grav
The two postulates of SR are:

1) the laws of physics are the same in every inertial frame
2) light is measured travelling isotropically at c in every inertial frame

I intend to derive SR by applying only the second postulate alone, that the speed of light is measured isotropically at c in every inertial frame. We will start with a reference frame where the light is always measured to travel isotropically at c. Whether this particular frame is considered absolute or just a preferred arbitrary frame is of no consequence since we will only be determining the results according to how the physics directly relates between observers, which is how SR would relate the physics since there is otherwise nothing else to relate it to within the philosophy of SR. All other frames are to be viewed from the perspective of this preferred frame. We will apply certain properties that might occur to observers as they move relative to that frame of reference, along with their clocks and rulers. The clocks of observers in motion to this frame may time dilate by a factor of z, lengths contract in the line of motion by a factor of Lx and perpendicularly to the line of motion by a factor of Ly. Each of these ratios are to be compared to that of the reference frame, so if any of these are found not to occur, then the value for that property will be 1.

Alice will always be considered the reference frame observer that remains stationary to that particular frame of reference for the purposes of this demonstration. We will have Bob and Carl travel away from Alice with a speed of v with a distance of d between them, Carl before Bob, as these measurements are taken by Alice. Light always travels isotropically to the reference frame at c, and we want to find what must be true for Bob and Carl to measure the same isotropic speed. Let's say a pulse is sent from Bob to Carl. According to Alice, the light travels a distance of d plus the extra distance Carl has travelled away from the pulse in the time it took for the pulse to reach Carl, so c t_BC = d + v t_BC, t_BC = d / (c - v). The time Alice will measure for a light pulse to travel from Carl to Bob while Bob moves toward the pulse in that time is c t_CB = d - v t_CB, so t_CB = d / (c + v).

If the clocks of Bob and Carl are synchronized to each other as Alice views them, then Bob and Carl will say that their own clocks are unsynchronized because the time it takes for the pulse to travel from Bob to Carl is different from the time it takes to travel from Carl to Bob. So Bob and Carl establish a new simultaneity convention between their own clocks. This is done by having Bob turn his own clock forward or having Carl turn his back by an amount which will produce a time lag between their readings of tl. Bob's clock now reads a greater time than Carl's according to Alice, but Bob and Carl say their clocks are synchronized. When the pulse travels from Bob to Carl, it travels from Bob when Bob's clock reads TB = tl and Carl's clock reads TC = 0, then reaches Carl when Carl's clock reads TC' = z t_BC, so with a difference in times as measured by Bob and Carl of t'_BC = TC' - TB = z t_BC - tl. When the pulse travels from Carl to Bob, Carl sends it when the clocks read TC = 0 and TB = tl, and Bob receives it when his clock reads TB' = tl + z t_CB, so in a time measured by them of t'_CB = TB' - TC = tl + z t_CB. Since they must measure these two times to be equal, then t'_BC = t'_CB, z t_BC - tl = tl + z t_CB, 2 tl = z t_BC - z t_CB = z d / (c - v) - z d / (c + v) = z d [(c + v) - (c - v)] / [(c - v) (c + v)] = 2 z d v / (c^2 - v^2), whereby tl = z d v / (c^2 - v^2). This is now the simultaneity difference between the clocks of Bob and Carl according to Alice. In the frame of Bob and Carl according to Alice, their rulers have been contracted in the line of motion by Lx, so if Alice measures a distance between them of d, then they will measure d' = d / Lx . So when the pulse is sent from Bob to Carl, it will be measured to have a speed of c' = d' / t'_BC = (d / Lx) / (z t_BC - tl) = (d / Lx) / [z d / (c - v) - z d v / (c^2 - v^2)] = [(c^2 - v^2) / (z Lx)] / [(c + v) - v] = c (1 - (v/c)^2) / (z Lx), whereby if c' = c, then z Lx = 1 - (v/c)^2 . Likewise, the speed measured from Carl to Bob gives the same result.

Now let's say that in the frame of Bob and Carl, an apparatus has been set up where light is allowed to travel across the lengths of two perpendicular arms of equal lengths d' and back. Since the pulses must have the isotropic speed of c along equal lengths of d', then the times to travel both arms are measured the same in the frame of the apparatus with c = d' / t' which is to be accepted as the usual definition of speed with distance measured over time measured, not a law that must be derived, but a given definition for speed, so since t' = d' / c where c and d' are measured the same along both arms, then so must be t' be the same along both arms. Also, because the pulses coincide in the same place upon separating and then coincide in the same place again when returning, then all observers in all frames must agree that the times to traverse both arms is the same for whatever time they measure between these two events. Let's say that one arm travels directly in the line of motion of the apparatus to the reference frame. From Alice's frame, the apparatus is contracted in the line of motion by Lx and perpendicularly by Ly, so the lengths of the arms are dx = Lx d' and dy = Ly d'. The time that Alice measures for the pulse to travel the arm in the line of motion and back is t_forward = dx / (c - v) and t_back = dx / (c + v), whereby tx = dx / (c - v) + dx / (c + v) = dx [(c + v) + (c - v)] / (c^2 - v^2) = 2 (Lx d') c / (c^2 - v^2). In the perpendicular direction, Alice measures a time of (c t_away)^2 = (v t_away)^2 + dy^2, so t_away = dy / sqrt(c^2 - v^2). The pulse travels in the same way along the same angle away and back, so t_perp = 2 dy / sqrt(1 - (v/c)^2) = 2 (Ly d') / sqrt(1 - (v/c)^2). In order for these times to be the same as Alice measures them, then 2 (Lx d') c / (c^2 - v^2) = 2 (Ly d') / sqrt(c^2 - v^2), and from this we gain Lx / Ly = sqrt(1 - (v/c)^2).

