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Omnibus
AlphaNumeric, the following is correct, regarding the Lorentz transformations:
QUOTE
If x=0 at t=0 and x'=0 at t'=0 and they are initially set to have t=t'=0, this doesn't mean that t'=t for all t or t' (which ever you are taking as the independent variable). For an instant their clocks will agree and then they will not. The gamma factor tells you that immediately. It's like saying :

"Let T = 2(t-t0). To set origins equal we put t0=0, so initially, at t=0 T=0. Therefore t=T for all t."

Obviously false but it's what you're doing.

You do realise these are exercises given to school children? I help teach 'Relativity and Motion' at my university and time dilation is included. Of course they don't do it with scalar equations, they aren't up to speed on using tensor calculus yet, they are only 1st years.

However, it does not help to explain away the fact the one particle in one state in k, described by F' = ma' is represented by two different states in K, namely, F = ma and F = beta^3ma, if we believe both the “Principle of Relativity” and the Lorentz transformations are physically correct. A single particle in a given state in a given system cannot be represented by a particle having two different states in another system, don’t you agree?
buttershug
QUOTE (Omnibus+Oct 25 2008, 02:04 PM)
AlphaNumeric, the following is correct, regarding the Lorentz transformations:

However, it does not help to explain away the fact the one particle in one state in k, described by F' = ma' is represented by two different states in K, namely, F = ma and F = beta^3ma, if we believe both the “Principle of Relativity” and the Lorentz transformations are physically correct. A single particle in a given state in a given system cannot be represented by a particle having two different states in another system, don’t you agree?

I've seen other papers comparing old techniques to new.
The authors of those are not saying the old way is correct. They are saying it's how it was done.

Do you agree that Newton believed that F=ma?
Do you agree that for high speeds it is not accurate?
Omnibus
QUOTE
Do you agree that Newton believed that F=ma?

Yes, I do.

QUOTE (->
 QUOTE Do you agree that Newton believed that F=ma?

Yes, I do.

Do you agree that for high speeds it is not accurate?

Yes I do but not because Eisntein's theory derives it. Einstein's theory cannot derive that inaccuracy because its first postulate (the "Principle of Relativity") is in conflict with the Lorentz transformations which invalidates it prior to any intention to claim anything following from it, as I've already shown.
buttershug
QUOTE (Omnibus+Oct 25 2008, 04:31 PM)
Yes, I do.

Yes I do but not because Eisntein's theory derives it. Einstein's theory cannot derive that inaccuracy because its first postulate (the "Principle of Relativity") is in conflict with the Lorentz transformations which invalidates it prior to any intention to claim anything following from it, as I've already shown.

And everyone has shown you that it is a misreading of the paper that makes you think that.

You should read up on that genious from India that had no formal training. He came up with lots of things that are accurrate but no one knows how he did it.
Not even him. But not knowing how he came up with it all doesn't make it wrong.
Omnibus
QUOTE
And everyone has shown you that it is a misreading of the paper that makes you think that.

That's called wishful thinking. You have no grounds to say that. These are empty words because if they were not you would've been able to shoot me down immediately with an argument instead of just an empty statement. Anyone can move the air with statements, such as yours above, about literally anything.

QUOTE (->
 QUOTE And everyone has shown you that it is a misreading of the paper that makes you think that.

That's called wishful thinking. You have no grounds to say that. These are empty words because if they were not you would've been able to shoot me down immediately with an argument instead of just an empty statement. Anyone can move the air with statements, such as yours above, about literally anything.

You should read up on that genious from India that had no formal training. He came up with lots of things that are accurrate but no one knows how he did it.
Not even him. But not knowing how he came up with it all doesn't make it wrong.

Training has noting to do with what we're discussing. The fact is, a you point out, that that Indian guy is right, while Einstein is wrong, as I've demonstrated conclusively.
buttershug
QUOTE (Omnibus+Oct 25 2008, 05:20 PM)
That's called wishful thinking. You have no grounds to say that. These are empty words because if they were not you would've been able to shoot me down immediately with an argument instead of just an empty statement. Anyone can move the air with statements, such as yours above, about literally anything.

Training has noting to do with what we're discussing. The fact is, a you point out, that that Indian guy is right, while Einstein is wrong, as I've demonstrated conclusively.

But the said that F<>ma at high speeds.

He made predicitons and they came true.

Ignore that part where you think he said that F=ma, what about the rest of what he said?
Omnibus
QUOTE
But the said that F<>ma at high speeds.

He may have said that but these are mere words. He has not derived that.

The ideas that m is velocity dependent were in the air at that time, and Einstein fixed himself in that crowd through cheating, pretending that such velocity dependence may be derived by first transforming F = ma from K into k by using one way of transforming it (using the "Principle of Relativity") and then, when F' = ma' became available, cheat and "forget" how that F' = ma' was actually obtained (and what it should inevitably be in K), and transform it back into K in a different way by using Lorentz transformations.

The result from that second way of transforming is something completely different from what should have been obtained if Einstein had not finagled.

If Einstein had not finagled and had not "forgotten" that F' = ma' had been obtained from F = ma and therefore, inevitably, that F' = ma' in k should become exactly F = ma in K (and not F = beta^3ma, as he erroneously obtains) he shouldn't even have published his paper because such crucial contradiction invalidates a theory and publishing an invalid theory is useless.
johanfprins
[QUOTE=AlphaNumeric,Oct 25 2008, 10:56 AM] Yeah. one of those 'bigots' who helped build the transistor and the GPS system. And whose predictions match all experiments.[/QUOTE]

You are not meaning Bardeen do you? He was a great man, but the BCS model is wrong.

QUOTE=AlphaNumeric,Oct 25 2008, 10:56 AM] All you say is "They tick the same" and then ignore experiments which show otherwise as 'misunderstood'. Simply denying an experiment doesn't mean it hasn't been done.
If x=0 at t=0 and x'=0 at t'=0 and they are initially set to have t=t'=0, this doesn't mean that t'=t for all t or t' (which ever you are taking as the independent variable). For an instant their clocks will agree and then they will not. The gamma factor tells you that immediately. It's like saying :

"Let T = 2(t-t0). To set origins equal we put t0=0, so initially, at t=0 T=0. Therefore t=T for all t."

Obviously false but it's what you're doing.[/QUOTE]

That is not what I am saying. Use the Lorentz transformation to transform the coordinates x(/)=0 and t(/) the time at X(/)=0: Then calculate the results using the TOTAL transformation: Guess what you get: x=VT: The Galilean transformation: Are you so stupid that you cannot do grade 5 algebra? It seems you are.

[QUOTE=AlphaNumeric,Oct 25 2008, 10:56 AM] You do realise these are exercises given to school children? I help teach 'Relativity and Motion' at my university and time dilation is included. [/QUOTE]

God help England!! Thank God you are not teaching my children. A bigger idiot than you never existed.

[QUOTE=AlphaNumeric,Oct 25 2008, 10:56 AM] Are you a physicist? I doubt it. And given the extremely vitriolic attitude you have towards them, or rather us, it would seem you failed to be any good at it when it came to crunch time and passing exams. [/QUOTE]

I do not attack real physicists: Just certifiable idiots like you!

[QUOTE=AlphaNumeric,Oct 25 2008, 10:56 AM] And for the record, I have a degree in mathematics, a masters in theoretical physics and am in my 3rd year of a string theory PhD. [/QUOTE]

For the record, I have a BSc in physics and Mathematics, An MSc in Physics, A DSc in Materials Science a very good citation index: I have taught Physics at Universities (including relativity theory) and worked in Industry. Therefore I have kept myself busy with reality not BS like "string theory". You are obviously going to be cuddled within a university career teaching BS to unsuspecting young people who, like you, are too stupid to understand reality from cloud cuckoo land.
Omnibus
johanfprins, what is x(/) versus X(/)?
ASTERIX*
Omnibus? .... or is it that twatfuck, Pentcho Valev? - who cares?

For everyone else who's not a complete knobspank :

Anti-Einstein Cranks.

More

Yet more

There's over 300k more links to these retarded abominations - feel free to browse yourself.

These links will serve to illuminate the type of groin-beating idiot who's started this thread.
Omnibus
QUOTE (ASTERIX*+Oct 25 2008, 08:04 PM)
Omnibus? .... or is it that twatfuck, Pentcho Valev? - who cares?

For everyone else who's not a complete knobspank :

Anti-Einstein Cranks.

More

Yet more

There's over 300k more links to these retarded abominations - feel free to browse yourself.

These links will serve to illuminate the type of groin-beating idiot who's started this thread.

Don't be a Nazi, don't give me 100 arguments. If you were right one should have been enough. For instance, just answer this one question--is it possible one electron whose state of motion in a given system k is described by F' = ma' to be in two different states of motion, namely F = ma and F = beta^3ma, in another system K, as Einstein derives ... unless beta = 1, that is, unless Lorentz transformations are invalid?
dimazin
QUOTE (ASTERIX*+Oct 26 2008, 01:04 AM)
Omnibus? .... or is it that twatfuck, Pentcho Valev? - who cares?

For everyone else who's not a complete knobspank :

Anti-Einstein Cranks.

More

Yet more

There's over 300k more links to these retarded abominations - feel free to browse yourself.

These links will serve to illuminate the type of groin-beating idiot who's started this thread.

