QUOTE
Relativity contains no internal contradiction. The only contradiction exists in your delusional misinterpretation.
Completely frivolous talk at odds with the facts.
Completely frivolous talk at odds with the facts.
QUOTE (Omnibus+Oct 27 2008, 10:40 PM)
Completely frivolous talk at odds with the facts.
What facts? You have no facts, just delusions based on ignorance.
What facts? You have no facts, just delusions based on ignorance.
[quoteI have 2 comments - you didn't read the page on the experimental verification of SR I posted did you?
Also, what exactly replaces relativity presumably reproducing the results of relativity in a different way in your crazy world? [quote]
I already told you, you don’t need to post links with purported experimental validations of the theory of relativity. The crucial contradiction I demonstrated invalidates it prior to any attempt for experimental verification. Therefore, there have never been nor can there ever be experimental proof for the validity of the theory of relativity. Any such claim is nothing else but crude manipulation. Therefore, any link or whatever claiming such validation has to be ignored out of hand.
As for what should replace the theory of relativity. Nothing. I already said that on several occasions. When a wheat field is weeded out no replacement for the weed is sought.
Also, what exactly replaces relativity presumably reproducing the results of relativity in a different way in your crazy world? [quote]
I already told you, you don’t need to post links with purported experimental validations of the theory of relativity. The crucial contradiction I demonstrated invalidates it prior to any attempt for experimental verification. Therefore, there have never been nor can there ever be experimental proof for the validity of the theory of relativity. Any such claim is nothing else but crude manipulation. Therefore, any link or whatever claiming such validation has to be ignored out of hand.
As for what should replace the theory of relativity. Nothing. I already said that on several occasions. When a wheat field is weeded out no replacement for the weed is sought.
Quite the contrary, delusions are on your part, as seen from frivolous statements such as the above. I've provided facts, you haven't
QUOTE (Omnibus+Oct 27 2008, 10:48 PM)
Quite the contrary, delusions are on your part, as seen from frivolous statements such as the above. I've provided facts, you haven't
Don't wet your pink tutu
Don't wet your pink tutu
prometheus, did you not read that I'm not commenting on the so-called relativistic equations? If you haven't, then read it now and don't bother wasting time asking me anything in connection with such pseudoscience. Relativistic equations are to be ignored out of hand because of the conclusively proven invalidity of the theory that creates them.
On the other hand, I'm waiting for you to tell me what is the equation for F valid for high velocities, known at the start of the discussed derivation?
Further, I'm waiting for your admission that even if you do provide me with such an equation (which is entirely impossible) then, again, the theory of relativity cannot be saved since then, again, the form of the equation in K produced through the Lorentz transformations will differ from that required by the first postulate.
On the other hand, I'm waiting for you to tell me what is the equation for F valid for high velocities, known at the start of the discussed derivation?
Further, I'm waiting for your admission that even if you do provide me with such an equation (which is entirely impossible) then, again, the theory of relativity cannot be saved since then, again, the form of the equation in K produced through the Lorentz transformations will differ from that required by the first postulate.
QUOTE (Omnibus+Oct 27 2008, 10:58 PM)
On the other hand, I'm waiting for you to tell me what is the equation for F valid for high velocities, known at the start of the discussed derivation?
You can derive it yourself, provided that you know elementary calculus:
p=mv/sqrt(1-(v/c)^2)
F=dp/dt
Let's see your level of mathematical ability. Did you graduate from high school, tutu-boy?
Now, who's the tutu boy, the one who proves these formulae are coming out of an incorrect theory and therefore must be ignored out of hand or the one insisting to have them considered, despite their uselessness?
@All,
Feedback is for the cowards, for those who don't have arguments and their only recourse is to come from behind and nip you.
Feedback is for the cowards, for those who don't have arguments and their only recourse is to come from behind and nip you.
QUOTE (Omnibus+Oct 28 2008, 12:27 AM)
Now, who's the tutu boy, the one who proves these formulae are coming out of an incorrect theory and therefore must be ignored out of hand or the one insisting to have them considered, despite their uselessness?
But reality proves them correct.
Forget the Einstien's paper. Reality trumps all.
and reality is F only closely equalls ma at relatively low speeds.
If they are so useless then how does the GPS work when their calculations are using those formulae you are dismissing.
But reality proves them correct.
Forget the Einstien's paper. Reality trumps all.
and reality is F only closely equalls ma at relatively low speeds.
If they are so useless then how does the GPS work when their calculations are using those formulae you are dismissing.
QUOTE (buttershug+Oct 27 2008, 07:33 PM)
But reality proves them correct.
Forget the Einstien's paper. Reality trumps all.
and reality is F only closely equalls ma at relatively low speeds.
If they are so useless then how does the GPS work when their calculations are using those formulae you are dismissing.
buttershug, don't ask me to forget Einstein's paper because that's is what we are discussing. Also, free yourself from that connection of F and ma. That's not the substantial part of the argument.
The important point of my argument is that the first postulate (the "Principle of Relativity") requires that whatever the mass m is of a given body in a given system that same body has to have that same mass m in any other system. Understand that this requirement applies to whatever velocity of the body you may think of. So, don't worry about the validity of whatever you would consider the law is at whatever velocity you may think of.
Also, understand, no matter what the experimental results are, no matter what GPS data you may get, the theory cannot derive any of it.
Further, you cannot be sure at all that what they tell you about GPS and all is true (it isn't). On the other hand, however, you see it with your own eyes that the theory is more than wanting, it is outright incorrect.
Forget the Einstien's paper. Reality trumps all.
and reality is F only closely equalls ma at relatively low speeds.
If they are so useless then how does the GPS work when their calculations are using those formulae you are dismissing.
buttershug, don't ask me to forget Einstein's paper because that's is what we are discussing. Also, free yourself from that connection of F and ma. That's not the substantial part of the argument.
The important point of my argument is that the first postulate (the "Principle of Relativity") requires that whatever the mass m is of a given body in a given system that same body has to have that same mass m in any other system. Understand that this requirement applies to whatever velocity of the body you may think of. So, don't worry about the validity of whatever you would consider the law is at whatever velocity you may think of.
Also, understand, no matter what the experimental results are, no matter what GPS data you may get, the theory cannot derive any of it.
Further, you cannot be sure at all that what they tell you about GPS and all is true (it isn't). On the other hand, however, you see it with your own eyes that the theory is more than wanting, it is outright incorrect.
QUOTE (Omnibus+Oct 27 2008, 08:58 PM)
The important point of my argument is that the first postulate (the "Principle of Relativity") requires that whatever the mass m is of a given body in a given system that same body has to have that same mass m in any other system. Understand that this requirement applies to whatever velocity of the body you may think of. So, don't worry about the validity of whatever you would consider the law is at whatever velocity you may think of.
This paragraph would seem to fairly summarize the trap you have blundered into. While a similar m may be measured the same by two observers in their two respective frames, you have somehow leaped to the conclusion that observer A, given the chance, would measure observer B's m to be the same as his own.
That is not what is meant by "m is invariant."
That m differs due to velocity is borne out daily in particle accelerators.
QUOTE (NoCleverName+Oct 27 2008, 08:12 PM)
This paragraph would seem to fairly summarize the trap you have blundered into. While a similar m may be measured the same by two observers in their two respective frames, you have somehow leaped to the conclusion that observer A, given the chance, would measure observer B's m to be the same as his own.
That is not what is meant by "m is invariant."
That m differs due to velocity is borne out daily in particle accelerators.
This has already been addressed. It doesn't matter what the experiments with accelerators or whatever other experiments demonstrate, let alone that such change of mass had been known long before the appearance of the theory of relativity. The point here is that it is specifically the theory of relativity that cannot derive it. Indeed, the first postulate (the "Principle of Relativity") requires that mass m in a given system must be represented by the same mass m in any other system (consult §10 of Einstein's 1905 paper to convince yourself that that's the case). On the other hand, Lorentz transformations derive that mass m in one system is represented in another system as beta^3m (consult §10 of Einstein's 1905 paper to convince yourself that that's the case). This is a crucial contradiction in the theory which invalidates it in its entirety and which has nothing to do with any experiment whatsoever. This contradiction also shows that there can never be an experiment that would prove the validity of that theory. This contradiction can only be resolved if beta = 1, that is, if the Lorentz transformations are invalid.
That is not what is meant by "m is invariant."
That m differs due to velocity is borne out daily in particle accelerators.
This has already been addressed. It doesn't matter what the experiments with accelerators or whatever other experiments demonstrate, let alone that such change of mass had been known long before the appearance of the theory of relativity. The point here is that it is specifically the theory of relativity that cannot derive it. Indeed, the first postulate (the "Principle of Relativity") requires that mass m in a given system must be represented by the same mass m in any other system (consult §10 of Einstein's 1905 paper to convince yourself that that's the case). On the other hand, Lorentz transformations derive that mass m in one system is represented in another system as beta^3m (consult §10 of Einstein's 1905 paper to convince yourself that that's the case). This is a crucial contradiction in the theory which invalidates it in its entirety and which has nothing to do with any experiment whatsoever. This contradiction also shows that there can never be an experiment that would prove the validity of that theory. This contradiction can only be resolved if beta = 1, that is, if the Lorentz transformations are invalid.
QUOTE (Omnibus+Oct 27 2008, 09:48 PM)
This has already been addressed. It doesn't matter what the experiments with accelerators or whatever other experiments demonstrate, let alone that such change of mass had been known long before the appearance of the theory of relativity. The point here is that it is specifically the theory of relativity that cannot derive it. Indeed, the first postulate (the "Principle of Relativity") requires that mass m in a given system must be represented by the same mass m in any other system (consult §10 of Einstein's 1905 paper to convince yourself that that's the case). On the other hand, Lorentz transformations derive that mass m in one system is represented in another system as beta^3m (consult §10 of Einstein's 1905 paper to convince yourself that that's the case). This is a crucial contradiction in the theory which invalidates it in its entirety and which has nothing to do with any experiment whatsoever. This contradiction also shows that there can never be an experiment that would prove the validity of that theory. This contradiction can only be resolved if beta = 1, that is, if the Lorentz transformations are invalid.
Nope. You've read it wrong and are clearly repeating the same mistake I mentioned above. Sure, m is m is systems k and K; but they are the same only when one measures within the respective system. The paper derives what m looks like from the perspective of the other system.
That's the relativity part. You seemed to be fixated on this "same mass" thing like someone looking at one of those optical illusions and can't let the "other interpretation pop into view". Sorry, man. You've blown it.
Nope. You've read it wrong and are clearly repeating the same mistake I mentioned above. Sure, m is m is systems k and K; but they are the same only when one measures within the respective system. The paper derives what m looks like from the perspective of the other system.
That's the relativity part. You seemed to be fixated on this "same mass" thing like someone looking at one of those optical illusions and can't let the "other interpretation pop into view". Sorry, man. You've blown it.
QUOTE (NoCleverName+Oct 27 2008, 09:18 PM)
Nope. You've read it wrong and are clearly repeating the same mistake I mentioned above. Sure, m is m is systems k and K; but they are the same only when one measures within the respective system. The paper derives what m looks like from the perspective of the other system.
That's the relativity part. You seemed to be fixated on this "same mass" thing like someone looking at one of those optical illusions and can't let the "other interpretation pop into view". Sorry, man. You've blown it.
What are you talking about? At least the other opponents know the elementary basis of the theory.
That's the relativity part. You seemed to be fixated on this "same mass" thing like someone looking at one of those optical illusions and can't let the "other interpretation pop into view". Sorry, man. You've blown it.
What are you talking about? At least the other opponents know the elementary basis of the theory.
QUOTE (Omnibus+Oct 27 2008, 10:23 PM)
What are you talking about? At least the other opponents know the elementary basis of the theory.
Yes they do; and that is why they know you are wrong.
Yes they do; and that is why they know you are wrong.
I think you might be less confused if instead of trying to look at one single particle from the standpoint of two separate frames you look at two identical particles each in their own frame. That way you can compare what you measure about your own particle with what you ascertain about the moving particle.
As the moving particle zips by, you would measure it's mass as different from your own copy. But then if you have that same particle brought to rest and put it beside you own copy you'd see they in fact have the same mass. However, you might notice the previously speeding particle is somewhat younger than yours.
As the moving particle zips by, you would measure it's mass as different from your own copy. But then if you have that same particle brought to rest and put it beside you own copy you'd see they in fact have the same mass. However, you might notice the previously speeding particle is somewhat younger than yours.
QUOTE (Omnibus+Oct 28 2008, 12:27 AM)
Now, who's the tutu boy, the one who proves these formulae are coming out of an incorrect theory and therefore must be ignored out of hand or the one insisting to have them considered, despite their uselessness?
So, you can't calculate the force. Just as I thought, you can't perform even a simple derivative. I suggest that you take the tutu off your head, it is constricting your temples.
Now, seriously speaking, if you don't know simple math, you can't continue to pretend that you are discussing physics. Most cranks in this forum get caught this way, you aren't the first .
So, you can't calculate the force. Just as I thought, you can't perform even a simple derivative. I suggest that you take the tutu off your head, it is constricting your temples.
Now, seriously speaking, if you don't know simple math, you can't continue to pretend that you are discussing physics. Most cranks in this forum get caught this way, you aren't the first .
QUOTE (Trout+Oct 27 2008, 09:56 PM)
So, you can't calculate the force. Just as I thought, you can't perform even a simple derivative. I suggest that you take the tutu off your head, it is constricting your temples.
Now, seriously speaking, if you don't know simple math, you can't continue to pretend that you are discussing physics. Most cranks in this forum get caught this way, you aren't the first .
You are the crank. Why? Because you ignore the fact that the theory of relativity is flawed and continue to use its formulae. This is how cranks are caught.
Now, seriously speaking, if you don't know simple math, you can't continue to pretend that you are discussing physics. Most cranks in this forum get caught this way, you aren't the first .
You are the crank. Why? Because you ignore the fact that the theory of relativity is flawed and continue to use its formulae. This is how cranks are caught.
QUOTE (bm1957+Oct 27 2008, 11:19 PM)
The conversation has moved on since you replied to me... but I would like to pick up on this point.
The reason you should read books is because you obviously don't know what the commonly accepted interpretation of relativity is. You've tried to say it's in error by setting up an almighty strawman which assumes that most physicists think that relativity says something which it quite obviously and trivially does not.
Now you obviously have a better understanding of the principles than someone like Omnibus, but you need more knowledge and formal teaching to really be any good... so go get the books out!!!
I note that you do not comment on my derivation using the Lorentz transformations that t=t(/). It is people like you and AlpaNumeric who have dragged, and are still dragging "real physics" down into the gutter!
I am always willing to read new books to gain new insights: But what Trout did was to direct me to books after what he stated is EXACTLY the same as what I have stated. He stated that the perspective of SR can be explained that you observe your TWIN brother as being shorter and he observes you as being shorter WHILE YOU ARE BOTH THE SAME LENGTH.
I have stated similarly that when your twin brother moves away from you at a CONSTANT SPEED you observe his clock to be slower while he observes your clock to be slower while IN ACTUAL REALITY THE TWO CLOCKS TICK AT EXACTLY THE SAME RATE.
Now if those books could not help Trout to realize that we are stating is the same thing in two different ways, why should I read books that he recommends?
The reason you should read books is because you obviously don't know what the commonly accepted interpretation of relativity is. You've tried to say it's in error by setting up an almighty strawman which assumes that most physicists think that relativity says something which it quite obviously and trivially does not.
Now you obviously have a better understanding of the principles than someone like Omnibus, but you need more knowledge and formal teaching to really be any good... so go get the books out!!!
I note that you do not comment on my derivation using the Lorentz transformations that t=t(/). It is people like you and AlpaNumeric who have dragged, and are still dragging "real physics" down into the gutter!
I am always willing to read new books to gain new insights: But what Trout did was to direct me to books after what he stated is EXACTLY the same as what I have stated. He stated that the perspective of SR can be explained that you observe your TWIN brother as being shorter and he observes you as being shorter WHILE YOU ARE BOTH THE SAME LENGTH.
I have stated similarly that when your twin brother moves away from you at a CONSTANT SPEED you observe his clock to be slower while he observes your clock to be slower while IN ACTUAL REALITY THE TWO CLOCKS TICK AT EXACTLY THE SAME RATE.
Now if those books could not help Trout to realize that we are stating is the same thing in two different ways, why should I read books that he recommends?
QUOTE (johanfprins+Oct 28 2008, 08:29 AM)
It is people like you and AlpaNumeric who have dragged, and are still dragging "real physics" down into the gutter!
Never mind there isn't an experiment which has shown special relativity wrong within its domain of application.
So it comes down to either you being wrong or nature being wrong. Wow, that's a tough one.....
Never mind there isn't an experiment which has shown special relativity wrong within its domain of application.
So it comes down to either you being wrong or nature being wrong. Wow, that's a tough one.....
QUOTE (johanfprins+Oct 28 2008, 03:29 AM)
I am always willing to read new books to gain new insights: But what Trout did was to direct me to books after what he stated is EXACTLY the same as what I have stated. He stated that the perspective of SR can be explained that you observe your TWIN brother as being shorter and he observes you as being shorter WHILE YOU ARE BOTH THE SAME LENGTH.
I have stated similarly that when your twin brother moves away from you at a CONSTANT SPEED you observe his clock to be slower while he observes your clock to be slower while IN ACTUAL REALITY THE TWO CLOCKS TICK AT EXACTLY THE SAME RATE.
Ah, the old confusion about the "Twin Paradox" ... feel free to read and learn some more.
Don't worry, I don't bother to understand it in detail, either (because there's only so much that can fit in this old head) ... but this has been dispensed with for over 100 years now.
I have stated similarly that when your twin brother moves away from you at a CONSTANT SPEED you observe his clock to be slower while he observes your clock to be slower while IN ACTUAL REALITY THE TWO CLOCKS TICK AT EXACTLY THE SAME RATE.
Ah, the old confusion about the "Twin Paradox" ... feel free to read and learn some more.
Don't worry, I don't bother to understand it in detail, either (because there's only so much that can fit in this old head) ... but this has been dispensed with for over 100 years now.
QUOTE (AlphaNumeric+Oct 28 2008, 11:07 AM)
Never mind there isn't an experiment which has shown special relativity wrong within its domain of application.
So it comes down to either you being wrong or nature being wrong. Wow, that's a tough one.....
AN,
You are again displaying your stupidity: I have NOT challenged Special Relativity anywhere. I accept the Lorentz transformations as valid and derive from them that two clocks moving with a constant speed relative to each other MUST tick at the same rate. Are you saying I am wrong to believe the Lorentz trnsformations?
What you say is that the experiments have proved the Lorentz transformations wrong! You are thus the one who does not believe in the validity of the equations which DEFINE special relativity!
So it comes down to either you being wrong or nature being wrong. Wow, that's a tough one.....
AN,
You are again displaying your stupidity: I have NOT challenged Special Relativity anywhere. I accept the Lorentz transformations as valid and derive from them that two clocks moving with a constant speed relative to each other MUST tick at the same rate. Are you saying I am wrong to believe the Lorentz trnsformations?
What you say is that the experiments have proved the Lorentz transformations wrong! You are thus the one who does not believe in the validity of the equations which DEFINE special relativity!
QUOTE (NoCleverName+Oct 28 2008, 11:15 AM)
Ah, the old confusion about the "Twin Paradox" ... feel free to read and learn some more.
Don't worry, I don't bother to understand it in detail, either (because there's only so much that can fit in this old head) ... but this has been dispensed with for over 100 years now.
If you cannot understand it then it is irresponsible of you to claim that it has been "dispensed with" over 100 years ago. A person like you should not partake in a scientific discussion because all you do is to muddy the water by accepting dogma which you cannot understand.
Furthermore, I am not considering the twin paradox which involves acceleration and deceleration. What I state is that when two clocks move at a constant speed relative to each other they MUST each tick away at exactly the same rate.
Don't worry, I don't bother to understand it in detail, either (because there's only so much that can fit in this old head) ... but this has been dispensed with for over 100 years now.
If you cannot understand it then it is irresponsible of you to claim that it has been "dispensed with" over 100 years ago. A person like you should not partake in a scientific discussion because all you do is to muddy the water by accepting dogma which you cannot understand.
Furthermore, I am not considering the twin paradox which involves acceleration and deceleration. What I state is that when two clocks move at a constant speed relative to each other they MUST each tick away at exactly the same rate.
QUOTE (johanfprins+Oct 28 2008, 05:20 AM)
If you cannot understand it then it is irresponsible of you to claim that it has been "dispensed with" over 100 years ago. A person like you should not partake in a scientific discussion because all you do is to muddy the water by accepting dogma which you cannot understand.
Oh, I can follow it and understand it. It's just that I don't commit its operation to "working, everyday knowledge" since I'm more of a computer guy and don't work in physics. Nevertheless, I can see how something like this has to be true from first principles.
Namely (as I stated above): What are the consequences of all sub-lightspeed observers measuring the speed of light to have the same speed d/t. Something's got to give with d and/or t, right?
Oh, I can follow it and understand it. It's just that I don't commit its operation to "working, everyday knowledge" since I'm more of a computer guy and don't work in physics. Nevertheless, I can see how something like this has to be true from first principles.
Namely (as I stated above): What are the consequences of all sub-lightspeed observers measuring the speed of light to have the same speed d/t. Something's got to give with d and/or t, right?
QUOTE (NoCleverName+Oct 28 2008, 11:26 AM)
Oh, I can follow it and understand it. It's just that I don't commit it's operation to "working, everyday knowledge" since I'm more of a computer guy and don't work in physics. Nevertheless, I can see how something like this has to be true from first principles.
Namely (as I stated above): What are the consequences of all sub-lightspeed observers measuring the speed of light to have the same speed d/t. Something's got to give with d and/or t, right?