Okay, here's where things get interesting. One might think that two observers measuring the same relative speed of each other would follow from the first postulate, since if the laws of physics is the same in all inertial frames, then with nothing else to relate the physics to except between the observers, then each must measure the same relative speed between them as the other does, but it actually follows from the second principle alone. Let's say that Bob and Carl pass Alice. Alice says that the time for Carl to pass and Bob to reach her is t = d / v, all as measured in her own frame. Now, from what is measured in the frame of Bob and Carl, when Carl passes Alice, his clock reads TC = 0 and Bob's reads TB = tl. When Bob passes Alice, his clock then reads TB' = tl + z t. Bob and Carl will read the difference in times that has passed between their clocks as t' = TB' - TC = tl + z t = z d v / (c^2 - v^2) + z d / v = [z d / (v (c^2 - v^2))] [v^2 + (c^2 - v^2)] = z d / (v (1 - (v/c)^2)) and the distance Alice has travelled of d' = d / Lx, giving a relative speed for Alice as Bob and Carl measure it of v' = d' / t' = (d / Lx) / [z d / (v (1 - (v/c)^2)] = v (1 - (v/c)^2) / (z Lx), but since we have already established earlier that z Lx = 1 - (v/c)^2, then we gain v' = v (1 - (v/c)^2) / (z Lx) = v, so the observers will measure the same relative speed of each other in both frames.

Now let's look at the addition of speeds. Let's say that according to Alice, who is stationary with the reference frame, Bob and Carl are travelling in one direction at v and Danielle is travelling past them in the other direction at u. According to Alice, it takes a time of t = d / (u + v) for Danielle to travel from Carl to Bob, so Bob and Carl measure their difference in times to be TC = 0 and TB' = tl + z t, so t' = TB' - TC = z t + tl = z d / (u + v) + z d v / (c^2 - v^2) = z d [(c^2 - v^2) + (u + v) v] / [(u + v) (c^2 - v^2)] = z (Lx d') [c^2 - v^2 + u v + v^2] / [(u + v) (c^2 - v^2)] = d' [c^2 + u v] / [(u + v) c^2] = d' [1 + u v / c^2] / (u + v). Therefore, the speed that Bob and Carl measure of Danielle is w = d' / t' = (u + v) / (1 + u v / c^2). If Danielle were to travel in the same direction as Bob and Carl at u, then Alice would measure a time for Danielle to travel from Bob to Carl of t = d / (u - v), whereby Bob and Carl would measure their difference in times to be TB = tl and TC' = z t, for a difference in times of t' = TC' - TB = z t - tl = z d / (u - v) - z d v / (c^2 - v^2) = [z d / ((u - v) (c^2 - v^2))] [(c^2 - v^2) - v (u - v)] = [z d / ((u - v) (c^2 - v^2))] [c^2 - u v] = [z (Lx d') / ((u - v) (1 - (v/c)^2)] [1 - u v / c^2] = [d' / (u - v)] [1 - u v / c^2]. The relative speed Bob and Carl measure for Danielle when travelling in the same direction, then, is w = d' / t' = (u - v) / (1 - u v / c^2).

Now let's say Bob and Danielle are both travelling in ships that they measure of a length of d' in their own frames. Let's determine what the length contraction Bob measures of Danielle's ship will be. Bob cannot measure the length of Danielle's ship at a distance or even directly by using his ruler while Danielle's ship is in motion to his, so he has to find another way. What he does is to find the difference in time that it takes for the front of Alice's ship to pass an antenna on his ship and then the back of her ship to pass the same antenna. At T=0 on his clock, the front of Alice's ship passes the antenna. According to Alice, the time that it takes for Danielle's ship to pass Bob's antenna is t = (Lx(u) d') / (u + v), where Lx(u) is the length contraction Alice measures of Danielle's ship. If t passes in Alice's frame, then z(v) t, where z(v) = z from before but now we are adding more speeds so must become more specific, passes for Bob and all observers must agree that this is Bob's reading when Danielle passes since the events of the readings upon his clock coincide in the same place as the front and back of Danielle's ship with Bob's antenna when the clock is placed in the same place as the antenna also. The length of Danielle's ship as Bob measures it, then, is d" = w t' = [(u + v) / (1 + u v / c^2)] [z(v) (Lx(u) d') / (u + v)] = z(v) Lx(u) d' / (1 + u v / c^2). The observed length contraction, then, is Lx(w) = d" / d' = z(v) Lx(u) / (1 + u v / c^2).

So now let's find out what Bob and Carl measure for the time dilation of Danielle's clock. We will place Carl in the front of the ship of proper length d' and Bob at the back. Alice says Danielle travels from Carl to Bob in a time of t = (Lx(v) d') / (u + v). When Danielle passes Carl, the readings upon the clocks according to Alice are TC=0 and TB = tl. When Danielle passes Bob, Bob's reading is TB' = tl + z(v) t, and again, all observers must agree since the events of the clock readings and Danielle directly passing the clocks coincide in the same places. Bob and Carl say the difference in times that has passed between their clocks is TB' - TC = tl + z(v) t = z(v) (Lx(v) d') v / (c^2 - v^2) + z(v) (Lx(v) d') / (u + v) = [z(v) Lx(v) d' / ((c^2 - v^2) (u + v))] [v (u + v) + (c^2 - v^2)] = d' [c^2 + u v] / (c^2 (u + v)) = d' (1 + u v / c^2) / (u + v) = d' / w. The amount of time that has passed upon Danielle's clock while travelling from Carl to Bob is t" = z(u) t = z(u) (Lx(v) d') / (u + v), so the time dilation Bob and Carl measure of Danielle's clock is z(w) = t" / t' = [z(u) (Lx(v) d') / (u + v)] / [d' / w] = [z(u) Lx(v)] [w / (u + v)] = z(u) Lx(v) / (1 + u v / c^2).