The Earth moves in space more quickly than the Sun! A satellite of the Earth moves in space more quickly than the Earth!

According to symmetry of the relativity a satellite observer should to see dilation of time on the Earth, but really he can see only narrowing of Earth time . Presence of angular speed in calculations proves it. The theory of the relativity is almost correct only for the observer of an axis of rotation of the Earth. Formulas of Newton are completely correct only for speed of motion in space, without taking into account speed of movement into the future.
ASTERIX*
QUOTE (Omnibus+Oct 26 2008, 02:10 AM)
Don't be a Nazi, don't give me 100 arguments. If you were right one should have been enough. For instance, just answer this one question--is it possible one electron whose state of motion in a given system k is described by F' = ma' to be in two different states of motion, namely F = ma and F = beta^3ma, in another system K, as Einstein derives ... unless beta = 1, that is, unless Lorentz transformations are invalid?

Oh sorry, was I the absolute wankstain who's suggested Einsteins seminal groundbreaking 2nd 1905 paper has no worth? - without Einsteins 'vision', would Lorenz's transform ever been applied to time dilation?

p.s you're a nobody assboil and we all know it.

johanfprins
QUOTE (Omnibus+Oct 25 2008, 04:04 PM)
However, it does not help to explain away the fact the one particle in one state in k, described by F' = ma' is represented by two different states in K, namely, F = ma and F = beta^3ma, if we believe both the “Principle of Relativity” and the Lorentz transformations are physically correct. A single particle in a given state in a given system cannot be represented by a particle having two different states in another system, don’t you agree?

I have got paragraph 10 in front of me: Einstein's first set of equations relates to the time JUST AFTER AN ELECTRON STARTS TO ACCELERATE: QUOTE: "if an electron is AT REST at a given epoch the motion of the electron ensues in the next instant of time according to equations: (F=ma). This is for an electron starting off from zero speed: Thus these equations are THUS only valid at low speed.

THEN: "Now, secondly, let the velocity of the electron at a given epoch by v. We seek the law of motion of the electron in the immediate ensuing instants of time." THE ELECTRON CAN NOW MOVE AT A HIGH SPEED. Einstein then went ahead and derived for this specific "epoch" or condition that F=(beta)^2ma.

Clearly in the first epoch the calculation is made when the initial speed of the electron is ZERO: THE ELECTRON THUS MOVES AT A SLOW SPEED!!

In the second epoch the electron starts off with a speed v>0 and this speed can thus be MUCH larger than ZERO!!!

So Einstein derived two formulas: One for low speed and one for high speed. And as it must the formula at high speed becomes the formula at low speed when the speed goes to zero!!

Obviously F=ma is valid at low speeds and E=(beta)^3ma is valid at high speeds within the SAME REFERENCE FRAME!!SO WHAT IS YOUR BEEF?
Steven Christopher
18 • Intentionally ignoring Steven's threads is sign of defeat
johanfprins
NOW LET US DO SOME "REAL PHYSICS": Consider the following gedanken experiment (If you do not know what this is Alphanumeric, ask your String Theory professors. And if they cannot explain it to you, stop doing your PhD).

We have two space travelers within their space ships far far away from any galaxy. They have switched off their engines and are coasting towards each other and eventually pass each other. When passing each other they synchronize their clocks. After they have passed each other, which clock is moving and which clock is not moving?
Omnibus
QUOTE
Obviously F=ma is valid at low speeds and E=(beta)^3ma is valid at high speeds within the SAME REFERENCE FRAME!!SO WHAT IS YOUR BEEF?

That’s not at all obvious because F = ma is the representation in K of F’ = ma’ in k. However, Einstein claims that F = beta^3ma is also the representation in K of F’ = ma’. This is impossible. One state of motion, that is F’ = ma’, of one electron in k must be represented by just one state of this one electron in K. This is a crucial contradiction which invalidates the Lorentz transformations as physically viable transformations (theory of relativity is invalidated due to internal contradiction still in §1 and §2 due to internal contradictions in it).

Omnibus
johanfprins, I asked you a question--what is x(/) and what is X(/) in your earlier posting?
Omnibus
QUOTE (johanfprins+Oct 26 2008, 09:35 AM)
NOW LET US DO SOME "REAL PHYSICS": Consider the following gedanken experiment (If you do not know what this is Alphanumeric, ask your String Theory professors. And if they cannot explain it to you, stop doing your PhD).

We have two space travelers within their space ships far far away from any galaxy. They have switched off their engines and are coasting towards each other and eventually pass each other. When passing each other they synchronize their clocks. After they have passed each other, which clock is moving and which clock is not moving?

I already told you, this is not the way to prove that the Lorentz transformations are non-physical. Jumping from frame to frame is a rookie approach which the zealots love to talk about and debunk. If you want to prove the physical (not mathematical; mathematically Lorentz transformations are internally consistent) inconsistency of the Lorentz transformations stay in one frame.
johanfprins
QUOTE (Omnibus+Oct 26 2008, 04:37 PM)

That’s not at all obvious because F = ma is the representation in K of F’ = ma’ in k.

Not when k and K move relative to each other at high speed v:

Dear Dear Omnibus,

AT low speeds F=ma is correct within any reference frame. So what Einstein did in his "second epoch" was to assume a high speed v within the reference frame K(x,y,z) when acceleration starts off: There must thus be another reference frame K(/) moving with the speed v relative to K. Within this reference frame the acceleration is starting off with an initial speed equal to zero and therefore within this reference frame you have that F=ma; just as it is within reference frame K when it is starting off from zero speed in K. But in K it now starts off with a speed v, and Einstein thus derives the formula when you start off with speed v by transforming from reference frame K(/), in which the acceleration starts of with zero speed to the reference frame K in which it starts off with a non-zero speed: He then obtains F=(beta)^3ma which is different from F=ma IN THE SAME REFERENCE FRAME IF THE ACCELERATION WOULD HAVE STARTED OFF FROM ZERO SPEED.

THUS ALTHOUGH YOU HAVE TWO FORMULAS WITHIN THE SAME REFERENCE FRAME, THE ONE (F=ma) IS FOR LOW SPEEDS WHILE THE OTHER (F-(beta)^3ma) IS FOR HIGH SPEEDS. FURTHERMORE THE LATTER EQUATION BECOMES THE FIRST WHEN V GOES TO ZERO: AS IT MUST!!!
Omnibus
QUOTE (johanfprins+Oct 26 2008, 09:55 AM)
Not when k and K move relative to each other at high speed v:

Dear Dear Omnibus,

AT low speeds F=ma is correct within any reference frame. So what Einstein did in his "second epoch" was to assume a high speed v within the reference frame K(x,y,z) when acceleration starts off: There must thus be another reference frame K(/) moving with the speed v relative to K. Within this reference frame the acceleration is starting off with an initial speed equal to zero and therefore within this reference frame you have that F=ma; just as it is within reference frame K when it is starting off from zero speed in K. But in K it now starts off with a speed v, and Einstein thus derives the formula when you start off with speed v by transforming from reference frame K(/), in which the acceleration starts of with zero speed to the reference frame K in which it starts off with a non-zero speed: He then obtains F=(beta)^3ma which is different from F=ma IN THE SAME REFERENCE FRAME IF THE ACCELERATION WOULD HAVE STARTED OFF FROM ZERO SPEED.

THUS ALTHOUGH YOU HAVE TWO FORMULAS WITHIN THE SAME REFERENCE FRAME, THE ONE (F=ma) IS FOR LOW SPEEDS WHILE THE OTHER (F-(beta)^3ma) IS FOR HIGH SPEEDS. FURTHERMORE THE LATTER EQUATION BECOMES THE FIRST WHEN V GOES TO ZERO: AS IT MUST!!!

That's incorrect.
johanfprins
QUOTE (Omnibus+Oct 26 2008, 04:44 PM)
johanfprins, I asked you a question--what is x(/) and what is X(/) in your earlier posting?

A typing error
johanfprins
QUOTE (Omnibus+Oct 26 2008, 04:57 PM)
That's incorrect.

It is NOT INCORRECT that Einstein derived first (his first epoch) the formula when acceleration starts of with zero speed relative to an inertial reference frame; while in his second epoch he derived the formula when acceleration starts off at a high speed v relative to the same inertial reference frame. Read his paper!!

You are now REALLY starting to waste every body's time
Omnibus
At the start of the derivation F = ma is the only known equation in K for any velocity v (never mind the blabber that “the electron is at rest at a given epoch”). Nothing more than that can be presumed. For instance, it cannot be presumed that m is v-dependent. It doesn’t matter that there may be experimental results indicating change of m with v; this change has to be derived, not used as an initial condition, thus committing the crucial error in reasoning called “petitio principii” (begging the question).

First postulate tells us that an observer at rest with k sees that F = ma in K as F’ = ma’ in k for any velocity v. Conversely, first postulate tell us that F’ = ma’ in k must be nothing else but F = ma in K for any velocity v.

Lorentz transformations applied to F’ = ma’ tell us, however, that for any velocity v the expression F = beta^3ma must be the equation in K for any velocity v.

