[Colin2B]

I am definitely learning a lot about the specific vernacular used to communicate these physics ideas to others on this forum...thanks for the feedback.

I'm glad to hear that, but how about learning to answer the question?  []

I use the same definitions for the variable "types" as Einstein within his book "Relativity: The Special and General Theory", i.e. the primed variable is with respect to the co-ordinate system K' and the unprimed variable is with respect to the co-ordinate system K. I use the Lorentz transformations in the same manner as well.

This doesn't tell us how you are interpreting what you are working with.
Nor does it answer the question 'which clock'
Let me take you back to your original post

In order to satisfy the first postulate, the relationship between their event times must remain unchanged in order to prevent one from ascertaining the motion of the inertial frame utilizing this scenario.

When you write this I have to assume you are working from the viewpoint of an observer in K', with a clock in K', moving at the same speed as K'. In this case that observer will say that c=c' and u=u'. So the race is the same as observed in K.

When you apply transforms to K' I have to assume you are working from the viewpoint of an observer in K, with a clock in K, observing events in K'. For this observer c=c', but u≠u' the value of u' has to be calculated using transforms. Hence the race is not the same as the race run in K.

We have to be careful how we use physical laws in relativity.
If we define that atomic clocks always show the 'correct' time, this only valid in inertial frames, not ones moving relative to one another.
The same is true if we define that a 1m platinum rod at temp x is always the same length. Relativity say only in inertial frames, not between frames moving relative to one another.

I think we can come to a common interpretation by "walking the dog" if you're up to it...wadda ya say?

Only if you can confirm which clock, which observer, and that you understand what I have written in this and my last post.

[PmbPhy]

EVENTS OCCURRING IN REST FRAME (v = 0)
- These events now occur within a reference frame considered to be at rest (rest frame).
- The event time for A is; Δτ = x/c.
- The event time for B is; Δt = x/u.
- The ratio of these event times gives us the relationship between their event times;
Δτ/Δt = (x/c)/(x/u) = u/c.

No wonder I had a problem understanding what it was you were doing. It was as if you actually went out of your way to take a very simple thing and say it in a very complex way. You should have simply said the following: Let Δx be the distance traveled by a photon and a particle. Let u be the speed of the particle (thus u < c). Let Δt be the time it takes the particle travel the distance Δx and ΔT the time it takes a photon to travel the same distance. Then Δx = uΔt = cΔT. Therefore ΔT/Δt = u/c.

See how simple that was? []

[jeffreyH]

EVENTS OCCURRING IN REST FRAME (v = 0)
- These events now occur within a reference frame considered to be at rest (rest frame).
- The event time for A is; Δτ = x/c.
- The event time for B is; Δt = x/u.
- The ratio of these event times gives us the relationship between their event times;
Δτ/Δt = (x/c)/(x/u) = u/c.

No wonder I had a problem understanding what it was you were doing. It was as if you actually went out of your way to take a very simple thing and say it in a very complex way. You should have simply said the following: Let x be the distance traveled by a photon and a particle. Let v be the speed of a particle where v < c. Let t be the time it takes the particle travel the distance x and T the time it takes a photon to travel the same distance. Then x = vt = cT. Therefore T/t = v/c.

See how simple that was?

I managed to get it. Probably because I'm still picking things up. I bet if I had a degree in physics it wouldn't have made any sense. That is why I try to read so much. So at least I may be able to explain things in terms that other members may readily understand. I am not always successful.

[Colin2B]

No wonder I had a problem understanding what it was you were doing. It was as if you actually went out of your way to take a very simple thing and say it in a very complex way. You should have simply said the following: Let x be the distance traveled by a photon and a particle. Let v be the speed of a particle where v < c. Let t be the time it takes the particle travel the distance x and T the time it takes a photon to travel the same distance. Then x = vt = cT. Therefore T/t = v/c.
See how simple that was?

Yes, but he makes it even more complicated in the following section where he considers the moving frame. He uses the same T and t for the values seen by the observer in the rest frame when viewing the moving frame, but T and t are what the observer within the moving frame sees and are seen as reduced by the observer in the rest frame. He needs new symbols for these transformed values as he is making the mistake of equating unequal values.
I find the entire analysis over complicated and it is very unclear. To be honest I haven't even looked to see whether the maths is correct as there are too many false assumptions before you even start on the maths. No, no, I'll be really honest, it's because I'm not a mathematician and I don't really enjoy the maths bit  []

[Waste of Time (OP)]

No wonder I had a problem understanding what it was you were doing. It was as if you actually went out of your way to take a very simple thing and say it in a very complex way. You should have simply said the following: Let x be the distance traveled by a photon and a particle. Let v be the speed of a particle where v < c. Let t be the time it takes the particle travel the distance x and T the time it takes a photon to travel the same distance. Then x = vt = cT. Therefore T/t = v/c.