I have added an additional comment while you also posted; pointing out that my statement does NOT relate to acceleration and deceleration. Furthermore, if you go back to one of the posts I did yesterday you will find a proof, derived from the the Lorentz transformations, that two clocks moving at a constant speed relative to each other will show the same time. As a computer expert you should be able to follow the mathematics more easily than our present mainstream "real physicists" like AlphaNumeric . To write an algorithm requires one to be able to handle logic. Most of our mainstream "real" theoretical "physicists" have ignored logic since the early 1930's.
Namely (as I stated above): What are the consequences of all sub-lightspeed observers measuring the speed of light to have the same speed d/t. Something's got to give with d and/or t, right?
I have added an additional comment while you also posted; pointing out that my statement does NOT relate to acceleration and deceleration. Furthermore, if you go back to one of the posts I did yesterday you will find a proof, derived from the the Lorentz transformations, that two clocks moving at a constant speed relative to each other will show the same time. As a computer expert you should be able to follow the mathematics more easily than our present mainstream "real physicists" like AlphaNumeric . To write an algorithm requires one to be able to handle logic. Most of our mainstream "real" theoretical "physicists" have ignored logic since the early 1930's.
QUOTE (johanfprins+Oct 28 2008, 05:32 AM)
I have added an additional comment while you also posted; pointing out that my statement does NOT relate to acceleration and deceleration.
Hmmm, two objects with a velocity difference and no acceleration to explain where that difference came from ... sure, right.
I believe the whole point of having acceleration is that it breaks the observed symmetry of the situation. That is, somewhere along the line somebody "changed frames". Again, this is dealt with (I think) in the Wiki so I'm not going to make a feeble attempt to rewrite that article here. Suffice it to say I can understand why it has to be true, even if I can't derive it myself.
Hmmm, two objects with a velocity difference and no acceleration to explain where that difference came from ... sure, right.
I believe the whole point of having acceleration is that it breaks the observed symmetry of the situation. That is, somewhere along the line somebody "changed frames". Again, this is dealt with (I think) in the Wiki so I'm not going to make a feeble attempt to rewrite that article here. Suffice it to say I can understand why it has to be true, even if I can't derive it myself.
QUOTE (johanfprins+Oct 28 2008, 08:29 AM)
I note that you do not comment on my derivation using the Lorentz transformations that t=t(/).
Like I said, the conversation had moved on. Regardless, you never showed me that you had used the transformations previously; if you were hoping for a retraction, you should have linked to where you did it before.
But you have shown that you are ignorant of current physics by insisting that physicists think something which they don't. So how can you comment on "real physics"? All you are commenting on is your perception of "current physics", which is demonstrably wrong.
Besides, who's getting abusive now?
But you have shown that you are ignorant of current physics by insisting that physicists think something which they don't. So how can you comment on "real physics"? All you are commenting on is your perception of "current physics", which is demonstrably wrong.
Besides, who's getting abusive now?
I am always willing to read new books to gain new insights: But what Trout did was to direct me to books after what he stated is EXACTLY the same as what I have stated.
But you believed that what you stated was at odds with "real science"; it's not. So read the books, and learn what "real science" really says. That's all.
I'm afraid this is your biggest genuine error. In truth, neither tick-rate is more less real than any other. You can say that they have the same tick-rate as measured locally (the same proper time), but this is utterly trivial and not worthy of discussion, IMO.
I'm afraid this is your biggest genuine error. In truth, neither tick-rate is more less real than any other. You can say that they have the same tick-rate as measured locally (the same proper time), but this is utterly trivial and not worthy of discussion, IMO.
Now if those books could not help Trout to realize that we are stating is the same thing in two different ways, why should I read books that he recommends?
To learn more about what "real science" says so you don't embarrass yourself by arguing that it says something wrong and you know how to put it right, which is what you've done.
bm1957m where does johanfprins
On the contrary, johanfprins shows that if x(/) = 0 then Lorentz transformations yield x = vt =/= 0 and conversely, if x = 0 then x(/) = vt(/) =/= 0.
The error is that he assumes x = x(/) which he has no grounds to do because x is in K while x(/) is in k and in order for his assumption to be true there must be t = t(/) but this is exactly what johanfprins needs to prove.
Lorentz transformations are mathematically internally consistent and their mathematical validity cannot be disputed. The only thing that can be disputed is their physical meaning and that’s done differently from what johanfprins tries to do.
I have shown one way to prove that Lorentz transformations lack physical meaning—by demonstrating that they contradict the first postulate (the “Principle of Relativity”) should one consider that postulate as valid.
bm1957m,
What are you talking about? What one equation. Any equation describing the motion of particle of a given mass in K transforms in any system according to the first postulate retaining the same mass. On the contrary, Lorentz transformations derive any equation such that the mass in one system becomes beta^3 times that mass in another system. Any equation. That’s a fatal conflict between the first postulate and the Lorentz transformations demonstrable for any equation. For any equation. Now who’s shown no understanding, you or I?
And they don't, because x is in K while x(/) is in k.
Like I said, the conversation had moved on. Regardless, you never showed me that you had used the transformations previously; if you were hoping for a retraction, you should have linked to where you did it before.
QUOTE
It is people like you and AlpaNumeric who have dragged, and are still dragging "real physics" down into the gutter!
But you have shown that you are ignorant of current physics by insisting that physicists think something which they don't. So how can you comment on "real physics"? All you are commenting on is your perception of "current physics", which is demonstrably wrong.
Besides, who's getting abusive now?
QUOTE (->
| QUOTE |
| It is people like you and AlpaNumeric who have dragged, and are still dragging "real physics" down into the gutter! |
But you have shown that you are ignorant of current physics by insisting that physicists think something which they don't. So how can you comment on "real physics"? All you are commenting on is your perception of "current physics", which is demonstrably wrong.
Besides, who's getting abusive now?
I am always willing to read new books to gain new insights: But what Trout did was to direct me to books after what he stated is EXACTLY the same as what I have stated.
But you believed that what you stated was at odds with "real science"; it's not. So read the books, and learn what "real science" really says. That's all.
QUOTE
IN ACTUAL REALITY THE TWO CLOCKS TICK AT EXACTLY THE SAME RATE.
I'm afraid this is your biggest genuine error. In truth, neither tick-rate is more less real than any other. You can say that they have the same tick-rate as measured locally (the same proper time), but this is utterly trivial and not worthy of discussion, IMO.
QUOTE (->
| QUOTE |
| IN ACTUAL REALITY THE TWO CLOCKS TICK AT EXACTLY THE SAME RATE. |
I'm afraid this is your biggest genuine error. In truth, neither tick-rate is more less real than any other. You can say that they have the same tick-rate as measured locally (the same proper time), but this is utterly trivial and not worthy of discussion, IMO.
Now if those books could not help Trout to realize that we are stating is the same thing in two different ways, why should I read books that he recommends?
To learn more about what "real science" says so you don't embarrass yourself by arguing that it says something wrong and you know how to put it right, which is what you've done.
QUOTE (Omnibus+Oct 28 2008, 04:33 AM)
You are the crank. Why? Because you ignore the fact that the theory of relativity is flawed and continue to use its formulae. This is how cranks are caught.
Only the person that says things that don't match observed results is the crank.
that's you by the way.
Only the person that says things that don't match observed results is the crank.
that's you by the way.
QUOTE (bm1957+Oct 28 2008, 11:44 AM)
Like I said, the conversation had moved on. Regardless, you never showed me that you had used the transformations previously; if you were hoping for a retraction, you should have linked to where you did it before.
I repeated the proof yesterday: Here it is again:
Consider the Lorentz transformation from the primed system top the non-primed system:
x=(beta)(x(/)+vt(/)) and t=(beta)(t(/)+(v/c^2)x(/))
The position of the clock in the primed system is at its origin: i.e. x(/)=0: Thus the equations become:
x=(beta)(vt(/)) and t=(beta)t(/): Now use the second equation to replace t(/) in the first equation: One then obtains that:
x=vt where x is the distance between the origins of the primed and unprimed systems.
Now consider the Lorentz transformation from the unprimed to the primed system:
x(/)=(beta)(x-vt) and t(/)=(beta)(t-(v/c^2)x). The clock in the unprimed system is at its origin so that x=0: Thus the equations become:
x(/)=(beta)(-vt) and t(/)=(beta)t: Now use t in the second equation to replace t in the first equation; and one obtains that:
-x(/) =vt(/) where -x(/) is now the distance between the origins of the two reference frames. At any instant in time (whether in the primed or unprimed reference frames) the distance between the origins has a single value: Thus -x(/)=x; and since v=v; one obtains that t(/)=t.
So first comment on this derivation before we proceed further
I repeated the proof yesterday: Here it is again:
Consider the Lorentz transformation from the primed system top the non-primed system:
x=(beta)(x(/)+vt(/)) and t=(beta)(t(/)+(v/c^2)x(/))
The position of the clock in the primed system is at its origin: i.e. x(/)=0: Thus the equations become:
x=(beta)(vt(/)) and t=(beta)t(/): Now use the second equation to replace t(/) in the first equation: One then obtains that:
x=vt where x is the distance between the origins of the primed and unprimed systems.
Now consider the Lorentz transformation from the unprimed to the primed system:
x(/)=(beta)(x-vt) and t(/)=(beta)(t-(v/c^2)x). The clock in the unprimed system is at its origin so that x=0: Thus the equations become:
x(/)=(beta)(-vt) and t(/)=(beta)t: Now use t in the second equation to replace t in the first equation; and one obtains that:
-x(/) =vt(/) where -x(/) is now the distance between the origins of the two reference frames. At any instant in time (whether in the primed or unprimed reference frames) the distance between the origins has a single value: Thus -x(/)=x; and since v=v; one obtains that t(/)=t.
So first comment on this derivation before we proceed further
QUOTE (Omnibus+Oct 28 2008, 01:48 AM)
This contradiction also shows that there can never be an experiment that would prove the validity of that theory. This contradiction can only be resolved if beta = 1, that is, if the Lorentz transformations are invalid.
No that contraction does not mean there can be no experiment. It might mean the paper is badly worded. But the other possiblility is that there is a difference in m and m'.
And it takes experiments to see which is correct. And you are wrong beta does not equal 1.
No that contraction does not mean there can be no experiment. It might mean the paper is badly worded. But the other possiblility is that there is a difference in m and m'.
And it takes experiments to see which is correct. And you are wrong beta does not equal 1.
QUOTE (NoCleverName+Oct 28 2008, 11:41 AM)
I believe the whole point of having acceleration is that it breaks the observed symmetry of the situation. That is, somewhere along the line somebody "changed frames". Again, this is dealt with (I think) in the Wiki so I'm not going to make a feeble attempt to rewrite that article here. Suffice it to say I can understand why it has to be true, even if I can't derive it myself.
As I said I have NOT considered acceleration or deceleration: So I am happy with this comment of yours
As I said I have NOT considered acceleration or deceleration: So I am happy with this comment of yours
QUOTE (johanfprins+Oct 28 2008, 06:06 AM)
As I said I have NOT considered acceleration or deceleration: So I am happy with this comment of yours
... but you must, since, regardless of the keyhole you are peering thru, it occurred.
... but you must, since, regardless of the keyhole you are peering thru, it occurred.
bm1957, johanfprins' notwithstanding, you have no grounds to say that:
because you have shown no error in my understanding.
QUOTE
Now you obviously have a better understanding of the principles than someone like Omnibus ...
because you have shown no error in my understanding.
QUOTE (johanfprins+Oct 28 2008, 11:02 AM)
I repeated the proof yesterday: Here it is again:
Consider the Lorentz transformation from the primed system top the non-primed system:
x=(beta)(x(/)+vt(/)) and t=(beta)(t(/)+(v/c^2)x(/))
The position of the clock in the primed system is at its origin: i.e. x(/)=0: Thus the equations become:
x=(beta)(vt(/)) and t=(beta)t(/): Now use the second equation to replace t(/) in the first equation: One then obtains that:
x=vt where x is the distance between the origins of the primed and unprimed systems.
Now consider the Lorentz transformation from the unprimed to the primed system:
x(/)=(beta)(x-vt) and t(/)=(beta)(t-(v/c^2)x). The clock in the unprimed system is at its origin so that x=0: Thus the equations become:
x(/)=(beta)(-vt) and t(/)=(beta)t: Now use t in the second equation to replace t in the first equation; and one obtains that:
-x(/) =vt(/) where -x(/) is now the distance between the origins of the two reference frames. At any instant in time (whether in the primed or unprimed reference frames) the distance between the origins has a single value: Thus -x(/)=x; and since v=v; one obtains that t(/)=t.
So first comment on this derivation before we proceed further
Calm down.
I never commented on your 'proof', I only stated that I didn't remember you using the lorentz transform when you stated that you had. Using it now doesn't show that you did before I said that.
I will comment, nonetheless.
You assume that x=0 and that x(/)=0, yet in your conclusion you make the implicit assumption that x and x(/) may take any values and generalise that t(/)=t thus always holds. I propose that your 'proof' simply shows that t(/)(0)=t(0), i.e., that the clocks are synchronised as they pass each other. Since this is a prerequisite to even perform the gedanken, I think all you've done is merely another triviality.
Consider the Lorentz transformation from the primed system top the non-primed system:
x=(beta)(x(/)+vt(/)) and t=(beta)(t(/)+(v/c^2)x(/))
The position of the clock in the primed system is at its origin: i.e. x(/)=0: Thus the equations become:
x=(beta)(vt(/)) and t=(beta)t(/): Now use the second equation to replace t(/) in the first equation: One then obtains that:
x=vt where x is the distance between the origins of the primed and unprimed systems.
Now consider the Lorentz transformation from the unprimed to the primed system:
x(/)=(beta)(x-vt) and t(/)=(beta)(t-(v/c^2)x). The clock in the unprimed system is at its origin so that x=0: Thus the equations become:
x(/)=(beta)(-vt) and t(/)=(beta)t: Now use t in the second equation to replace t in the first equation; and one obtains that:
-x(/) =vt(/) where -x(/) is now the distance between the origins of the two reference frames. At any instant in time (whether in the primed or unprimed reference frames) the distance between the origins has a single value: Thus -x(/)=x; and since v=v; one obtains that t(/)=t.
So first comment on this derivation before we proceed further
Calm down.
I never commented on your 'proof', I only stated that I didn't remember you using the lorentz transform when you stated that you had. Using it now doesn't show that you did before I said that.
I will comment, nonetheless.
You assume that x=0 and that x(/)=0, yet in your conclusion you make the implicit assumption that x and x(/) may take any values and generalise that t(/)=t thus always holds. I propose that your 'proof' simply shows that t(/)(0)=t(0), i.e., that the clocks are synchronised as they pass each other. Since this is a prerequisite to even perform the gedanken, I think all you've done is merely another triviality.
buttershug,
How come that’s me, by the way? The crank is the person who says things that don’t match observed results such as the fact that first postulate requires m to be the same m in every system while Lorentz transformation derive that m in one system is beta^3m in another system, denying that therefore first postulate and the Lorentz transformations are in conflict, is the crank. Denying the obvious makes one a crank.
QUOTE
Only the person that says things that don't match observed results is the crank.
that's you by the way.
that's you by the way.
How come that’s me, by the way? The crank is the person who says things that don’t match observed results such as the fact that first postulate requires m to be the same m in every system while Lorentz transformation derive that m in one system is beta^3m in another system, denying that therefore first postulate and the Lorentz transformations are in conflict, is the crank. Denying the obvious makes one a crank.
QUOTE (Omnibus+Oct 28 2008, 12:10 PM)
bm1957, johanfprins' notwithstanding, you have no grounds to say that:
because you have shown no error in my understanding.
Omnibus,
You have expicitly refused to engage in any conversation about relativity, repeatedly coming back to the point that you have issue with a single equation. When anybody tries to explain your misunderstanding, you refuse to accept it, stating that their explanation is invalid because you don't understand that equation.
You have shown no understanding; you have not even given yourself a platform to show any understanding because you refuse to enter any technical dialogue, so the assumption that somebody who is trying to discuss physics understands the principles better than you is valid, IMHO.
because you have shown no error in my understanding.
Omnibus,
You have expicitly refused to engage in any conversation about relativity, repeatedly coming back to the point that you have issue with a single equation. When anybody tries to explain your misunderstanding, you refuse to accept it, stating that their explanation is invalid because you don't understand that equation.
You have shown no understanding; you have not even given yourself a platform to show any understanding because you refuse to enter any technical dialogue, so the assumption that somebody who is trying to discuss physics understands the principles better than you is valid, IMHO.
QUOTE (bm1957+Oct 28 2008, 01:30 PM)
Omnibus,
You have expicitly refused to engage in any conversation about relativity, repeatedly coming back to the point that you have issue with a single equation. When anybody tries to explain your misunderstanding, you refuse to accept it, stating that their explanation is invalid because you don't understand that equation.
You have shown no understanding; you have not even given yourself a platform to show any understanding because you refuse to enter any technical dialogue, so the assumption that somebody who is trying to discuss physics understands the principles better than you is valid, IMHO.
Amen!
You have expicitly refused to engage in any conversation about relativity, repeatedly coming back to the point that you have issue with a single equation. When anybody tries to explain your misunderstanding, you refuse to accept it, stating that their explanation is invalid because you don't understand that equation.
You have shown no understanding; you have not even given yourself a platform to show any understanding because you refuse to enter any technical dialogue, so the assumption that somebody who is trying to discuss physics understands the principles better than you is valid, IMHO.
Amen!
bm1957m where does johanfprins
QUOTE
... assume that x=0 and that x(/)=0 ...
On the contrary, johanfprins shows that if x(/) = 0 then Lorentz transformations yield x = vt =/= 0 and conversely, if x = 0 then x(/) = vt(/) =/= 0.
The error is that he assumes x = x(/) which he has no grounds to do because x is in K while x(/) is in k and in order for his assumption to be true there must be t = t(/) but this is exactly what johanfprins needs to prove.
Lorentz transformations are mathematically internally consistent and their mathematical validity cannot be disputed. The only thing that can be disputed is their physical meaning and that’s done differently from what johanfprins tries to do.
I have shown one way to prove that Lorentz transformations lack physical meaning—by demonstrating that they contradict the first postulate (the “Principle of Relativity”) should one consider that postulate as valid.
QUOTE (bm1957+Oct 28 2008, 01:24 PM)
You assume that x=0 and that x(/)=0, yet in your conclusion you make the implicit assumption that x and x(/) may take any values and generalize that t(/)=t thus always holds. I propose that your 'proof' simply shows that t(/)(0)=t(0), i.e., that the clocks are synchronized as they pass each other. Since this is a prerequisite to even perform the gedanken, I think all you've done is merely another triviality.
I only now saw your reply. The position of the clock in the primed system is at x(/)=0 and at x not equal to zero (but to the distance between the origins) relative to the unprimed system. Furthermore, the clock in the unprimed system is at x=0 and at x(/) not equal to zero (but equal to the distance between the origins measured along the negative x-axis of the primed system) relative to the primed system. You can REALLY not be so stupid can you?
The distance between the two clocks cannot have two values at any time and the positions of the clocks within their reference frames are at the origins of the reference frames. Thus when the distance has a certain value and the time is t(/) on the clock within the primed reference frame and the corresponding time on the clock within the unprimed reference frame is t then t (/) MUST BE equal to t (as mathematically proved above). Whether it is trivial or not is NOT at issue. It is a fact! If it is found experimentally not to be so, the Lorentz transformations are violated since this fact follows by correct algebra from these equations.
Of course the clocks have been synchronized by setting t(/)(0)=t(0)=0. This is implicit in the derivation! But according to the Lorentz transformation they MUST stay synchronized for any distance d between them, unless you are using rules for algebra which I have never encountered.
I only now saw your reply. The position of the clock in the primed system is at x(/)=0 and at x not equal to zero (but to the distance between the origins) relative to the unprimed system. Furthermore, the clock in the unprimed system is at x=0 and at x(/) not equal to zero (but equal to the distance between the origins measured along the negative x-axis of the primed system) relative to the primed system. You can REALLY not be so stupid can you?
The distance between the two clocks cannot have two values at any time and the positions of the clocks within their reference frames are at the origins of the reference frames. Thus when the distance has a certain value and the time is t(/) on the clock within the primed reference frame and the corresponding time on the clock within the unprimed reference frame is t then t (/) MUST BE equal to t (as mathematically proved above). Whether it is trivial or not is NOT at issue. It is a fact! If it is found experimentally not to be so, the Lorentz transformations are violated since this fact follows by correct algebra from these equations.
Of course the clocks have been synchronized by setting t(/)(0)=t(0)=0. This is implicit in the derivation! But according to the Lorentz transformation they MUST stay synchronized for any distance d between them, unless you are using rules for algebra which I have never encountered.
bm1957m,
QUOTE
Omnibus,
You have expicitly refused to engage in any conversation about relativity, repeatedly coming back to the point that you have issue with a single equation. When anybody tries to explain your misunderstanding, you refuse to accept it, stating that their explanation is invalid because you don't understand that equation.
You have shown no understanding; you have not even given yourself a platform to show any understanding because you refuse to enter any technical dialogue, so the assumption that somebody who is trying to discuss physics understands the principles better than you is valid, IMHO.
You have expicitly refused to engage in any conversation about relativity, repeatedly coming back to the point that you have issue with a single equation. When anybody tries to explain your misunderstanding, you refuse to accept it, stating that their explanation is invalid because you don't understand that equation.
You have shown no understanding; you have not even given yourself a platform to show any understanding because you refuse to enter any technical dialogue, so the assumption that somebody who is trying to discuss physics understands the principles better than you is valid, IMHO.
What are you talking about? What one equation. Any equation describing the motion of particle of a given mass in K transforms in any system according to the first postulate retaining the same mass. On the contrary, Lorentz transformations derive any equation such that the mass in one system becomes beta^3 times that mass in another system. Any equation. That’s a fatal conflict between the first postulate and the Lorentz transformations demonstrable for any equation. For any equation. Now who’s shown no understanding, you or I?