This is a crucial contradiction. On the one hand, we know without presuming that m is v dependent (as we should approach the problem if we are to reason correctly), that in K for any velocity v the law is F = ma. On the other, we are told that using transformations, F = beta^3ma is the correct law in K for any velocity v. Suppose, we don’t know anything about the experimental results, why should we prefer F = beta^3ma over F = ma? Because the latter is more beautiful maybe? That’s an aesthetic argument, not scientific. Scientifically, if we are not to presume anything (as we should act as scientists) we must require that beta = 1 because only in this way the two expressions will show the one and the same state of motion of one electron in one system K.
Omnibus
QUOTE (johanfprins+Oct 26 2008, 09:59 AM)
A typing error

Can you give the argument again without the typo?
johanfprins
Omnibus,

If you are not able to understand simple English, even though Einstein's style is ponderous, I cannot help you. I do not believe, like Alphanumeric, that people should be written off as cranks when they disagree with you. One should try and argue logic on both sides. I am sorry, but after many attempts, you come back with the same refrain: in total contradiction to what Einstein has written and done in his paper, and also in total contradiction to scientifically sound and logical arguments which have been raised by myself and many others on this thread. I do not want to accuse you of being a crank; but I am now starting to experience you as being one.

Thus I am not going to even try and argue with you. If you want to believe that Einstein's model should be banned from physics: So be it. Agitate for it by demonstrating with a placard alongside all the people who claim "The end is nigh!!Goodbye
Omnibus
QUOTE (johanfprins+Oct 26 2008, 10:53 AM)
Omnibus,

If you are not able to understand simple English, even though Einstein's style is ponderous, I cannot help you. I do not believe, like Alphanumeric, that people should be written off as cranks when they disagree with you. One should try and argue logic on both sides. I am sorry, but after many attempts, you come back with the same refrain: in total contradiction to what Einstein has written and done in his paper, and also in total contradiction to scientifically sound and logical arguments which have been raised by myself and many others on this thread. I do not want to accuse you of being a crank; but I am now starting to experience you as being one.

Thus I am not going to even try and argue with you. If you want to believe that Einstein's model should be banned from physics: So be it. Agitate for it by demonstrating with a placard alongside all the people who claim "The end is nigh!!Goodbye

You are not going to argue more because you don't have an argument. Accept that Einstein is in crucial error and let's move on.
bm1957
QUOTE (johanfprins+Oct 26 2008, 03:35 PM)
We have two space travelers within their space ships far far away from any galaxy. They have switched off their engines and are coasting towards each other and eventually pass each other. When passing each other they synchronize their clocks. After they have passed each other, which clock is moving and which clock is not moving?

It's relative. Either one, or both; take your pick. What's your point?
Trout
QUOTE (Omnibus+Oct 26 2008, 04:03 PM)
You are not going to argue more because you don't have an argument. Accept that Einstein is in crucial error and let's move on.

Bzzt, no. You are simply unable to learn the difference between proper and relativistic mass and you blame Einstein and relativity for your ignorance. You are not the first to do this, you will not be the last.
Omnibus
QUOTE (Trout+Oct 26 2008, 06:02 PM)
Bzzt, no. You are simply unable to learn the difference between proper and relativistic mass and you blame Einstein and relativity for your ignorance. You are not the first to do this, you will not be the last.

This has already been addressed. "Proper" ("rest") and "relativistic" mass can only be derived if the crucial contradiction I am demonstrating were not there. It exists, however, and no concepts such as "rest", "relativistic" or whatever else mass, let alone mass-energy relationship, popularly known as E = mc^2, can ever be derived based on the "theory" of relativity. Not that mass-energy relationship does not exist (it does; it even exists classically in the Poynting vector). What is shown here is that it is specifically the "theory" of relativity that cannot derive it, as well as anything else claimed to be derived by that "theory".
Omnibus
Trout, you might have come here first to discuss the issue before slapping feedback. As seen you're the one who's in error. Now what?
Trout
QUOTE (Omnibus+Oct 26 2008, 11:19 PM)
It exists, however, and no concepts such as "rest", "relativistic" or whatever else mass, let alone mass-energy relationship, popularly known as E = mc^2, can ever be derived based on the "theory" of relativity.

This is yet another misconception of yours . I'll give you another neg next week, don't worry about that .
Omnibus
Just saying "This is yet another misconception of yours" doesn't cut the mustard. Where's your argument? You can give me as many negs as you wish, who can stop you. That won't do any good, though, until you come up with an argument.
eyeque
That's right, Unfortunatly i freak out about the neggs and do my block because they besmerch my identity.
Omnibus
QUOTE (eyeque+Oct 26 2008, 07:07 PM)
That's right, Unfortunatly i freak out about the neggs and do my block because they besmerch my identity.

Trout
QUOTE (Omnibus+Oct 27 2008, 12:03 AM)
Just saying "This is yet another misconception of yours" doesn't cut the mustard. Where's your argument?

My argument is that you don't have an argument.
Omnibus
QUOTE (Trout+Oct 26 2008, 07:09 PM)
My argument is that you don't have an argument.

Mildly funny.
prometheus
QUOTE (Omnibus+Oct 26 2008, 11:19 PM)
...let alone mass-energy relationship, popularly known as E = mc^2, can ever be derived based on the "theory" of relativity.

E = mc^2 certainly can be derived from SR. All you have to do is work out the norm of the momentum four vector thusly:

User posted image: User posted image

QUOTE (Omnibus+Oct 26 2008, 11:19 PM)
Not that mass-energy relationship does not exist (it does; it even exists classically in the Poynting vector).

I have no idea how you work this out. Poynting vectors come up when you try to prove that the electric and magnetic fields are conserved, that is they don't disappear somewhere and reappear in others.

Your arguing technique is not good either. You complain when people say "you're wrong" and don't write a long post to explain why, yet you write gibberish like the nonsense about Poynting vectors and claim it as a proof. It is not even close to disproving relativity, something which people have been trying to do for 100 years and failed. Also, I find it rather annoying that you put theory in quotes every time you write "theory of relativity" as if this were somehow a point against it. There are many theories of physics that are unphysical, yet they are interesting because they provide insight into a more complicated theory for example. SR is an example of the theory that is valid above a certain scale, below which there is something more fundamental.

You have yet to explain why the predictions of SR are borne out in experiments so well - do you by any chance have your own pet theory that conflicts with special relativity that you're waiting to spring on us?
johanfprins
QUOTE (bm1957+Oct 26 2008, 11:23 PM)
It's relative. Either one, or both; take your pick. What's your point?

My point is the following: If a moving clock ticks slower, which clock is ticking slower? Each clock must be ticking slower than the other: But this is am absurdity. One cannot be shorter than your brother AND you brother shorter than you. Thus the two clocks must in actual reality tick at the same rate. Time-dilation is NOT an actual slow-down of a clock when it moves relative to another clock. Thus Haefele and Keating's experiment MUST be flawed: They obtained what they wanted to obtain.

The fact is that an observer experiences the clock moving relative to him/her as ticking slower when he/she has to communicate with the other observer by means of electromagnetic waves. Similarly the other observer experiences the clock of the original observer to tick slower when communicating with this observer. This is NOT A REAL SLOWING DOWN OF THE CLOCKS, BUT IS CAUSED BY THE FACT THAT BOTH OBSERVERS EXPERIENCE LIGHT SPEED TO BE REFERENCED RELATIVE TO HIM/HER. BOTH OBSERVERS EXPERIENCE HIS/HER REFERENCE FRAME AS THE STATIONARY REFERENCE FRAME IN OUR UNIVERSE. BUT IN ACTUAL FACT THE CLOCKS ARE TICKING AWAY AT EXACTLY THE SAME RATE.
bm1957
QUOTE (johanfprins+Oct 27 2008, 08:49 AM)

My point is the following: If a moving clock ticks slower, which clock is ticking slower? Each clock must be ticking slower than the other: But this is am absurdity. One cannot be shorter than your brother AND you brother shorter than you. Thus the two clocks must in actual reality tick at the same rate. Time-dilation is NOT an actual slow-down of a clock when it moves relative to another clock. Thus Haefele and Keating's experiment MUST be flawed: They obtained what they wanted to obtain.

The fact is that an observer experiences the clock moving relative to him/her as ticking slower when he/she has to communicate with the other observer by means of electromagnetic waves. Similarly the other observer experiences the clock of the original observer to tick slower when communicating with this observer. This is NOT A REAL SLOWING DOWN OF THE CLOCKS, BUT IS CAUSED BY THE FACT THAT BOTH OBSERVERS EXPERIENCE LIGHT SPEED TO BE REFERENCED RELATIVE TO HIM/HER. BOTH OBSERVERS EXPERIENCE HIS/HER REFERENCE FRAME AS THE STATIONARY REFERENCE FRAME IN OUR UNIVERSE. BUT IN ACTUAL FACT THE CLOCKS ARE TICKING AWAY AT EXACTLY THE SAME RATE.

Are you arguing that the clocks' tick rate as measured locally is identical? Because that's pretty obvious.

This is no argument that the change is not real though. If simultaneity could be agreed on, then yes, you would have a point. But the fact that simultaneity is relative means that your argument holds no water I'm afraid.

It certainly sounds like you don't understand what's happening properly, I would suggest that you swallow a humple pill for a second and at least be open to the fact that AN might be right. If you stop with the insults, he might even be kind enough to try explaining it to you.
johanfprins
QUOTE (bm1957+Oct 27 2008, 11:38 AM)
Are you arguing that the clocks' tick rate as measured locally is identical? Because that's pretty obvious.

