WIKI and Time Dilation: The Possible Error in Relative Velocity

[chinglu1998]

If you have the context, why did you accuse me of the "mistake" of saying that the wiki article said "the clock at rest with the observer"? The context made clear I was talking about your statement, not the wiki article. If you're not a troll, please acknowledge that this was an unfounded accusation.

Then it's simple, in the observer's frame the light source is at rest so there is no aberration, therefore the light just travels on a vertical path of length L from the bottom to the top, so the time is $$\Delta t = \frac{L}{c}$$ to go from bottom to top.

What?

The observer's frame does not have the light source, the clock does. This is an error.

Are you going to calculate from the clock frame?

 

[JesseM]

What?

The observer's frame does not have the light source, the clock does. This is an error.

Didn't you just say "the clock at rest wrt to the observer"? So the clock (and light source) are at rest in the same frame, right? If you mean something different by "the clock at rest wrt to the observer" then you are speaking in an extremely bizarre way and you need to explain your weird terminology.

 

[chinglu1998]

Didn't you just say "the clock at rest wrt to the observer"? So the clock (and light source) are at rest in the same frame, right? If you mean something different by "the clock at rest wrt to the observer" then you are speaking in an extremely bizarre way and you need to explain your weird terminology.

Whatever, are you going to calculate from the frame at rest with the clock? I am waiting to see what the moving frame is doing.

 

[JesseM]

Whatever, are you going to calculate from the frame at rest with the clock? I am waiting to see what the moving frame is doing.

OK, if you're too confused to know whether the observer is at rest relative to the clock or moving relative to the clock, then just change the word "observer" to "clock" in my previous post #244 and there you have it:

Then it's simple, in the clock's frame the light source is at rest so there is no aberration, therefore the light just travels on a vertical path of length L from the bottom to the top, so the time is $$\Delta t = \frac{L}{c}$$ to go from bottom to top.

 

[espen180]

I think he basically means to find a light cone such that every position on the ellipsoid is also the position of some event on the light cone, then translate the positions and times of these events into the primed frame using the LT. Of course it's easier to work backwards--assume a light cone in the primed frame starting from x'=y'=z'=t'=0, consider the set of events at t'=r/c which all satisfy x'² + y'² + z'² = r², then translate these events to the unprimed frame and show their positions form an ellipsoid.

Okay.

I don't get why this is being discussed though. I would think this phenomenon is a consequence of not including the relativity of simultaneity. The OP has been asking for mathematics, so I will now give him mathematics.

Frame S : $$A^{\mu}=(ct,x,y,z)$$

Frame S' : $$B^{\nu}=(ct^\prime,x^\prime,y^\prime,z^\prime)$$

We assume the origins coincide at t=t'=0 and that the orientation of axes are equal. We also assume that S' frame moves with velocity v in the x-direction in the S frame. We apply the Lorentz transform:

$$B^{\nu}=\Gamma^{\nu}_{\mu}A^{\mu}= \left( \begin{array}{cccc}\gamma & -\gamma\frac{v}{c} & 0 & 0 \\ -\gamma\frac{v}{c} & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{array} \right)\left(\begin{array}{c} ct \\ x \\ y \\ z \end{array}\right)=\left( \begin{array}{c} \gamma\left(ct-\frac{vx}{c}\right) \\ \gamma (x-vt) \\ y \\ z \end{array}\right)$$

So we have

$$t^\prime = \gamma \left(t-\frac{vx}{c^2}\right)$$

$$x^\prime = \gamma (x-vt)$$

$$y^\prime = y$$

$$z^\prime = z$$

The metric of flat space-time is $$\eta_{\mu\nu}=\mathrm{diag}(1,-1,-1,-1)$$ which by the definition of the metric gives us for frame S:

$$ds^2=\eta_{\mu\nu}dA^{\mu}dA^{\nu}=c^2dt^2-dx^2-dy^2-dz^2$$

and for frame S':

$$ds^2=\gamma^2c^2dt^2-\gamma^2\frac{v^2}{c^2}dx^2-\gamma^2dx^2+\gamma^2v^2dt^2-dy^2-dz^2=\left( \gamma^2(c^2+v^2) \right)dt^2-\left( \gamma^2\left(\frac{v^2}{c^2}+1\right) \right)dx^2-dy^2-dz^2$$

The OP can check this for himself if he wishes.