Quote: see above:You assume that x=0 and that x(/)=0, yet in your conclusion you make the implicit assumption that x and x(/) may take any values and generalize that t(/)=t thus always holds. I propose that your 'proof' simply shows that t(/)(0)=t(0), i.e., that the clocks are synchronized as they pass each other. Since this is a prerequisite to even perform the gedanken, I think all you've done is merely another triviality.
I only now saw your reply. The position of the clock in the primed system is at x(/)=0 and at x not equal to zero (but to the distance between the origins) relative to the unprimed system. Furthermore, the clock in the unprimed system is at x=0 and at x(/) not equal to zero (but equal to the distance between the origins measured along the negative x-axis of the primed system) relative to the primed system. You can REALLY not be so stupid can you?
The distance between the two clocks cannot have two values at any time and the positions of the clocks within their reference frames are at the origins of the reference frames. Thus when the distance has a certain value and the time is t(/) on the clock within the primed reference frame and the corresponding time on the clock within the unprimed reference frame is t then t (/) MUST BE equal to t (as mathematically proved above). Whether it is trivial or not is NOT at issue. It is a fact! If it is found experimentally not to be so, the Lorentz transformations are violated since this fact follows by correct algebra from these equations.
Of course the clocks have been synchronized by setting t(/)(0)=t(0)=0. This is implicit in the derivation! But according to the Lorentz transformation they MUST stay synchronized for any distance d between them, unless you are using rules for algebra which I have never encountered.
I only now saw your reply. The position of the clock in the primed system is at x(/)=0 and at x not equal to zero (but to the distance between the origins) relative to the unprimed system. Furthermore, the clock in the unprimed system is at x=0 and at x(/) not equal to zero (but equal to the distance between the origins measured along the negative x-axis of the primed system) relative to the primed system. You can REALLY not be so stupid can you?
The distance between the two clocks cannot have two values at any time and the positions of the clocks within their reference frames are at the origins of the reference frames. Thus when the distance has a certain value and the time is t(/) on the clock within the primed reference frame and the corresponding time on the clock within the unprimed reference frame is t then t (/) MUST BE equal to t (as mathematically proved above). Whether it is trivial or not is NOT at issue. It is a fact! If it is found experimentally not to be so, the Lorentz transformations are violated since this fact follows by correct algebra from these equations.
Of course the clocks have been synchronized by setting t(/)(0)=t(0)=0. This is implicit in the derivation! But according to the Lorentz transformation they MUST stay synchronized for any distance d between them, unless you are using rules for algebra which I have never encountered.
QUOTE
The distance between the two clocks cannot have two values at any time
And they don't, because x is in K while x(/) is in k.
QUOTE (bm1957+Oct 28 2008, 07:30 AM)
Omnibus,
You have expicitly refused to engage in any conversation about relativity, repeatedly coming back to the point that you have issue with a single equation. When anybody tries to explain your misunderstanding, you refuse to accept it, stating that their explanation is invalid because you don't understand that equation.
You have to admit: it's a pretty clever tactic to have the position "I don't need to understand any of it because according to me, it's all invalid in the first place." In other words, a role reversal in which the crank claims the establishment are cranks so it's pointless arguing about the "crank predictions" of the establishment.
Love it.
At this point, it's time to throw Omnibus under the bus
You have expicitly refused to engage in any conversation about relativity, repeatedly coming back to the point that you have issue with a single equation. When anybody tries to explain your misunderstanding, you refuse to accept it, stating that their explanation is invalid because you don't understand that equation.
You have to admit: it's a pretty clever tactic to have the position "I don't need to understand any of it because according to me, it's all invalid in the first place." In other words, a role reversal in which the crank claims the establishment are cranks so it's pointless arguing about the "crank predictions" of the establishment.
Love it.At this point, it's time to throw Omnibus under the bus
QUOTE (Omnibus+Oct 28 2008, 11:56 AM)
bm1957m,
What are you talking about? What one equation. Any equation describing the motion of particle of a given mass in K transforms in any system according to the first postulate retaining the same mass. On the contrary, Lorentz transformations derive any equation such that the mass in one system becomes beta^3 times that mass in another system. Any equation. That’s a fatal conflict between the first postulate and the Lorentz transformations demonstrable for any equation. For any equation. Now who’s shown no understanding, you or I?
The first postulate is given as "old" thinking.
A conflict is given.
The conflict needs to be solved by reality.
The experimental data shows to resolve the conflict is to throw out the first postulate.
(it was only put into the theory of relativity for the purposes of throwing it out.)
Einstien knew he would throw it out when he wrote the paper.
What are you talking about? What one equation. Any equation describing the motion of particle of a given mass in K transforms in any system according to the first postulate retaining the same mass. On the contrary, Lorentz transformations derive any equation such that the mass in one system becomes beta^3 times that mass in another system. Any equation. That’s a fatal conflict between the first postulate and the Lorentz transformations demonstrable for any equation. For any equation. Now who’s shown no understanding, you or I?
The first postulate is given as "old" thinking.
A conflict is given.
The conflict needs to be solved by reality.
The experimental data shows to resolve the conflict is to throw out the first postulate.
(it was only put into the theory of relativity for the purposes of throwing it out.)
Einstien knew he would throw it out when he wrote the paper.
QUOTE (NoCleverName+Oct 28 2008, 07:01 AM)
You have to admit: it's a pretty clever tactic to have the position "I don't need to understand any of it because according to me, it's all invalid in the first place." In other words, a role reversal in which the crank claims the establishment are cranks so it's pointless arguing about the "crank predictions" of the establishment. Love it.
At this point, it's time to throw Omnibus under the bus
Never mind. Don't bother.
Love it.At this point, it's time to throw Omnibus under the bus
Never mind. Don't bother.
QUOTE (buttershug+Oct 28 2008, 07:01 AM)
The first postulate is given as "old" thinking.
A conflict is given.
The conflict needs to be solved by reality.
The experimental data shows to resolve the conflict is to throw out the first postulate.
(it was only put into the theory of relativity for the purposes of throwing it out.)
Einstien knew he would throw it out when he wrote the paper.
You are not serious, are you? Throw out the first postulate and yet insists that the theory of relativity predicts outcome of experiments ... If you continue like this I'll ignore you, the way I ignored NoCleverName.
A conflict is given.
The conflict needs to be solved by reality.
The experimental data shows to resolve the conflict is to throw out the first postulate.
(it was only put into the theory of relativity for the purposes of throwing it out.)
Einstien knew he would throw it out when he wrote the paper.
You are not serious, are you? Throw out the first postulate and yet insists that the theory of relativity predicts outcome of experiments ... If you continue like this I'll ignore you, the way I ignored NoCleverName.
QUOTE (Omnibus+Oct 28 2008, 12:08 PM)
You are not serious, are you? Throw out the first postulate and yet insists that the theory of relativity predicts outcome of experiments ... If you continue like this I'll ignore you, the way I ignored NoCleverName.
Are you serious?
You throw out the results of experiments?
And it wasn't a true first postulate in the TOR.
How do you show a paradigm shift without showing the old as well as the new? I would like to know how you do that.
Are you serious?
You throw out the results of experiments?
And it wasn't a true first postulate in the TOR.
How do you show a paradigm shift without showing the old as well as the new? I would like to know how you do that.
QUOTE (Omnibus+Oct 28 2008, 08:08 AM)
You are not serious, are you? Throw out the first postulate and yet insists that the theory of relativity predicts outcome of experiments ... If you continue like this I'll ignore you, the way I ignored NoCleverName.
Oh, I'm sooo hurt. How can I go on?
P.S. Do you seriously think I care whether or not you are someday convinced or forced to retract your views? I know well you are simply playing the time-honored role of troll. And you know well that (particularly in this forum) you can run your mouth forever in the blissful knowledge that there are never any consequences to you personally, that you can get all the psychic pleasure you want. "Just say no" works on the Internet. Undoubtedly you are here because you have a nothing job and no control over your life.
Seen it before; will see it again.
Oh, I'm sooo hurt. How can I go on?
P.S. Do you seriously think I care whether or not you are someday convinced or forced to retract your views? I know well you are simply playing the time-honored role of troll. And you know well that (particularly in this forum) you can run your mouth forever in the blissful knowledge that there are never any consequences to you personally, that you can get all the psychic pleasure you want. "Just say no" works on the Internet. Undoubtedly you are here because you have a nothing job and no control over your life.
Seen it before; will see it again.
QUOTE (johanfprins+Oct 28 2008, 12:55 PM)
I only now saw your reply. The position of the clock in the primed system is at x(/)=0 and at x not equal to zero (but to the distance between the origins) relative to the unprimed system. Furthermore, the clock in the unprimed system is at x=0 and at x(/) not equal to zero (but equal to the distance between the origins measured along the negative x-axis of the primed system) relative to the primed system. You can REALLY not be so stupid can you?
Oh dear you really haven't understood me at all. I won't suggest that you must be stupid to have misunderstood me that badly, but I will think it.
I was never suggesting that x=x(/)=0 within the same co-ordinate system. My point is that if the clocks are moving (which I assume they are), then they are only at the origin of their respective co-ordinate systems for a moment of time. x=0 only holds at t=t(0), surely? So your manipulation of the transforms is only valid for t=t(0), and you can't use the deductions to make any hypothesis about what may or may not be true generally.
What you appearing to be trying to do is moving the frame with the clock at v, to validate x=0 always holding, but then making deductions about behaviour relative to a completely different frame. I'm no expert, but that doesn't seem valid to me.
How about you specify your gedanken again so we can agree on what is actually happening and what is being measured with respect to what before we carry on analysing the algebra, let's start from scratch. Happy with that?
t (as measured in the unprimed system) will always be equal to t(/) (as measured in the primed system); this is trivial and is confirmed by definition, it doesn't even need experiment. You think your algebra shows that t(/) (as measured in the unprimed frame) is equal to t (as measured in the unprimed frame). It doesn't.
Oh dear
you really haven't understood me at all. I won't suggest that you must be stupid to have misunderstood me that badly, but I will think it.I was never suggesting that x=x(/)=0 within the same co-ordinate system. My point is that if the clocks are moving (which I assume they are), then they are only at the origin of their respective co-ordinate systems for a moment of time. x=0 only holds at t=t(0), surely? So your manipulation of the transforms is only valid for t=t(0), and you can't use the deductions to make any hypothesis about what may or may not be true generally.
What you appearing to be trying to do is moving the frame with the clock at v, to validate x=0 always holding, but then making deductions about behaviour relative to a completely different frame. I'm no expert, but that doesn't seem valid to me.
How about you specify your gedanken again so we can agree on what is actually happening and what is being measured with respect to what before we carry on analysing the algebra, let's start from scratch. Happy with that?
QUOTE
The distance between the two clocks cannot have two values at any time and the positions of the clocks within their reference frames are at the origins of the reference frames. Thus when the distance has a certain value and the time is t(/) on the clock within the primed reference frame and the corresponding time on the clock within the unprimed reference frame is t then t (/) MUST BE equal to t (as mathematically proved above). Whether it is trivial or not is NOT at issue. It is a fact! If it is found experimentally not to be so, the Lorentz transformations are violated since this fact follows by correct algebra from these equations.
t (as measured in the unprimed system) will always be equal to t(/) (as measured in the primed system); this is trivial and is confirmed by definition, it doesn't even need experiment. You think your algebra shows that t(/) (as measured in the unprimed frame) is equal to t (as measured in the unprimed frame). It doesn't.
QUOTE (Omnibus+Oct 28 2008, 12:56 PM)
bm1957m,
What are you talking about? What one equation. Any equation describing the motion of particle of a given mass in K transforms in any system according to the first postulate retaining the same mass. On the contrary, Lorentz transformations derive any equation such that the mass in one system becomes beta^3 times that mass in another system. Any equation. That’s a fatal conflict between the first postulate and the Lorentz transformations demonstrable for any equation. For any equation.
Ok, I meant transform instead of equation. Sorry.
My reading of the first postulate is that it refers to a local observer, is this not the case?
The Lorentz transformation is to transform quantities for a non-local observer, is it not?
If you use the Lorentz transform for a local observer, beta=1 so there is no conflict, is there?
Still you.
What are you talking about? What one equation. Any equation describing the motion of particle of a given mass in K transforms in any system according to the first postulate retaining the same mass. On the contrary, Lorentz transformations derive any equation such that the mass in one system becomes beta^3 times that mass in another system. Any equation. That’s a fatal conflict between the first postulate and the Lorentz transformations demonstrable for any equation. For any equation.
Ok, I meant transform instead of equation. Sorry.
My reading of the first postulate is that it refers to a local observer, is this not the case?
The Lorentz transformation is to transform quantities for a non-local observer, is it not?
If you use the Lorentz transform for a local observer, beta=1 so there is no conflict, is there?
QUOTE
Now who’s shown no understanding, you or I?
Still you.
bm1957,
How is that trivial using Lorentz transformations? The opposite is true, according to these transformations: t(/) =/= t because t(/) = beta(t – vx/c^2).
That’s exactly the reason why johanfprins cannot assume that x = x(/).
Lorentz transformations cannot be abolished mathematically.
QUOTE
t (as measured in the unprimed system) will always be equal to t(/) (as measured in the primed system); this is trivial and is confirmed by definition
How is that trivial using Lorentz transformations? The opposite is true, according to these transformations: t(/) =/= t because t(/) = beta(t – vx/c^2).
That’s exactly the reason why johanfprins cannot assume that x = x(/).
Lorentz transformations cannot be abolished mathematically.
QUOTE (bm1957+Oct 28 2008, 02:23 PM)
I was never suggesting that x=x(/)=0 within the same co-ordinate system. My point is that if the clocks are moving (which I assume they are), then they are only at the origin of their respective co-ordinate systems for a moment of time. x=0 only holds at t=t(0), surely? So your manipulation of the transforms is only valid for t=t(0), and you can't use the deductions to make any hypothesis about what may or may not be true generally.
This is amazing: The main point of relativity is that two entities can ONLY move relative to each other! Within their own coordinate systems they are STATIONARY!!
Thus the clock which is stationary within the unprimed coordinate system STAYS at x=0, AND the clock which is stationary within the primed reference system STAYS at x(/)=0. You claim above that in addition to moving relative to each other they also move relative to their own coordinate systems. Can you not even see that you are talking BS.
When the origins pass the clocks are synchronized by setting t(/)(0)=t=0: I am considering the situation after a time t on the clock which stays stationary at position x=0 while the other clock moves relative to it; and at the time t(/) on the clock which stays stationary at a position x(/)=0 while the other clock is moving from it along the x(/)-axis. Specifically I consider the situation when the clocks have moved a distance d from each other: Now let me go through the derivation again:
Consider the Lorentz transformation from the primed system top the non-primed system:
x=(beta)(x(/)+vt(/)) and t=(beta)(t(/)+(v/c^2)x(/))
The position of the clock in the primed system is at its origin: i.e. x(/)=0; while relative to the unprimed system at it is at position x=d: Thus the equations become:
d=(beta)(vt(/)) and t=(beta)t(/): Now use the second equation to replace t(/) in the first equation: One then obtains that:
d=vt where d is the distance between the origins of the primed and unprimed systems.
Now consider the Lorentz transformation from the unprimed to the primed system:
x(/)=(beta)(x-vt) and t(/)=(beta)(t-(v/c^2)x). The clock in the unprimed system is at its origin so that x=0; while relative to the primed system it is at position x(/)=-d: Thus the equations become:
-d=(beta)(-vt) and t(/)=(beta)t: Now use t in the second equation to replace t in the first equation; and one obtains that:
d =vt(/) where d is now the distance between the origins of the two reference frames. Since d=vt and d=vt(/) and v=v; one obtains that t(/)=t.
This is amazing: The main point of relativity is that two entities can ONLY move relative to each other! Within their own coordinate systems they are STATIONARY!!
Thus the clock which is stationary within the unprimed coordinate system STAYS at x=0, AND the clock which is stationary within the primed reference system STAYS at x(/)=0. You claim above that in addition to moving relative to each other they also move relative to their own coordinate systems. Can you not even see that you are talking BS.
When the origins pass the clocks are synchronized by setting t(/)(0)=t=0: I am considering the situation after a time t on the clock which stays stationary at position x=0 while the other clock moves relative to it; and at the time t(/) on the clock which stays stationary at a position x(/)=0 while the other clock is moving from it along the x(/)-axis. Specifically I consider the situation when the clocks have moved a distance d from each other: Now let me go through the derivation again:
Consider the Lorentz transformation from the primed system top the non-primed system:
x=(beta)(x(/)+vt(/)) and t=(beta)(t(/)+(v/c^2)x(/))
The position of the clock in the primed system is at its origin: i.e. x(/)=0; while relative to the unprimed system at it is at position x=d: Thus the equations become:
d=(beta)(vt(/)) and t=(beta)t(/): Now use the second equation to replace t(/) in the first equation: One then obtains that:
d=vt where d is the distance between the origins of the primed and unprimed systems.
Now consider the Lorentz transformation from the unprimed to the primed system:
x(/)=(beta)(x-vt) and t(/)=(beta)(t-(v/c^2)x). The clock in the unprimed system is at its origin so that x=0; while relative to the primed system it is at position x(/)=-d: Thus the equations become:
-d=(beta)(-vt) and t(/)=(beta)t: Now use t in the second equation to replace t in the first equation; and one obtains that:
d =vt(/) where d is now the distance between the origins of the two reference frames. Since d=vt and d=vt(/) and v=v; one obtains that t(/)=t.
QUOTE (Omnibus+Oct 28 2008, 01:43 PM)
How is that trivial using Lorentz transformations?
It is trivial because they are identical clocks, so they both have the same proper time.
And when v=0, this is equivalent to t(/)=t, which is essentially what he has shown. That the local measurement of a clock rate (from a co-moving frame) will not change.
And when v=0, this is equivalent to t(/)=t, which is essentially what he has shown. That the local measurement of a clock rate (from a co-moving frame) will not change.
That’s exactly the reason why johanfprins cannot assume that x = x(/).
Huh? You're way out of your depth. Please answer my question about the first postulate and your interpretation of it.
Now I think you're just taking the p!ss. What does that even mean?
No, you’re confused. Lorentz transformations and first postulate are two different ways of transforming a given equation. They must yield the same result (which they don’t, as I’ve shown). Read, for instance, §3 of the 1905, especially that part:
“We now have to prove that any ray of light, measured in the moving system, is propagated with the velocity c, if, as we have assumed, this is the case in the stationary system; for we have not as yet furnished the proof that the principle of the constancy of the velocity of light is compatible with the principle of relativity.”
There, in §3 Einstein thinks he has demonstrated the compatibility of the two principles when applied to a spherical wave (he hasn’t but that’s a separate discussion). As we saw, however, in §10 the discrepancy is blatant. He hasn’t noticed it because otherwise he would not have even published the paper. Publishing an invalid theory is worthless.
My reading of the first postulate is that it refers to a local observer, is this not the case?
By local do you mean in the same frame?
If so then you can't have movement and the equation means nothing.
With this you’re confirming exactly what myself and johnfprins have been insisting all along—time in each system is as the time in any other system when observed from within that system. Therefore, Lorentz transformations are only illusions about that time in each system. That’s not what the zealots think, though, because in that case (which is the truth) Physics has no use for the Lorentz transformations. The Lorentz transformations afficionados claim, ridiculous as it is, that the time in k, from within k, is actually running at a different rate than the time in K.
That the latter is not the case cannot be proven through only observing Lorentz transformations as only mathematical constructs, as johnfprins tries to do, mindless of their physical meaning. Lorentz transformations are without a doubt mathematically consistent and there’s no way to disprove their mathematical validity. Lorentz transformations have no physical meaning and that can be shown in various ways. I have shown one crucial way to prove their physical invalidity.
It is trivial because they are identical clocks, so they both have the same proper time.
QUOTE
The opposite is true, according to these transformations: t(/) =/= t because t(/) = beta(t – vx/c^2).
And when v=0, this is equivalent to t(/)=t, which is essentially what he has shown. That the local measurement of a clock rate (from a co-moving frame) will not change.
QUOTE (->
| QUOTE |
| The opposite is true, according to these transformations: t(/) =/= t because t(/) = beta(t – vx/c^2). |
And when v=0, this is equivalent to t(/)=t, which is essentially what he has shown. That the local measurement of a clock rate (from a co-moving frame) will not change.
That’s exactly the reason why johanfprins cannot assume that x = x(/).
Huh? You're way out of your depth. Please answer my question about the first postulate and your interpretation of it.
QUOTE
Lorentz transformations cannot be abolished mathematically.
Now I think you're just taking the p!ss. What does that even mean?
QUOTE
Ok, I meant transform instead of equation. Sorry.
My reading of the first postulate is that it refers to a local observer, is this not the case?
The Lorentz transformation is to transform quantities for a non-local observer, is it not?
If you use the Lorentz transform for a local observer, beta=1 so there is no conflict, is there?
My reading of the first postulate is that it refers to a local observer, is this not the case?
The Lorentz transformation is to transform quantities for a non-local observer, is it not?
If you use the Lorentz transform for a local observer, beta=1 so there is no conflict, is there?
No, you’re confused. Lorentz transformations and first postulate are two different ways of transforming a given equation. They must yield the same result (which they don’t, as I’ve shown). Read, for instance, §3 of the 1905, especially that part:
“We now have to prove that any ray of light, measured in the moving system, is propagated with the velocity c, if, as we have assumed, this is the case in the stationary system; for we have not as yet furnished the proof that the principle of the constancy of the velocity of light is compatible with the principle of relativity.”