See how simple that was?

If that is a format that you understand better, then yes...go with it. The original format makes perfect sense to me, but then again I wrote it which makes me biased to understanding.

[Waste of Time (OP)]

Only if you can confirm which clock, which observer, and that you understand what I have written in this and my last post.

Just as Einstein defines his primed variables (such as t'), so I define mine. With regards to the Lorentz transformation t' = (t - vx/c^2)γ, which clock and observer does Einstein use for t'?

[Waste of Time (OP)]

When I asked "I think we can come to a common interpretation by "walking the dog" if you're up to it...wadda ya say?", I meant to start at the beginning and step through the progression (changing as necessary in small pieces) of the scenario ensuring that you can follow my logic (as you best understand) until we get to the conclusion. My original writing style was how I think and write, which is obviously not best for a general audience that is used to another style.

[PmbPhy]

If that is a format that you understand better, then yes...go with it. The original format makes perfect sense to me, but then again I wrote it which makes me biased to understanding.

I'm sorry to see that you don't understand what we're trying to get through to you yet. Here you made the mistake of interpreting my response as a complaint of format whereas it was actually a complaint against your overly complicated and very unclear description of the problem. The best way to communicate something to someone else is to make it as clear and as simple as you can. You didn't do that. You should have first stated the problem very clearly in words. After that's done, and only after it's done, do you translate it into the language of spacetime physics. Do you understand this now? I'll try to do this with your opening post.

First off I've asked you several times what "There are two parallel linear events (A and B) with uniform velocity along the positive x-axis." means and I've yet to get a response. You didn't even state what the worldlines are which are supposed to be parallel. I'm going to assume that you're referring to the following worldlines;
Worldline A: Worldline connecting origin with event A
Worldline B: Worldline connecting origin with event B

Worldline A is the worldline of a photon which is emitted from the origin and moves in the +x-direction and ends up at event A. That means that it's a line which is 45 degrees with respect to the +x-axis (and of course its also a line which is 45 degrees with the ct-axis).

Worldline B is the worldline of a particle which moves at a speed less than the speed of light and ends up at event B. That means that it's a line which is greater than 45 degrees with respect to the +x-axis.

This means that it is a line which is 45 degrees with respect to the +x-axis (and of course it’s also a line which is 45 degrees with the ct-axis).

Therefore it follows that these two worldlines are not parallel. So what in the world do you mean by “parallel events”?

[PmbPhy]

   In order to satisfy the first postulate, the relationship between their event times must remain unchanged in order to prevent one from ascertaining the motion of the inertial frame utilizing this scenario.

Okay. Now I see where you're going with all of this. This statement is not true. The first postulate states that the laws of physics are the same in all frames of reference. That means that no experiment can be done which would tell the observer that he's in motion relative to an absolute frame of reference. It does not mean that the relationship you established is invariant. This is the error which led you to the error in the paper you posted in post #1. You forget, those relationships are well tested so if you thought that you found a mistake in them then you can be certain that you're wrong.

[Waste of Time (OP)]

First off I've asked you several times what "There are two parallel linear events (A and B) with uniform velocity along the positive x-axis." means and I've yet to get a response. You didn't even state what the worldlines are which are supposed to be parallel. I'm going to assume that you're referring to the following worldlines;
Worldline A: Worldline connecting origin with event A
Worldline B: Worldline connecting origin with event B

Worldline A is the worldline of a photon which is emitted from the origin and moves in the +x-direction and ends up at event A. That means that it's a line which is 45 degrees with respect to the +x-axis (and of course its also a line which is 45 degrees with the ct-axis).

Worldline B is the worldline of a particle which moves at a speed less than the speed of light and ends up at event B. That means that it's a line which is greater than 45 degrees with respect to the +x-axis.

This means that it is a line which is 45 degrees with respect to the +x-axis (and of course it’s also a line which is 45 degrees with the ct-axis).

Therefore it follows that these two worldlines are not parallel. So what in the world do you mean by “parallel events”?