Light travels along null geodesics, meaning ds=0, and these equations then represent light cones in the two frames. Now, note the following:

(1) If we look at 3D slices of constant t in S, we get spheres.

(2) If we look at slices of constant t' in S', we get spheres (Remember that $$x^\prime \neq x$$).

(3) If we look at slices of constant t in S' or at constant t' in S, we get ellipsoids.

If the OP doubts this, by all means try to disprove it.

These three points confirm that the light sphere/light ellipsoid discussion is simply nonphysical. It is a result of combining measurements from different frames, which is illegal. The OP should now be convinced that his earlier claims about Euclidean spaces and whatnot are flawed.

 

[JesseM]

These three points confirm that the light sphere/light ellipsoid discussion is simply nonphysical. It is a result of combining measurements from different frames, which is illegal. The OP should now be convinced that his earlier claims about Euclidean spaces and whatnot are flawed.

Well, you can think about it just in terms of considering a particular set of events--events which happen to be simultaneous and form a sphere in one frame, while they are non-simultaneous and their positions form an ellipsoid in the other. It's not illegal to consider how the same set of physical events looks in two different frames. But the original claim about the stationary frame being treated as "Euclidean" and the other as "Minkowski" still doesn't make sense, since even once you have labeled one frame as "stationary", you are still free to pick a set of events that form a sphere in the non-stationary frame and an ellipsoid in the stationary one.

 

[espen180]

Well, you can think about it just in terms of considering a particular set of events--events which happen to be simultaneous and form a sphere in one frame, while they are non-simultaneous and their positions form an ellipsoid in the other. It's not illegal to consider how the same set of physical events looks in two different frames. But the original claim about the stationary frame being treated as "Euclidean" and the other as "Minkowski" still doesn't make sense, since even once you have labeled one frame as "stationary", you are still free to pick a set of events that form a sphere in the non-stationary frame and an ellipsoid in the stationary one.

Ah, of course. It was not my intention to counterargue this.

I think I might have misinterpreted the problem. My impression was that the OP's claim was that a light sphere in one frame would be measured as an ellipsoid in another.

 

[JesseM]

Ah, of course. It was not my intention to counterargue this.

I think I might have misinterpreted the problem. My impression was that the OP's claim was that a light sphere in one frame would be measured as an ellipsoid in another.

Well, the OP's argument is fairly unintelligible. He seems to agree that if we take a surface of simultaneity in any frame then the slice of the light cone which lies on that surface will form a sphere in the coordinates of that frame, but still insists that when approaching a relativity problem it is critical to treat one frame as "stationary" and the other as "moving" (i.e. he insists these are something more than arbitrary labels for the sake of convenience in verbal discussions), because in the stationary frame we deal with "Euclidean space" where light forms a sphere, but then we must treat the moving frame as a "Minkowski space" where it forms an ellipsoid. Why it isn't equally valid to consider a set of events which are simultaneous in the moving frame (and thus form a sphere) but non-simultaneous in the stationary frame (and form an ellipsoid), he never explains.

 

[Dale]

x'1 = ( -r - v(r/c) )³
x'2 = ( r - v(r/c) )³
Obviously x'1 ` x'2.

So what?

For these events t'1 ` t'2 in the primed frame.
The same applies in the unprimed frame, for two events on the light cone if t1 ` t2 then x1 ` x2 in the unprimed frame also.

So this does not distinguish the primed frame from the unprimed frame in any way. They each have the same metric, in each frame simultaneous events on a light cone form a sphere, non-simultaneous events do not. There is no distinction.

 

[Dale]

You people are so willing to accept the WIKI calculations and yet so reluctant to calculate from the clock frame. Why?

We aren't reluctant, it is just trivial. In the clock's frame the clock is not time dilated (Lorentz factor is 1). Plus, it has taken hundreds of posts to get you to even occasionally specify what you are measuring velocities relative to.

 

[Dale]

Sowhat are the times when the clokc frame is taken as stationary.

Why are all you people terrified of answering this?

There you go sliding back into your old habits again. You didn't specify what the clock frame is stationary wrt.

Why are you so terrified of the fact that velocities are relative?

 

[darkhorror]

I have the context.

I want to see your calculation with the clock at rest wrt to the observer. See i even followed your rules, as if that was the issue.

I want to see the times with the clock at rest.

What?

The observer's frame does not have the light source, the clock does. This is an error.

Are you going to calculate from the clock frame?