Now let's compare what we have gained for the time dilation and length contraction Bob and Carl measure of Danielle and dive into a little math logic for this part of the demonstration. We have found that Lx(w) = z(v) Lx(u) / (1 + u v / c^2) and z(w) = z(u) Lx(v) / (1 + u v / c^2), whereby (1 + u v / c^2) = z(v) Lx(u) / Lx(w) = z(u) Lx(v) / z(w), so by rearranging we gain [z(v) / Lx(v)] [Lx(u) / z(u)] [z(w) / Lx(w)] = 1. The values here that are represented by u and v are what Alice measures for the relative speeds of Danielle and of Bob and Carl, respectively. w represents the relative speed that is measured by Bob and Carl of Danielle. Now, from Alice's perspective, u and v can have any arbitrary values for the relative speeds to the reference frame and w will be determined by what those values are. u and v can have the same value and still be arbitrary, so let's say that u = v. In that case, [z(v) / Lx(v)] [Lx(u) / z(u)] = 1 so [z(w) / Lx(w)] = 1 also. Since w can still have any arbitrary value with any arbitrary values of u and v where u=v, then [z(w) / Lx(w)] = 1 for any arbitrary value whatsoever, therefore z(w) / Lx(w) always equals 1 for any relative speed of w. If that is the case, then [z(v) / Lx(v)] [Lx(u) / z(u)] [z(w) / Lx(w)] = [z(v) / Lx(v)] [Lx(u) / z(u)] = 1 for any arbitrary values of u and v even when the speeds are not equal, and the only way they can do that when changing the speed of u slightly while keeping the speed of v the same, for instance, is if z(v) / Lx(v) = 1 and z(u) / Lx(u) = 1 always also.

So since we had z(v) Lx(v) = 1 - (v/c)^2 and Lx(v) / Ly(v) = sqrt(1 - (v/c)^2) as found at the beginning of the demonstration, and now we have z(v) / Lx(v) = 1, whereby z(v) = Lx(v), then z(v) Lx(v) = z(v)^2 = 1 - (v/c)^2, giving z(v) = Lx(v) = sqrt(1 - (v/c)^2), as well as Lx(v) / Ly(v) = sqrt(1 - (v/c)^2), giving Ly(v) = 1, so no contraction takes place perpendicularly to the line of motion. And there we have it. We have determined that all of the basic principles of SR can be determined from the second postulate alone.
Bivalves
Could you summarize that nonsense down to only 27 trillion words please, as whilst reading the first bit I've noticed the geology changing outside.

Trout
QUOTE (grav+Feb 6 2010, 06:44 AM)
We have determined that all of the basic principles of SR can be determined from the second postulate alone.

You haven't determined anything, you are simply blabbing nonsense.
Starting with the second postulate ONLY derive the covariance of Maxwell's wave equation. This is a very good example of "proving" the first postulate . You can't do it.
grav
QUOTE (Trout+Feb 6 2010, 12:55 AM)
You haven't determined anything, you are simply blabbing nonsense.
Starting with the second postulate ONLY derive the covariance of Maxwell's wave equation. This is a very good example of "proving" the first postulate . You can't do it.

Maxwell's equations come before SR and determine that light travels isotropically at c, so that is where the second postulate comes from to begin with. Here is what I am presenting. Normally, by using only the second postulate alone, one might only be able to get as far as to determine that z Lx = 1 - (v/c)^2 and Lx / Ly = sqrt(1 - (v/c)^2). It is then only by applying the first postulate that the precise values are usually obtained, say with Doppler, where Bob moves away from Alice, so Bob measures a Doppler shift of D = (1 - v/c) /z and Alice measures D = z / (1 + v/c), so by including that the laws of physics are the same in every inertial frame, then they will get identical results between them, whereby (1 - v/c)/ z = z / (1 + v/c), so z^2 = 1 - (v/c)^2, and we receive z = Lx = sqrt(1 - (v/c)^2) and Ly = 1 from this. However, by using the math logic in the second to last paragraph of the OP, as it relates to the two previous paragraphs where Bob and Carl find the time dilaltion and length contraction of Danielle, the precise values can then be determined, all from involving only the second postulate alone.
Trout
QUOTE (grav+Feb 6 2010, 06:20 PM)
Maxwell's equations come before SR and determine that light travels isotropically at c, so that is where the second postulate comes from to begin with. Here is what I am presenting.

This is not what I asked you. You were asked to show the covariance. you don't even understand what you are asked to do.
Please top repeating the same tripe and pay attention to what you are challenged to do.
grav
QUOTE (Trout+Feb 6 2010, 02:58 PM)
This is not what I asked you. You were asked to show the covariance. you don't even understand what you are asked to do.
Please top repeating the same tripe and pay attention to what you are challenged to do.

There is no need to do that as it is not relevant to this thread. I have made no claim that SR is incorrect or that it will not apply to any field in the way it usually does. All I have claimed is that the basic principles of SR can be derived from the second postulate alone. If your observations were as astute as your demands, then you might have noticed that the first postulate is still necessary for what I have demonstrated because there is otherwise nothing that says that different materials might not contract to varying degrees in different frames unless the physics is the same in every inertial frame. As such, I have already been upped by an attentive poster in this other thread.
Trout
QUOTE (grav+Feb 6 2010, 10:38 PM)
There is no need to do that as it is not relevant to this thread. I have made no claim that SR is incorrect or that it will not apply to any field in the way it usually does.

Sure there is need, you claim that you have a theory competing with SR that is (in your dreams) "more intuitive" (every crank's claim) but you are unable to calculate anything with it.

QUOTE
All I have claimed is that the basic principles of SR can be derived from the second postulate alone. If your observations were as astute as your demands, then you might have noticed that the first postulate is still necessary for what I have demonstrated because there is otherwise nothing that says that different materials might not contract to varying degrees in different frames unless the physics is the same in every inertial frame. As such, I have already been upped by an attentive poster in this other thread.