I have no idea where you are getting this information or interpretation. When I say parallel, I mean parallel. A photon travels parallel to the x-asix, and an electron travels parallel to the x-axis. Simple geometry, like your car travels parallel to the surface of the road. No need for world lines or tilting through any degrees. You don't understand my original scenario...I get it. No need to keep stating the same thing...I get it.

[Waste of Time (OP)]

Paraphrasing is the cause for much of this confusion (all my mistakes), so let's return to Einstein's definitions as the source instead of my failed attempts.

In accordance with Einstein's "Relativity: The Special and General Theory", the Lorentz transformations will transform an event (event as referenced within "Relativity: The Special and General Relativity) that occurs within K (x,y,z,t) to a system K' (x',y',z',t') or vice versa. Refer to Part 1, Chapter 11 of his book.

Within this co-ordinate system scenario given by Einstein, we define an experiment based upon two events;
(A) As seen by an observer in K, a photon propagates (c) from (x = 0) to (x) in a time interval (τ).
(B) As seen by an observer in K, an electron travels in uniform motion (u<c) from (x = 0) to (x), identical length along the x-axis as the photon, in a time interval (t).

Does this clarify the beginning of this scenario?

[PmbPhy]

I have no idea where you are getting this information or interpretation.

Well, I've asked you directly what that meant in post #7 but you never answered me. Therefore I took a guess.

In post #7 I asked

What is "parallel linear events" supposed to mean?

When I say parallel, I mean parallel.

That's the problem. You might think that such a phrase makes sense but in reality it doesn't. You spoke of parallel events and since events are but mere points in spacetime its meaningless to speak of two points being parallel. Also when you're speaking of events its natural to speak in terms of spacetime diagrams and worldlines. That was the motivation for my response.

A photon travels parallel to the x-asix, and an electron travels parallel to the x-axis. Simple geometry, like your car travels parallel to the surface of the road. No need for world lines or tilting through any degrees. You don't understand my original scenario...I get it. No need to keep stating the same thing...I get it.

Then you made an error when you spoke of two parallel linear events (A and B) and that caused a problem since I was unable to understand what you meant by it. I now see the problem. What you said was meaningless. Do you understand why? In post number 1 you wrote about "two parallel linear events" (why you called them 'linear' is beyond me) when in fact there was nothing parallel about them.

By the description that you just posted you weren't talking about parallel events. You were talking about parallel lines. One line terminated on event A and the other line terminated on event B. Each line originated at the origin.

Do you now understand why you can't say that two events are parallel?

In any case your conclusion is incorrect.

[PmbPhy]

   The following scenario utilizes events that are applicable to Einstein’s Special Relativity Theory. What follows requires further analysis regarding application of the first postulate WRT SRT. For information with images, see http://gsjournal.net/Science-Journals/Research%20Papers-Relativity%20Theory/Download/5916.

Okay. Given your responses which clarify what you were trying to do I went over this carefully and now see that your conclusion is wrong. The Lorentz transformation (LT) works for any two events whatsoever so long as they can be observed in reality. After all, that's exactly what the LT is for. The problem is your erroneous belief that

In order to satisfy the first postulate, the relationship between their event times must remain unchanged in order to prevent one from ascertaining the motion of the inertial frame utilizing this scenario.

This assertion is quite wrong and based on a misunderstanding of the first postulate of special relativity. Here's why: The result of the LT ends up being

Δτ/Δt = Δτ’(1+ β)/Δt’(1+ βx’/cΔt’), x’ = uΔt’

Δτ/Δt = Δτ’(1+ β)/Δt’(1+ β(u/c))

This means that the ratio as measured in frame K has a different value than that measured in K'. The transformation of the ratio is a function of the relative speed of the two frames. It's okay for observers in one frame to measure his velocity relative to another inertial frame. Therefore your assumption that the first postulate is wrong is incorrect.

[Colin2B]

Okay. Given your responses which clarify what you were trying to do I went over this carefully and now see that your conclusion is wrong.

While you were composing yours, I was composing mine, you beat me to it by moments. I approach it from a different direction, but we reach the same conclusions. Perhaps you can confirm my logic as I see  HeyBert creating an invalid conclusion by failing to perform an audit trail of observers and frames.
Perhaps between the two responses he will be able to understand how he should be viewing his scenarios.