So does the observer's frame have a clock or not? You first say the clock(light clock aka a light source) is at rest wrt the observer, then you say the observer's frame does not have a light source.

Obviously if the clock is at rest wrt the observer it's time isn't dilated in that frame of reference. So what are you asking?

 

[chinglu1998]

So does the observer's frame have a clock or not? You first say the clock(light clock aka a light source) is at rest wrt the observer, then you say the observer's frame does not have a light source.

Obviously if the clock is at rest wrt the observer it's time isn't dilated in that frame of reference. So what are you asking?

No, the light source is with the clock and the observer is moving relative to the clock.


So, the question is, what time interval will each frame conclude.

 

[chinglu1998]

OK, if you're too confused to know whether the observer is at rest relative to the clock or moving relative to the clock, then just change the word "observer" to "clock" in my previous post #244 and there you have it:

I am not confused.

Assume yoy are calculating from the view of the clock and the observer is moving relative to the clock.

What are the time intervals for the 2 frames.

 

[chinglu1998]

Ah, of course. It was not my intention to counterargue this.

I think I might have misinterpreted the problem. My impression was that the OP's claim was that a light sphere in one frame would be measured as an ellipsoid in another.

This is off track, but LT calculates an ellipsoid. However, in the frame, it is a sphere.

 

[chinglu1998]

Okay.

I don't get why this is being discussed though. I would think this phenomenon is a consequence of not including the relativity of simultaneity. The OP has been asking for mathematics, so I will now give him mathematics.

Frame S : $$A^{\mu}=(ct,x,y,z)$$

Frame S' : $$B^{\nu}=(ct^\prime,x^\prime,y^\prime,z^\prime)$$

We assume the origins coincide at t=t'=0 and that the orientation of axes are equal. We also assume that S' frame moves with velocity v in the x-direction in the S frame. We apply the Lorentz transform:

$$B^{\nu}=\Gamma^{\nu}_{\mu}A^{\mu}= \left( \begin{array}{cccc}\gamma & -\gamma\frac{v}{c} & 0 & 0 \\ -\gamma\frac{v}{c} & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{array} \right)\left(\begin{array}{c} ct \\ x \\ y \\ z \end{array}\right)=\left( \begin{array}{c} \gamma\left(ct-\frac{vx}{c}\right) \\ \gamma (x-vt) \\ y \\ z \end{array}\right)$$

So we have

$$t^\prime = \gamma \left(t-\frac{vx}{c^2}\right)$$

$$x^\prime = \gamma (x-vt)$$

$$y^\prime = y$$

$$z^\prime = z$$

The metric of flat space-time is $$\eta_{\mu\nu}=\mathrm{diag}(1,-1,-1,-1)$$ which by the definition of the metric gives us for frame S:

$$ds^2=\eta_{\mu\nu}dA^{\mu}dA^{\nu}=c^2dt^2-dx^2-dy^2-dz^2$$

and for frame S':

$$ds^2=\gamma^2c^2dt^2-\gamma^2\frac{v^2}{c^2}dx^2-\gamma^2dx^2+\gamma^2v^2dt^2-dy^2-dz^2=\left( \gamma^2(c^2+v^2) \right)dt^2-\left( \gamma^2\left(\frac{v^2}{c^2}+1\right) \right)dx^2-dy^2-dz^2$$

The OP can check this for himself if he wishes.

Light travels along null geodesics, meaning ds=0, and these equations then represent light cones in the two frames. Now, note the following:

(1) If we look at 3D slices of constant t in S, we get spheres.

(2) If we look at slices of constant t' in S', we get spheres (Remember that $$x^\prime \neq x$$).

(3) If we look at slices of constant t in S' or at constant t' in S, we get ellipsoids.

If the OP doubts this, by all means try to disprove it.

These three points confirm that the light sphere/light ellipsoid discussion is simply nonphysical. It is a result of combining measurements from different frames, which is illegal. The OP should now be convinced that his earlier claims about Euclidean spaces and whatnot are flawed.

If you are able to prove that LT maps a sphere to a sphere, have at it. I will easily refute that argument.

 

[chinglu1998]

Well, the OP's argument is fairly unintelligible. He seems to agree that if we take a surface of simultaneity in any frame then the slice of the light cone which lies on that surface will form a sphere in the coordinates of that frame, but still insists that when approaching a relativity problem it is critical to treat one frame as "stationary" and the other as "moving" (i.e. he insists these are something more than arbitrary labels for the sake of convenience in verbal discussions), because in the stationary frame we deal with "Euclidean space" where light forms a sphere, but then we must treat the moving frame as a "Minkowski space" where it forms an ellipsoid. Why it isn't equally valid to consider a set of events which are simultaneous in the moving frame (and thus form a sphere) but non-simultaneous in the stationary frame (and form an ellipsoid), he never explains.