There, in §3 Einstein thinks he has demonstrated the compatibility of the two principles when applied to a spherical wave (he hasn’t but that’s a separate discussion). As we saw, however, in §10 the discrepancy is blatant. He hasn’t noticed it because otherwise he would not have even published the paper. Publishing an invalid theory is worthless.
QUOTE (bm1957+Oct 28 2008, 12:38 PM)
My reading of the first postulate is that it refers to a local observer, is this not the case?
By local do you mean in the same frame?
If so then you can't have movement and the equation means nothing.
QUOTE (bm1957+Oct 28 2008, 02:49 PM)
And when v=0, this is equivalent to t(/)=t, which is essentially what he has shown. That the local measurement of a clock rate (from a co-moving frame) will not change.
That is NOT what I have proved: See the simplified proof above: I have proved that two clocks moving with a constant speed relative to each other will stay synchronized by ticking at the same rate!
That is NOT what I have proved: See the simplified proof above: I have proved that two clocks moving with a constant speed relative to each other will stay synchronized by ticking at the same rate!
QUOTE (johanfprins+Oct 28 2008, 08:55 AM)
That is NOT what I have proved: See the simplified proof above: I have proved that two clocks moving with a constant speed relative to each other will stay synchronized by ticking at the same rate!
That this is so is trivial.
Now, bring the clocks together (in the same frame, at rest). What do they say?
That this is so is trivial.
Now, bring the clocks together (in the same frame, at rest). What do they say?
QUOTE
It is trivial because they are identical clocks, so they both have the same proper time.
With this you’re confirming exactly what myself and johnfprins have been insisting all along—time in each system is as the time in any other system when observed from within that system. Therefore, Lorentz transformations are only illusions about that time in each system. That’s not what the zealots think, though, because in that case (which is the truth) Physics has no use for the Lorentz transformations. The Lorentz transformations afficionados claim, ridiculous as it is, that the time in k, from within k, is actually running at a different rate than the time in K.
That the latter is not the case cannot be proven through only observing Lorentz transformations as only mathematical constructs, as johnfprins tries to do, mindless of their physical meaning. Lorentz transformations are without a doubt mathematically consistent and there’s no way to disprove their mathematical validity. Lorentz transformations have no physical meaning and that can be shown in various ways. I have shown one crucial way to prove their physical invalidity.
QUOTE (NoCleverName+Oct 28 2008, 02:59 PM)
That this is so is trivial.
Now, bring the clocks together (in the same frame, at rest). What do they say?
To do that you have to accelerate the clocks. If you accelerate both equally until they reach each other (i.e. the clock in the unprimed system accelerate with acceleration a and the clock in the primed system decelerate with a deceleration a) you must find that the clocks will still read the same when they reach other. Only when one clock is accelerated and the other not will one, according to general relativity, measure different times.
Now, bring the clocks together (in the same frame, at rest). What do they say?
To do that you have to accelerate the clocks. If you accelerate both equally until they reach each other (i.e. the clock in the unprimed system accelerate with acceleration a and the clock in the primed system decelerate with a deceleration a) you must find that the clocks will still read the same when they reach other. Only when one clock is accelerated and the other not will one, according to general relativity, measure different times.
QUOTE (johanfprins+Oct 28 2008, 01:45 PM)
This is amazing: The main point of relativity is that two entities can ONLY move relative to each other! Within their own coordinate systems they are STATIONARY!!
Thus the clock which is stationary within the unprimed coordinate system STAYS at x=0, AND the clock which is stationary within the primed reference system STAYS at x(/)=0. You claim above that in addition to moving relative to each other they also move relative to their own coordinate systems. Can you not even see that you are talking BS.
When the origins pass the clocks are synchronized by setting t(/)(0)=t=0: I am considering the situation after a time t on the clock which stays stationary at position x=0 while the other clock moves relative to it; and at the time t(/) on the clock which stays stationary at a position x(/)=0 while the other clock is moving from it along the x(/)-axis. Specifically I consider the situation when the clocks have moved a distance d from each other: Now let me go through the derivation again:
Yes, I can see where I was talking BS, I must have got myself confused.
But, I have seen where your blunder is, and it is certainly a whopper!!!
Here you are trying to calculate the position of the unprimed clock a seen by the primed clock. The position of the primed clock relative to itself is irrelevant and certainly x(/)<>0. In fact, x(/)=-d=vt (I think).
You have also stated that you are considering the situation after time t on the unprimed clock, hence t=t and x=x, your transforms are unnecessary and are calculating what the primed clock sees, contrary to the situation you claimed you were considering.
Here you are trying to calculate the position of the unprimed clock a seen by the primed clock. The position of the primed clock relative to itself is irrelevant and certainly x(/)<>0. In fact, x(/)=-d=vt (I think).
You have also stated that you are considering the situation after time t on the unprimed clock, hence t=t and x=x, your transforms are unnecessary and are calculating what the primed clock sees, contrary to the situation you claimed you were considering.
d=(beta)(vt(/)) and t=(beta)t(/): Now use the second equation to replace t(/) in the first equation: One then obtains that:
d=vt where d is the distance between the origins of the primed and unprimed systems.
Invalid due to above. Happy that d=vt (if t is as measured on the unprimed clock and d is the distance measured by the unprimed clock.)
Again; here you are trying to calculate the position of the primed clock as seen by the unprimed clock. The position of the unprimed clock relative to itself is irrelevant and x<>0. In fact, x=d=vt (I'm pretty sure of that now.)
Again; here you are trying to calculate the position of the primed clock as seen by the unprimed clock. The position of the unprimed clock relative to itself is irrelevant and x<>0. In fact, x=d=vt (I'm pretty sure of that now.)
: Thus the equations become:
-d=(beta)(-vt) and t(/)=(beta)t: Now use t in the second equation to replace t in the first equation; and one obtains that:
d =vt(/) where d is now the distance between the origins of the two reference frames. Since d=vt and d=vt(/) and v=v; one obtains that t(/)=t.
Again, further manipulation is invalid. Also, happy that d=vt(/) (if t(/) is as measured on the primed clock and d is the distance measured by the primed clock.)
So, you end up with t(/)=t, where t(/) is presumably the time the unprimed clock sees on the primed clock and t is the time that the primed clock sees on the unprimed clock.
It looks like you've used the transforms incorrectly, and through poor choice of variables ended up showing that the time A sees at B is the same as the time B sees at A.
You certainly haven't come close to showing that the time A sees at A is equal to the time A sees at B, if that is what you were trying to do.
Thus the clock which is stationary within the unprimed coordinate system STAYS at x=0, AND the clock which is stationary within the primed reference system STAYS at x(/)=0. You claim above that in addition to moving relative to each other they also move relative to their own coordinate systems. Can you not even see that you are talking BS.
When the origins pass the clocks are synchronized by setting t(/)(0)=t=0: I am considering the situation after a time t on the clock which stays stationary at position x=0 while the other clock moves relative to it; and at the time t(/) on the clock which stays stationary at a position x(/)=0 while the other clock is moving from it along the x(/)-axis. Specifically I consider the situation when the clocks have moved a distance d from each other: Now let me go through the derivation again:
Yes, I can see where I was talking BS, I must have got myself confused.
But, I have seen where your blunder is, and it is certainly a whopper!!!
QUOTE
Consider the Lorentz transformation from the primed system top the non-primed system:
x=(beta)(x(/)+vt(/)) and t=(beta)(t(/)+(v/c^2)x(/))
The position of the clock in the primed system is at its origin: i.e. x(/)=0; while relative to the unprimed system at it is at position x=d: Thus the equations become:
x=(beta)(x(/)+vt(/)) and t=(beta)(t(/)+(v/c^2)x(/))
The position of the clock in the primed system is at its origin: i.e. x(/)=0; while relative to the unprimed system at it is at position x=d: Thus the equations become:
Here you are trying to calculate the position of the unprimed clock a seen by the primed clock. The position of the primed clock relative to itself is irrelevant and certainly x(/)<>0. In fact, x(/)=-d=vt (I think).
You have also stated that you are considering the situation after time t on the unprimed clock, hence t=t and x=x, your transforms are unnecessary and are calculating what the primed clock sees, contrary to the situation you claimed you were considering.
QUOTE (->
| QUOTE |
| Consider the Lorentz transformation from the primed system top the non-primed system: x=(beta)(x(/)+vt(/)) and t=(beta)(t(/)+(v/c^2)x(/)) The position of the clock in the primed system is at its origin: i.e. x(/)=0; while relative to the unprimed system at it is at position x=d: Thus the equations become: |
Here you are trying to calculate the position of the unprimed clock a seen by the primed clock. The position of the primed clock relative to itself is irrelevant and certainly x(/)<>0. In fact, x(/)=-d=vt (I think).
You have also stated that you are considering the situation after time t on the unprimed clock, hence t=t and x=x, your transforms are unnecessary and are calculating what the primed clock sees, contrary to the situation you claimed you were considering.
d=(beta)(vt(/)) and t=(beta)t(/): Now use the second equation to replace t(/) in the first equation: One then obtains that:
d=vt where d is the distance between the origins of the primed and unprimed systems.
Invalid due to above. Happy that d=vt (if t is as measured on the unprimed clock and d is the distance measured by the unprimed clock.)
QUOTE
Now consider the Lorentz transformation from the unprimed to the primed system:
x(/)=(beta)(x-vt) and t(/)=(beta)(t-(v/c^2)x). The clock in the unprimed system is at its origin so that x=0; while relative to the primed system it is at position x(/)=-d
x(/)=(beta)(x-vt) and t(/)=(beta)(t-(v/c^2)x). The clock in the unprimed system is at its origin so that x=0; while relative to the primed system it is at position x(/)=-d
Again; here you are trying to calculate the position of the primed clock as seen by the unprimed clock. The position of the unprimed clock relative to itself is irrelevant and x<>0. In fact, x=d=vt (I'm pretty sure of that now.)
QUOTE (->
| QUOTE |
| Now consider the Lorentz transformation from the unprimed to the primed system: x(/)=(beta)(x-vt) and t(/)=(beta)(t-(v/c^2)x). The clock in the unprimed system is at its origin so that x=0; while relative to the primed system it is at position x(/)=-d |
Again; here you are trying to calculate the position of the primed clock as seen by the unprimed clock. The position of the unprimed clock relative to itself is irrelevant and x<>0. In fact, x=d=vt (I'm pretty sure of that now.)
: Thus the equations become:
-d=(beta)(-vt) and t(/)=(beta)t: Now use t in the second equation to replace t in the first equation; and one obtains that:
d =vt(/) where d is now the distance between the origins of the two reference frames. Since d=vt and d=vt(/) and v=v; one obtains that t(/)=t.
Again, further manipulation is invalid. Also, happy that d=vt(/) (if t(/) is as measured on the primed clock and d is the distance measured by the primed clock.)
So, you end up with t(/)=t, where t(/) is presumably the time the unprimed clock sees on the primed clock and t is the time that the primed clock sees on the unprimed clock.
It looks like you've used the transforms incorrectly, and through poor choice of variables ended up showing that the time A sees at B is the same as the time B sees at A.
You certainly haven't come close to showing that the time A sees at A is equal to the time A sees at B, if that is what you were trying to do.
QUOTE (Omnibus+Oct 28 2008, 03:04 PM)
Therefore, Lorentz transformations are only illusions about that time in each system. That’s not what the zealots think, though, because in that case (which is the truth) Physics has no use for the Lorentz transformations. The Lorentz transformations afficionados claim, ridiculous as it is, that the time in k, from within k, is actually running at a different rate than the time in K.
.
I have decided at a point not to respond to Omnibus anymore: However, what he claims above is correct. The only difference is that "relativistic illusions are real" for the observer who encounters them. This sounds funny but is not really. Consider a cannonball being launched vertically within a coordinate system that moves relative to an observer. The observer will see the cannonball following a parabolic path while within the reference frame in which it had been launched, it moves straight up and straight down. The parabolic path that the observer experiences is real: If he gets in the way of that path he will be killed.
The fact that "movement" leads to relativistic "illusions" have already been noted by Zeno: The equations describing these illusions however models real physical effects. Therefore the Lorentz transformations cannot be discarded even though the time within each inertial reference frame is actually ticking away at the same rate. Fortunately, this fact can be derived from the Lorentz transformations as I have illustrated above. If this could not have been done; then I would agree with Omnibus that these equations will have to be discarded.
.
I have decided at a point not to respond to Omnibus anymore: However, what he claims above is correct. The only difference is that "relativistic illusions are real" for the observer who encounters them. This sounds funny but is not really. Consider a cannonball being launched vertically within a coordinate system that moves relative to an observer. The observer will see the cannonball following a parabolic path while within the reference frame in which it had been launched, it moves straight up and straight down. The parabolic path that the observer experiences is real: If he gets in the way of that path he will be killed.
The fact that "movement" leads to relativistic "illusions" have already been noted by Zeno: The equations describing these illusions however models real physical effects. Therefore the Lorentz transformations cannot be discarded even though the time within each inertial reference frame is actually ticking away at the same rate. Fortunately, this fact can be derived from the Lorentz transformations as I have illustrated above. If this could not have been done; then I would agree with Omnibus that these equations will have to be discarded.
bm1957, you are confused. Drop posting for a while and give it some thought. How come:
How is the position of the primed clock x(/)<>0 since johanfprons tells you, as the initial condition, that it is x(/) = 0. What are you talking about?
QUOTE
Here you are trying to calculate the position of the unprimed clock a seen by the primed clock. The position of the primed clock relative to itself is irrelevant and certainly x(/)<>0. In fact, x(/)=-d=vt (I think).
How is the position of the primed clock x(/)<>0 since johanfprons tells you, as the initial condition, that it is x(/) = 0. What are you talking about?
QUOTE (Omnibus+Oct 28 2008, 11:28 AM)
buttershug,
How come that’s me, by the way? The crank is the person who says things that don’t match observed results such as the fact that first postulate requires m to be the same m in every system while Lorentz transformation derive that m in one system is beta^3m in another system, denying that therefore first postulate and the Lorentz transformations are in conflict, is the crank. Denying the obvious makes one a crank.
Now,now. You pulled your tutu over your eyes and you refuse to see the reality.
First and foremost, the first postulate does not "requires m to be the same m in every system". So, this is your first mistake.
Secondly, m_0, i.e. proper mass is what is indeed frame invariant. Relativistic mass is not frame invariant (it is dependent on gamma=1/sqrt(1-(v/c)^2), i.e. it is speed dependent and we already know that in relativity speed is, well....relative!). This is your second mistake.
So, have you given more thought about taking up underwater tennis as a hobby? Denying relativity is not going very well for you.
How come that’s me, by the way? The crank is the person who says things that don’t match observed results such as the fact that first postulate requires m to be the same m in every system while Lorentz transformation derive that m in one system is beta^3m in another system, denying that therefore first postulate and the Lorentz transformations are in conflict, is the crank. Denying the obvious makes one a crank.
Now,now. You pulled your tutu over your eyes and you refuse to see the reality.
First and foremost, the first postulate does not "requires m to be the same m in every system". So, this is your first mistake.
Secondly, m_0, i.e. proper mass is what is indeed frame invariant. Relativistic mass is not frame invariant (it is dependent on gamma=1/sqrt(1-(v/c)^2), i.e. it is speed dependent and we already know that in relativity speed is, well....relative!). This is your second mistake.
So, have you given more thought about taking up underwater tennis as a hobby? Denying relativity is not going very well for you.
QUOTE (Omnibus+Oct 28 2008, 01:53 PM)
No, you’re confused. Lorentz transformations and first postulate are two different ways of transforming a given equation. They must yield the same result (which they don’t, as I’ve shown). Read, for instance, §3 of the 1905, especially that part:
“We now have to prove that any ray of light, measured in the moving system, is propagated with the velocity c, if, as we have assumed, this is the case in the stationary system; for we have not as yet furnished the proof that the principle of the constancy of the velocity of light is compatible with the principle of relativity.”
There, in §3 Einstein thinks he has demonstrated the compatibility of the two principles when applied to a spherical wave (he hasn’t but that’s a separate discussion). As we saw, however, in §10 the discrepancy is blatant. He hasn’t noticed it because otherwise he would not have even published the paper. Publishing an invalid theory is worthless.
Will you answer my questions?
“We now have to prove that any ray of light, measured in the moving system, is propagated with the velocity c, if, as we have assumed, this is the case in the stationary system; for we have not as yet furnished the proof that the principle of the constancy of the velocity of light is compatible with the principle of relativity.”
There, in §3 Einstein thinks he has demonstrated the compatibility of the two principles when applied to a spherical wave (he hasn’t but that’s a separate discussion). As we saw, however, in §10 the discrepancy is blatant. He hasn’t noticed it because otherwise he would not have even published the paper. Publishing an invalid theory is worthless.
Will you answer my questions?
QUOTE
My reading of the first postulate is that it refers to a local observer, is this not the case?
The Lorentz transformation is to transform quantities for a non-local observer, is it not?
If you use the Lorentz transform for a local observer, beta=1 so there is no conflict, is there?
The Lorentz transformation is to transform quantities for a non-local observer, is it not?
If you use the Lorentz transform for a local observer, beta=1 so there is no conflict, is there?
QUOTE (buttershug+Oct 28 2008, 01:55 PM)
By local do you mean in the same frame?
Yes.
I disagree.
Yes.
QUOTE
If so then you can't have movement and the equation means nothing.
I disagree.
johafprins, it's immaterial that someone may claim that for him in his own system that's how the things that are happening in the other system look like. No zealot would agree that that's what Lorentz transformations provide because Physics has no use for it. What Physics looks for is to determine the real truth from within a system. Lorentz transformations don't provide that real truth but your way of proving it isn't doing it. Like I said, when you derive that x = vt and x(/) = vt(/) you unfoundedly accept that x = x(/). That is not true because x is in K while x(/) is in k and exactly due to the Lorentz transformations x and x(/) must be different.
Lorentz transformations cannot be abolished within their own mathematical framework because they are mathematically consistent beyond any doubt. Trying to abolish them mathematically only turns you into an easy prey for the zealots who only wait for such prey to come their way to gladly shoot it down. They love 'critics' like you.
Lorentz transformations cannot be abolished within their own mathematical framework because they are mathematically consistent beyond any doubt. Trying to abolish them mathematically only turns you into an easy prey for the zealots who only wait for such prey to come their way to gladly shoot it down. They love 'critics' like you.
QUOTE (Omnibus+Oct 28 2008, 02:04 PM)
With this you’re confirming exactly what myself and johnfprins have been insisting all along—time in each system is as the time in any other system when observed from within that system. Therefore, Lorentz transformations are only illusions about that time in each system.
That's exactly what any 'real' scientist will tell you, and I don't believe that is what johan is trying to show.
SO wrong.
SO wrong.
The Lorentz transformations afficionados claim, ridiculous as it is, that the time in k, from within k, is actually running at a different rate than the time in K.
More nonsense.
No, no and no. Like I said, both transformations are just different ways of doing the same thing for the same observer. Therefore, whatever the outcome from these transformations is, it refers to exactly the same thing. Read the paragraph I quoted, read also §6 of the 1905 paper and understand carefully why Einstein says: “Evidently the two systems of equations found for system k must express exactly the same thing, since both systems of equations are equivalent to the Maxwell-Hertz equations for system K.”
Here’s your whopper:
d=vt is “relative to the unprimed system”.
v(t/) is “relative to the primed system”, thus you have to denote it as d(/) = vt(/) and not use the Einsteineque manner of sneaking the notation which suits you but doesn’t correspond to the reality of the situation.
Therefore, d =/= d(/).
In order for d to be equal to d(/) you must have t = t(/) but this is exactly what you were supposed to prove.
Have you heard of rhetorical questions?
Have you heard of rhetorical questions?
How is the position of the primed clock x(/)<>0 since johanfprons tells you, as the initial condition, that it is x(/) = 0. What are you talking about?
The position of the primed clock is at x(/)=0. But, in the equation he uses, x(/) represents the position of the primed clock with respect to the unprimed clock, which is certainly not equla to 0.
Like I said, you are confused. Primed quantities are always referring to the primed system. Take some time to learn these things before posting.
That's exactly what any 'real' scientist will tell you, and I don't believe that is what johan is trying to show.
QUOTE
That’s not what the zealots think, though, because in that case (which is the truth) Physics has no use for the Lorentz transformations.
SO wrong.
QUOTE (->
| QUOTE |
| That’s not what the zealots think, though, because in that case (which is the truth) Physics has no use for the Lorentz transformations. |
SO wrong.
The Lorentz transformations afficionados claim, ridiculous as it is, that the time in k, from within k, is actually running at a different rate than the time in K.
QUOTE
That the latter is not the case cannot be proven through only observing Lorentz transformations as only mathematical constructs, as johnfprins tries to do, mindless of their physical meaning. Lorentz transformations are without a doubt mathematically consistent and there’s no way to disprove their mathematical validity. Lorentz transformations have no physical meaning and that can be shown in various ways. I have shown one crucial way to prove their physical invalidity.
More nonsense.
QUOTE
My reading of the first postulate is that it refers to a local observer, is this not the case?
The Lorentz transformation is to transform quantities for a non-local observer, is it not?
If you use the Lorentz transform for a local observer, beta=1 so there is no conflict, is there?
The Lorentz transformation is to transform quantities for a non-local observer, is it not?
If you use the Lorentz transform for a local observer, beta=1 so there is no conflict, is there?
No, no and no. Like I said, both transformations are just different ways of doing the same thing for the same observer. Therefore, whatever the outcome from these transformations is, it refers to exactly the same thing. Read the paragraph I quoted, read also §6 of the 1905 paper and understand carefully why Einstein says: “Evidently the two systems of equations found for system k must express exactly the same thing, since both systems of equations are equivalent to the Maxwell-Hertz equations for system K.”
QUOTE (bm1957+Oct 28 2008, 03:23 PM)
Yes, I can see where I was talking BS, I must have got myself confused.