Looking at the thread it is clear that you've been told the same thing as here: you are wrong and you don't know what you are doing. You've been told the same thing in the third forum, here. Why do you feel compelled to spam half the internet with your tripe?
grav
QUOTE (Trout+)
Sure there is need, you claim that you have a  theory competing with SR that is (in your dreams) "more intuitive" (every crank's claim) but you are unable to calculate anything with it.
I think you have the wrong thread for that one. I have only derived SR in its current form here. There is no competing theory.

QUOTE (Trout+)
Looking at the thread it is clear that you've been told the same thing as here: you are wrong and you don't know what you are doing. You've been told the same thing in the third forum, here. Why do you feel compelled to spam half the internet with your tripe?
There is a difference between told you are wrong and being proven wrong. I have only been proven wrong once in the thread I already linked to, although of course, that is all it takes, and I have conceded this thread accordingly.
Trout
QUOTE (grav+Feb 7 2010, 12:14 AM)
There is a difference between told you are wrong and being proven wrong. I have only been proven wrong once in the thread I already linked to, although of course, that is all it takes, and I have conceded this thread accordingly.

So, do you need to be proven wrong in all forums before you admit that you are wrong?
grav
QUOTE (Trout+Feb 6 2010, 07:27 PM)
So, do you need to be proven wrong in all forums before you admit that you are wrong?

I've already admitted it in both of my last couple of posts and posted that I was wrong in the other forums as well and even included a link to the thread that showed where I was wrong, so what's the problem there?
brucep
QUOTE (grav+Feb 7 2010, 12:14 AM)
There is a difference between told you are wrong and being proven wrong. I have only been proven wrong once in the thread I already linked to, although of course, that is all it takes, and I have conceded this thread accordingly.

rpenner and Trout have shown you why it's useless. Useless = Wrong in my book. Trout has asked you to apply your model, in a useful way, several times yet you seem to think applying your model to problem solving isn't important. Show us how you would derive E/m = dt/dTau and P/m = ds/dTau as conserved quantities over path segments, ie constants of the motion, using your model with the absolute aether rest frame. With these conserved quantities and the invariant mass you can derive the energy equation. I'll link you to the first chapter in Edwin Taylor's Exploring black holes which you can look at beginning at section 5 Energy in Special Relativity andt help you understand my query.

http://www.eftaylor.com/pub/chapter1.pdf
grav
QUOTE (brucep+Feb 6 2010, 08:47 PM)
rpenner and Trout have shown you why it's useless. Useless = Wrong in my book. Trout has asked you to apply your model, in a useful way, several times yet you seem to think applying your model to problem solving isn't important. Show us how you would derive E/m = dt/dTau and P/m = ds/dTau as conserved quantities over path segments, ie constants of the motion, using your model with the absolute aether rest frame. With these conserved quantities and the invariant mass you can derive the energy equation. I'll link you to the first chapter in Edwin Taylor's Exploring black holes which you can look at beginning at section 5 Energy in Special Relativity andt help you understand my query.

http://www.eftaylor.com/pub/chapter1.pdf

Sorry, brucep. I think you have the wrong thread also. There is no alternative model proposed here.
grav
deleted post
brucep
QUOTE (grav+Feb 7 2010, 02:57 AM)
Sorry, brucep. I think you have the wrong thread also. There is no alternative model proposed here.

Ok. In this thread you told Trout you admitted you're wrong. I assume you meant your aether model is wrong?
Trout
QUOTE (brucep+Feb 7 2010, 03:43 AM)
Ok. In this thread you told Trout you admitted you're wrong. I assume you meant your aether model is wrong?

grav has two crackpot threads going on simultaneously. He admitted that this particular thread is wrong (which doesn't stop him from continuing the masquarade in another forum) , he has not admitted that his "OAR" is wrong. I looked on the BAUT forum and the "OAR" thread was locked by the moderators because grav was caught cheating (and because his claims are false, of course).
grav
QUOTE (brucep+Feb 6 2010, 09:43 PM)
Ok. In this thread you told Trout you admitted you're wrong. I assume you meant your aether model is wrong?

No, just for this presentation. It requires both postulates to derive SR.
grav
QUOTE (Trout+Feb 6 2010, 10:15 PM)
grav has two crackpot threads going on simultaneously. He admitted that this particular thread is wrong (which doesn't stop him from continuing the masquarade in another forum) , he has not admitted that his "OAR" is wrong. I looked on the BAUT forum and the "OAR" thread was locked by the moderators because grav was caught cheating (and because his claims are false, of course).

I was not caught "cheating", although the thread was closed due to the accusation. That is just a common tactic in these types of debates, not unlike your own, as a matter of fact. I simply editted a post where I had forgotten to drop a 'z' down. If I had editted it only to match the correct results as was implied, then I would have had to go back through the entire thread and edit all of my posts where I had produced the exact same results. Also, if you think my claims are false in the other thread here, then simply show me where an error lies and I will admit that I was wrong about that too.
brucep
QUOTE (Trout+Feb 7 2010, 04:15 AM)
grav has two crackpot threads going on simultaneously. He admitted that this particular thread is wrong (which doesn't stop him from continuing the masquarade in another forum) , he has not admitted that his "OAR" is wrong. I looked on the BAUT forum and the "OAR" thread was locked by the moderators because grav was caught cheating (and because his claims are false, of course).

Thanks Trout

How do you cheat in a forum, probably involves telling whoopers. Like pretending a preferred frame aether theory is more convenient to use than relativity.
Trout
QUOTE (grav+Feb 7 2010, 04:47 AM)
I was not caught "cheating", although the thread was closed due to the accusation. That is just a common tactic in these types of debates, not unlike your own, as a matter of fact. I simply editted a post where I had forgotten to drop a 'z' down. If I had editted it only to match the correct results as was implied, then I would have had to go back through the entire thread and edit all of my posts where I had produced the exact same results. Also, if you think my claims are false in the other thread here, then simply show me where an error lies and I will admit that I was wrong about that too.