Are you talking behind my back?

I am saying the ellipsoid represents the simultaneity in the context of the other frame according to LT.

You did not look at my equation did you. If you are fairly good at math, you can graph the intersection I talked about.

The problem is that simultaneity in the other frame has a image in the stationary frame and it is different thatn the light sphere in the stationary frame. Now, that is to be expected from the R of S.

However, since it is a complete image of an ellipsoid in the stationary frame, then simultaneity in the other frame occurs over an interval of time according to LT in the stationary system.

So, SR is mapping intervals of time to 1 time. That direction is functionally OK. But, whne you apply the inverse of this, you have one time to many which is not a math function. This therefore, violates the rules of mathematics.

 

[DrGreg]

If you are able to prove that LT maps a sphere to a sphere, have at it. I will easily refute that argument.

Undoubtedly the LT maps a static sphere to a static ellipsoid. But it maps a sphere that is expanding at a rate of c to an expanding sphere that is expanding at a rate of c. That seems to be the point you are missing.

 

[chinglu1998]

So what?

For these events t'1 ` t'2 in the primed frame.
The same applies in the unprimed frame, for two events on the light cone if t1 ` t2 then x1 ` x2 in the unprimed frame also.

So this does not distinguish the primed frame from the unprimed frame in any way. They each have the same metric, in each frame simultaneous events on a light cone form a sphere, non-simultaneous events do not. There is no distinction.

They may have the same metric, so what.

What I showed is a differenece not detected by the metric and hence the metric is flawed and blind to clear geometric differentials.

You show the metric is invariant. I show the LT difference in geometry.

 

[chinglu1998]

Undoubtedly the LT maps a static sphere to a static ellipsoid. But it maps a sphere that is expanding at a rate of c to an expanding sphere that is expanding at a rate of c. That seems to be the point you are missing.

Wrong. The intersection of the light postulate with the equation I applied shows simultaneity in the other frames occurs over an interval of time in the stationary frame.

People are used to comparing two points for the R of S and that looks fine.

However, on a continuous function, simultaneity from the view of the stationary frame for the other frame is a continuous interval whereas it is one time in the other frame.

The mapping of an interval of time to one time in another frame is logically inconsistent with nature. In other words, my clocks would expire an interval of time and yours would need to freeze under the rules of LT.

 

[JesseM]

I am not confused.

Well, if you're not confused, you keep changing your story. Now you say the observer is moving relative to the clock:

Assume yoy are calculating from the view of the clock and the observer is moving relative to the clock.

But before you said the clock was stationary wrt the observer (in [post=3072796]post 227[/post]):

Geez, please take the clock as the stationary frame wrt to the observer.

And in [post=3072845]post 240[/post] you said the clock is "at rest wrt to the observer":

I want to see your calculation with the clock at rest wrt to the observer.

So, do you agree that "the observer is moving relative to the clock" contradicts your earlier statements like "the clock at rest wrt to the observer"? Or are you using some bizarro terminology where the two statements can both be true?

What are the time intervals for the 2 frames.

Duh, if the observer is moving relative to the clock that's exactly the scenario calculated in the wiki article, you can find the correct "time intervals for the 2 frames" there (divide the time intervals by two if you just want the time for the light to go from bottom mirror to top as opposed to the time for a two-way trip).

 

[chinglu1998]

Well, if you're not confused, you keep changing your story. Now you say the observer is moving relative to the clock:

We are doing a 2nd calculation.Is this an intellectual challenge?

But before you said the clock was stationary wrt the observer (in [post=3072796]post 227[/post]):

We are doing a new problem.

And in [post=3072845]post 240[/post] you said the clock is "at rest wrt to the observer":

So, do you agree that "the observer is moving relative to the clock" contradicts your earlier statements like "the clock at rest wrt to the observer"? Or are you using some bizarro terminology where the two statements can both be true?

This is not reasonable to try to trap me in like this when I clearly said let's switch the context.

Duh, if the observer is moving relative to the clock that's exactly the scenario calculated in the wiki article, you can find the correct "time intervals for the 2 frames" there (divide the time intervals by two if you just want the time for the light to go from bottom mirror to top as opposed to the time for a two-way trip).