I am glad you realize that it is obvious. But what do YOU mean by "locally"? Do you mean "locally" when the clocks are stationary relative to each other at the same position or the local time at each clock while they are moving with a constant speed relative to each other?
bm1957
QUOTE (johanfprins+Oct 27 2008, 10:46 AM)
I am glad you realize that it is obvious. But what do YOU mean by "locally"? Do you mean "locally" when the clocks are stationary relative to each other at the same position or the local time at each clock while they are moving with a constant speed relative to each other?

I am using the word with usually accepted meaning of the word. The local time is the time measured near to the clock, regardless of what relative speed it is moving at. (After all, since speed is relative, speed cannot change how time is measured locally.)

A moving observer will observe the tick rate to be slower, and in any symetrical system, when the clock and observer are brought back together, they will still be synchronised. This is standard relativity and no scientist will argue with you on that point.

I'll say again, you should really learn exactly what is predicted by relativity before you start arguing against it.
johanfprins
QUOTE (bm1957+Oct 27 2008, 12:42 PM)
A moving observer will observe the tick rate to be slower, and in any symetrical system, when the clock and observer are brought back together, they will still be synchronised. This is standard relativity and no scientist will argue with you on that point.

Really! Then why did AlphaNumeric become so abusive when I said that relative to their own reference frames (i.e. relative to their "local positions") the clocks are ticking at the same rate? Although the fact that an observer experiences a moving clock to tick slower, is "real" for the observer, does not mean that this slower rate is the rate at which the clock is actually ticking.

After being abused by Alphanumeric, you suddenly maintain that this is "accepted special relativity"! It cannot be the accepted interpretation if Stephen Hawking writes in his book "A Brief History of Time" that: page 22: "....each observer must have his own measure of time, as recorded by a clock carried by him, and that identical clocks carried by different observers would not necessarily agree". In other words there are different "local times" on identical clocks. Whose interpretation is the official one? Yours or Stephen Hawking's? I think I will go with Hawking. After all I have NEVER encountered a "non-local" observer .
johanfprins
QUOTE (bm1957+Oct 27 2008, 12:42 PM)
I'll say again, you should really learn exactly what is predicted by relativity before you start arguing against it.

Where have I argued "against" special relativity? Nowhere ever!! I think that if you want to make accusations you should be willing to substantiate them.
bm1957
QUOTE (johanfprins+Oct 27 2008, 12:36 PM)
Really! Then why did AlphaNumeric become so abusive when I said that relative to their own reference frames (i.e. relative to their "local positions") the clocks are ticking at the same rate?

Probably because that statement is a) very poorly worded (actually doesn't make sense) and b) is pretty trivial if you take the most likely interpretation of what you said. I'm guessing he interpreted it differently to how you meant it.

QUOTE
Although the fact that an observer experiences a moving clock to tick slower, is "real" for the observer, does not mean that this slower rate is the rate at which the clock is actually ticking.

You need to read more about the relativity of simultaneity. There is no absolute 'real' ticking rate. The rate the observer sees is just as 'real' as the rate that a local observer would measure.

QUOTE (->
 QUOTE Although the fact that an observer experiences a moving clock to tick slower, is "real" for the observer, does not mean that this slower rate is the rate at which the clock is actually ticking.

You need to read more about the relativity of simultaneity. There is no absolute 'real' ticking rate. The rate the observer sees is just as 'real' as the rate that a local observer would measure.

After being abused by Alphanumeric, you suddenly maintain that this is "accepted special relativity"!

It's so simple, that I'm assuming AN mis-interpreted you, thinking that you couldn't possibly be saying what I think you're saying, because it is so trivial. But AN would have to confirm or deny that... I'm only speculating.

QUOTE
It cannot be the accepted interpretation if Stephen Hawking writes in his book "A Brief History of Time" that: page 22:  "....each observer must have his own measure of time, as recorded by a clock carried by him, and that identical clocks carried by different observers would not necessarily agree".

That is entirely consistent with what I am saying.

QUOTE (->
 QUOTE It cannot be the accepted interpretation if Stephen Hawking writes in his book "A Brief History of Time" that: page 22:  "....each observer must have his own measure of time, as recorded by a clock carried by him, and that identical clocks carried by different observers would not necessarily agree".

That is entirely consistent with what I am saying.

In other words there are different "local times" on identical clocks.

Again, if you understood relativity of simultaneity, you would understand what this means. The difficulty is in defining when you are going to compare the clocks to see if they are still in synch. If both clocks make identical journeys and come back together, they will still be in synch. If you try to communicate without bringing the clocks back together, you must start calculating transmission times and correcting for relative velocities. I suggest we don't consider that yet. If one clock accelerates away, turns round and comes back (a la the twin paradox), when they come back together, the clocks will be out of synch. The one which changed direction will have 'lost' time. This is standard relativity.

QUOTE
Whose interpretation is the official one? Yours or Stephen Hawking's? I think I will go with Hawking. After all I have NEVER encountered a "non-local" observer :rolleyes: .

My interpretation is the same as that of Hawking's. That you don't understand Hawking's is an issue I cannot rectify.

If you observe a satellite (say, by using GPS), you are a non-local observer. You will have to make corrections (which the GPS does for you), otherwise there will be discrepencies in time measurements (and you will end up in a field)

QUOTE (->
 QUOTE Whose interpretation is the official one? Yours or Stephen Hawking's? I think I will go with Hawking. After all I have NEVER encountered a "non-local" observer :rolleyes: .

My interpretation is the same as that of Hawking's. That you don't understand Hawking's is an issue I cannot rectify.

If you observe a satellite (say, by using GPS), you are a non-local observer. You will have to make corrections (which the GPS does for you), otherwise there will be discrepencies in time measurements (and you will end up in a field)

Where have I argued "against" special relativity? Nowhere ever!! I think that if you want to make accusations you should be willing to substantiate them.

When you stated that the experiments which flew clocks around the world were inherently flawed. The results were in agreement with GR and SR calculations, so if you argue that the results of the experiment were flawed, you are arguing against SR, no? If I'm mistaken then I'll retract the statement.
johanfprins
QUOTE (bm1957+Oct 27 2008, 02:31 PM)
Again, if you understood relativity of simultaneity, you would understand what this means. The difficulty is in defining when you are going to compare the clocks to see if they are still in synch. If both clocks make identical journeys and come back together, they will still be in synch. If you try to communicate without bringing the clocks back together, you must start calculating transmission times and correcting for relative velocities. I suggest we don't consider that yet. If one clock accelerates away, turns round and comes back (a la the twin paradox), when they come back together, the clocks will be out of synch. The one which changed direction will have 'lost' time. This is standard relativity.

My argument did not involve acceleration. All I said, and you now ALSO say, BUT WHICH YOU CONSIDER AS supposedly "trivial" is that although an observer experiences a moving clock to be slower than his own clock when he/she communicates with it, the other clock is actually ticking at the same rate. In addition I used the Lorentz transformations to prove that it must be so.

I also mentioned quite clearly in subsequent postings that acceleration is not taken into account. Thus if two clocks are synchronized and can be brought together in future without one having to accelerate and decelerate more than the other (it is trivial to construct an appropriate thought-experiment) they MUST have the same reading. But I got attacked, and am still being attacked by you. Whether YOU consider the point trivial or not, does not imply that I am NOT understanding SR.

What I have pointed out is that Hafeale and Keating calculated a part of the time difference they measured to the constant motion NOT all to acceleration. Therefore I distrust their results. Why do you have to become abusive?
bm1957
QUOTE (johanfprins+Oct 27 2008, 01:49 PM)
My argument did not involve acceleration. All I said, and you now ALSO say, BUT WHICH YOU CONSIDER AS supposedly "trivial" is that although an observer experiences a moving clock to be slower than his own clock when he/she communicates with it, the other clock is actually ticking at the same rate. In addition I used the Lorentz transformations to prove that it must be so.

I am saying that you cannot say that. Relativity of simultaneity requires that the tick-rate observed by a non-local observer is just as real and valid as the local tick-rate.

Of course it is trivial that 2 identical clocks will always have the same tick-rate as measured locally. That is surely implied by the definition 'identical clocks'?

I certainly don't remember you using Lorentz transformations to prove anything (I won't be trawling back through the thread to check, so I could be wrong), all I remember you using is a variation on d=v/t

QUOTE
I also mentioned quite clearly in subsequent postings that acceleration is not taken into account. Thus if two clocks are synchronized and can be brought together in future without one having to accelerate and decelerate more than the other (it is trivial to construct an appropriate thought-experiment) they MUST have the same reading.

Agreed. You also argued that this was not the current understanding. I imagine this is why you found yourself in argument with AN.

QUOTE (->
 QUOTE I also mentioned quite clearly in subsequent postings that acceleration is not taken into account. Thus if two clocks are synchronized and can be brought together in future without one having to accelerate and decelerate more than the other (it is trivial to construct an appropriate thought-experiment) they MUST have the same reading.

Agreed. You also argued that this was not the current understanding. I imagine this is why you found yourself in argument with AN.

But I got attacked, and am still being attacked by you. Whether YOU consider the point trivial or not, does not imply that I am NOT understanding SR.

Sorry if you feel attacked, that was not my purpose. If you present a case and argue that it flies in the face of the current interpretation of relativity when it doesn't, then you either misunderstand relativity or don't know the accepted interpretation. Perhaps it is the accepted interpretation which you are not familiar with?