In other words....you chaeted! And you got caught.
grav
QUOTE (brucep+Feb 6 2010, 11:31 PM)
Thanks Trout

How do you cheat in a forum, probably involves telling whoopers. Like pretending a preferred frame aether theory is more convenient to use than relativity.

Who ever said anything about a preferred frame aether theory? There can be an arbitrary preferred frame in SR only, which is very convenient to use, as I have already stated in the other thread. An aether frame would be more than just an arbitrary preferred frame, it would be considered absolute. And why are we discussing this here instead of there?
grav
QUOTE (Trout+Feb 6 2010, 11:39 PM)
In other words....you chaeted! And you got caught.

Well, if you consider editting a typo cheating, then I see you've cheated numerous times in this thread already. Granted, it was a math typo, but than again, mathematics is the language of physics. My results were already scattered throughout the thread, and math does not lie. The results were only according to what the math says, not me.
brucep
QUOTE (grav+Feb 8 2010, 01:53 AM)
Who ever said anything about a preferred frame aether theory? There can be an arbitrary preferred frame in SR only, which is very convenient to use, as I have already stated in the other thread. An aether frame would be more than just an arbitrary preferred frame, it would be considered absolute. And why are we discussing this here instead of there?

Plonk!!!!!!!!!
Trout
QUOTE (grav+Feb 8 2010, 02:01 AM)
Well, if you consider editting a typo cheating, then I see you've cheated numerous times in this thread already. Granted, it was a math typo, but than again, mathematics is the language of physics. My results were already scattered throughout the thread, and math does not lie. The results were only according to what the math says, not me.

You had the same mistake in several posts throughout the thread. You editted the mistake only in one post. Now, you are outright lying about it and cheating in this thread. Plonk!
grav
QUOTE (brucep+Feb 7 2010, 08:06 PM)
Plonk!!!!!!!!!

So what is that supposed to mean? Reread the other thread again. It's all still there in black and white.
grav
QUOTE (Trout+Feb 7 2010, 08:40 PM)
You had the same mistake in several posts throughout the thread. You editted the mistake only in one post. Now, you are outright lying about it and cheating in this thread.  Plonk!

I wrote the same thing I wrote in every other post, that the simultaneity difference, or time lag, observed between the clocks in a frame that is moving relative to an observer is tl = d' v / c^2, where d' is the distance between the clocks according to the frame of the clocks, so d' = d / z = y d, where d is the distance between the clocks according to the observing frame, whereby tl = y d v / c^2, exactly as it should be. It was stated from the very first post of the thread and the relativity of simultaneity was demonstrated fully in post #21 where it was demonstrated how the result was derived using train and tunnel observers. So as was similarly requested in the thread, I will ask the same thing of you. If you disagree with that result, that tl = d' v / c^2 or tl = y d v / c^2, then please, please, by all means, enlighten me as to what you believe it should really be.
brucep
QUOTE (grav+Feb 8 2010, 03:15 AM)
So what is that supposed to mean? Reread the other thread again. It's all still there in black and white.

grav
QUOTE (brucep+Feb 7 2010, 09:29 PM)

So what then? Are you now also disagreeing that a preferred frame in SR, in other words from the point of view of an observer that always remains stationary, is the most convenient way to perform calculations?
grav
Okay, folks. Looks like all has not yet been lost for this presentation after all. I may have thrown in the towel to soon.

As also seen in this thread, Einstein assumes homogeneity in his calculations, and that's also exactly what I have used in my presentation rather than the first postulate. If we still add it as a second postulate to my presentation, then SR must also, so it then has three. Okay, so regardless of the materials that a ship is composed of, it is only the relation between the length of the rulers to the ship and so forth that they measure, and we can have varying values for z and Lx and still have observers measure their own ships the same length as well as measuring the speed of light isotropically at c. For instance, let's say that in a particular frame, an observer has a meter stick and measures his ship to have some length, as well as measuring the speed of light isotropically at c, gaining some value for z and Lx as viewed from another frame. Another observer in the same frame as the first, however, carries a meter stick that is twice as long as the other observer, but still measures their own ship the same length because their ship is really twice as long as the first's ship also, and if the second observer's clock ticks at half the rate of the first observer's clock, then the second observer will also measure light travelling isotropically at c. But for the second observer, an observer in another frame will measure twice the value of L than for the first observer, and half the value for z, still gaining z Lx = 1 - (v/c)^2, but no definite values determined from that. So we find we can have varying values of z and Lx to another frame.

Okay, but here's what happened. One can see in the presentation where Danielle was introduced in the last four paragraphs and then the observations of her were used to determine the precise values of z and Lx. Now, notice that Danielle measures the length of her ship d' to be the same as the distance that Bob and Carl measure between themselves. One can place them also in the front and back of a ship for that. Then notice that Alice also measures Bob and Carl's ship to be d as well as measuring Danielle's ship to be the same. This makes all the difference. It is a given here that Danielle's ship and Bob and Carl's ship are identical, but what does that mean? Well, first off it means that if Danielle's ship and Bob and Carl's ship were in the same frame, then they could do a direct comparison between the two ships to determine whether they were actually the same length or not, thereby the lengths of their rulers must also be the same. But Danielle is travelling in the opposite direction than Bob and Carl, so a direct comparison cannot be made, and all we can say about those observers is that they measure their own ships to have the same length as each other by using their own rulers.