We are trying to calculate the clock at rest and the observer is moving and calculate the time intervals between the frames in the view of the clock frame. Can you do this?

 

[JesseM]

Why it isn't equally valid to consider a set of events which are simultaneous in the moving frame (and thus form a sphere) but non-simultaneous in the stationary frame (and form an ellipsoid), he never explains.

Are you talking behind my back?

I am saying the ellipsoid represents the simultaneity in the context of the other frame according to LT.

That's what I just said. If we look at a set of events which are simultaneous in the moving frame (and thus form a sphere) and use the LT to translate them to the stationary frame, we get a set of events whose positions form the ellipsoid you talked about (and which are non-simultaneous in the stationary frame). So since we have a sphere in the moving frame and an ellipsoid in the stationary frame, doesn't this mean that according to your terminology we have a "Minkowski space" in the stationary frame and a "Euclidean space" in the moving frame?

You did not look at my equation did you. If you are fairly good at math, you can graph the intersection I talked about.

Yes, I did look at it, but you still aren't answering my question from the end of [post=3072773]post 224[/post] of what you think the point of this exercise in when it just confirms what I already said to you in [post=3072718]post 214[/post], namely "if he makes a different choice then the positions of the events will form an ellipsoid in his frame but a sphere in the moving frame." Anyway, in answer to your question the intersection of the ellipsoid with the light cone can be found by assigning time coordinates to each point on the ellipsoid with the equation $$t = \frac{\gamma r}{c} + \frac{v(x - (\gamma vr/c))}{c^2}$$, and then if you translate these events into the primed frame they all satisfy $$x'^2 + y'^2 + z'^2 = r^2$$ and $$t' = r/c$$.

The problem is that simultaneity in the other frame has a image in the stationary frame and it is different thatn the light sphere in the stationary frame.

Yes, and likewise a simultaneous set of points on the light cone in the stationary frame has an image in the moving frame and it is different than the light sphere in the moving frame. The situation is 100% symmetrical, so "stationary" and "moving" have no meaning except as arbitrary verbal labels.

So, SR is mapping intervals of time to 1 time. That direction is functionally OK. But, whne you apply the inverse of this, you have one time to many which is not a math function. This therefore, violates the rules of mathematics.

Are you on drugs? There is nothing in the "rules of mathematics" which says a set of events which all have the same t-coordinate can't be mapped to a set of events which have multiple different t-coordinates. Maybe in some hazy confused way you're thinking of the fact that a "function" should have a unique output for each input, but each individual event in one frame does indeed map to a unique event with unique coordinates in the other frame.

 

[chinglu1998]

That's what I just said. If we look at a set of events which are simultaneous in the moving frame (and thus form a sphere) and use the LT to translate them to the stationary frame, we get a set of events whose positions form the ellipsoid you talked about (and which are non-simultaneous in the stationary frame). So since we have a sphere in the moving frame and an ellipsoid in the stationary frame, doesn't this mean that according to your terminology we have a "Minkowski space" in the stationary frame and a "Euclidean space" in the moving frame?

Have you graphed the two light spheres, they are in different places and different shapes. So, "Minkowski space" and "Euclidean space" do not overlay when brought together in one frame even though the light sphere is supposed to be one object. Do you understand the problem with this? But, I am encouraged you at least understand the basics on what I am doing given you are bringing back continuous simultaneity to the stationary frame from the moving frame. If you assume the moving frame light sphere has acquired a radius r, this will become obvious.

 

[Dale]

What I showed is a differenece not detected by the metric

How is what you showed a difference at all? In both frames, events on the light cone that are not simultaneous are not at the same distance. And in both frames, events on the light cone that are simultaneous are at the same distance. So both frames are the same in that respect, not different.

 

[JesseM]

We are doing a 2nd calculation.Is this an intellectual challenge?



We are doing a new problem.



This is not reasonable to try to trap me in like this when I clearly said let's switch the context.

Why don't you stop lying, none of you're posts said "let's switch the context". Instead you repeatedly made nasty accusations that I was making "mistakes" and "errors" when I just took you at your word, and then you pretended like you hadn't said what you had in fact said. Read again:

Geez, please take the clock as the stationary frame wrt to the observer.

In that case there is only one frame to consider, the frame where both the clock and observer are at rest

Where in the WIKI article is the clock at rest with the observer? You make many mistakes.