But, I have seen where your blunder is, and it is certainly a whopper!!!
It looks like you've used the transforms incorrectly, and through poor choice of variables ended up showing that the time A sees at B is the same as the time B sees at A.
You certainly haven't come close to showing that the time A sees at A is equal to the time A sees at B, if that is what you were trying to do.
Thanks for admitting your BS: This indicates that we might be able to come to an agreement based on logic:
Unfortunately you then proceeded with some more BS. I did NOT make a mistake; The Lorentztransformations are coordinate transformations from one coordinate system (unprimed) with its origin at x=0 to another coordinate system whose origin at a time t>0 (at a clock at its origin: clock 1) after synchronization is at a position x>0. Similarly one also has a coordinate transformation from the primed coordinate system whose origin relative to its own system MUST BE and REMAINS at x(/)=0, while after synchronization at a time t(/) on its clock at its origin the coordinate of the clock (clock 2) within the unprimed system relative to the primed coordinate system is x(/)<0 (i.e. the unprimed system is moving in the negative dirction relative to the primed system): Thus we have the following coordinate situations for:
1. the clock in the primed system: Relative to the primed system at any time
t(/) one has that x(/)=0 : Relative to the unprimed system its corresponding coordinates must be t and x: To calculate these coordinates one has to transform the coordinates x(/)=0, t(/) from the primed system into the corresponding coordinates x, t of the unprimed system using the Lorentz transformations: One ends up with a relationship between the distance d between the clocks (which is obviously also the distance between the origins of the two systems) d=vt:
2. The clock in the unprimed primed system: Relative to the unprimed system at any time t) one has that its postion coordinate is x=0 : Relative to the primed system its corresponding coordinates must be t(/) and x(/): To calculate these coordinates one has to transform the coordinates t, x=0 from the unprimed system into the corresponding coordinates x(/) and t(/) within the primed system using the Lorentz transformations: One ends up with a relationship between the distance d between the clocks (which is obviously also the distance between the origins of the two systems) -d=-vt(/):
Both transformations are correct and must thus correlate for physics to be consistent: This leads to t=t(/). So where is my whopper!?
Obviously I have not tried to prove "that the time A sees at A is equal to the time A sees at B". What I have proved is that the time A sees at A is the same as the time that B sees at B. you will now probably say it is trivial: Even if it is why are you not able to follow the proof?
But, I have seen where your blunder is, and it is certainly a whopper!!!
It looks like you've used the transforms incorrectly, and through poor choice of variables ended up showing that the time A sees at B is the same as the time B sees at A.
You certainly haven't come close to showing that the time A sees at A is equal to the time A sees at B, if that is what you were trying to do.
Thanks for admitting your BS: This indicates that we might be able to come to an agreement based on logic:
Unfortunately you then proceeded with some more BS. I did NOT make a mistake; The Lorentztransformations are coordinate transformations from one coordinate system (unprimed) with its origin at x=0 to another coordinate system whose origin at a time t>0 (at a clock at its origin: clock 1) after synchronization is at a position x>0. Similarly one also has a coordinate transformation from the primed coordinate system whose origin relative to its own system MUST BE and REMAINS at x(/)=0, while after synchronization at a time t(/) on its clock at its origin the coordinate of the clock (clock 2) within the unprimed system relative to the primed coordinate system is x(/)<0 (i.e. the unprimed system is moving in the negative dirction relative to the primed system): Thus we have the following coordinate situations for:
1. the clock in the primed system: Relative to the primed system at any time
t(/) one has that x(/)=0 : Relative to the unprimed system its corresponding coordinates must be t and x: To calculate these coordinates one has to transform the coordinates x(/)=0, t(/) from the primed system into the corresponding coordinates x, t of the unprimed system using the Lorentz transformations: One ends up with a relationship between the distance d between the clocks (which is obviously also the distance between the origins of the two systems) d=vt:
2. The clock in the unprimed primed system: Relative to the unprimed system at any time t) one has that its postion coordinate is x=0 : Relative to the primed system its corresponding coordinates must be t(/) and x(/): To calculate these coordinates one has to transform the coordinates t, x=0 from the unprimed system into the corresponding coordinates x(/) and t(/) within the primed system using the Lorentz transformations: One ends up with a relationship between the distance d between the clocks (which is obviously also the distance between the origins of the two systems) -d=-vt(/):
Both transformations are correct and must thus correlate for physics to be consistent: This leads to t=t(/). So where is my whopper!?
Obviously I have not tried to prove "that the time A sees at A is equal to the time A sees at B". What I have proved is that the time A sees at A is the same as the time that B sees at B. you will now probably say it is trivial: Even if it is why are you not able to follow the proof?
QUOTE (johanfprins+Oct 28 2008, 02:25 PM)
I have decided at a point not to respond to Omnibus anymore: However, what he claims above is correct. The only difference is that "relativistic illusions are real" for the observer who encounters them. This sounds funny but is not really. Consider a cannonball being launched vertically within a coordinate system that moves relative to an observer. The observer will see the cannonball following a parabolic path while within the reference frame in which it had been launched, it moves straight up and straight down. The parabolic path that the observer experiences is real: If he gets in the way of that path he will be killed.
So if you can accept that, why do you not accept that time dilation is a real effect?
So if you can accept that, why do you not accept that time dilation is a real effect?
QUOTE (Omnibus+Oct 28 2008, 02:29 PM)
bm1957, you are confused. Drop posting for a while and give it some thought. How come:
You ask a question but don't want me to post? Hmm...
The position of the primed clock is at x(/)=0. But, in the equation he uses, x(/) represents the position of the primed clock with respect to the unprimed clock, which is certainly not equla to 0.
You ask a question but don't want me to post? Hmm...
QUOTE
How is the position of the primed clock x(/)<>0 since johanfprons tells you, as the initial condition, that it is x(/) = 0. What are you talking about?
The position of the primed clock is at x(/)=0. But, in the equation he uses, x(/) represents the position of the primed clock with respect to the unprimed clock, which is certainly not equla to 0.
QUOTE
Both transformations are correct and must thus correlate for physics to be consistent: This leads to t=t(/). So where is my whopper!?
Here’s your whopper:
d=vt is “relative to the unprimed system”.
v(t/) is “relative to the primed system”, thus you have to denote it as d(/) = vt(/) and not use the Einsteineque manner of sneaking the notation which suits you but doesn’t correspond to the reality of the situation.
Therefore, d =/= d(/).
In order for d to be equal to d(/) you must have t = t(/) but this is exactly what you were supposed to prove.
QUOTE (bm1957+Oct 28 2008, 03:56 PM)
So if you can accept that, why do you not accept that time dilation is a real effect?
I have not said that time dilation is not a real effect: It is a "real relativistic effect" in the sense that an observer A concludes that time dilates at the position of an observer B while in actual fact the time on the clock observed by observer B is changing at the same rate as the time on A's clock. Similarly it is a "real relativistic effect" in the sense that an observer B concludes that time dilates at the position of an observer A while in actual fact the time on the clock observed by observer A is changing at the same rate as the time on B's clock.
These perceptions are created by relative movement and must be taken into account when actions within one system can be observed from the other system. However, as Feynman said, if you cannot "look out", each system will be the same: i.e. a clock in system A will tick away at the same rate as in system B. What my derivation proves is that when you have an interaction from A observed from B and the opposite interaction in B observed from A, you get the exact same time rate as will be the case when you cannot "look out". This is exactly as it must be. This is another feather in the cap for Special Relativity!
Omnibus thus have some valid points: Relativistic effects are in effect "illusions" but these "illusions" have real physical effects for the observer who is experiencing them: And in order to model these REAL PHYSICAL effects, the Lorentz transformations have withstood the test of time!
I have not said that time dilation is not a real effect: It is a "real relativistic effect" in the sense that an observer A concludes that time dilates at the position of an observer B while in actual fact the time on the clock observed by observer B is changing at the same rate as the time on A's clock. Similarly it is a "real relativistic effect" in the sense that an observer B concludes that time dilates at the position of an observer A while in actual fact the time on the clock observed by observer A is changing at the same rate as the time on B's clock.
These perceptions are created by relative movement and must be taken into account when actions within one system can be observed from the other system. However, as Feynman said, if you cannot "look out", each system will be the same: i.e. a clock in system A will tick away at the same rate as in system B. What my derivation proves is that when you have an interaction from A observed from B and the opposite interaction in B observed from A, you get the exact same time rate as will be the case when you cannot "look out". This is exactly as it must be. This is another feather in the cap for Special Relativity!
Omnibus thus have some valid points: Relativistic effects are in effect "illusions" but these "illusions" have real physical effects for the observer who is experiencing them: And in order to model these REAL PHYSICAL effects, the Lorentz transformations have withstood the test of time!
QUOTE
bm1957, you are confused. Drop posting for a while and give it some thought. How come:
You ask a question but don't want me to post? Hmm...
You ask a question but don't want me to post? Hmm...
Have you heard of rhetorical questions?
QUOTE (->
| QUOTE |
| bm1957, you are confused. Drop posting for a while and give it some thought. How come: You ask a question but don't want me to post? Hmm... |
Have you heard of rhetorical questions?
How is the position of the primed clock x(/)<>0 since johanfprons tells you, as the initial condition, that it is x(/) = 0. What are you talking about?
The position of the primed clock is at x(/)=0. But, in the equation he uses, x(/) represents the position of the primed clock with respect to the unprimed clock, which is certainly not equla to 0.
Like I said, you are confused. Primed quantities are always referring to the primed system. Take some time to learn these things before posting.
QUOTE (Omnibus+Oct 28 2008, 02:49 PM)
No, no and no.
You are answering no to all questions?
In that case, I think you are beyond help, I give up.
You are answering no to all questions?
In that case, I think you are beyond help, I give up.
QUOTE (bm1957+Oct 28 2008, 04:00 PM)
The position of the primed clock is at x(/)=0. But, in the equation he uses, x(/) represents the position of the primed clock with respect to the unprimed clock, which is certainly not equla to 0.
I cannot remember that I did this: As outlined above the coordinates of the primed clock is x(/)=0 ant t(/) relative to the primed coordinate system while it is x and t relative to the unprimed system, The Lorentz transformations then connect these coordinates.
Similarly the coordinates of the unprimed clock is x=0 and t relative to the unprimed system and x(/) and t(/) relative to the primed system.
I cannot remember that I did this: As outlined above the coordinates of the primed clock is x(/)=0 ant t(/) relative to the primed coordinate system while it is x and t relative to the unprimed system, The Lorentz transformations then connect these coordinates.
Similarly the coordinates of the unprimed clock is x=0 and t relative to the unprimed system and x(/) and t(/) relative to the primed system.
Hi Omnibus, johanfprins, vergon, RealityCheck, prometheus, fairy, buttershug, NoCleverName, Trout et al,
I am about to post a long answer but considering this single thread has lasted over 24 web pages with a lot of heat and no light... I have decided to "speak up"... I apologize for the length of this post but only because I can tell you what this answer actually is and end this confusion for those who can actually understand the argument and read the references clearly... I also apologize for any typos in advance as there always will be some...
I have not been "intimately" involved with this discussion but it seems to me that it gets way off the track with side issues and this is the result of an "ill defined" problem. With all due respect... what I said in my initial post on this thread on page 1 still stands.
Good Elf on this thread earlier.
Einstein's initial paper included "relativistic mass" (though not explicitly stated he used longitudinal and transverse mass... not good!) and that point is implicit in section 10 (the most disputed section). This is not a good idea and Einstein himself stated this later (.. as I quoted in that original post of mine). This is "doubly confirmed" by the quote by Wheeler and Taylor that followed it. We should only use "invariant mass" (Do not use "rest mass" which implies the idea of "relativistic mass"). This says "nothing" about the Prof. LB Okun... Lev Okun was the head of the Laboratory of Elementary-Particle Theory at the Institute of Theoretical and Experimental Physics, in Moscow, USSR (So... he is no "slouch"). I have found a link to his presentation in Physics Today June 1989.
Lev Okun, The Concept of Mass, Physics Today, June 1989.
He is still presenting major papers on this and other subjects at prestigious events and has stood his ground regarding this "principle" calling it a "pedagogical virus".... I tend to agree. So please look into this matter carefully... this link above is pretty clear and provides a lot of philosophical "meat". Because nobody has stated just what this problem is and what various equations are supposed to mean... I will assist and try to provide the answer to just one "simple" problem. Feel free to criticize but keep within the "fenceline" of my problem and please do not confound it with some other undefined problems with "no fencelines"... OK?
Ignore those conventions in the link above and pay the price as you all play games with the various definitions of what is being observed. In my opinion is is a "bad move" to deal with any "relativistic mass", we should only deal with "invariant mass" on which "everyone" who understands what is going on agrees. I can see that this has not been agreed on and "all hell is released" and unless somebody states that the mass is the "invariant mass" then the equation F = ma will not hold in each and every reference frame. Let me say it for you... "m" is the invariant mass... the one and only mass... OK!... and you have sometimes been referring to "relativistic mass" which "cannot be agreed upon" as Einstein stated and is no longer a "correct theoretical concept". Forget it! There is "relativistic energy" but that is quite different (see below).
Einstein did nothing wrong from the historical context of his time, though anyone who uses these "old" ideas today needs to be very clear that he/she is no longer at liberty to bend rules since they have already been "set in stone". What you are all doing is trying to create some self-consistent argument with the advantage of hindsight... This is not a "textbook perfect" case and there are inevitable inconsistency and Levy Okun, John A Wheeler, Edwin Floriman Taylor and Albert Einstein (also many others) have warned everyone of the consequences. The Fourmilab commentary on Einstein's Paper should also be heeded and should have also alerted you all to "problems" in the original paper (specifically that section) and nobody should be directly quoting the controversial parts of Einstein's Paper directly... read the footnotes.... specifically the one related to para 10 where we are speaking of that transverse and longitudinal mass.
Most technical papers have errors in them, but the general truth of Einstein's paper is still there despite some problems. Relativity exists and it is not possible to avoid the consequences of time dilation and length contraction (as some suggest... this is total nonsense)... these are real measurable phenomena and they affect "real clocks" and real "measuring rods", though perhaps not in the way people think because of this "smokescreen" of directionless nonsense created by an inadequate educational system whose desire it is to make everything seem to be "self evident" and through the confusing influence of Okun's "pedagogical virus".
Just to expand and clear up on the "historical process" of Einstein's Special Theory... Over the following years since 1905 a number of matters were "elucidated" and "emphasized" by Einstein. This include the fact that the second "Principle" (... of two basic Special Relativity "Principles") was not required .... the constancy of the speed of light is already part of Maxwell's Equations and is not specifically required as an assumption.
see paragraph 2 of "ON THE ELECTRODYNAMICS OF MOVING BODIES" By A. Einstein June 30, 1905
Many people do not realize that the earlier Maxwell's Equations are "already unified" as being entirely consistent with the Special Theory of Relativity and they remain so to this day. There are some "wild statements" to the contrary in this Forum (and in other places)... they are entirely wrong. Einstein has not changed anything related to that. Another aspect was the problem with "relativistic mass" and the need to understand and use only "invariant mass"... that topic we are currently "unfortunately" discussing. Please note... The "meaning of invariant mass" is that "matter" has "rest energy" equal to mc² in addition to any "potential and or kinetic energy" and is measured in the rest frame of the matter to which it applies. You measure that rest mass "comparatively" using a spring in that freely falling inertial system and measure the relative accelerations applied "equally" to two different masses (one a test mass) that are contra-accelerated using the masses attached to the ends of the spring... m'a' = m"a" where a"/a' = m'/m" (easy peezy!). Note there are no "absolute masses" because we do not know the absolute energy content of any mass, we only know their relative masses (aside from Burkard Heim who had some interesting ideas but I tend to disagree because I think the principle of Relativity "overrides" this "absolute mass" calculation concept). This is because all energy is "relative" to some arbitrary "zero point". We measure "rest mass" in comparison to a standard mass. There is absolutely no way (till now) of determining the mass of a particle from first principles. Probably it would be a violation of Einstein's Theory right there... and if you prove it you can probably put your hand out for "The Gong".
Every inertial "rest" frame measures the same "invariant mass" and any comparisons "between frames" should be in terms of this "invariant mass" that is measured in any "rest frame" the masses are to be internally consistent. This does not mean that you will not find "difficulties" with many important papers in the field and with references on the Internet where there is "absolute nonsense and even lies". It is the nature of scientific papers to be "presented" in the most turgid way possible to confuse the "unwashed multitudes". The energy content of a particle is composed of a rest energy mC² plus the kinetic and potential energy. Of course those potential and kinetic energies are also "relative to something somewhere" and depends on how you measure them and with respect to"what"? I try and think of contained "systems" rather than the separate particles.
There were other very minor technical errors in the paper as well and you can read them as commentary at the bottom of the Fourmilab "re-translation".
ON THE ELECTRODYNAMICS OF MOVING BODIES By A. Einstein June 30, 1905 (referenced above)
One matter in the para 10 is there as a caution (as I mentioned)... footnote 9. In case people have not read it... it states "explicitly"... "The definition of force here given is not advantageous, as was first shown by M. Planck. It is more to the point to define force in such a way that the laws of momentum and energy assume the simplest form. Editor's Notes". This does not change the fact that Einstein "is still" correct. Things are done differently nowadays and what you are arguing about is "hindsight". Einstein never even studied Maxwell's Equations at his Institute that he graduated from.... he had to interpret all that from his own personal readings at a time when there was no "Internet"... So unlike all of you who have been versed in these matters with 100 years of thoughtful contemplation, Einstein needed to work things out for himself, and in these matters he may have made some problems with his own definitions. As noted previously ... he was not a good teacher... But he was a good scientist. So much for the history lesson.
The next point to bring up is far more complex, subtle and technical and also more critical to this discussion. The idea of "relativistic acceleration". This is not a term related to relativistic mass. What this really means (and we need to all agree on this) is the observation of remote virtual forces developing in an arbitrary "non-relativistically accelerating" particle moving at very high velocity (with a relative frame speed a significant portion of the speed of light) viewed by a non-accelerating observer in/from an inertial frame of reference. It must also be stated that only one of these two frames can be accelerated if we are to use "simple" Special Relativity and the "actual" accelerated frame is clearly the one in which an acceleration is actually felt... other frames in which we may "see" accelerations are no guarantee that they are truly accelerating because our frame may be the one accelerating relative to that "external" frame (accelerations are vectors and can sum to unusual values relative to various frames). You can easily tell if a frame is being accelerated "for real" if a spring balance in the frame shows an extension F = mg where g is an acceleration. If there is no "extension" ... there is no "inertial" acceleration. No acceleration does not mean that the mass is undefined... we have discussed that point already above. There might be "practical problems" doing things in general but the principle must be considered first and foremost in any "consistent theory".
For instance falling freely in a gravity field tells us the object "falling freely" is not "currently" suffering the time dilation due to Special Relativity while an object being actually accelerated such as yourselves on the surface of the earth are being accelerated (have a look at a spring balance and the way it extends) and are therefore the ones undergoing the "time dilation" (I am assuming quite low velocity here). The previous history of a particle is important only if the clock synchronization "in the one frame" occurs prior to an acceleration phase or "boost". Subsequently the time dilation between the frames is "cumulative" depending on the duration the particle remains at a high velocity... that is a property of spacetime.
This also means that clocks in the lower floors of buildings run slower "relative" to clocks in the upper stories of the same building due to it's slightly lowered gravitational acceleration rate being further from the center of the earth.
This can be easily measured and is not something that is able to be simply argued in a "debate"... it is just a "fact". A "remote observation" of the kinematics of a relatively moving frame of reference alone is no guarantee that an "observed acceleration" is due to inertial forces or due to frame dragging so it is not possible to say for sure if the clocks are "actually" running slower in the moving frame relative to a "stay at home clock" that has been accelerated, then brought back together and compared in the same observer frame once more. Only the accelerated clock will be time dilated due to Special Relativistic Motion when compared with a clock that was initially synchronized with in the one frame before the trip (as what might occur on a round trip to some star system at a very high relativistic velocity). The ideal is assumed that the clock left behind will be in free fall in it's own "unaccelerated" observer frame. In the case of an observer left behind on the earth and the twin goes off at one gee acceleration then I would expect to see little or no difference on return since the General Principle of Equivalence states that the inertial and gravitational accelerations are absolutely equivalent.
The "actual" ponderable forces (like the forces acting in gravity) in any accelerating frame of reference can be considered as piecemeal "inertial" and "in frame" and might be in any generalized direction... and for all short "boosts" this internal acceleration is "low" and should be "non-relativistic" though the relative base velocity of the frame can be very high (relative to the observer frame). The "history" of the frame and any relative velocity with respect to some other frame may mean nothing when considering the "small" vector acceleration so the idea that simply taking the second derivative of the relative position vectors between frames will provide any measure of the acceleration in the accelerated frame is misleading and avoids the question and also disguises it. After all... this is an "accelerating frame" by definition and while we are speaking about is Special Relativity then the best way to treat this as Special Relativity is to consider the accelerating frame as being "frozen" at two different sequential points in time (events) in their frame then compared with the "permanently" inertial (unaccelerated) frame of the observer and the transformations "correcting" between frames should indicate that the two snapshots of the "accelerating" frame will be seen "accelerating" relative to the first un-accelerated frame only by a "difference".
Instantaneously each "pseudo-piecewise-inertial frame" is moving with constant (though high) relative velocity (though each snapshot has a different "low" constant velocity changed by an "internal acceleration"). This acceleration is remotely observed and we are trying not to determine the acceleration as it "appears" by just remotely observing this dilated phenomenon but to estimate the effect of what the "test pilot" in a rocket moving with such a particle may "feel" or what he may discover to be the acceleration in that non-inertial frame. The answer is "obviously".... F' = τF (translated... the "force" or the acceleration measured in the observer rest frame (primed) of the distant accelerated frame must be identical to the observations of the kinematics measured in the accelerated frame via a set of transformations (τ). We need to have synchronized clocks and the origin of the frames need to be set so a single sighting can be made of the accelerating particle in each subsequent "snapshot".