However, in the second to last paragraph of the presentation, we set Danille's speed equal to Bob and Carl's speed as measured by Alice, whereas u=v, but in opposite directions. Now, as an extension of the given that their ships are identical, Alice measuring the lengths of both ships as d, that means that if Danielle were travelling in the same direction at v as Bob and Carl also at v, Alice could also do a direct comparison between the ships to tell if they are actually the same length. But Danielle is travelling in the opposite direction instead, so what do we do here? We simply require that all observations be homogeneous in any direction from any position in the same frame. That is, if Danielle's ship is observed to have a certain length when travelling to the right at v according to Alice, then if Danielle's ship were to travel to the left at v instead, it would still be observed to have the same length, and so would be identical. So if the length of Danielle's ship is identical to the length of Bob and Carl's ship when travelling in opposite directions at v, then it would be the same as a direct comparison when both ships are travelling in the same direction at v. Therefore, if we assume that all observations are homogeneous in this way, and are comparing the length of identical ships and rulers, and therefore identically ticking clocks which together measure the speed of light as isotropical at c, then we can determine the precise values of z = Lx = sqrt(1 - (v/c)^2), as has been done in the demonstration.
Trout
QUOTE (grav+Feb 8 2010, 03:27 AM)
I wrote the same thing I wrote in every other post,

You are a liar and a cheat. You persist in lying and cheating.

QUOTE
hat the simultaneity difference, or time lag, observed between the clocks in a frame that is moving relative to an observer is tl = d' v / c^2, where d' is the distance between the clocks according to the frame of the clocks, so d' = d / z = y d, where d is the distance between the clocks according to the observing frame, whereby tl = y d v / c^2,

d'=d/gamma (standard length contraction) so you can NEVER obtain tl = y d v / c^2 from tl = d' v / c^2,, through your crackpot method. You are still trying to cheat.
grav
QUOTE (Trout+Feb 7 2010, 10:29 PM)
You are a liar and a cheat. You persist in lying and cheating.
About what? Neither of you have even provided any replies that are useful enough to comment on, other than "PLONK!". Surely you two are not the only significant voices on this forum, or I will very disappointed.

QUOTE
d'=d/gamma (standard length contraction) so you can NEVER obtain tl = y d v / c^2 from  tl = d' v / c^2,, through your crackpot method. You are still trying to cheat.
Are you just looking at how you've seen d' described before without thinking about what it means and how I might be using it in a different way? The standard for d is what the observer sees of their own ruler and d' for what they see of a ruler in another frame, but I have used d for what the observer sees of the ruler in the other frame and d' for what an observer in the other frame sees of the same ruler, so it is d' = y d in the demonstration. It was just easier to consider all of Alice's observations of the other frame as the unprimed and that of the other frame of their measurements in their own frame as primed. Although, now that I see it can produce some confusion, maybe you have made a point that I should have done it the other way around instead, so I'll give you that.
Trout
QUOTE (grav+Feb 8 2010, 05:09 AM)

Are you just looking at how you've seen d' described before without thinking about what it means and how I might be using it in a different way?

Nonsense, d'=d/gamma. ALWAYS. You don't know the very basics abouut length contraction.
grav
QUOTE (Trout+Feb 7 2010, 11:49 PM)
Nonsense, d'=d/gamma. ALWAYS. You don't know the very basics abouut length contraction.

Not when I have defined d as the observations made of the moving frame and d' as the observations made by a moving observer in the moving frame, but you have made a good point, as I said, so maybe I should have used d' and d" or something to stay more in line with the standard interpretation in order to avoid possible confusion, which may have led to the problems in the other thread you referred to now that you mention it. It's too late to change my presentation now but I'll remember that in the future. Thanks.
Trout
QUOTE (grav+Feb 8 2010, 06:10 AM)
Not when I have defined d as the observations made of the moving frame and d' as the observations made by a moving observer in the moving frame,

What in the word ALWAYS don't you understand?
grav
QUOTE (Trout+Feb 8 2010, 12:19 AM)
What in the word ALWAYS don't you understand?

It depends upon the circumstances and how it is specifically defined. For example, if observer B passes observer A at a relative speed of v and they synchronize clocks upon passing, while a time of t then passes for observer A, t' might refer to the time dilation that observer A measures upon observer B's clock, which would be t' = t / y, or it might refer to the time that observer B observes upon his own clock directly when applying the Lorentz transforms with t' = y (t - v x / c^2).
Trout
QUOTE (grav+Feb 18 2010, 08:00 AM)
It depends upon the circumstances and how it is specifically defined.

No, it doesn't. What in ALWAYS don't you understand?
You STILL don't know the very basics about length contraction.
grav
QUOTE (Trout+Feb 18 2010, 09:20 AM)
No, it doesn't. What in ALWAYS don't you understand?
You STILL don't know the very basics about length contraction.

Well, I just showed you an example of how it is used differently with the Lorentz transform itself, where t and t' are the readings each observer reads upon their own clocks for the same event as compared to t' with time dialtion for what is read upon the other's clock. So if it is ALWAYS to be used a particular way, then which one is correct?

With this thread, observations made by Alice in the preferred frame are unprimed and observations made by Bob and Carl are primed. But if you wish, I can use d(A>BC) for what Alice observes of the distance between Bob and Carl and d(BC>BC) for the distance Bob and Carl observe between themselves to make things more clear. In that case, d(BC>BC) = d(A>BC) / sqrt(1 - (v/c)^2).
Trout
QUOTE (grav+Feb 18 2010, 03:46 PM)
Well, I just showed you an example of how it is used differently with the Lorentz transform itself,

No, it isn't. The derivation of length contraction in SR is ALWAYS d'=d/gamma.
In your fringe theory it might be different.

QUOTE
I can use d(A>BC) for what Alice observes of the distance between Bob and Carl and d(BC>BC) for the distance Bob and Carl observe between themselves to make things more clear. In that case, d(BC>BC) = d(A>BC) / sqrt(1 - (v/c)^2).

No, you can't. Until you learn SR properly , you can forget about fabricating your fringe theories.
grav
QUOTE (Trout+Feb 18 2010, 09:52 AM)
No, it isn't. The derivation of length contraction in SR is ALWAYS d'=d/gamma.
In your fringe theory it might be different.