Then in the only post of yours that used the word "context", you said that the context was still about "the clock at rest wrt to the observer":

I have the context.

I want to see your calculation with the clock at rest wrt to the observer. See i even followed your rules, as if that was the issue.

Then it's simple, in the observer's frame the light source is at rest so there is no aberration, therefore the light just travels on a vertical path of length L from the bottom to the top, so the time is $$\Delta t = \frac{L}{c}$$ to go from bottom to top.

What?

The observer's frame does not have the light source, the clock does. This is an error.

So in both cases, you said the observer was at rest relative to the clock, I responded to that, and then you accused me of a mistake/error. There was nothing like "oh sorry, I know I said the observer was at rest relative to the clock but now I want to change the context and talk about an observer moving relative to the clock" (when I pointed out you had changed your story, your response in post #248 was "whatever"). So you are either confused or trolling, I'm thinking the latter looks pretty likely now. If you want to show you're not a troll you can acknowledge that you were incorrect to accuse me of mistakes in the two posts above, if you don't do that I will be pretty confident that you are trolling and will no longer respond to your posts (and will probably also report this thread to the mods in hopes they will lock it so others aren't fooled into wasting time responding as if your questions were asked in earnest).

We are trying to calculate the clock at rest and the observer is moving and calculate the time intervals between the frames in the view of the clock frame. Can you do this?

To calculate the "time intervals between the frames", one of the time intervals we must calculate is the time in the observer's frame. What the hell does it mean to calculate the time in the observer's frame "in the view of the clock frame"? Are you asking to apply the LT to the times in the clock frame? Are you asking to figure out the clock frame's view of the two clocks the observer uses to locally measure the time-coordinates of the light leaving the bottom mirror and the light hitting the top one? If the question is serious and not a trollish provocation, you need to explain precisely what it is you want calculated here.

 

[chinglu1998]

How is what you showed a difference at all? In both frames, events on the light cone that are not simultaneous are not at the same distance. And in both frames, events on the light cone that are simultaneous are at the same distance. So both frames are the same in that respect, not different.

Not true. The stationary frame will show the moving light sphere located at origin vt when applying LT to its own light sphere at time t.

The other frame will see the origin at -vt.

We have a difference.

Bit, we are talking about the difference in the context of one frame. That has been established in this thread as fact.

Another thing established in this thread is that light aberration is absolute based on the frame that contains the light source. Hence for the WIKI example, t'=tγ regardless of which frame is taken as stationary.

 

[chinglu1998]

Why don't you stop lying, none of you're posts said "let's switch the context". Instead you repeatedly made nasty accusations that I was making "mistakes" and "errors" when I just took you at your word, and then you pretended like you hadn't said what you had in fact said. Read again:

I am so sorry, you are never wrong.

Anyway, I made it clear I wanted to calculate time dilation from the context of the clock frame and that is a fact. You take the time to validate that. I said it several time.

Then in the only post of yours that used the word "context", you said that the context was still about "the clock at rest wrt to the observer":

You are in error. This is just a few.
Geez, please take the clock as the stationary frame wrt to the observer. #227
You people are so willing to accept the WIKI calculations and yet so reluctant to calculate from the clock frame. Why? #230
So what are the times when the clock frame is taken as stationary.#239

To calculate the "time intervals between the frames", one of the time intervals we must calculate is the time in the observer's frame. What the hell does it mean to calculate the time in the observer's frame "in the view of the clock frame"? Are you asking to apply the LT to the times in the clock frame? Are you asking to figure out the clock frame's view of the two clocks the observer uses to locally measure the time-coordinates of the light leaving the bottom mirror and the light hitting the top one? If the question is serious and not a trollish provocation, you need to explain precisely what it is you want calculated here.

This seems so complicated. You know which frame the WIKI use to calculate? Yea, use the other and do not forget light aberration.

 

[Dale]

Not true. The stationary frame will show the moving light sphere located at origin vt when applying LT to its own light sphere at time t.

The other frame will see the origin at -vt.

We have a difference.

I can't parse that sentence at all. But, if something is stationary in the unprimed frame then it is moving at velocity v in the primed frame, and if something is stationary in the primed frame it is moving at velocity -v in the unprimed frame. How is that a difference?

Bit, we are talking about the difference in the context of one frame. That has been established in this thread as fact.