I recommend again that you read about the relativity of simultaneity. I believe that if you do it may resolve a number of your issues.

QUOTE
What I have pointed out is that Hafeale and Keating calculated a part of the time difference they measured to the constant motion NOT all to acceleration. Therefore I distrust their results.

No, you vaguely dismissed the experiment as evidence that time dilation is real. You did this alongside an argument that tick-rates don't change due to motion.

The experiments showed a time difference, which is evidence that time dilation is a real effect.

If you only meant to question the accuracy of the results, and are happy to accept that there really was a time difference, due to relative velocity, acceleration and gravitational effects, then I retract my statement that you were arguing against SR. (Although only the relative velocity effects are SR, the others are both GR)

QUOTE (->
 QUOTE What I have pointed out is that Hafeale and Keating calculated a part of the time difference they measured to the constant motion NOT all to acceleration. Therefore I distrust their results.

No, you vaguely dismissed the experiment as evidence that time dilation is real. You did this alongside an argument that tick-rates don't change due to motion.

The experiments showed a time difference, which is evidence that time dilation is a real effect.

If you only meant to question the accuracy of the results, and are happy to accept that there really was a time difference, due to relative velocity, acceleration and gravitational effects, then I retract my statement that you were arguing against SR. (Although only the relative velocity effects are SR, the others are both GR)

Why do you have to become abusive?

Have I become abusive?
DavidD
if realativity exist then it is beta coefiecent 1/(1-v/c)
Alphanumeric is total stupid crank like you all btw
Trout
QUOTE (johanfprins+Oct 27 2008, 07:49 AM)

My point is the following: If a moving clock ticks slower, which clock is ticking slower? Each clock must be ticking slower than the other: But this is am absurdity. One cannot be shorter than your brother AND you brother shorter than you.

Of course you can: you and your brother walk away from each other on a street. From time to time you look over your respective shoulders: each one apperas smaller and smaller to each one of you. Mutual time dilation operates the same exact way. You are simply regurgitating the arguments of Herbert Dingle.
Omnibus
prometheus, you wrote:

QUOTE
...let alone mass-energy relationship, popularly known as E = mc^2, can ever be derived based on the "theory" of relativity.

E = mc^2 certainly can be derived from SR. All you have to do is work out the norm of the momentum four vector thusly:

User posted image: User posted image

The equation you have written is arrived at based on the Lorentz transformations, does it not? Lorentz transformations, however, are in conflict with the first postulate of the theory of relativity (the “Principle of Relativity”), as I have shown (and will repeat briefly below, for your convenience).

Therefore, the theory of relativity cannot derive E = mc^2.

Again, in order for the equation you have written to be physically valid you have to resolve the contradiction I have shown between the first postulate (the “Principle of Relativity”) and the Lorentz transformations.

QUOTE (->
 QUOTE ...let alone mass-energy relationship, popularly known as E = mc^2, can ever be derived based on the "theory" of relativity. E = mc^2 certainly can be derived from SR. All you have to do is work out the norm of the momentum four vector thusly:User posted image: User posted image

The equation you have written is arrived at based on the Lorentz transformations, does it not? Lorentz transformations, however, are in conflict with the first postulate of the theory of relativity (the “Principle of Relativity”), as I have shown (and will repeat briefly below, for your convenience).

Therefore, the theory of relativity cannot derive E = mc^2.

Again, in order for the equation you have written to be physically valid you have to resolve the contradiction I have shown between the first postulate (the “Principle of Relativity”) and the Lorentz transformations.

Your arguing technique is not good either. You complain when people say "you're wrong" and don't write a long post to explain why, yet you write gibberish like the nonsense about Poynting vectors and claim it as a proof. It is not even close to disproving relativity, something which people have been trying to do for 100 years and failed. Also, I find it rather annoying that you put theory in quotes every time you write "theory of relativity" as if this were somehow a point against it. There are many theories of physics that are unphysical, yet they are interesting because they provide insight into a more complicated theory for example. SR is an example of the theory that is valid above a certain scale, below which there is something more fundamental.

You have yet to explain why the predictions of SR are borne out in experiments so well - do you by any chance have your own pet theory that conflicts with special relativity that you're waiting to spring on us?

You have a point about my putting the ‘theory’ in quotes because it may appear to some that it is somehow a point against it. It is not. I apologize for that and I’ll restrain from putting it in quotes in the future.

Neither is the fact that the Poynting vector containing the mass-energy connection an argument against the theory in question. Sorry if this is how it is getting across. If you’re curious, though, you may convince yourself that it is not gibberish and nonsense that it contains the mass-energy connection by observing the dimensions it is expressed in. Poynting vector, is not the subject of discussion here, so I will not continue with it.

Further, as I have shown, and I, as I said, will repeat it briefly below for your convenience, the theory of relativity is incorrect in its entirety. There is nothing in it of use for physics neither is there any need for it to be substituted by any improved or new theory, let alone my pet theory. The weed is removed from a wheat field without the need to replace it by a new weed.

Caloric and phlogiston theories, as incorrect as they are, might have been useful in a certain way for Science unlike the theory of relativity which has only incurred damage—there has never been another instance in Science to have a theory based on internal contradictions gain so much ground due to imposing it by methods having nothing to do with Science. Even unphysical theories such as Astrology of Clairvoyance are not based on internal contradictions. Bible has internal contradictions but faith is not science, is it?

Now, here’s briefly my argument, as promised.

As is known at the start of the derivation, the second Newton’s law F = ma is valid in K for any velocity of the electron (not only for slow motion of the electron as Einstein wants us to believe). First postulate (“Principle of Relativity”) requires F = ma in K to be presented as F’ = ma’ in k and vice versa—first postulate requires that F’ = ma’ in k to be presented in K as nothing else but F = ma. Therefore, presenting F’ = ma’ in k as F = Ma where M =/= m, as Einstein has done by applying Lorentz transformations, is in conflict with the first postulate requiring, as said, F’ = ma’ in k to be presented only by F = ma.

This is a crucial contradiction between the first postulate and the Lorentz transformations which requires abandoning the theory at once and not continuing to use further the F = Ma which it does not derive (to obtain, for instance relativistic mass, E = mc^2 etc.), as Einstein does.

Thus, before telling me that I “have yet to explain why the predictions of SR are borne out in experiments so well” you have to explain how can the demonstrated contradiction be resolved. And if it cannot be resolved (as it obviously can’t) I do not need to explain “why the predictions of SR are borne out by experiments so well” because there simply cannot be such predictions at all and the claim that there are is nothing else but unscrupulous propaganda.
Trout
QUOTE (Omnibus+Oct 27 2008, 03:34 PM)
prometheus, you wrote:

The equation you have written is arrived at based on the Lorentz transformations, does it not? Lorentz transformations, however, are in conflict with the first postulate of the theory of relativity (the “Principle of Relativity”), as I have shown (and will repeat briefly below, for your convenience).

No, they are not, find a different hobby. Stamp collecting would be good.
buttershug
QUOTE (Omnibus+Oct 27 2008, 03:34 PM)
prometheus, you wrote:

This is a crucial contradiction between the first postulate and the Lorentz transformations which requires abandoning the theory at once and not continuing to use further the F = Ma which it does not derive (to obtain, for instance relativistic mass, E = mc^2 etc.), as Einstein does.

Thus, before telling me that I “have yet to explain why the predictions of SR are borne out in experiments so well” you have to explain how can the demonstrated contradiction be resolved. And if it cannot be resolved (as it obviously can’t) I do not need to explain “why the predictions of SR are borne out by experiments so well” because there simply cannot be such predictions at all and the claim that there are is nothing else but unscrupulous propaganda.

Einstein was pointing out the contradiction.
You are faced with two choices.
Throw out the old equation.
Throw out the experimental data that does not support the old equation.

Why do you throw out the observered data?
johanfprins
QUOTE (bm1957+Oct 27 2008, 03:12 PM)
I certainly don't remember you using Lorentz transformations to prove anything (I won't be trawling back through the thread to check, so I could be wrong), all I remember you using is a variation on d=v/t

Lorentz transformations from x(/) to x

x=beta(x(/)+vt(/)) and t=beta(t(/)+(v/c^2)x(/))

The clock within the primed system is situated at x(/)=0 is it not?
Thus put x(/) equal to zero and guess what:

x= beta(vt(/)) and t=beta(t(/)): now use second equation to substitute for t(/) in the first: and you get:

x=vt where x is the distance between the origins of the non-primed and primed systems.

Now do the reverse transformation: Which you should be able to do: And if you can understand the algebra involved you will get that

x(/)=-vt(/) where now -x(/) is the distance between the origins of the primed and non-primed systems as measured along the negative x(/) axis. NOW SURELY ONE MUST HAVE A SINGLE DISTANCE BETWEEN THE ORIGINS! i.e. HAVE THAT :x=-x(/): AND that v=v: SO THAT ONE OBTAINS THAT t=t(/)
QED.

Did I use the Lorentz transformations or not?