Right, I have made it that way with my hypothesis, where Alice's observations are unprimed and Bob and Carl's are primed, but I can see how it might be confusing at times when considering the convention for time dilation as you say, so from this point on I will denote them with A>BC and BC>BC instead.
Trout
QUOTE (grav+Feb 18 2010, 04:00 PM)
Right, I have made it that way with my hypothesis, where Alice's observations are unprimed and Bob and Carl's are primed, but I can see how it might be confusing at times when considering the convention for time dilation as you say, so from this point on I will denote them with A>BC and BC>BC instead.

It is not a matter of notation, it is a matter of your inability to understand the very basics of SR.
grav
QUOTE (Trout+Feb 18 2010, 10:01 AM)
It is not a matter of notation, it is a matter of your inability to understand the very basics of SR.

I understand SR fine. If you do not think so, then point to a place where I have shown such a misunderstanding.
Trout
QUOTE (grav+Feb 18 2010, 04:04 PM)
I understand SR fine. If you do not think so, then point to a place where I have shown such a misunderstanding.

How about the two threads you opened in this forum? They both show very deep misconceptions about SR. You should really take the advice and enroll in a SR 101 class.

Let me prove to you that you do not understand basic relativity.

You have two frames in relative motion with speed v. The Lorentz transform between frame F and frame F' is:

x'=g(x-vt)
t'=g(t-vx/c^2)

Let's say that you have a ruler of length dx, as measured in F. Its length, as measured in F' is determined by marking its two endpoints simultaneously in F':

dt'=0

But :

dx'=g(dx-v*dt)

dt'=g(dt-vdx/c^2)

Since dt'=0 it follows that dt= vdx/c^2. Substitute the expression for dt into the expression for dx' and you get

dx'=dx/g (length contraction)

A similar approach is used for determining time dilation, you need to look at two events that are co-located in frame F (dx=0) as seen from the perspective of the observer in frame F' . When you substitute dx=0 into the expression of dt' you get

dt'=gdt (time dilation)

Now, if you want to find out the expressions from the perspective of an observer located in frame F, you simply need to use the inverse Lorentz transforms

x=g(x'+vt')
t=g(t'+vx'/c^2)

Find dt as a function of dt'
Find dx as a function of dx'

If you do the above correctly, you will find the errors in your two threads.
grav
QUOTE (Trout+Feb 18 2010, 10:41 AM)
How about the two threads you opened in this forum? They both show very deep misconceptions about SR. You should really take the advice and enroll in a SR 101 class.

Let me prove to you that you do not understand basic relativity.

You have two frames in relative motion with speed v. The Lorentz transform between frame F and frame F' is:

x'=g(x-vt)
t'=g(t-vx/c^2)

Let's say that you have a ruler of length dx, as measured in F. Its length, as measured in F' is determined by marking its two endpoints simultaneously in F':

dt'=0

But :

dx'=g(dx-v*dt)

dt'=g(dt-vdx/c^2)

Since dt'=0 it follows that dt= vdx/c^2. Substitute the expression for dt into the expression for dx' and you get

dx'=dx/g (length contraction)

A similar approach is used for determining time dilation, you need to look at two events that are co-located in frame F (dx=0) as seen from the perspective of the observer in frame F' . When you substitute dx=0 into the expression of dt' you get

dt'=gdt (time dilation)

Now, if you want to find out the expressions from the perspective of an observer located in frame F, you simply need to use the inverse Lorentz transforms

x=g(x'+vt')
t=g(t'+vx'/c^2)

Find dt as a function of dt'
Find dx as a function of dx'

If you do the above correctly, you will find the errors in your two threads.

Right, so you received dt' = g dt when the event occurs with dx = 0 in the unprimed frame, where dt' = g dt is not the usual standard for time dilation either, but is normally expressed as dt' = dt / g = sqrt(1 - (v/c)^2) dt for what one observer observes of another's clock. However, if we make dx' = 0 for the event taking place in the primed frame, then we receive dx' = g (dx - v dt), so dx = v dt. Then dt' = g (dt - v dx / c^2) = g (dt - v^2 dt / c^2) = g dt (1 - (v/c)^2) = g dt / g^2 = dt / g = sqrt(1 - (v/c)^2) dt.
Trout
QUOTE (grav+Feb 18 2010, 05:03 PM)
Right, so you received dt' = g dt when the event occurs with dx = 0 in the unprimed frame, where dt' = g dt is not the usual standard for time dilation either, but is normally expressed as dt' = dt / g = sqrt(1 - (v/c)^2) dt for what one observer observes of another's clock.

You have serious problems, you really need to take that class.

QUOTE
However, if we make dx' = 0 for the event taking place in the primed frame, then we receive dx' = g (dx - v dt), so dx = v dt. Then dt' = g (dt - v dx / c^2) = g (dt - v^2 dt / c^2) = g dt (1 - (v/c)^2) = g dt / g^2 = dt / g = sqrt(1 - (v/c)^2) dt.

SR NEVER claims such a stupidity.
grav
QUOTE (Trout+Feb 18 2010, 11:05 AM)
You have serious problems, you really need to take that class.

SR NEVER claims such a stupidity.

Why, do you see anything wrong with what I just did? Whether it it comes out to dt' = dt / g or dt' = g dt just depends upon which frame the event occurs, that's all. The reason is that it is not the time dilation that occurs directly between the primed and unprimed observers at all, but only the time each directly reads upon their own clocks when the event occurs according to their frame. By the way, if it were the time dilation observed between the observers, then by convention, the part you highlighted, where dt' = sqrt(1 - (v/c)^2) dt, dt for one's own frame and dt' for what is observed upon the clock of the other, is correct.
Trout
QUOTE (grav+Feb 18 2010, 05:17 PM)
Why, do you see anything wrong with what I just did?