My reply to HeyBert

In accordance with Einstein's "Relativity: The Special and General Theory", the Lorentz transformations will transform an event (event as referenced within "Relativity: The Special and General Relativity) that occurs within K (x,y,z,t) to a system K' (x',y',z',t') or vice versa. Refer to Part 1, Chapter 11 of his book.

The fact that you are starting to repeat your previous analysis even after we - PmbPhy and I - have posted replies, means you have either not read them or not understood them.

Just as Einstein defines his primed variables (such as t'), so I define mine. With regards to the Lorentz transformation t' = (t - vx/c^2)γ, which clock and observer does Einstein use for t'?

This confirms that you have not thought this through, nor have you understood the replies we have posted which answered this. Copying the formulae without understanding which observer, which clock, which frame means you are creating logical inconsistencies that result in wrong answers.

In order to cut this short:

When the race occurs within the stationary laboratory frame (v = 0), no calculation is needed (inverse Lorentz transformation reduces to Galilean format) and the ratio of the photon time span to the electron time span WRT the stationary laboratory frame is simply;

Photon:    τ = τ'

Electron:   t = t'

Ratio #1:  τ/t = τ'/t'

Wrong
This experiment is performed in the stationary laboratory frame K and for the observer and clocks in this frame, let's call him A, the experiment delivers 2 results T and t. It tells us nothing about K', T' or t'. Just so we are really clear about the observer and frame let's call these results T_KA - this means T as measured in frame K by A. Similarly t_KA

The way we find out about T' and t' is to allow A to travel in frame K' which is moving at v relative to K, and use the clocks in this frame. He performs the same experiment as in K but now gets results T' and t'. However, when he compares T and T' he finds they are the same, T=T', similarly t=t'. This means that by performing this experiment he is unable to determine whether he is moving relative to K, he also confirms that the laws of physics are the same in these 2 frames of reference.

When the race occurs within the inertial frame at velocity (v), the calculation of the ratio of the photon time span to the electron time span WRT the stationary laboratory frame IAW the inverse Lorentz transformation is;

Photon:     τ=(τ'+vx'/c^2)γ
                Since it is a photon, the second postulate gives x' = cτ'
                τ=(τ'+v(cτ')/c^2)γ
                τ=(τ'+v(τ')/c)γ
                τ=τ'(1+v/c)γ

 τ=(τ'+vx'/c^2)γ  wrong
 Since it is a photon, the second postulate gives x' = cτ' correct
 τ=(τ'+v(cτ')/c^2)γ  wrong
 τ=(τ'+v(τ')/c)γ  wrong
 τ=τ'(1+v/c)γ  wrong

At this point I am not interested in whether the transform formulae are correct or not, I am still looking at observers, frames and clocks, and the logic of their relationships. Until we understand this, the formulae are of little use.

Let us look at your final formula:

τ=τ'(1+v/c)γ
We know from the experiments performed by A in K and K' that T=T' so:
T=T(1+v/c)γ

Really? Do you believe that?  1=(1+v/c)γ,  where γ = (1-v^2/c^2)^-1/2

All this comes about because you are copying formulae without considering observers, frames or clocks.

Let's look at the observation of K' from K.

A second observer, let's call him B, sits in K and observes A conducting his experiment in K'. Although he can perform the same experiment as A performed in K, and similarly derive T and t, what does he see when he observes the experiment in K'.

B sees that K' is moving at v and knows that because of this the clocks in K' appear to be running slow and any time T' or t' he observes has to be adjusted by the Lorentz transforms.
So we can call the value of T' that B observes in K' as T_K'B.
Let's not worry about the exact maths but say that:
T_K'B=ƒT' (where ƒ is the transform and ƒ is not 0 or 1)
But we know that T'=T so
T_K'B=ƒT
Hence T_K'B≠T
In other words
T_K'B≠T_KA≠T
which is why your formula τ=(τ'+vx'/c^2)γ is wrong, you assumed the wrong value for T, you should have been deriving T_K'B not T_KA.

This is why your ratios don't work. You have not considered in which frames and for which observers and clocks your defined terms are valid.

To be fair you were misled by the paper you posted.
I don't intend to go through the detail of the calculations or your round trip scenario, but leave you to work through the correct versions.

[PmbPhy]

While you were composing yours, I was composing mine, you beat me to it by moments. I approach it from a different direction, but we reach the same conclusions. Perhaps you can confirm my logic as I see  HeyBert creating an invalid conclusion by failing to perform an audit trail of observers and frames.
Perhaps between the two responses he will be able to understand how he should be viewing his scenarios.