I have an interruption but will come back later to some of the other points you have raised.
dimazin
QUOTE (DavidD+Oct 27 2008, 01:16 PM)
if realativity exist then it is beta coefiecent 1/(1-v/c)
Alphanumeric is total stupid crank like you all btw

Trout
QUOTE (johanfprins+Oct 27 2008, 04:51 PM)
Lorentz transformations from x(/) to x

x=beta(x(/)+vt(/)) and t=beta(t(/)+(v/c^2)x(/))

The clock within the primed system is situated at x(/)=0 is it not?
Thus put x(/) equal to zero and guess what:

x= beta(vt(/)) and t=beta(t(/)): now use second equation to substitute for t(/) in the first: and you get:

x=vt where x is the distance between the origins of the non-primed and primed systems.

Now do the reverse transformation: Which you should be able to do: And if you can understand the algebra involved you will get that

x(/)=-vt(/) where now -x(/) is the distance between the origins of the primed and non-primed systems as measured along the negative x(/) axis. NOW SURELY ONE MUST HAVE A SINGLE DISTANCE BETWEEN THE ORIGINS! i.e. HAVE THAT :x=-x(/): AND that v=v: SO THAT ONE OBTAINS THAT t=t(/)
QED.

Did I use the Lorentz transformations or not?

I have an interruption but will come back later to some of the other points you have raised.

You are repeating a classical error by Dingle. Here is the correct resolution.
johanfprins
QUOTE (bm1957+Oct 27 2008, 03:12 PM)
I recommend again that you read about the relativity of simultaneity. I believe that if you do it may resolve a number of your issues.

I know the relativity of simultaneity very well. Two simultaneous events at the same position within an inertial reference frame is observed as simultaneous events by an observer moving at a high speed relative to this reference frame. Only when the events are at separate positions within the inertial reference frame does an observer moving at a high speed relative to this reference frame conclude that the events are not simultaneous.

BUT for two events to be simultaneous within a single inertial reference frame demands a Newtonian reference frame within which the time is the same at all position points: i.e. space and time are not interrelated. It is only the observer moving at a high speed who interprets that the times at different position points are different and thus experiences a space-time manifold (which does not exist within the reference frame within which the events are simultaneous). This is the same as when the moving observer observes a clock ticking away at a slower rate while in ACTUAL FACT it is not doing so at all!!

Thus if Hafeale and Keating in their clock experiments did measure an actual time-dilation totally attributable to relative motion (no acceleration being taken into account) they most probably cooked their results in the same expert fashion that first year physics students have done during laboratory practice for many many years!
johanfprins
QUOTE (Trout+Oct 27 2008, 03:40 PM)
Of course you can: you and your brother walk away from each other on a street. From time to time you look over your respective shoulders: each one appears smaller and smaller to each one of you. Mutual time dilation operates the same exact way. You are simply regurgitating the arguments of Herbert Dingle.

Very funny Although the perspective changes your length and your brother' slength do not change the fact that your lengths are exactly the same. OK relativity theory has to do with perspective: But this is exactly what I am saying. The fact that you experience the other clock as ticking slower, is not ACTUALLY happening within the refrence frame of the clock:
Trout
QUOTE (johanfprins+Oct 27 2008, 05:53 PM)
Very funny Although the perspective changes your length and your brother' slength do not change the fact that your lengths are exactly the same. OK relativity theory has to do with perspective: But this is exactly what I am saying. The fact that you experience the other clock as ticking slower, is not ACTUALLY happening

...which is exactly what happens with proper time.
Relativistic time, as well as relativistic length , on the other change exactly like the perspective height for you and your brother in the example I gave you. I can recommend a few very good books on the subject, you should take a break from posting nonsense and read them.
Trout
QUOTE (johanfprins+Oct 27 2008, 05:45 PM)

Thus if Hafeale and Keating in their clock experiments did measure an actual time-dilation totally attributable to relative motion (no acceleration being taken into account) they most probably cooked their results in the same expert fashion that first year physics students have done during laboratory practice for many many years!

They didn't. The HK experiment is fully explained within the framework of General Relativity. Judging by the fact that you don't even understand Special Relativity, the explanation of the HK experiment is totally beyond you.
johanfprins
QUOTE (Trout+Oct 27 2008, 08:05 PM)
...which is exactly what happens with proper time.
Relativistic time, as well as relativistic length , on the other change exactly like the perspective height for you and your brother in the example I gave you. I can recommend a few very good books on the subject, you should take a break from posting nonsense and read them.

Why should I read books you recommend if what I am saying is according to you "exactly" correct? It will be helpful if you could be more specific when you claim that I am "posting nonsense". You will see above that I have always been willing to engage in logic. Maybe it is not your strong point: I have found in the past that people who want to recommend "books" instead of being able to argue logic, usually do not know Arthur from Martha.

It is claimed in all the books on relativity that the clock of a twin moving away at a constant speed ACTUALLY ticks slower than the clock of the other twin. This is obviously BS.
johanfprins
QUOTE (Trout+Oct 27 2008, 08:09 PM)
They didn't. The HK experiment is fully explained within the framework of General Relativity. Judging by the fact that you don't even understand Special Relativity, the explanation of the HK experiment is totally beyond you.

Really. In their paper I saw them using the Lorentz transformations of special relativity. If I missed a point and they did interpret their results totally in terms of accelerations, I stand corrected: However, I do not think that they did. I did not see a solution of Einstein's equations for general relativity. Maybe YOU could quickly show us here how they did it?
Omnibus
QUOTE (Trout+Oct 27 2008, 11:29 AM)
No, they are not, find a different hobby. Stamp collecting would be good.

Yes, they are. I've proven it while you just believe they are not. Start collecting stamps instead.
Omnibus
buttershug, the following is incorrect:

QUOTE
Einstein was pointing out the contradiction.
You are faced with two choices.
Throw out the old equation.
Throw out the experimental data that does not support the old equation.

Why do you throw out the observered data?

What Einstein did is try to convince us that at the start of his derivation the old equation, that is, the equation F = ma is only valid for a slow motion of the electron. That is not true. At the start of the derivation F = ma is known to be applicable to any velocity. There is no way you won't agree with that.

Therefore, its corresponding equation in k is only F' = ma' and nothing else. F' = ma' maps into K as F = ma and not as F = Ma, where M =/= m, OK?

Thus, there is no new equation to compare with the observed data.
Trout
QUOTE (johanfprins+Oct 27 2008, 06:15 PM)

It is claimed in all the books on relativity that the clock of a twin moving away at a constant speed ACTUALLY ticks slower than the clock of the other twin. This is obviously BS.

Why is it BS? Because you are incapable of understanding it? There is ample experimental proof of the phaenomenon.
Trout
QUOTE (Omnibus+Oct 27 2008, 07:02 PM)
buttershug, the following is incorrect:

What Einstein did is try to convince us that at the start of his derivation the old equation, that is, the equation F = ma is only valid for a slow motion of the electron. That is not true. At the start of the derivation F = ma is known to be applicable to any velocity. There is no way you won't agree with that.

Bzzt, wrong.
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.
NoCleverName
QUOTE (johanfprins+Oct 27 2008, 02:15 PM)
It is claimed in all the books on relativity that the clock of a twin moving away at a constant speed ACTUALLY ticks slower than the clock of the other twin. This is obviously BS.

It would seem that the basis of your misunderstanding starts with the belief that there is a one, true and universal "ACTUALLY".

Now, I am not a trained physicist, but even I "know" that "ACTUALLY" over there is not the same as "ACTUALLY" for me.

But, notwithstanding that point, it seems to me that relativity largely deals with the proposition that "if the speed of light is a constant and universal speed limit and if everyone in any sub-lightspeed frame determines that this is so, then how must the concept of "time" and "distance" be adjusted to allow this to happen?"
Omnibus
QUOTE
Bzzt, wrong.
The formula valid for any velocity is F=dp/dt
where p=momentum
In relativity, momentum is give 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.

Don't tell me what formula relativity gives because I've already proven conclusively that relativity is invalid prior to any formula that would come out of it. Therefore, calculus won't help.

You have to first prove that you can get F = Ma out of the law valid for any velocity, namely, out of F = ma (where M =/= m). Obviously, Einstein himself couldn't do that, let alone someone with a handle Trout.
Omnibus
I was just typing that and NoCleverName beat me to it. I'll post it anyway:

johanfprins, I told you to restrain from trying to disprove Lorentz transformations within their own framework because in this way you become an easy prey for the zealots. Case in point, this assertion, leading to what you think is the proof that t = t’ is unfounded:
QUOTE
NOW SURELY ONE MUST HAVE A SINGLE DISTANCE BETWEEN THE ORIGINS! i.e. HAVE THAT :x=-x(/)

The condition you have nonchalantly assumed, namely that x = x(/) is untenable because x is the coordinate in K while x(/) is the coordinate in k. In order for this condition to be fulfilled you must have that t = t(/) but this is exactly what you have to prove.

Lorentz transformations can be criticized only based on their non-physicality, not by trying to prove they are inconsistent (because the are mathematically consistent without a doubt) and the arguments showing that are different from what you’re trying to push.
Steven Christopher
shut up fuk tards
theres glass in the sky the moon is bowl-shaped and the earth is inverted
8,000 mile wide universe for you fuked up little brains

buttershug
QUOTE (Omnibus+Oct 27 2008, 07:02 PM)
buttershug, the following is incorrect:

What Einstein did is try to convince us that at the start of his derivation the old equation, that is, the equation F = ma is only valid for a slow motion of the electron. That is not true. At the start of the derivation F = ma is known to be applicable to any velocity. There is no way you won't agree with that.