The idea is that the position vectors and times are is not so "obvious" since the propagation times of an observation "signal flash" back to a single observer must be corrected for propagation times. For instance what does the word "simultaneously" mean to any remotely observed event in an accelerating non-inertial accelerating reference frame moving "through" an inertial observer reference frame like a jet from a distant black hole or rapidly retreating Hubble shifted quasar being frame dragged away from us? Clearly it does not mean "measured at the same time" as it does in Newtonian Physics (... the speed of propagation of light would be infinite) because light takes time to propagate from distant points in the rest frame of the observer, in and from the distant universe. The observer time of observation needs correction "back" to the time in the rest frame where the event occurred.
t = T - ct'
t = true observer frame time of a properly synchronized remote event.
T = observed "simultaneous time" but "uncorrected" in the observer frame.
ct' = time of propagation "directly" to the observer origin in the observer frame (as a ray length S = ct'... the minus sign indicates that the event occurred "before" it was observed at the origin of this inertial frame).
This T is the time an observer at the origin of the observer rest frame would think is the time after T0 = 0 (time when all clocks in the inertial frame and the clock of the "subsequently" accelerated frame were all set to zero in the one observer rest frame) that is then an event seen at a remote position of the rest frame. The time t is what should be recorded as an estimate of the time in the frame at the remote position of the event at the end of the optical vector S (where the flash is seen in the rest frame). In reality for "cosmic events" the observed position of these events needs "something more" to determine how far away an event is. Currently we use simple optical parallax due to the diameter of the orbit of the earth for nearby targets or Hubble shift to determine distance by measuring the red shift in the spectral bands of the moving source's light. Otherwise we would not have a clue at all, it would all still be a complete mystery (which in all probability it probably still is).
Alternatively assuming that we have a "johnny on the spot" everywhere of interest in the observer frame where interesting events are occurring then the equations everyone knows "apply" if that "Johnny on the spot" carries a synchronized clock and "knows where he is in the rigid inertial frame xyz coordinate system". This requires the "Einstein clock synchronization convention" that all should be aware of to achieve this "minimum requirement". Otherwise a correction "in-frame" is required. Distant events on the other side of the Universe must be corrected for global spacetime curvature which we "earth observers" really do not entirely understand or normally actually see yet (... check the WMAP Data for instance). There are exceptions where we can actually "see" spacetime curvature for instance where gravitational lensing has occurred in the vacuum of space in "Einstein's Rings". However Special Relativity only applies over a "piecemeal flatspace" and we know spacetime itself is not only curved but even measurably still "stretching".
If we understand this then the acceleration "always" depends on mass through the equation F = ma or F' = ma' (for equivalent masses measured in different frames) and to have any meaning these forces must be convertible and they are absolutely independent of any relative velocity between frames and small accelerations... but most importantly to get these values they must be measured within the relatively moving frame itself and the mass "estimated" must be the "invariant mass".
I am about to post a long answer but considering this single thread has lasted over 24 web pages with a lot of heat and no light... I have decided to "speak up"... I apologize for the length of this post but only because I can tell you what this answer actually is and end this confusion for those who can actually understand the argument and read the references clearly... I also apologize for any typos in advance as there always will be some...
I have not been "intimately" involved with this discussion but it seems to me that it gets way off the track with side issues and this is the result of an "ill defined" problem. With all due respect... what I said in my initial post on this thread on page 1 still stands.
Good Elf on this thread earlier.
Einstein's initial paper included "relativistic mass" (though not explicitly stated he used longitudinal and transverse mass... not good!) and that point is implicit in section 10 (the most disputed section). This is not a good idea and Einstein himself stated this later (.. as I quoted in that original post of mine). This is "doubly confirmed" by the quote by Wheeler and Taylor that followed it. We should only use "invariant mass" (Do not use "rest mass" which implies the idea of "relativistic mass"). This says "nothing" about the Prof. LB Okun... Lev Okun was the head of the Laboratory of Elementary-Particle Theory at the Institute of Theoretical and Experimental Physics, in Moscow, USSR (So... he is no "slouch"). I have found a link to his presentation in Physics Today June 1989.
Lev Okun, The Concept of Mass, Physics Today, June 1989.
He is still presenting major papers on this and other subjects at prestigious events and has stood his ground regarding this "principle" calling it a "pedagogical virus".... I tend to agree. So please look into this matter carefully... this link above is pretty clear and provides a lot of philosophical "meat". Because nobody has stated just what this problem is and what various equations are supposed to mean... I will assist and try to provide the answer to just one "simple" problem. Feel free to criticize but keep within the "fenceline" of my problem and please do not confound it with some other undefined problems with "no fencelines"... OK?
Ignore those conventions in the link above and pay the price as you all play games with the various definitions of what is being observed. In my opinion is is a "bad move" to deal with any "relativistic mass", we should only deal with "invariant mass" on which "everyone" who understands what is going on agrees. I can see that this has not been agreed on and "all hell is released" and unless somebody states that the mass is the "invariant mass" then the equation F = ma will not hold in each and every reference frame. Let me say it for you... "m" is the invariant mass... the one and only mass... OK!... and you have sometimes been referring to "relativistic mass" which "cannot be agreed upon" as Einstein stated and is no longer a "correct theoretical concept". Forget it! There is "relativistic energy" but that is quite different (see below).
Einstein did nothing wrong from the historical context of his time, though anyone who uses these "old" ideas today needs to be very clear that he/she is no longer at liberty to bend rules since they have already been "set in stone". What you are all doing is trying to create some self-consistent argument with the advantage of hindsight... This is not a "textbook perfect" case and there are inevitable inconsistency and Levy Okun, John A Wheeler, Edwin Floriman Taylor and Albert Einstein (also many others) have warned everyone of the consequences. The Fourmilab commentary on Einstein's Paper should also be heeded and should have also alerted you all to "problems" in the original paper (specifically that section) and nobody should be directly quoting the controversial parts of Einstein's Paper directly... read the footnotes.... specifically the one related to para 10 where we are speaking of that transverse and longitudinal mass.
Most technical papers have errors in them, but the general truth of Einstein's paper is still there despite some problems. Relativity exists and it is not possible to avoid the consequences of time dilation and length contraction (as some suggest... this is total nonsense)... these are real measurable phenomena and they affect "real clocks" and real "measuring rods", though perhaps not in the way people think because of this "smokescreen" of directionless nonsense created by an inadequate educational system whose desire it is to make everything seem to be "self evident" and through the confusing influence of Okun's "pedagogical virus".
Just to expand and clear up on the "historical process" of Einstein's Special Theory... Over the following years since 1905 a number of matters were "elucidated" and "emphasized" by Einstein. This include the fact that the second "Principle" (... of two basic Special Relativity "Principles") was not required .... the constancy of the speed of light is already part of Maxwell's Equations and is not specifically required as an assumption.
see paragraph 2 of "ON THE ELECTRODYNAMICS OF MOVING BODIES" By A. Einstein June 30, 1905
Many people do not realize that the earlier Maxwell's Equations are "already unified" as being entirely consistent with the Special Theory of Relativity and they remain so to this day. There are some "wild statements" to the contrary in this Forum (and in other places)... they are entirely wrong. Einstein has not changed anything related to that. Another aspect was the problem with "relativistic mass" and the need to understand and use only "invariant mass"... that topic we are currently "unfortunately" discussing. Please note... The "meaning of invariant mass" is that "matter" has "rest energy" equal to mc² in addition to any "potential and or kinetic energy" and is measured in the rest frame of the matter to which it applies. You measure that rest mass "comparatively" using a spring in that freely falling inertial system and measure the relative accelerations applied "equally" to two different masses (one a test mass) that are contra-accelerated using the masses attached to the ends of the spring... m'a' = m"a" where a"/a' = m'/m" (easy peezy!). Note there are no "absolute masses" because we do not know the absolute energy content of any mass, we only know their relative masses (aside from Burkard Heim who had some interesting ideas but I tend to disagree because I think the principle of Relativity "overrides" this "absolute mass" calculation concept). This is because all energy is "relative" to some arbitrary "zero point". We measure "rest mass" in comparison to a standard mass. There is absolutely no way (till now) of determining the mass of a particle from first principles. Probably it would be a violation of Einstein's Theory right there... and if you prove it you can probably put your hand out for "The Gong".
Every inertial "rest" frame measures the same "invariant mass" and any comparisons "between frames" should be in terms of this "invariant mass" that is measured in any "rest frame" the masses are to be internally consistent. This does not mean that you will not find "difficulties" with many important papers in the field and with references on the Internet where there is "absolute nonsense and even lies". It is the nature of scientific papers to be "presented" in the most turgid way possible to confuse the "unwashed multitudes". The energy content of a particle is composed of a rest energy mC² plus the kinetic and potential energy. Of course those potential and kinetic energies are also "relative to something somewhere" and depends on how you measure them and with respect to"what"? I try and think of contained "systems" rather than the separate particles.
There were other very minor technical errors in the paper as well and you can read them as commentary at the bottom of the Fourmilab "re-translation".
ON THE ELECTRODYNAMICS OF MOVING BODIES By A. Einstein June 30, 1905 (referenced above)
One matter in the para 10 is there as a caution (as I mentioned)... footnote 9. In case people have not read it... it states "explicitly"... "The definition of force here given is not advantageous, as was first shown by M. Planck. It is more to the point to define force in such a way that the laws of momentum and energy assume the simplest form. Editor's Notes". This does not change the fact that Einstein "is still" correct. Things are done differently nowadays and what you are arguing about is "hindsight". Einstein never even studied Maxwell's Equations at his Institute that he graduated from.... he had to interpret all that from his own personal readings at a time when there was no "Internet"... So unlike all of you who have been versed in these matters with 100 years of thoughtful contemplation, Einstein needed to work things out for himself, and in these matters he may have made some problems with his own definitions. As noted previously ... he was not a good teacher... But he was a good scientist. So much for the history lesson.
The next point to bring up is far more complex, subtle and technical and also more critical to this discussion. The idea of "relativistic acceleration". This is not a term related to relativistic mass. What this really means (and we need to all agree on this) is the observation of remote virtual forces developing in an arbitrary "non-relativistically accelerating" particle moving at very high velocity (with a relative frame speed a significant portion of the speed of light) viewed by a non-accelerating observer in/from an inertial frame of reference. It must also be stated that only one of these two frames can be accelerated if we are to use "simple" Special Relativity and the "actual" accelerated frame is clearly the one in which an acceleration is actually felt... other frames in which we may "see" accelerations are no guarantee that they are truly accelerating because our frame may be the one accelerating relative to that "external" frame (accelerations are vectors and can sum to unusual values relative to various frames). You can easily tell if a frame is being accelerated "for real" if a spring balance in the frame shows an extension F = mg where g is an acceleration. If there is no "extension" ... there is no "inertial" acceleration. No acceleration does not mean that the mass is undefined... we have discussed that point already above. There might be "practical problems" doing things in general but the principle must be considered first and foremost in any "consistent theory".
For instance falling freely in a gravity field tells us the object "falling freely" is not "currently" suffering the time dilation due to Special Relativity while an object being actually accelerated such as yourselves on the surface of the earth are being accelerated (have a look at a spring balance and the way it extends) and are therefore the ones undergoing the "time dilation" (I am assuming quite low velocity here). The previous history of a particle is important only if the clock synchronization "in the one frame" occurs prior to an acceleration phase or "boost". Subsequently the time dilation between the frames is "cumulative" depending on the duration the particle remains at a high velocity... that is a property of spacetime.
This also means that clocks in the lower floors of buildings run slower "relative" to clocks in the upper stories of the same building due to it's slightly lowered gravitational acceleration rate being further from the center of the earth.
This can be easily measured and is not something that is able to be simply argued in a "debate"... it is just a "fact". A "remote observation" of the kinematics of a relatively moving frame of reference alone is no guarantee that an "observed acceleration" is due to inertial forces or due to frame dragging so it is not possible to say for sure if the clocks are "actually" running slower in the moving frame relative to a "stay at home clock" that has been accelerated, then brought back together and compared in the same observer frame once more. Only the accelerated clock will be time dilated due to Special Relativistic Motion when compared with a clock that was initially synchronized with in the one frame before the trip (as what might occur on a round trip to some star system at a very high relativistic velocity). The ideal is assumed that the clock left behind will be in free fall in it's own "unaccelerated" observer frame. In the case of an observer left behind on the earth and the twin goes off at one gee acceleration then I would expect to see little or no difference on return since the General Principle of Equivalence states that the inertial and gravitational accelerations are absolutely equivalent.
The "actual" ponderable forces (like the forces acting in gravity) in any accelerating frame of reference can be considered as piecemeal "inertial" and "in frame" and might be in any generalized direction... and for all short "boosts" this internal acceleration is "low" and should be "non-relativistic" though the relative base velocity of the frame can be very high (relative to the observer frame). The "history" of the frame and any relative velocity with respect to some other frame may mean nothing when considering the "small" vector acceleration so the idea that simply taking the second derivative of the relative position vectors between frames will provide any measure of the acceleration in the accelerated frame is misleading and avoids the question and also disguises it. After all... this is an "accelerating frame" by definition and while we are speaking about is Special Relativity then the best way to treat this as Special Relativity is to consider the accelerating frame as being "frozen" at two different sequential points in time (events) in their frame then compared with the "permanently" inertial (unaccelerated) frame of the observer and the transformations "correcting" between frames should indicate that the two snapshots of the "accelerating" frame will be seen "accelerating" relative to the first un-accelerated frame only by a "difference".
Instantaneously each "pseudo-piecewise-inertial frame" is moving with constant (though high) relative velocity (though each snapshot has a different "low" constant velocity changed by an "internal acceleration"). This acceleration is remotely observed and we are trying not to determine the acceleration as it "appears" by just remotely observing this dilated phenomenon but to estimate the effect of what the "test pilot" in a rocket moving with such a particle may "feel" or what he may discover to be the acceleration in that non-inertial frame. The answer is "obviously".... F' = τF (translated... the "force" or the acceleration measured in the observer rest frame (primed) of the distant accelerated frame must be identical to the observations of the kinematics measured in the accelerated frame via a set of transformations (τ). We need to have synchronized clocks and the origin of the frames need to be set so a single sighting can be made of the accelerating particle in each subsequent "snapshot".
The idea is that the position vectors and times are is not so "obvious" since the propagation times of an observation "signal flash" back to a single observer must be corrected for propagation times. For instance what does the word "simultaneously" mean to any remotely observed event in an accelerating non-inertial accelerating reference frame moving "through" an inertial observer reference frame like a jet from a distant black hole or rapidly retreating Hubble shifted quasar being frame dragged away from us? Clearly it does not mean "measured at the same time" as it does in Newtonian Physics (... the speed of propagation of light would be infinite) because light takes time to propagate from distant points in the rest frame of the observer, in and from the distant universe. The observer time of observation needs correction "back" to the time in the rest frame where the event occurred.
t = T - ct'
t = true observer frame time of a properly synchronized remote event.
T = observed "simultaneous time" but "uncorrected" in the observer frame.
ct' = time of propagation "directly" to the observer origin in the observer frame (as a ray length S = ct'... the minus sign indicates that the event occurred "before" it was observed at the origin of this inertial frame).
This T is the time an observer at the origin of the observer rest frame would think is the time after T0 = 0 (time when all clocks in the inertial frame and the clock of the "subsequently" accelerated frame were all set to zero in the one observer rest frame) that is then an event seen at a remote position of the rest frame. The time t is what should be recorded as an estimate of the time in the frame at the remote position of the event at the end of the optical vector S (where the flash is seen in the rest frame). In reality for "cosmic events" the observed position of these events needs "something more" to determine how far away an event is. Currently we use simple optical parallax due to the diameter of the orbit of the earth for nearby targets or Hubble shift to determine distance by measuring the red shift in the spectral bands of the moving source's light. Otherwise we would not have a clue at all, it would all still be a complete mystery (which in all probability it probably still is).
Alternatively assuming that we have a "johnny on the spot" everywhere of interest in the observer frame where interesting events are occurring then the equations everyone knows "apply" if that "Johnny on the spot" carries a synchronized clock and "knows where he is in the rigid inertial frame xyz coordinate system". This requires the "Einstein clock synchronization convention" that all should be aware of to achieve this "minimum requirement". Otherwise a correction "in-frame" is required. Distant events on the other side of the Universe must be corrected for global spacetime curvature which we "earth observers" really do not entirely understand or normally actually see yet (... check the WMAP Data for instance). There are exceptions where we can actually "see" spacetime curvature for instance where gravitational lensing has occurred in the vacuum of space in "Einstein's Rings". However Special Relativity only applies over a "piecemeal flatspace" and we know spacetime itself is not only curved but even measurably still "stretching".
If we understand this then the acceleration "always" depends on mass through the equation F = ma or F' = ma' (for equivalent masses measured in different frames) and to have any meaning these forces must be convertible and they are absolutely independent of any relative velocity between frames and small accelerations... but most importantly to get these values they must be measured within the relatively moving frame itself and the mass "estimated" must be the "invariant mass".
QUOTE (Trout+)
You can derive it yourself, provided that you know elementary calculus:
p=mv/sqrt(1-(v/c)^2)
F=dp/dt
Let's see your level of mathematical ability. Did you graduate from high school, tutu-boy?
Trout has mentioned (above) that we should deal with change in momentum... that is correct up to a point but to say that F = ma is not true (see reference below) then state that F=dp/dt makes a mockery of convention... what is the difference since dp/dt = mdv/dt = ma? What else could it mean since there is actually only one mass not two?... but once again errors result when there is no common paradigm. One moment Trout admonishes the relativistic mass and the next moment he is using it?...
p=mv/sqrt(1-(v/c)^2)
F=dp/dt
Let's see your level of mathematical ability. Did you graduate from high school, tutu-boy?
Trout has mentioned (above) that we should deal with change in momentum... that is correct up to a point but to say that F = ma is not true (see reference below) then state that F=dp/dt makes a mockery of convention... what is the difference since dp/dt = mdv/dt = ma? What else could it mean since there is actually only one mass not two?... but once again errors result when there is no common paradigm. One moment Trout admonishes the relativistic mass and the next moment he is using it?...
QUOTE (Trout Today at 5:54 AM+)
The formula valid for any velocity is F=dp/dt
where p=momentum
In relativity, momentum is given by p=m_0*v/sqrt(1-(v/c)^2)
where m_0=proper mass, v=particle speed, c=speed of light
So, if you know any calculus, you can now go back and calculate the force.
Let's see if you can calculate a simple derivative
.. Here Trout is quoting a "proper mass"... m_0 ... why the extra "m"... m is the "invariant mass" = "proper mass" and he calls it m_0??? Thus he shows p = mv = m_0*v/sqrt(1-(v/c)^2)... two different things meaning one thing? Which is it to be??
johanfprins is trying to say that although Lorentz transformations don’t give the real truth about spatial and temporal dimensions from within the moving system when observed by the stationary observer, they can be used by the stationary observer for practical purposes as engineering empirical equations. He thinks Lorentz equations somehow help the stationary observer to figure out from his own perspective for his own purposes what the story is from within the moving system.
Thus, although the moving observer sees things proper in his own frame, the deformed image provided by the Lorentz transformations can be of some use, johanfprins thinks, for the stationary observer who actually knows the image is deformed but he (the stationary observer) can always obtain the correct image by undoing them by applying the inverse Lorentz transformations.
This begs the question, why use equations to get a deformed image and then undo them to get the correct final answer?
where p=momentum
In relativity, momentum is given by p=m_0*v/sqrt(1-(v/c)^2)
where m_0=proper mass, v=particle speed, c=speed of light
So, if you know any calculus, you can now go back and calculate the force.
Let's see if you can calculate a simple derivative
.. Here Trout is quoting a "proper mass"... m_0 ... why the extra "m"... m is the "invariant mass" = "proper mass" and he calls it m_0??? Thus he shows p = mv = m_0*v/sqrt(1-(v/c)^2)... two different things meaning one thing? Which is it to be??
QUOTE (Trout Today at 7:31 AM+)
Actually F=ma only for the trivial case v=0. In this case a=0 so F is trivially equal to zero. It is therefore safe to say that F is never equal to ma.
Not so safe... So acceleration in all kinematics is not even close? It "apparently" only applies when V = 0. Clearly if V = 0 then a = 0 and p =0 etc... his trivial case I presume. I entirely disagree. F = ma in all accelerated system "piecemealwise". It does not depend on any history so it actually means instantaneous acceleration of "proper mass". The change in velocity can be vanishingly small or "moderate" up to many gees... just not too extreme like a million gees for instance "instantaneously" and lasting over a very short time. If you let this "time" drag out then you must integrate along a special curve for a possible varying vector acceleration function. I do not want to go there it is too complicated for this discussion.
This would mean with some qualification that this change in momentum or force (... is related to the change in relative velocity between two snapshot frames... one at the beginning of a short boost and the other indicating the state of the frame after the short boost). Presumably since "p" in the first of Trout's equations is a relativistically altered momentum... this altered value must be what is "seen" in the observer frame happening to a particle in the "high velocity" frame... as V -> C p -> ∞. While in Trout's second equation the F quoted there to be consistent is what is measured in the observer frame (unprimed) from the unaccelerated frame (it must be an accelerated frame now since we have "forces" acting). We know this is correct where there is only a small relative velocity... this is Newtonian Mechanics... that's fine. As V -> C what does this equation F = dp/dt do? F = dp/dt = m dv/dt as V -> C dv/dt -> 0 (it's acceleration drops to zero as observed in the remote frame). Wrong... the acceleration "in frame" may be anything at all and is not specifically dependent on the relative frame velocity or even it's direction.