Another thing established in this thread is that light aberration is absolute based on the frame that contains the light source. Hence for the WIKI example, t'=tγ regardless of which frame is taken as stationary.

Stationary wrt what? The clock or the observer? You really have a very bad habit of not specifying what you are measuring velocity relative to. I don't think that this thread has even established what you are asking, let alone what the answer is.

 

[chinglu1998]

If something is stationary in the unprimed frame then it is moving at velocity v in the primed frame, and if something is stationary in the primed frame it is moving at velocity -v in the unprimed frame. How is that a difference?

Stationary wrt what? The clock or the observer? You really have a very bad habit of not specifying what you are measuring velocity relative to. I don't think that this thread has even established what you are asking, let alone what the answer is.

Let's calculate time dilation from the clock frame using the light aberration argument of the WIKI artucle.

 

[Dale]

Let's calculate time dilation from the clock frame using the light aberration argument of the WIKI artucle.

1

 

[JesseM]

I am so sorry, you are never wrong.

Sarcastic non-answer doesn't cut it for me to see you as a non-troll. Do you acknowledge you changed your story in those specific posts I quoted, where first you said you wanted the clock and the observer at rest wrt each other, then when I dealt with that scenario you accused me of mistakes/errors? Yes or no? If you don't give a serious answer here (and if the answer is 'no', an explanation of what specific mistake I made about what scenario you had described in those posts), then as I said I'm writing you off as a troll and reporting the thread.

Anyway, I made it clear I wanted to calculate time dilation from the context of the clock frame and that is a fact. You take the time to validate that. I said it several time.

Yeah but I keep asking you what the hell that means and you never answer.

Then in the only post of yours that used the word "context", you said that the context was still about "the clock at rest wrt to the observer":

You are in error.

What am I in error about?

This is just a few.
Geez, please take the clock as the stationary frame wrt to the observer. #227
You people are so willing to accept the WIKI calculations and yet so reluctant to calculate from the clock frame. Why? #230
So what are the times when the clock frame is taken as stationary.#239

Hmm, none of this has anything whatsoever to do about my comment Then in the only post of yours that used the word "context", you said that the context was still about "the clock at rest wrt to the observer". That comment was about post #240, which was in fact "the only post of yours that used the word "context", and in that post you said the context "the clock at rest wrt to the observer". I never denied that you wanted to calculate something "from the clock frame", I just pointed out you had changed your story about whether the observer was moving or stationary relative to the clock frame, and I still don't understand what it means to calculate the observer's time interval "from the clock frame".

This seems so complicated. You know which frame the WIKI use to calculate?

Yeah, the wiki used both frames, and you are on drugs if you think it only used the observer's frame. This statement from the wiki is clearly discussing the perspective of the clock frame:

In the frame where the clock is at rest (diagram at right), the light pulse traces out a path of length 2L and the period of the clock is 2L divided by the speed of light:

$$\Delta t = \frac{2L}{c}$$

If you still think the wiki calculates everything from the observer's frame, why don't you explain in detail how you derive the equation $$\Delta t = \frac{2L}{c}$$ "from the observer's frame", and maybe this will finally allow us to understand what the hell you are talking about when you ask to calculate the observer's time interval "from the clock frame".

 

[darkhorror]

No, the light source is with the clock and the observer is moving relative to the clock.


So, the question is, what time interval will each frame conclude.

In the observer's frame the clock will be running slower compared to the observer's time.

In the clock's frame the observer's time will be moving slower when compared to the clock.

 

[chinglu1998]

In the observer's frame the clock will be running slower compared to the observer's time.

In the clock's frame the observer's time will be moving slower when compared to the clock.

Can we show with with the light aberration argument of WIKI?

 

[chinglu1998]

Sarcastic non-answer doesn't cut it for me to see you as a non-troll. Do you acknowledge you changed your story in those specific posts I quoted, where first you said you wanted the clock and the observer at rest wrt each other, then when I dealt with that scenario you accused me of mistakes/errors? Yes or no? If you don't give a serious answer here (and if the answer is 'no', an explanation of what specific mistake I made about what scenario you had described in those posts), then as I said I'm writing you off as a troll and reporting the thread.

Fine, I'm done. I showed you at least three times where I wanted to look at the calculation from the view of the clock frame. There are many more. Then, you continue to call me a liar even after I have provided evidence in a thread you are supposed to be reading.


I do not want to be at a place where I have answered truthfully, provided evidence of what I said and am still called names by someone out of anger.