Therefore, its corresponding equation in k is only F' = ma' and nothing else. F' = ma' maps into K as F = ma and not as F = Ma, where M =/= m, OK?

Thus, there is no new equation to compare with the observed data.

There was no way for Newton to know otherwise, but F does not exactly equal ma at any speed except zero.

I'm glad the people that make the GPS system understood what you don't.
Trout
QUOTE (Omnibus+Oct 27 2008, 08:22 PM)
Don't tell me what formula relativity gives because I've already proven conclusively that relativity is invalid prior to any formula that would come out of it. Therefore, calculus won't help.

Realy? You should send your works to Stockholm and ask them to mail you your Physics Nobel Prize. Pronto.
All you have proven is that you are ignorant. If you don't like stamp collecting as a hobby I suggest you take up underwater pingpong.
bm1957
QUOTE (johanfprins+Oct 27 2008, 07:15 PM)
Why should I read books you recommend if what I am saying is according to you "exactly" correct?

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!!!
prometheus
QUOTE (Omnibus+Oct 27 2008, 03:34 PM)
As is known at the start of the derivation, the second Newton’s law F = ma...

This is not the most general form of Newtons second law. It is F = dp/dt. The reason you get F = ma is because Newton defined p = mv so F = m dv/dt = ma. For an object at low velocity this is ok but not for an object with arbitrary velocity. More on this in a minute.

QUOTE (Omnibus+Oct 27 2008, 03:34 PM)
...is valid in K for any velocity of the electron (not only for slow motion of the electron as Einstein wants us to believe).

If you want to prove your argument you have to justify this statement. Why is F = ma valid for any velocity? (It's not by the way).

QUOTE (Omnibus+Oct 27 2008, 03:34 PM)
First postulate (“Principle of Relativity”) requires F = ma in K to be presented as F’ = ma’ in k and vice versa—first postulate requires that F’ = ma’ in k to be presented in K as nothing else but F = ma.

Apart from this being word salad and showing precisely nothing, the premise is wrong. For the record, this is the first postulate of relativity: "If a system of coordinates K is chosen so that, in relation to it, physical laws hold good in their simplest form, the same laws hold good in relation to any other system of coordinates K' moving in uniform translation relatively to K." The physical law that is invariant in any inertial frame is F = dp/dt not F = ma because the two are not the same in full generality.

QUOTE (Omnibus+Oct 27 2008, 03:34 PM)

Thus, before telling me that I “have yet to explain why the predictions of SR are borne out in experiments so well” you have to explain how can the demonstrated contradiction be resolved. And if it cannot be resolved (as it obviously can’t) I do not need to explain “why the predictions of SR are borne out by experiments so well” because there simply cannot be such predictions at all and the claim that there are is nothing else but unscrupulous propaganda.

I've demonstrated the flaw in your logic - to repeat F = dp/dt is not equal to ma in general. So now, why are the predictions of SR in such close agreement with what we observe in nature. For your appreciation.
Trout
QUOTE (buttershug+Oct 27 2008, 08:42 PM)
There was no way for Newton to know otherwise, but F does not exactly equal ma at any speed except zero.

I'm glad the people that make the GPS system understood what you don't.

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.
Omnibus
QUOTE (buttershug+Oct 27 2008, 03:42 PM)
There was no way for Newton to know otherwise, but F does not exactly equal ma at any speed except zero.

I'm glad the people that make the GPS system understood what you don't.

Never mind GPS or any other experiment. That's not the point. The point is whether or not the theory in question can derive what it is claiming. It can't, as shown. The derivation it pretends to have carried out is in fact based on the crucial error in reasoning called 'petitio principii', that is, having the question contain the answer together with selectively "forgetting" what it had previously derived.

Therefore, there is a double violation of the scientific principles:

1) First, foisting on us the obviously incorrect idea that F = ma is only valid for slow motion of the electron which implicitly presumes velocity dependence of m which is exactly what it needs to prove.

2) Then, conveniently "forgetting" the fact that the first first postulate requires that F = ma in K must be F' = ma' in k and that F' = ma' in k must be F = ma in K. Thus, after "forgetting" or ignoring this crucial fact it unperturbed continues to use a different law in K, namely, F = Ma (where M =/= m) without heeding the fact that it contradicts its first postulate, for further derivations (of 'relativistic; mass, E = mc^2 etc.)

This is reprehensible manipulation, having nothing to do with Science and anyone in his right mind who calls himself a scientist should feel offended rather than turn a blind eye and continue the hopeless and unbecoming activity of defending it.

Mind you also that if, indeed, F = Ma is the law in K for high velocities (where M =/= m) then, according to the first postulate, Einstein should start in k from F' = Ma' and not from F' = ma', as he has done. But then the law in K becomes F = beta^3Ma after applying the Lorentz transformations which again is different from the law in K required by the first postulate F = Ma. So, there's no escape, low velocities or high, there's always a contradiction between the first postulate and the Lorentz transformations which conclusively invalidates the theory of relativity in its entirety.
Trout
QUOTE (Omnibus+Oct 27 2008, 09:32 PM)

2) Then, conveniently "forgetting" the fact that the first first postulate requires that F = ma in K must be F' = ma' in k and that F' = ma' in k must be F = ma in K.

Bzzt, no.
The first postulate says "If F=dp/dt in K then F'=dp'/dt' in K' "
You really need to find a different hobby, did you give some more thought to underwater pingpong?
Omnibus
QUOTE
Realy? You should send your works to Stockholm and ask them to mail you your Physics Nobel Prize. Pronto.
All you have proven is that you are ignorant. If you don't like stamp collecting as a hobby I suggest you take up underwater pingpong

This doesn’t deserve response because no technical arguments are provided.

Omnibus
QUOTE
Bzzt, no.
The first postulate says "If F=dp/dt in K then F'=dp'/dt' in K' "
You really need to find a different hobby, did you give some more thought to underwater pingpong?

Before sending me to read whatever, you have to send Einstein to do some reading because you know better than him that what he used in §10 in paper, F = ma in K then F’ = ma’ is incorrect.

Omnibus
QUOTE
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.

That’s not what Einstein says and if you want to defend him do some reading first.

Trout
QUOTE (Omnibus+Oct 27 2008, 09:40 PM)
Before sending me to read whatever, you have to send Einstein to do some reading because you know better than him that what he used in §10 in paper, F = ma in K then F’ = ma’ is incorrect.

The calendar reads 2008, so we teach students to use F=dp/dt when deriving the relativistic force. If you are still stuck in the 1905 timewarp, this is your problem. The more I think, the more I like you playing underwater pingpong in a tutu
Omnibus
Prometeus,
QUOTE
The reason you get F = ma …

Einstein does, not me.

QUOTE (->
 QUOTE The reason you get F = ma …

Einstein does, not me.

If you want to prove your argument you have to justify this statement. Why is F = ma valid for any velocity? (It's not by the way).

Way are you saying F = ma is not valid for any velocity. What other law do you know at the start of the argument that would be valid for high velocity?

QUOTE
Apart from this being word salad and showing precisely nothing, the premise is wrong. For the record, this is the first postulate of relativity: "If a system of coordinates K is chosen so that, in relation to it, physical laws hold good in their simplest form, the same laws hold good in relation to any other system of coordinates K' moving in uniform translation relatively to K." The physical law that is invariant in any inertial frame is F = dp/dt not F = ma because the two are not the same in full generality.

What is first postulate and how it is applied is known by Einstein better than anybody and that requires F = ma in K to be presented as F’ = ma’ in k and vice versa—first postulate requires that F’ = ma’ in k to be presented in K as nothing else but F = ma. You can check §10 and convince yourself that that’s exactly the case. If that’s a word salad for you, that’s only your problem.

QUOTE (->
 QUOTE Apart from this being word salad and showing precisely nothing, the premise is wrong. For the record, this is the first postulate of relativity: "If a system of coordinates K is chosen so that, in relation to it, physical laws hold good in their simplest form, the same laws hold good in relation to any other system of coordinates K' moving in uniform translation relatively to K." The physical law that is invariant in any inertial frame is F = dp/dt not F = ma because the two are not the same in full generality.

What is first postulate and how it is applied is known by Einstein better than anybody and that requires F = ma in K to be presented as F’ = ma’ in k and vice versa—first postulate requires that F’ = ma’ in k to be presented in K as nothing else but F = ma. You can check §10 and convince yourself that that’s exactly the case. If that’s a word salad for you, that’s only your problem.

I've demonstrated the flaw in your logic - to repeat F = dp/dt is not equal to ma in general. So now, why are the predictions of SR in such close agreement with what we observe in nature. For your appreciation.

You did not need to provide a link to experimental confirmations of the theory of relativity because there are no such. Because of the crucial argument I’m providing, invalidating categorically the theory of relativity, any claim for experimental validation must be considered as nothing else but unscrupulous propaganda.

And, don’t remain wit the impression that you have demonstrated a flaw in my logic by claiming that F = dp/dt is not equal to ma in general because you have to tell me what was F known to be equal to at the start of the derivation. I am telling you and you try to prove me wrong hat at the beginning of the discussed derivation F = ma has been the only known law describing F.