There is no "hint" of what might be seen in the observer frame. Clearly F = dp/dt in all "Newtonian systems" which means this is the standard momentum "in frame" not in the observer frame unless interframe velocity V -> 0, then it will be "close" (ignoring all de Broglie effects). Trout's equation B is a non-sequitur from equation A. The "in frame" acceleration is still F = ma where "a" is an acceleration determined "in frame" relative only to the first or second snapshot and not to the observation of the relative V from the observer frame. Trout ... never wants to be explicit but if you want to answer a real problem you must be very explicit. Goal posts will be rearranged and balls will be "hidden" in case anyone should take a "free kick". This is the problem I am trying to give an answer to so I stick whith this.
This equation F = dp/dt holds "only" if the mass here is the "intrinsic or invariant mass" or Trout's "proper mass" (measured in the "instantaneous" rest frame... it does not matter much if it is snapshot 1 or snapshot 2 since this is going to be the "instantaneous" acceleration) and "a" (F = ma) and "a' " (F' = ma') are accelerations and forces measured in the accelerated frame and unaccelerated "observer" frame of masses in those frames. Without some primed frame (observer frame) this is a meaningless set of useless and inconsistent equations and would be marked wrong in any examination by anyone who understands the subject matter. What any rest frame observer wants to know is what "a" is in the accelerated frame and not what "a' " is since as V -> C you will not be able to observe "a" directly since "a" is the proper acceleration or intrinsic acceleration in the relatively moving and accelerating frame. If you disagree then the question is "what else could it mean?"
As V -> C (the "initial" relative velocity between the inertial frame and the accelerated frame) the "acceleration" of "everything" in the accelerated frame observationally -> 0, everything becomes "relatively frozen" on the light cone wall. It is "difficult" to measure the acceleration of a rocket when V -> C especially if we are unsure of how space is "curved" between the observer and the observed due to non-inertial influences. We know "non-inertial influences" dominate the Universe... called Hubble expansion.
What the observer wants to know is if an "astronaut" in a rocket that is "still" undergoing acceleration as it approaches the speed of light and is "still feeling comfortable" and what acceleration "a" he "feels" in his frame is still currently within human limits. While relative V can approach C, since this V is only the relative velocity it has absolutely no effect on this intrinsic value of "a" in the accelerated frame which can vary instantaneously and may be arbitrarily aligned. An estimate of the instantaneous acceleration is possible only after correction between the frames piecemealwise.... Agreed? The imprecision in this figure is going to increase as V -> C the measurables of time and position are going to become more and more "difficult" according to the orthogonality of these measurables.
Fourier Transforms and Uncertainty
What it is not possible to say is that you simply take the second differential of the position vector to obtain the acceleration because the velocity V between the observer frame and the accelerated frame is not applicable and are "distorted" subject to length contraction and time dilation as seen from the observer frame. Common sense and Einstein's first Principle tells us that... that the laws of physics are the same in all frames (that is the meaning according to Einstein that is to be preserved in Special Relativity despite the minor problems stated previously). Even assuming a flatspace this second differential of the position vector does not give the correct acceleration unless the acceleration is in the same direction as the velocity (any "relativistic" distant receding particle) will always tend to appear to travel directly away from the origin of the observers inertial frame given enough time and space. If the "boost" is "nearly" in that direction (like a rocket boost) then the observed accelerations will tend to the value...
a = a' [ 1 - U²/C²]³⁄² .... this is what you are all looking for in the simplest form for the simplest problem possible.
So F = ma = F' = ma' etc... If U ≈0 then a = a' in the limit but this evaluation works at all low relative velocity (low compared with the speed of light). This has that caution in it so do not generalize this idea beyond its domain of applicability. This "evaluation" got us to the moon and I am sure it will still be good a million years from now regardless of how we want to calculate kinematics. Full details of the derivation are here...
Acceleration in Special Relativity
Ask a few questions if you want me to explain the finer details...
I would point to the composition of relativistic velocities as a starting point here...
How Do You Add Velocities in Special Relativity?
The rest of the derivation above is "simple" in that given the simplified problem it is the limit taken in regard to some elementary "fractions" and this treatment does not require any special differentiation or integration as suggested by some if we are trying to determine "instantaneous proper intrinsic relativistic acceleration"... a "lot of words" in case there is any doubt as to what this really means.
It is important to realize that this is only where the acceleration is in the same direction as the inter-frame velocity and that velocity is chosen as the X direction. A more general treatment for arbitrary acceleration in an arbitrary direction should be found here...
Relativistic force transformation - Valery P. Dmitriyev 13Jul 2005
It is not "simple" but it is "almost" straightforward but more necessarily complicated... Warning: there are some minor typos in this treatment but the overall results are correct.
Some additional "sticky" points... A "method" is required to allow us to compare the two accelerated frame "snapshots" with the observer frame and that is not so easily done using standard relativity using the relative velocity between the two "snapshot" frames where the time between the snapshots is "arbitrarily short" but not zero. Remember that in the remote observer inertial frame a second of time may be quite different to a "second" in the accelerated frame whose relative velocity may be approaching the speed of light (and the internal "proper velocity" is approaching "infinity" while the "proper acceleration" may remain "internally constant" lets say 1 gee). Remember the "subjective experience of time" in both frames is identical and there is no "increase in any mass" as V -> C. What high relativistic speed provides is the ability to cross vast distances of many light years of space in a short period of ship's time at the expense of dilation of the travelers time relative to his "stay at home twin"... though if that twin undergoes the same history of acceleration gravitationally as our traveler experiences purely relativistically, maybe there will be no time difference at all on a round trip at constant magnitude of 1 gee acceleration. The simplistic relativity equation for time dilation assumes no acceleration of the "stay at home" twin and large change in velocity for the accelerated traveler. In that simplified case it is the accelerated/deaccelerated particle that suffers the time dilation (check it out with a spring balance). It is definitely not a symmetrically interchangeable effect.
To complete this problem easily as shown in the references above you need the law of composition of relativistic velocities and then apply limiting conditions to the arbitrarily "boosted" particle. It is also important to realize that this relationship between the high velocity particle and the low velocity particle or observer frame is not "universal" but is dependent on important interactions between them to realize their full potential (such as the transfer of "energy"). No interaction then there is no energy. One type of interaction is for the space ship scenario to be a "round trip" where the identical twins begin and end the trip in the same inertial reference frame. Other pairs of particles do not include this high - low velocity relationship so it appears quite different ... what is a constant in all "similarly" accelerated frames at very different relative velocities is the same proper relativistic acceleration (... read simple in-frame acceleration) in all frames and it remains that same acceleration independent of any of the individual relative velocity relationships. That is Einstein's Relativity Principle.
While this "base" velocity "distorts" what is being "visually seen" in/from the observer or "traveler" frame this "base" velocity has absolutely no independent influence on the forces or the acceleration in the accelerated frame because these forces are independent of any "relative velocity" between any quasi-stationary "snapshot" of an accelerated frame (treated as an piecemeal unaccelerated frame) and the observer frame and it is solely dependent on this "absolute acceleration" which is felt only in the accelerating frame of which we have two "snapshots" some "short time and distance apart" viewed by/from the observer frame as a "pair" of relativistically moving "objects" at different times and positions "corrected" using the SR transformations.
We first determine the positions and times of events in the two relativistic snapshot frames from the corrected measurements from the observer frame. Once we know the relative positions and times of events between these two snapshots in the snapshot frames of reference, and because the difference between the two frames have only a small acceleration we can estimate this instantaneous acceleration (quite accurately) using Newtonian approximation (assuming the first snapshot was in a frame of "instantaneous rest"). That is where you can use the equation F = dp/dt or even F = ma (... the one we and NASA all know and love!).
We need a value for "a" (above) given a value of "a' " determined from measurements in the observer inertial frame... the measured relative in frame acceleration. The relative "speed" between the observing frame and the accelerating frame is not about to alter any factor in this estimate since it is an "in frame boost". This "boost" does not need to have any relationship whatsoever to the prevailing velocity between frames... not even to its direction. It is "hard" to distinguish a small acceleration in an arbitrary direction of a far off extreme relativistically moving object due to the imprecision and also due to the fewer number of executing events being recorded from the relativistic progression of observations. In the extreme case the object moving away just marginally under the speed of light becomes totally unobserved and even "static" as it recedes away from the observer suffering more and more extreme time dilation until a point is reached where almost no events are occurring in the observed phenomenon resulting in an "observational paradox" of observing "nothing happening" in the limiting case. with "nothing happening" there is no significant temporal progress in what is being observed... it is "frozen in time".
If "Special Relativity" alone was the main determinant in the kinematics of distant objects... none of these distant bodies would "appear" to recede faster than C/2. This is very easy to prove as a body travels at near the speed of light traversing one light second in distance in just marginally over one second of observer frame time, the consecutive flashes from this object can only propagate at a speed of C so the arrival frequency of consecutive flashes is one flash per two seconds while the particle "visibly recedes" at an uncorrected speed in the observer frame of C/2. However this is not the "true" velocity of the near luminally moving particle whose velocity is causing it to "streak ahead" in real terms of that perceived position at nearly the full speed of light that would be recorded by an army of "johnny's on the spot" at rest in the observer frame in close proximity and along the outward path of the particle and successive "johnny's on the spot" recording the particles progress with correctly synchronized clocks and comparing "notes" later to obtain that velocity of nearly C.
In fact there is another seeming paradox of many cosmic bodies appearing to recede at nearly C in the far off distance of the Universe (up to 99%C). This is "impossible" if only a single perspective measurement is made on a body subject only to the influence of Special Relativity (or observation of the frequency of spectral lines)... something "else" is going on to provide these velocities and it might be due to more than just "simple" Hubble Shift... and there are many theories of which I am not about to speak this time.
Clearly this force or acceleration does not transform according to normal Special Relativity because it is something felt "entirely" in the accelerating frame, it is not "between frames" in the way the velocity or the time (events) are transformed "between" relatively moving inertial frames. The accelerations are not entirely covariant forms like the the other system measurables. The acceleration is measured entirely in the accelerated frame and is "mg" for any "rocket", for instance, where g is a constant acceleration of the rocket and each and every point in an ideal "rigid" rocket will undergo that same acceleration.
So I think this answers the original question the way Einstein would have liked to have answered it. I am also using hindsight but I am quoting references and I have respected opinion backing this point of view. Okun's cautions were appropriate and through it all Einstein still shapes up according to his "principles". Sorry about the length but I look behind me and it is a monster of 26 pages of disjointed ideas all competing with each other without this unifying context. Little bits and pieces of other peoples opinions are partially OK but nobody bothered to define a problem in full.
Cheers
Not so safe... So acceleration in all kinematics is not even close? It "apparently" only applies when V = 0. Clearly if V = 0 then a = 0 and p =0 etc... his trivial case I presume. I entirely disagree. F = ma in all accelerated system "piecemealwise". It does not depend on any history so it actually means instantaneous acceleration of "proper mass". The change in velocity can be vanishingly small or "moderate" up to many gees... just not too extreme like a million gees for instance "instantaneously" and lasting over a very short time. If you let this "time" drag out then you must integrate along a special curve for a possible varying vector acceleration function. I do not want to go there it is too complicated for this discussion.
This would mean with some qualification that this change in momentum or force (... is related to the change in relative velocity between two snapshot frames... one at the beginning of a short boost and the other indicating the state of the frame after the short boost). Presumably since "p" in the first of Trout's equations is a relativistically altered momentum... this altered value must be what is "seen" in the observer frame happening to a particle in the "high velocity" frame... as V -> C p -> ∞. While in Trout's second equation the F quoted there to be consistent is what is measured in the observer frame (unprimed) from the unaccelerated frame (it must be an accelerated frame now since we have "forces" acting). We know this is correct where there is only a small relative velocity... this is Newtonian Mechanics... that's fine. As V -> C what does this equation F = dp/dt do? F = dp/dt = m dv/dt as V -> C dv/dt -> 0 (it's acceleration drops to zero as observed in the remote frame). Wrong... the acceleration "in frame" may be anything at all and is not specifically dependent on the relative frame velocity or even it's direction.
There is no "hint" of what might be seen in the observer frame. Clearly F = dp/dt in all "Newtonian systems" which means this is the standard momentum "in frame" not in the observer frame unless interframe velocity V -> 0, then it will be "close" (ignoring all de Broglie effects). Trout's equation B is a non-sequitur from equation A. The "in frame" acceleration is still F = ma where "a" is an acceleration determined "in frame" relative only to the first or second snapshot and not to the observation of the relative V from the observer frame. Trout ... never wants to be explicit but if you want to answer a real problem you must be very explicit. Goal posts will be rearranged and balls will be "hidden" in case anyone should take a "free kick". This is the problem I am trying to give an answer to so I stick whith this.
This equation F = dp/dt holds "only" if the mass here is the "intrinsic or invariant mass" or Trout's "proper mass" (measured in the "instantaneous" rest frame... it does not matter much if it is snapshot 1 or snapshot 2 since this is going to be the "instantaneous" acceleration) and "a" (F = ma) and "a' " (F' = ma') are accelerations and forces measured in the accelerated frame and unaccelerated "observer" frame of masses in those frames. Without some primed frame (observer frame) this is a meaningless set of useless and inconsistent equations and would be marked wrong in any examination by anyone who understands the subject matter. What any rest frame observer wants to know is what "a" is in the accelerated frame and not what "a' " is since as V -> C you will not be able to observe "a" directly since "a" is the proper acceleration or intrinsic acceleration in the relatively moving and accelerating frame. If you disagree then the question is "what else could it mean?"
As V -> C (the "initial" relative velocity between the inertial frame and the accelerated frame) the "acceleration" of "everything" in the accelerated frame observationally -> 0, everything becomes "relatively frozen" on the light cone wall. It is "difficult" to measure the acceleration of a rocket when V -> C especially if we are unsure of how space is "curved" between the observer and the observed due to non-inertial influences. We know "non-inertial influences" dominate the Universe... called Hubble expansion.
What the observer wants to know is if an "astronaut" in a rocket that is "still" undergoing acceleration as it approaches the speed of light and is "still feeling comfortable" and what acceleration "a" he "feels" in his frame is still currently within human limits. While relative V can approach C, since this V is only the relative velocity it has absolutely no effect on this intrinsic value of "a" in the accelerated frame which can vary instantaneously and may be arbitrarily aligned. An estimate of the instantaneous acceleration is possible only after correction between the frames piecemealwise.... Agreed? The imprecision in this figure is going to increase as V -> C the measurables of time and position are going to become more and more "difficult" according to the orthogonality of these measurables.
Fourier Transforms and Uncertainty
What it is not possible to say is that you simply take the second differential of the position vector to obtain the acceleration because the velocity V between the observer frame and the accelerated frame is not applicable and are "distorted" subject to length contraction and time dilation as seen from the observer frame. Common sense and Einstein's first Principle tells us that... that the laws of physics are the same in all frames (that is the meaning according to Einstein that is to be preserved in Special Relativity despite the minor problems stated previously). Even assuming a flatspace this second differential of the position vector does not give the correct acceleration unless the acceleration is in the same direction as the velocity (any "relativistic" distant receding particle) will always tend to appear to travel directly away from the origin of the observers inertial frame given enough time and space. If the "boost" is "nearly" in that direction (like a rocket boost) then the observed accelerations will tend to the value...
a = a' [ 1 - U²/C²]³⁄² .... this is what you are all looking for in the simplest form for the simplest problem possible.
So F = ma = F' = ma' etc... If U ≈0 then a = a' in the limit but this evaluation works at all low relative velocity (low compared with the speed of light). This has that caution in it so do not generalize this idea beyond its domain of applicability. This "evaluation" got us to the moon and I am sure it will still be good a million years from now regardless of how we want to calculate kinematics. Full details of the derivation are here...
Acceleration in Special Relativity
Ask a few questions if you want me to explain the finer details...
I would point to the composition of relativistic velocities as a starting point here...
How Do You Add Velocities in Special Relativity?
The rest of the derivation above is "simple" in that given the simplified problem it is the limit taken in regard to some elementary "fractions" and this treatment does not require any special differentiation or integration as suggested by some if we are trying to determine "instantaneous proper intrinsic relativistic acceleration"... a "lot of words" in case there is any doubt as to what this really means.
It is important to realize that this is only where the acceleration is in the same direction as the inter-frame velocity and that velocity is chosen as the X direction. A more general treatment for arbitrary acceleration in an arbitrary direction should be found here...
Relativistic force transformation - Valery P. Dmitriyev 13Jul 2005
It is not "simple" but it is "almost" straightforward but more necessarily complicated... Warning: there are some minor typos in this treatment but the overall results are correct.
Some additional "sticky" points... A "method" is required to allow us to compare the two accelerated frame "snapshots" with the observer frame and that is not so easily done using standard relativity using the relative velocity between the two "snapshot" frames where the time between the snapshots is "arbitrarily short" but not zero. Remember that in the remote observer inertial frame a second of time may be quite different to a "second" in the accelerated frame whose relative velocity may be approaching the speed of light (and the internal "proper velocity" is approaching "infinity" while the "proper acceleration" may remain "internally constant" lets say 1 gee). Remember the "subjective experience of time" in both frames is identical and there is no "increase in any mass" as V -> C. What high relativistic speed provides is the ability to cross vast distances of many light years of space in a short period of ship's time at the expense of dilation of the travelers time relative to his "stay at home twin"... though if that twin undergoes the same history of acceleration gravitationally as our traveler experiences purely relativistically, maybe there will be no time difference at all on a round trip at constant magnitude of 1 gee acceleration. The simplistic relativity equation for time dilation assumes no acceleration of the "stay at home" twin and large change in velocity for the accelerated traveler. In that simplified case it is the accelerated/deaccelerated particle that suffers the time dilation (check it out with a spring balance). It is definitely not a symmetrically interchangeable effect.
To complete this problem easily as shown in the references above you need the law of composition of relativistic velocities and then apply limiting conditions to the arbitrarily "boosted" particle. It is also important to realize that this relationship between the high velocity particle and the low velocity particle or observer frame is not "universal" but is dependent on important interactions between them to realize their full potential (such as the transfer of "energy"). No interaction then there is no energy. One type of interaction is for the space ship scenario to be a "round trip" where the identical twins begin and end the trip in the same inertial reference frame. Other pairs of particles do not include this high - low velocity relationship so it appears quite different ... what is a constant in all "similarly" accelerated frames at very different relative velocities is the same proper relativistic acceleration (... read simple in-frame acceleration) in all frames and it remains that same acceleration independent of any of the individual relative velocity relationships. That is Einstein's Relativity Principle.
While this "base" velocity "distorts" what is being "visually seen" in/from the observer or "traveler" frame this "base" velocity has absolutely no independent influence on the forces or the acceleration in the accelerated frame because these forces are independent of any "relative velocity" between any quasi-stationary "snapshot" of an accelerated frame (treated as an piecemeal unaccelerated frame) and the observer frame and it is solely dependent on this "absolute acceleration" which is felt only in the accelerating frame of which we have two "snapshots" some "short time and distance apart" viewed by/from the observer frame as a "pair" of relativistically moving "objects" at different times and positions "corrected" using the SR transformations.
We first determine the positions and times of events in the two relativistic snapshot frames from the corrected measurements from the observer frame. Once we know the relative positions and times of events between these two snapshots in the snapshot frames of reference, and because the difference between the two frames have only a small acceleration we can estimate this instantaneous acceleration (quite accurately) using Newtonian approximation (assuming the first snapshot was in a frame of "instantaneous rest"). That is where you can use the equation F = dp/dt or even F = ma (... the one we and NASA all know and love!).
We need a value for "a" (above) given a value of "a' " determined from measurements in the observer inertial frame... the measured relative in frame acceleration. The relative "speed" between the observing frame and the accelerating frame is not about to alter any factor in this estimate since it is an "in frame boost". This "boost" does not need to have any relationship whatsoever to the prevailing velocity between frames... not even to its direction. It is "hard" to distinguish a small acceleration in an arbitrary direction of a far off extreme relativistically moving object due to the imprecision and also due to the fewer number of executing events being recorded from the relativistic progression of observations. In the extreme case the object moving away just marginally under the speed of light becomes totally unobserved and even "static" as it recedes away from the observer suffering more and more extreme time dilation until a point is reached where almost no events are occurring in the observed phenomenon resulting in an "observational paradox" of observing "nothing happening" in the limiting case. with "nothing happening" there is no significant temporal progress in what is being observed... it is "frozen in time".
If "Special Relativity" alone was the main determinant in the kinematics of distant objects... none of these distant bodies would "appear" to recede faster than C/2. This is very easy to prove as a body travels at near the speed of light traversing one light second in distance in just marginally over one second of observer frame time, the consecutive flashes from this object can only propagate at a speed of C so the arrival frequency of consecutive flashes is one flash per two seconds while the particle "visibly recedes" at an uncorrected speed in the observer frame of C/2. However this is not the "true" velocity of the near luminally moving particle whose velocity is causing it to "streak ahead" in real terms of that perceived position at nearly the full speed of light that would be recorded by an army of "johnny's on the spot" at rest in the observer frame in close proximity and along the outward path of the particle and successive "johnny's on the spot" recording the particles progress with correctly synchronized clocks and comparing "notes" later to obtain that velocity of nearly C.
In fact there is another seeming paradox of many cosmic bodies appearing to recede at nearly C in the far off distance of the Universe (up to 99%C). This is "impossible" if only a single perspective measurement is made on a body subject only to the influence of Special Relativity (or observation of the frequency of spectral lines)... something "else" is going on to provide these velocities and it might be due to more than just "simple" Hubble Shift... and there are many theories of which I am not about to speak this time.
