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All that I show is that the current format of the Lorentz temporal transformation complies with the first postulate for round-trip events, but not one-way events. We can either choose to accept this as a limitation of STR, or choose a format for the temporal transformation that does comply with the first postulate for one-way events.

So what you are saying is that light is intrinsically linked to gravitation.

- There are two parallel linear events

- Event A is an electromagnetic event (velocity = c)...

- Event B is an inertial body event (velocity = u) ...

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.EVENTS OCCURRING IN INERTIAL FRAME (v)- These events now occur within a reference frame considered to be in uniform motion (inertial frame).- The SRT event time for A is; Δτ = (Δτ’+vx’/c^2)γ. Applying the second postulate;Δτ = (Δτ’+vx’/c^2)γ, x’ = cΔτ’Δτ = (Δτ’+v(cΔτ’)/c^2)γΔτ = (Δτ’+vΔτ’/c)γΔτ = Δτ’(1+ β)γ- The SRT event time for B is; Δt = (Δt’+vx’/c^2)γ.- The ratio of these event times gives us their the relationship between their event times;Δτ/Δt = [Δτ’(1+ β)γ]/[(Δt’+vx’/c^2)γ]Δτ/Δt = [Δτ’(1+ β)]/[(Δt’+vx’/c^2)]Δτ/Δt = [Δτ’(1+ β)]/[Δt’(1+vx’/(c^2)Δt’)]Δτ/Δt = [Δτ’(1+ β)]/[Δt’(1+ βx’/cΔt’)]

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.

Quote from: jeffreyH on 17/02/2015 00:37:02So what you are saying is that light is intrinsically linked to gravitation.I thought variability of light was only seen in accelerating frames eg influence of gravity, not inertial frames.Did I miss something in the arguments?

Quote from: Colin2B on 17/02/2015 13:28:31Quote from: jeffreyH on 17/02/2015 00:37:02So what you are saying is that light is intrinsically linked to gravitation.I thought variability of light was only seen in accelerating frames eg influence of gravity, not inertial frames.Did I miss something in the arguments?No you didn't miss anything. It just gave me a thought which I am following up on. I followed his train of thought with difficulty but saw something interesting. It might be garbage but I am following it up.

At first sight it appears to me that this could be a confusion between who is observing and from which frame, can you make this clearer please?. The introduction of your 'unity' as a physical law could well be compounding this.

Quote from: HeyBert In order to satisfy the first postulate, the relationship between their event ..I don't understand. Who are "they" when you say "their event"?

Perhaps we can look at it this way as my original choice of words may have not been understood clearly by all...(τ and t indicate time spans, or time intervals measured by a clock).

Does the ratio of these two scenarios mathematically equal each other (same race happening in two different inertial frames)? If so, then how? If not, then what would stop us from determining the velocity of the inertial frame at velocity (v) by conducting such a "race" experiment and measuring how the ratio changes with increasing velocity of the inertial frame.

My calculations, where γ = (1-v^2/c^2)^-1/2;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'

Quote from: HeyBert on 19/02/2015 04:10:36My calculations, where γ = (1-v^2/c^2)^-1/2;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'You keep posting things like this without defining them. What you said tells us nothing about what those quantities are/mean.

OK, things are staring to look clearer, but before we confuse ourselves with maths let's be really clear that we are talking about the same things.Quote from: HeyBert on 19/02/2015 02:32:17Perhaps we can look at it this way as my original choice of words may have not been understood clearly by all...(τ and t indicate time spans, or time intervals measured by a clock).Which clock? Sorry to labour this but there are 2 clocks, one in the rest frame and one in the moving frame. For the observer in the moving frame the moving clock measures the same time intervals as when he made the same measurements in the rest frame so T/t is the same. So by doing a physics experiment he sees no difference between the frames. From his point of view he is stationary (because he is moving with the frame) and no Lorentz transforms are required.For the observer in the rest frame observing the moving frame, he sees the clock in the moving frame measuring time more slowly than the one he has next to him in the rest frame. So the T/t is not the same and this difference is calculated using the Lorentz transforms.Quote from: HeyBert on 19/02/2015 02:32:17Does the ratio of these two scenarios mathematically equal each other (same race happening in two different inertial frames)? If so, then how? If not, then what would stop us from determining the velocity of the inertial frame at velocity (v) by conducting such a "race" experiment and measuring how the ratio changes with increasing velocity of the inertial frame.As we can see, it depends who is doing the observing. For the observer who was stationary and is now moving there is no difference and this is what the postulate is all about. The observer cannot tell by the race experiment whether he is moving or not. He can however, look out of the window and see he is moving away from the rest frame, so would perceive himself as moving.For the observer who is stationary observing the moving frame, no the ratios are not the same. That's what relativity is all about, frames moving relative to one another.Again sorry to labour this, but it think there is a confusion of observers.I hate to say it but I don't think you are comparing light with light! []

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 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.

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.

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

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.

Quote from: HeyBert 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?

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?

QuoteOnly 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'?

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

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.

QuoteFirst 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 AWorldline B: Worldline connecting origin with event BWorldline 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.

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 AWorldline B: Worldline connecting origin with event BWorldline 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.

What is "parallel linear events" supposed to mean?

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.

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 ...sorry, you cannot view external links. To see them, please REGISTER or LOGIN.

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.

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.

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'?

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'

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

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.

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.

...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."

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.

Quote from: HeyBertIt appears from your remark that (T' = T) ...QuoteTo 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?

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

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

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.

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'.

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.?

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;

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.?

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)^-1/2Referencing Einstein’s book (Part 1, Chapter 11), we merely apply the transformation to two events (i.e. moving electrons) in K from (x = 0) to (x) in unequal times (τ) and (t).WRT K’: τ’ = (τ-vx/c^2)γ t’ = (t-vx/c^2)γEvaluating 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/tPART 2 (Round-Trip Events)Referencing Einstein’s book (Part 1, Chapter 11), we merely apply the transformation to two events (i.e. moving electrons) in K from (x = 0) to (x) in unequal times (ϖ) and (T), then from (x) to (x = 0) in these same times (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’ = 2TγEvaluating the ratio of these times at either (v or v = 0) gives the results WRT to K’ or K as;* ϖ’/T' = ϖ/T

And what law states I have to answer your questions anyways?

Quote from: HeyBertIt appears from your remark that (T' = T) ...To whom are you speaking to? I don't see Colin making any such comment.Quote from: HeyBert...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.

Working things out on another forum right now. Learned to frame the problem in a "relativity friendly" format from another user and have good positive communication going. Thanks for the offer though, and I'll let you know If I come back to this or identify where my logic went astray.

Yeah. I know. I saw that and I saw that you're making mistakes there too. I've already shown you how to do this right so what's your objection to it?