Now, suppose you prove me wrong (impossible, but suppose for the sake of argument you do) and that for high velocities we have F = Ma, rather than F = ma. If, indeed, F = Ma is the law in K for high velocities (where M =/= m) then, according to the first postulate, Einstein should start in k from F' = Ma' and not from F' = ma', as he has done. But then the law in K becomes F = beta^3Ma after applying the Lorentz transformations which again is different from the law in K required by the first postulate F = Ma. So, there's no escape, low velocities or high, there's always a contradiction between the first postulate and the Lorentz transformations which conclusively invalidates the theory of relativity in its entirety.
Omnibus
QUOTE
The calendar reads 2008, so we teach students to use F=dp/dt when deriving the relativistic force. If you are still stuck in the 1905 timewarp, this is your problem. The more I think, the more I like you playing underwater pingpong in a tutu

Is that all you have to say? Then, hear it again:

Before sending me to read whatever, you have to send Einstein to do some reading because you know better than him that what he used in §10 in paper, F = ma in K then F’ = ma’ is incorrect.

We’re not discussing here someone’s teaching whatever whoever. What we’re discussing here is Einstein and the fatal contradiction in his paper, invalidating it in its entirety.
Trout
QUOTE (Omnibus+Oct 27 2008, 10:02 PM)
Is that all you have to say? Then, hear it again:

Before sending me to read whatever, you have to send Einstein to do some reading because you know better than him that what he used in §10 in paper, F = ma in K then F’ = ma’ is incorrect.

The fact that Einstein used a different formula in 1905 is inconsequential:

1. the formula is not used in the derivation of the theory , it is just a consequence of the theory, this is why it shows up in paragraph 10.

2. Since 1905, both Einstein and many others have corrected the derivation of force.
He used that derivation in 1905. We no longer use that derivation. We no longer use the terms longitudinal and transverse mass.
We are in 2008.
Since 1905 we have long learned that we need to use p=mv/sqrt(1-(v/c)^2) and F=dp/dt. This news has not traveled to your end of the world, that is too bad. Make sure that you don't get your tutu wet when you play underwater pingpong.
prometheus
A simple question: what is p? Both a mathematical definition and a wordy definition please.
Trout
QUOTE (prometheus+Oct 27 2008, 10:08 PM)
A simple question: what is p? Both a mathematical definition and a wordy definition please.

Relativistic momentum: p=mv/sqrt(1-(v/c)^2)
Omnibus
QUOTE (Trout+Oct 27 2008, 05:04 PM)
He used that in 1905.
We are in 2008. Since 1905 we have long learned that we need to use F=mv/sqrt(1-(V/c)^2). This news has not traveled to your end of the world, that is too bad. Make sure that you don't get your tutu wet when you play underwater pingpong.

Not so. Aristotelian geocentricism has survived even more, thousands of years, despite it being wrong. Using longevity as a proof for the validity of a theory is as flawed an argument as anything else flawed you're allowing yourself to waste everybody's time here with. Theory of relativity is wrong, it has never been nor can it ever be proven correct. I've provided one crucial argument as to why that's the case.
Trout
QUOTE (Omnibus+Oct 27 2008, 10:12 PM)
Not so. Aristotelian geocentricism has survived even more, thousands of years, despite it being wrong. Using longevity as a proof for the validity of a theory is as flawed an argument as anything else flawed you're allowing yourself to waste everybody's time here with.

I am not using longevity as proof of correctness. Quite the opposite, I showed you how the physics world evolved from 1905. There is nothing that I can do for you if you are still stuck in the 1905 timewarp, playing underwater pingpong dressed in a pink tutu.

QUOTE
Theory of relativity is wrong, it has never been nor can it ever be proven correct. I've provided one crucial argument as to why that's the case.

I suggest that you send your "discovery" to the Karolinska Intitutet in Stockholm. I can give you the exact address. make sure that you include a self-addressed, properly stamped envelope so they can mail you the Nobel prize.
prometheus
QUOTE (Trout+Oct 27 2008, 10:10 PM)
Relativistic momentum: p=mv/sqrt(1-(v/c)^2)

PS, I know the answer to this. I want to see omnibus' version.
Omnibus
QUOTE
The fact that Einstein used a different formula in 1905 is inconsequential:

That’s wishful thinking. Admit first the obvious, namely, that Einstein’s 1905 paper is badly flawed and then we’ll move to something else. The easy escape you’re seeking, by pronouncing Einstein’s 1905 paper as inconsequential, exposes your weakness of arguments as much as the rest of the nonsense you feel the need to spread here.
Trout
QUOTE (Omnibus+Oct 27 2008, 10:17 PM)
That’s wishful thinking. Admit first the obvious, namely, that Einstein’s 1905 paper is badly flawed and then we’ll move to something else. The easy escape you’re seeking, by pronouncing Einstein’s 1905 paper as inconsequential, exposes your weakness of arguments as much as the rest of the nonsense you feel the need to spread here.

Not the whole paper, just the derivation in paragraph 10.This has been known (and corrected) for about 99 years. Are you reading and comprehension challenged?
prometheus
QUOTE (Omnibus+Oct 27 2008, 09:58 PM)
You did not need to provide a link to experimental confirmations of the theory of relativity because there are no such. Because of the crucial argument I’m providing, invalidating categorically the theory of relativity, any claim for experimental validation must be considered as nothing else but unscrupulous propaganda.

So you're suggesting that physics departments the world over in competition with each other are somehow colluding to cover up the incorrectness or relativity? For the record

1) I have shown there is a flaw in your logic
2) There are rafts of experimental evidence that support the predictions of relativity.

Is nature wrong, or is it you?

Omnibus
QUOTE
PS, I know the answer to this. I want to see omnibus' version.

You know very well my answer to that. I've said it on several occasions--any so-called relativistic formula must be ignored out of hand because of the crucial contradiction I've demonstrated, invalidating said theory in its entirety.
Trout
QUOTE (Omnibus+Oct 27 2008, 10:22 PM)
You know very well my answer to that. I've said it on several occasions--any so-called relativistic formula must be ignored out of hand because of the crucial contradiction I've demonstrated, invalidating said theory in its entirety.

1. You demonstrated only your ignorance. Plenty.
2. Theories can only be disproven via experiment. You didn't know that? Tsk,tsk.
Omnibus
QUOTE
1) I have shown there is a flaw in your logic

No, you haven’t. All you have said is that F = ma isn’t valid for high velocities but you never showed what the expression for F is for high velocities at the start of the derivation. That’s first. Second, as I already wrote to you, even if you have shown me such an unlikely thing, then, again, you would be in error.

Therefore, again, no, you have not shown there is a flaw in my logic.

QUOTE (->
 QUOTE 1) I have shown there is a flaw in your logic

No, you haven’t. All you have said is that F = ma isn’t valid for high velocities but you never showed what the expression for F is for high velocities at the start of the derivation. That’s first. Second, as I already wrote to you, even if you have shown me such an unlikely thing, then, again, you would be in error.

Therefore, again, no, you have not shown there is a flaw in my logic.

2) There are rafts of experimental evidence that support the predictions of relativity.

No, there have never been, neither can there ever be experiments validating said theory because there are crucial contradictions in it, as the one I’ve already shown, which render it completely wrong. Anyone claiming experimental evidence in support of the theory of relativity is nothing else but an unscrupulous manipulator.

Scientific reasoning requires that the theory in question must be abandoned for good.
prometheus
QUOTE (Omnibus+Oct 27 2008, 10:22 PM)
You know very well my answer to that. I've said it on several occasions--any so-called relativistic formula must be ignored out of hand because of the crucial contradiction I've demonstrated, invalidating said theory in its entirety.

Can you please reproduce it. I genuinely haven't seen it and I'm far too busy to read 22 pages of this thread.
Omnibus
[quote] 1. You demonstrated only your ignorance. Plenty.
[quote]
Completely frivolous talk.

[quote]2. Theories can only be disproven via experiment. You didn't know that? Tsk,tsk. [/quote]
Wrong. The above applies to theories which don’t contain internal contradictions. A theory such as the theory of relativity is disproved by exposing the internal contradiction which invalidates it prior to any experiment. It also makes carrying out of experiments to check the validity of the theory unnecessary.
prometheus
QUOTE (Omnibus+Oct 27 2008, 10:30 PM)
No, there have never been, neither can there ever be experiments validating said theory because there are crucial contradictions in it, as the one I’ve already shown, which render it completely wrong. Anyone claiming experimental evidence in support of the theory of relativity is nothing else but an unscrupulous manipulator.

Scientific reasoning requires that the theory in question must be abandoned for good.

I 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?
Trout
QUOTE (Omnibus+Oct 27 2008, 10:34 PM)
Wrong. The above applies to theories which don’t contain internal contradictions. A theory such as the theory of relativity is disproved by exposing the internal contradiction

Omnibus
QUOTE (prometheus+Oct 27 2008, 05:31 PM)
Can you please reproduce it. I genuinely haven't seen it and I'm far too busy to read 22 pages of this thread.

Did you read what I wrote or you like to play games? I'm taking you seriously and I expect from you the same.
prometheus
QUOTE (Omnibus+Oct 27 2008, 10:37 PM)
Did you read what I wrote or you like to play games? I'm taking you seriously and I expect from you the same.

I've read what you've put in the last 2 pages or so. At no point have you defined p. If you have then point it out to me and I will gladly apologise for my oversight.
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