You mixed frames, so you got garbage. Take that class, stop wasting your time posting hilarious mistakes.
grav
QUOTE (Trout+Feb 18 2010, 11:21 AM)
You mixed frames, so you got garbage. Take that class, stop wasting your time posting hilarious mistakes.

I didn't mix anything. The times do not represent what each observer reads upon the clock of the other, only upon their own clocks when the event takes place according to their frame, so it is not the time dilation between them.
Trout
QUOTE (grav+Feb 18 2010, 05:29 PM)
I didn't mix anything.

Trout
QUOTE (grav+Feb 18 2010, 05:03 PM)
Right, so you received dt' = g dt when the event occurs with dx = 0 in the unprimed frame, where dt' = g dt is not the usual standard for time dilation either, but is normally expressed as dt' = dt / g = sqrt(1 - (v/c)^2) dt for what one observer observes of another's clock.

grav
QUOTE (Trout+Feb 18 2010, 12:22 PM)

Apparently you still don't see it, so let me make things clearer by using dt(A>A) for the time that A reads upon A's own clock and dt(B>B) for the time that B reads upon B's own clock. With the Lorentz transforms, each uses their own set of coordinates for what they see of another observer C with dx(A>C) and dt(A>A) for A and dx(B>C) and dt(B>B) for B. So if dx(A>C) = 0 where C is at rest to A, then when an event occurs at C, we gain dt(B>B) = g dt(A>A) for how the times each reads upon their own clocks relate to each other as you had. If C is at rest in the B frame, then dx(B>C) = 0 and we gain dt(B>B) = dt(A>A) / g. This is not the same as the time dilation that occurs directly between the observers, which would be dt(A>B) = dt(A>A) / g and dt(B>A) = dt(B>B) / g.
Trout
QUOTE (grav+Feb 18 2010, 06:42 PM)
which would be dt(A>B) = dt(A>A) / g and dt(B>A) = dt(B>B) / g.

Not. Are you home schooled?
grav
QUOTE (Trout+Feb 18 2010, 12:43 PM)
Not. Are you home schooled?

So what's the matter with that now? If you don't like division, then we can simply rearrange with dt(A>A) = g dt(A>B) and dt(B>B) = g dt(B>A). Is that better?
Bivalves
QUOTE (Trout+Feb 18 2010, 06:43 PM)
Are you home schooled?

Possibly, if his home was for incredulously deranged fucked-up mental atrocities, with mammoth genetic issues and suck.

Trout
QUOTE (grav+Feb 18 2010, 06:52 PM)
So what's the matter with that now? If you don't like division, then we can simply rearrange with dt(A>A) = g dt(A>B) and dt(B>B) = g dt(B>A).

I don't know why you persist in your errors. I tried to teach you how to do the derivation correctly. Sorry, I failed, you are not the teachable kind.

QUOTE
Is that better?

You can't get it right if your life depended on it.
grav
QUOTE (Bivalves+Feb 18 2010, 12:54 PM)
Possibly, if his home was for incredulously deranged fucked-up mental atrocities, with mammoth genetic issues and suck.

No, but that's only what I've been finding here so far. Is everyone here like this? If so, I don't see how this forum has managed to keep any decent members at all, and probably hasn't for long, leaving only this rather special club.
Trout
QUOTE (grav+Feb 18 2010, 06:42 PM)
and we gain dt(B>B) = dt(A>A) / g.

NEVER.

QUOTE
This is not the same as the time dilation that occurs directly between the observers, which would be dt(A>B) = dt(A>A) / g and dt(B>A) = dt(B>B) / g.

NOT.

Using your notation, the correct formulas are:

dt(A>B) = dt(A>A) * g and dt(B>A) = dt(B>B) * g

See the difference?

I tried to teach you how to derive the correct formulas. Sorry, I failed.
grav
QUOTE (Trout+Feb 18 2010, 01:08 PM)
NEVER.

NOT.

Using your notation, the correct formulas are:

dt(A>B) = dt(A>A) * g and dt(B>A) = dt(B>B) * g

See the difference?

I tried to teach you how to derive the correct formulas. Sorry, I failed.

Did you forget that g = 1 / sqrt(1 - (v/c)^2)?
Trout
QUOTE (grav+Feb 18 2010, 07:19 PM)
Did you forget that g = 1 / sqrt(1 - (v/c)^2)?

No, I didn't . Why? What does it have to do with your inability to comprehend either length contraction or time dilation?
grav
QUOTE (Trout+Feb 18 2010, 01:22 PM)
No, I didn't . Why?

Because if dt(A>A) is the time that A observes passing upon A's own clock and dt(A>B) is the time that A observes passing upon B's clock, then since A sees B's clock time dilating, then dt(A>B) = sqrt(1 - (v/c)^2) dt(A>A) = dt(A>A) / g, or dt(A>A) = g dt(A>B).
grav
Actually, I have come to realize that all this talk of dt and dx has confounded the Lorentz transformations also. t and x are the coordinates of an event that takes place according to a frame, so just remains t and x, t being the time that passes on the observer's clock from the time the two observers pass each other and synchronize their clocks to T=0 and x being the distance that an event takes place from the origin of the frame where the observer is located. After observers A and B synchronize their clocks upon passing each other, if an event occurs at x(A>C) = 0, then it occurs at the origin of A at some time of t(A>A), giving t(B>B) = g t(A>A) for when the event takes place according to the clock of the B frame. Likewise, if an event occurs at the origin of B where x(B>C) = 0, then we get t(A>A) = g t(B>B).
Trout
QUOTE (grav+Feb 18 2010, 07:26 PM)
Because if dt(A>A) is the time that A observes passing upon A's own clock and dt(A>B) is the time that A observes passing upon B's clock, then since A sees B's clock time dilating, then dt(A>B) = sqrt(1 - (v/c)^2) dt(A>A) = dt(A>A) / g, or dt(A>A) = g dt(A>B).

You are not fixable, I give up.
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