Sure thing. I'll do my best. However it will have to wait since I have to go to the store and get an ink cartridge for my printer.

[Waste of Time (OP)]

It appears from your remark that (T' = T) means that you are assuming I am discussing what K measures in their own frame vs what K' measures in their own frame. This was not at all what I was discussing. Of course they will not measure anything different, that is merely a result of the first postulate. Since all the following remarks are based upon this view, I can see how you came to your conclusion.

[PmbPhy]

It appears from your remark that (T' = T) ...

To whom are you speaking to? I don't see Colin making any such comment.

...means that you are assuming I am discussing what K measures in their own frame vs what K' measures in their own frame. This was not at all what I was discussing.

That is exactly what it appears that you're discussing. If it isn't then your posts are extremely deceptive. When someone places a prime on a variable it means that the quantity that has the prime on it is measured in the primed frame and vice versa. I.e. A' is measured in frame K', B is measured in frame K, etc.

You also speak of measuring observables in the "rest frame (v = 0)" where variables have no prime on them and in the "inertial frame (v)" where variables have a prime on them. By the way, this is a bad name for a frame since all frames used in special relativity are inertial frames. Then you apply the LT which means you're relating variables between inertial frames. That's why your comment

...means that you are assuming I am discussing what K measures in their own frame vs what K' measures in their own frame. This was not at all what I was discussing."

is quite misleading if its at all true. In fact you yourself defined the primed and unprimed variables in post #17 where you wrote

For all the primed variables, the clarification is as follows. IAW Einstein's book "Relativity: The Special and General Theory", the primed variable is with respect to the co-ordinate system K' and the unprimed variable is with respect to the co-ordinate system K.

Of course they will not measure anything different, that is merely a result of the first postulate.

The ratio is not invariant as you claim that it is. Are you unable to understand that? If not then you have a serious problem understanding special relativity. I explained why you're wrong above. Did you not read it?

[Waste of Time (OP)]

It appears from your remark that (T' = T) ...

To whom are you speaking to? I don't see Colin making any such comment.

I was referring to his comment about "Really? Do you believe that?  1=(1+v/c)γ,  where γ = (1-v^2/c^2)^-1/2"

Can you see it?

[Waste of Time (OP)]

It is obvious that my original presentations suffer many flaws due to the wording and the related mathematical results as a result of this wording. What if I strip out all the extraneous referencing to observers, rest frames, etc.?

[Waste of Time (OP)]

PART 1 (One-Way Events)

WRT Einstein’s book (Relativity: The Special and General Theory), Part 1, Chapter 11 gives the Lorentz transformation (LT) as;

t’ = (t-vx/c^2)γ, where γ = (1-v^2/c^2)^-1/2

Referencing Einstein’s book (Part 1, Chapter 11), we merely apply the transformation to two events originating in K that travel from (x = 0) to (x) in unequal times (τ) and (t).

WRT K’:
     τ’ = (τ-vx/c^2)γ

     t’ = (t-vx/c^2)γ

Evaluating the ratio of these times at (v) gives the results WRT K’ as;
*    τ’/t’ = (τ-vx/c^2)/(t-vx/c^2)

Evaluating this ratio at (v = 0) gives the results WRT K as;
     τ’/t’ = (τ-0*x/c^2)/(t-0*x/c^2)
*    τ’/t’ = τ/t

PART 2 (Round-Trip Events)

Referencing Einstein’s book (Part 1, Chapter 11), we merely apply the transformation to two events originating in K that travel from (x = 0) to (x) in unequal times (ϖ) and (T), then are reflected back such that they return to their points of origin (x = 0) in this same time (ϖ) and (T).

WRT K’:
     2ϖ’ = (ϖ-vx/c^2)γ + (ϖ+vx/c^2)γ
     2ϖ’ = ϖγ-(vx/c^2)γ+ϖγ+(vx/c^2)γ
*    2ϖ’ = 2ϖγ

     2T’ = (T-vx/c^2)γ + (T+vx/c^2)γ
     2T’ = Tγ-(vx/c^2)γ+Tγ+(vx/c^2)γ
*    2T’ = 2Tγ

Evaluating the ratio of these times at either (v or v = 0) gives the results WRT to K’ or K as;
*    ϖ’/T' = ϖ/T