Clearly this force or acceleration does not transform according to normal Special Relativity because it is something felt "entirely" in the accelerating frame, it is not "between frames" in the way the velocity or the time (events) are transformed "between" relatively moving inertial frames. The accelerations are not entirely covariant forms like the the other system measurables. The acceleration is measured entirely in the accelerated frame and is "mg" for any "rocket", for instance, where g is a constant acceleration of the rocket and each and every point in an ideal "rigid" rocket will undergo that same acceleration.
So I think this answers the original question the way Einstein would have liked to have answered it. I am also using hindsight but I am quoting references and I have respected opinion backing this point of view. Okun's cautions were appropriate and through it all Einstein still shapes up according to his "principles". Sorry about the length but I look behind me and it is a monster of 26 pages of disjointed ideas all competing with each other without this unifying context. Little bits and pieces of other peoples opinions are partially OK but nobody bothered to define a problem in full.
Cheers
Hi Good Elf,
You have not added anything I did not already know or which clashes with anything I have posted here. But thanks for posting it.
You have not added anything I did not already know or which clashes with anything I have posted here. But thanks for posting it.
QUOTE (johanfprins+Oct 28 2008, 02:55 PM)
1. the clock in the primed system: Relative to the primed system at any time
t(/) one has that x(/)=0 : Relative to the unprimed system its corresponding coordinates must be t and x
No.
The co-ordinates for the primed clock (relative to the unprimed clock) are x(/) and t(/), which must be calculated using the transforms. You are confusing yourself by using x(/) to represent two different things: the primed clock's location in the primed frame (which is false) and the primed clock's location relative to the unprimed clock.
If that is what you havetried to prove, then you have not done it properly and youhave used poor notation. However, it is trivial andis exactly what every scientist knows is true.
What you have concluded is t=t(/) This is confusing because by convention, t(/) (primed t) is used to represent the time at frame B relative to another frame, A.
If you aren't trying to show that t=t', which is what I thought you were saying, then I have no disagreement with your conclusion, but it is trivial. Your manipulation of the transforms is false, but that's moot if you aren't trying to show that t=t'
t(/) one has that x(/)=0 : Relative to the unprimed system its corresponding coordinates must be t and x
No.
The co-ordinates for the primed clock (relative to the unprimed clock) are x(/) and t(/), which must be calculated using the transforms. You are confusing yourself by using x(/) to represent two different things: the primed clock's location in the primed frame (which is false) and the primed clock's location relative to the unprimed clock.
QUOTE
Obviously I have not tried to prove "that the time A sees at A is equal to the time A sees at B". What I have proved is that the time A sees at A is the same as the time that B sees at B. you will now probably say it is trivial: Even if it is why are you not able to follow the proof?
If that is what you havetried to prove, then you have not done it properly and youhave used poor notation. However, it is trivial andis exactly what every scientist knows is true.
What you have concluded is t=t(/) This is confusing because by convention, t(/) (primed t) is used to represent the time at frame B relative to another frame, A.
If you aren't trying to show that t=t', which is what I thought you were saying, then I have no disagreement with your conclusion, but it is trivial. Your manipulation of the transforms is false, but that's moot if you aren't trying to show that t=t'
QUOTE (johanfprins+Oct 28 2008, 03:13 PM)
I have not said that time dilation is not a real effect: It is a "real relativistic effect" in the sense that an observer A concludes that time dilates at the position of an observer B while in actual fact the time on the clock observed by observer B is changing at the same rate as the time on A's clock. Similarly it is a "real relativistic effect" in the sense that an observer B concludes that time dilates at the position of an observer A while in actual fact the time on the clock observed by observer A is changing at the same rate as the time on B's clock.
These perceptions are created by relative movement and must be taken into account when actions within one system can be observed from the other system. However, as Feynman said, if you cannot "look out", each system will be the same: i.e. a clock in system A will tick away at the same rate as in system B. What my derivation proves is that when you have an interaction from A observed from B and the opposite interaction in B observed from A, you get the exact same time rate as will be the case when you cannot "look out". This is exactly as it must be. This is another feather in the cap for Special Relativity!
Omnibus thus have some valid points: Relativistic effects are in effect "illusions" but these "illusions" have real physical effects for the observer who is experiencing them: And in order to model these REAL PHYSICAL effects, the Lorentz transformations have withstood the test of time!
In that case I'm not really sure what your original point was or who you thought you were in disagreement with.
These perceptions are created by relative movement and must be taken into account when actions within one system can be observed from the other system. However, as Feynman said, if you cannot "look out", each system will be the same: i.e. a clock in system A will tick away at the same rate as in system B. What my derivation proves is that when you have an interaction from A observed from B and the opposite interaction in B observed from A, you get the exact same time rate as will be the case when you cannot "look out". This is exactly as it must be. This is another feather in the cap for Special Relativity!
Omnibus thus have some valid points: Relativistic effects are in effect "illusions" but these "illusions" have real physical effects for the observer who is experiencing them: And in order to model these REAL PHYSICAL effects, the Lorentz transformations have withstood the test of time!
In that case I'm not really sure what your original point was or who you thought you were in disagreement with.
QUOTE (Omnibus+Oct 28 2008, 03:14 PM)
Have you heard of rhetorical questions?
Like I said, you are confused. Primed quantities are always referring to the primed system. Take some time to learn these things before posting.
Bleurgh...
Like I said, you are confused. Primed quantities are always referring to the primed system. Take some time to learn these things before posting.
Bleurgh...
QUOTE (bm1957+Oct 28 2008, 09:17 AM)
You are answering no to all questions?
In that case, I think you are beyond help, I give up.
Read my answer carefully. Also, read and try to understand what Einstein has in mind--I gave you two citations to ruminate over. With your questions you're trying to parse something that cannot be parsed.
In that case, I think you are beyond help, I give up.
Read my answer carefully. Also, read and try to understand what Einstein has in mind--I gave you two citations to ruminate over. With your questions you're trying to parse something that cannot be parsed.
QUOTE (johanfprins+Oct 28 2008, 03:19 PM)
I cannot remember that I did this: As outlined above the coordinates of the primed clock is x(/)=0 ant t(/) relative to the primed coordinate system while it is x and t relative to the unprimed system, The Lorentz transformations then connect these coordinates.
Similarly the coordinates of the unprimed clock is x=0 and t relative to the unprimed system and x(/) and t(/) relative to the primed system.
You didn't do that.
What I showed is what x(/) should have represented for the equation to be valid. You let x(/)=0, which is not what is required for the equation to be valid. That is where you blundered.
Similarly the coordinates of the unprimed clock is x=0 and t relative to the unprimed system and x(/) and t(/) relative to the primed system.
You didn't do that.
What I showed is what x(/) should have represented for the equation to be valid. You let x(/)=0, which is not what is required for the equation to be valid. That is where you blundered.
QUOTE (Omnibus+Oct 28 2008, 03:37 PM)
Read my answer carefully. Also, read and try to understand what Einstein has in mind--I gave you two citations to ruminate over. With your questions you're trying to parse something that cannot be parsed.
I asked you three direct yes/no questions.
Are you saying that you were being flippant when you said no, no, no?
Please just answer the first two honestly. If you answer correctly, I can explain your misunderstanding. If you answer incorrectly I can try to explain why you are incorrect. If you disagree at any point then we can discuss it.
Why do you have a problem with that?
I asked you three direct yes/no questions.
Are you saying that you were being flippant when you said no, no, no?
Please just answer the first two honestly. If you answer correctly, I can explain your misunderstanding. If you answer incorrectly I can try to explain why you are incorrect. If you disagree at any point then we can discuss it.
Why do you have a problem with that?
QUOTE
In that case I'm not really sure what your original point was or who you thought you were in disagreement with.
johanfprins is trying to say that although Lorentz transformations don’t give the real truth about spatial and temporal dimensions from within the moving system when observed by the stationary observer, they can be used by the stationary observer for practical purposes as engineering empirical equations. He thinks Lorentz equations somehow help the stationary observer to figure out from his own perspective for his own purposes what the story is from within the moving system.
Thus, although the moving observer sees things proper in his own frame, the deformed image provided by the Lorentz transformations can be of some use, johanfprins thinks, for the stationary observer who actually knows the image is deformed but he (the stationary observer) can always obtain the correct image by undoing them by applying the inverse Lorentz transformations.
This begs the question, why use equations to get a deformed image and then undo them to get the correct final answer?
Hi johanfprins et al,
You have a problem, not I. Your questions contain ambiguities such as “local obser” and “non-local observer”. There is no such a thing. These questions are senseless and you have to do some reading and thinking in order to understand what really is going on here.
Just a little help. Forget local and non-local and imagine you are in K and know a certain law of Physics in K. Both first postulate and Lorentz transformations give how another observe, an observer at rest with k will view the law you’re seeing in K. Obviously, the observer in k cannot see one and the same law in two different ways.
Does that help?
You need more than hitting the book, ginger Trout. You need to think. Because Physics books are not the Bible where you can get away on pure faith.
Theory of relativity cannot prove that m is variable. Therefore, the formula m=m(v)=m_0/sqrt(1-(v/c)^2) you’re using is illegitimate. You can slap any formula you wish, as you’ve done, but that doesn’t mean it’s valid.
The above formula is not valid because it contradicts the first postulate of the theory of relativity which requires that the mass m in a given system to be the same mass in all other systems. According to your formula m in one system is in another system m(v) =/= m for v =/= 0. That’s a fatal conflict with the first postulate.
Well, it’s another story if you start insisting that the first postulate isn’t valid. That’ll be very interesting. Go ahead, say it.
Like I said, forget about "local" and "non-local". They are confusing you. We're not discussing quantum mechanics here. Read very carefully what I told you, especially the last post which I hope will really help you understand what's really going on here.
You need the help not I. So don’t ask me “does that help”.
The above only shows your muddled thinking and unwillingness to understand things. I gave you two quotes from Einstein for rumination but you still refuse to understand that both first postulate and the Lorentz transformations, both of them are ways to transform a law from one frame into another. Both of them must give the same result. Read the quotes from Einstein i gave you and don’t fantasize.
This isn't your blunder. The transformations you present are correct. bm1957 is confused, he doesn't know what he's talking about as all his recent post prove.
Your blunder is that you accept x = x(/).
This is untenable because x is in K while x(/) is in k.
In order for x to be equal to x(/) the time t must be equal ti t(/) which is exactly what you were supposed to prove.
Yes, that's indeed your understanding but that's not what Einstein has in mind.
Recall from §6
"Evidently the two systems of equations found for system k must express exactly the same thing, since both systems of equations are equivalent to the Maxwell-Hertz equations for system K. "
or, say from §3
"We now have to prove that any ray of light, measured in the moving system, is propagated with the velocity c, if, as we have assumed, this is the case in the stationary system; for we have not as yet furnished the proof that the principle of the constancy of the velocity of light is compatible with the principle of relativity."
More importantly, try to understand your blunder when claiming that x = x(/). That unfounded acceptance kills your argument which otherwise is based on correct application of the Lorentz transformations.
QUOTE (johanfprins+)
Hi Good Elf,
You have not added anything I did not already know or which clashes with anything I have posted here. But thanks for posting it.
You have not added anything I did not already know or which clashes with anything I have posted here. But thanks for posting it.
If people agree with me I am very pleased. We would then be on the same track. I am not looking for "trouble" and if this assists anyone at all I am gratified and pleased to have done it. 27 pages of "stuff" is too deeply "mixed up and turgid" for my taste and even though my reply is "long" it is not nearly as long as what is currently in progress here... 
Thanks again... and Cheers
Thanks again... and Cheers
QUOTE (GoodElf+)
Trout has mentioned (above) that we should deal with change in momentum... that is correct up to a point but to say that F = ma is not true (see reference below) then state that F=dp/dt makes a mockery of convention... what is the difference since dp/dt = mdv/dt = ma? What else could it mean since there is actually only one mass not two?... but once again errors result when there is no common paradigm. One moment Trout admonishes the relativistic mass and the next moment he is using it?...
If you don't know physics, just say so.
You can't write dp/dt = mdv/dt because "m" is not a constant, it is variable by virtue of the fact that m=m(v)=m_0/sqrt(1-(v/c)^2) and "v" is also variable : v=v(t).
It is time for you to stop pretending.. Hit the books Little Elf.
If you don't know physics, just say so.
You can't write dp/dt = mdv/dt because "m" is not a constant, it is variable by virtue of the fact that m=m(v)=m_0/sqrt(1-(v/c)^2) and "v" is also variable : v=v(t).
It is time for you to stop pretending.. Hit the books Little Elf.
QUOTE
Why do you have a problem with that?
You have a problem, not I. Your questions contain ambiguities such as “local obser” and “non-local observer”. There is no such a thing. These questions are senseless and you have to do some reading and thinking in order to understand what really is going on here.
Just a little help. Forget local and non-local and imagine you are in K and know a certain law of Physics in K. Both first postulate and Lorentz transformations give how another observe, an observer at rest with k will view the law you’re seeing in K. Obviously, the observer in k cannot see one and the same law in two different ways.
Does that help?
QUOTE (Omnibus+Oct 28 2008, 03:48 PM)
johanfprins is trying to say that although Lorentz transformations don’t give the real truth about spatial and temporal dimensions from within the moving system when observed by the stationary observer, they can be used by the stationary observer for practical purposes as engineering empirical equations. He thinks Lorentz equations somehow help the stationary observer to figure out from his own perspective for his own purposes what the story is from within the moving system.
Thus, although the moving observer sees things proper in his own frame, the deformed image provided by the Lorentz transformations can be of some use, johanfprins thinks, for the stationary observer who actually knows the image is deformed but he (the stationary observer) can always obtain the correct image by undoing them by applying the inverse Lorentz transformations.
This begs the question, why use equations to get a deformed image and then undo them to get the correct final answer?
I just gave you an accidental pos! Lucky you
Do you know what 'begs the question' means? I think not.
You're trying to bring down relativity and you admit that you don't know what local and non-local mean?
*Holds head in hands, head drops and hits the desk repeatedly*
Thus, although the moving observer sees things proper in his own frame, the deformed image provided by the Lorentz transformations can be of some use, johanfprins thinks, for the stationary observer who actually knows the image is deformed but he (the stationary observer) can always obtain the correct image by undoing them by applying the inverse Lorentz transformations.
This begs the question, why use equations to get a deformed image and then undo them to get the correct final answer?
I just gave you an accidental pos! Lucky you
Do you know what 'begs the question' means? I think not.
QUOTE
You have a problem, not I. Your questions contain ambiguities such as “local obser” and “non-local observer”. There is no such a thing. These questions are senseless and you have to do some reading and thinking in order to understand what really is going on here.
You're trying to bring down relativity and you admit that you don't know what local and non-local mean?
*Holds head in hands, head drops and hits the desk repeatedly*
QUOTE (bm1957+Oct 28 2008, 02:56 PM)
I just gave you an accidental pos! Lucky you
Baaad
Baaad
QUOTE
If you don't know physics, just say so.
You can't write dp/dt = mdv/dt because "m" is not a constant, it is variable by virtue of the fact that m=m(v)=m_0/sqrt(1-(v/c)^2) and "v" is also variable : v=v(t).
It is time for you to stop pretending.. Hit the books Little Elf.
You can't write dp/dt = mdv/dt because "m" is not a constant, it is variable by virtue of the fact that m=m(v)=m_0/sqrt(1-(v/c)^2) and "v" is also variable : v=v(t).
It is time for you to stop pretending.. Hit the books Little Elf.
You need more than hitting the book, ginger Trout. You need to think. Because Physics books are not the Bible where you can get away on pure faith.
Theory of relativity cannot prove that m is variable. Therefore, the formula m=m(v)=m_0/sqrt(1-(v/c)^2) you’re using is illegitimate. You can slap any formula you wish, as you’ve done, but that doesn’t mean it’s valid.
The above formula is not valid because it contradicts the first postulate of the theory of relativity which requires that the mass m in a given system to be the same mass in all other systems. According to your formula m in one system is in another system m(v) =/= m for v =/= 0. That’s a fatal conflict with the first postulate.
Well, it’s another story if you start insisting that the first postulate isn’t valid. That’ll be very interesting. Go ahead, say it.
QUOTE (Omnibus+Oct 28 2008, 03:56 PM)
Just a little help. Forget local and non-local and imagine you are in K and know a certain law of Physics in K. Both first postulate and Lorentz transformations give how another observe, an observer at rest with k will view the law you’re seeing in K. Obviously, the observer in k cannot see one and the same law in two different ways.
Does that help?
This is your mistake.
The first postulate says that the observer in k will see that same law in k as you are seeing in K.
The transforms tell you how you will see the law in k and how the observer in k will see the law in K.
Does that help?
Does that help?
This is your mistake.
The first postulate says that the observer in k will see that same law in k as you are seeing in K.
The transforms tell you how you will see the law in k and how the observer in k will see the law in K.
Does that help?
QUOTE
You're trying to bring down relativity and you admit that you don't know what local and non-local mean?
Like I said, forget about "local" and "non-local". They are confusing you. We're not discussing quantum mechanics here. Read very carefully what I told you, especially the last post which I hope will really help you understand what's really going on here.
QUOTE (Omnibus+Oct 28 2008, 04:11 PM)
Like I said, forget about "local" and "non-local". They are confusing you. We're not discussing quantum mechanics here. Read very carefully what I told you, especially the last post which I hope will really help you understand what's really going on here.
Read my reply above, it explains your gross misconception.
Read my reply above, it explains your gross misconception.
QUOTE (bm1957+Oct 28 2008, 04:40 PM)
What I showed is what x(/) should have represented for the equation to be valid. You let x(/)=0, which is not what is required for the equation to be valid. That is where you blundered.
I did not blunder anywhere:
First epoch: Coordinates of clock stationary in unprimed system:
Relative to unprimed system: x=0, t
Relative to primed system: x(/), t(/)
Coordinates x(/) t(/) calculated by Lorentz transformation from x=0, t: i.e. transforming correctly from unprimed to primed system
Second epoch: Coordinates of clock stationary in primed system:
Relative to primed system: x(/)=0, t(/)
Relative to unprimed system: x, t
Coordinates x, t calculated by Lorentz transformation from x(/)=0, t(/): i.e. ftransforming correctly rom primed to unprimed system
So where did I let x(/)=0 in such a a manner that the transformations are wrong?
You are becoming like Omnibus: Just repeating the same wrong conclusion!
I did not blunder anywhere:
First epoch: Coordinates of clock stationary in unprimed system:
Relative to unprimed system: x=0, t
Relative to primed system: x(/), t(/)
Coordinates x(/) t(/) calculated by Lorentz transformation from x=0, t: i.e. transforming correctly from unprimed to primed system
Second epoch: Coordinates of clock stationary in primed system:
Relative to primed system: x(/)=0, t(/)
Relative to unprimed system: x, t
Coordinates x, t calculated by Lorentz transformation from x(/)=0, t(/): i.e. ftransforming correctly rom primed to unprimed system
So where did I let x(/)=0 in such a a manner that the transformations are wrong?
You are becoming like Omnibus: Just repeating the same wrong conclusion!
QUOTE
The first postulate says that the observer in k will see that same law in k as you are seeing in K.
The transforms tell you how you will see the law in k and how the observer in k will see the law in K.
The transforms tell you how you will see the law in k and how the observer in k will see the law in K.
You need the help not I. So don’t ask me “does that help”.
The above only shows your muddled thinking and unwillingness to understand things. I gave you two quotes from Einstein for rumination but you still refuse to understand that both first postulate and the Lorentz transformations, both of them are ways to transform a law from one frame into another. Both of them must give the same result. Read the quotes from Einstein i gave you and don’t fantasize.
QUOTE
So where did I let x(/)=0 in such a a manner that the transformations are wrong?
This isn't your blunder. The transformations you present are correct. bm1957 is confused, he doesn't know what he's talking about as all his recent post prove.
Your blunder is that you accept x = x(/).
This is untenable because x is in K while x(/) is in k.
In order for x to be equal to x(/) the time t must be equal ti t(/) which is exactly what you were supposed to prove.
QUOTE (bm1957+Oct 28 2008, 05:09 PM)
The first postulate says that the observer in k will see that same law in k as you are seeing in K.
The transforms tell you how you will see the law in k and how the observer in k will see the law in K.
You are finally getting there:
Your first sentence: That is why the clocks are ticking away at the same rate (as I have derived above)
Your second sentence: This is why a "stationary" observer will experience a time-dilation for a clock moving relative to him/her, although there is not an actual time-dilation within the moving reference frame.
The transforms tell you how you will see the law in k and how the observer in k will see the law in K.
You are finally getting there:
Your first sentence: That is why the clocks are ticking away at the same rate (as I have derived above)
Your second sentence: This is why a "stationary" observer will experience a time-dilation for a clock moving relative to him/her, although there is not an actual time-dilation within the moving reference frame.
QUOTE
Your first sentence: That is why the clocks are ticking away at the same rate (as I have derived above)
Your second sentence: This is why a "stationary" observer will experience a time-dilation for a clock moving relative to him/her, although there is not an actual time-dilation within the moving reference frame.
Your second sentence: This is why a "stationary" observer will experience a time-dilation for a clock moving relative to him/her, although there is not an actual time-dilation within the moving reference frame.
Yes, that's indeed your understanding but that's not what Einstein has in mind.
Recall from §6
"Evidently the two systems of equations found for system k must express exactly the same thing, since both systems of equations are equivalent to the Maxwell-Hertz equations for system K. "
or, say from §3
"We now have to prove that any ray of light, measured in the moving system, is propagated with the velocity c, if, as we have assumed, this is the case in the stationary system; for we have not as yet furnished the proof that the principle of the constancy of the velocity of light is compatible with the principle of relativity."
More importantly, try to understand your blunder when claiming that x = x(/). That unfounded acceptance kills your argument which otherwise is based on correct application of the Lorentz transformations.
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