Undergrad WIKI and Time Dilation: The Possible Error in Relative Velocity

[JesseM]

No, I mean t = √(x²+y²+z²)/c.

That is simply the equation for all events on the light cone in the unprimed frame. If we determine which of these events also satisfy (x-vrγ/c)²/(rγ)² + y²/r² + z²/r² = 1, and find the corresponding time for each position, then we can use the LT to translate these positions and times into the primed frame and show that in the primed frame they all occur at the same time and for a sphere.

We are operating from the context of the unprimed frame proving t' = r/c in the context of the primed frame.

And how do you want to prove t'=r/c in the context of the primed frame, except by first finding the positions and times of all the events on the ellipsoid in the unprimed frame, then translate those events into the primed frame using the LT? If this is not the procedure you have in mind, you need to either explain more clearly or just show the full derivation you have in mind.

 

[Doc Al]

Are you saying SR cannot use light aberration to check times from the clock frame? That would be a CPT violation.

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

Why are all you people terrified of answering this?

You're not making any sense!

Let me guess at what you're trying to ask for. Can the primed observer, who sees the clock as moving and the light rays slanted, use aberration to figure out the angle that the light makes in the unprimed frame? Sure. He'll find those light rays will be perpendicular, not slanted. (I'll leave that as an exercise for you.) And he'll find that Δt' = γΔt, just as expected.

 

[chinglu1998]

That is simply the equation for all events on the light cone in the unprimed frame. If we determine which of these events also satisfy (x-vrγ/c)²/(rγ)² + y²/r² + z²/r² = 1, and find the corresponding time for each position, then we can use the LT to translate these positions and times into the primed frame and show that in the primed frame they all occur at the same time and for a sphere.

And how do you want to prove t'=r/c in the context of the primed frame, except by first finding the positions and times of all the events on the ellipsoid in the unprimed frame, then translate those events into the primed frame using the LT? If this is not the procedure you have in mind, you need to either explain more clearly or just show the full derivation you have in mind.

You can prove this yourself. At least you did not have to find it like I did, so it is easier for you.

 

[JesseM]

I have the context.

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.

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

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.

 

[JesseM]

You can prove this yourself. At least you did not have to find it like I did, so it is easier for you.

All I'm asking is whether the method I suggested to prove it was the one you had in mind. If not, I don't understand what type of proof you're looking for. In any case, are you ever going to respond to [post=3072773]post #224[/post], where I asked at the end why you think I need to do this exercise, given that it would only be a mathematical demonstration of what I already said in words?

 

[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.