WIKI and Time Dilation: The Possible Error in Relative Velocity

[IsometricPion]

So, t' = tγ.

This means when the unprimed frame is at rest and observering the to clock, it is moving and beating faster than it rest clock.

As with the calculation in my previous post, $$\Delta{}t'=\gamma{}\Delta{}t\Rightarrow{}\Delta{}n'=\frac{\Delta{}n}{\gamma}$$, so in each frame clocks in motion tick slower than clocks at rest (no matter which frame is taken to be at rest).

 

[chinglu1998]

But you never explained why you think we can't also assume the same Euclidean laws hold in the moving frame, or why doing so would mean that "complete and total breakdown of SR occurs."

No, because I was talking about the moving frame, as I think was pretty clear from the bolded statements above. Do you disagree that in the moving frame, "the position of the light at any specific time T' satisfies x'² + y'² + z'² = c² T'²"? Isn't this "Euclidean geometry"?

I see you point. We did not communicate.

I meant when calculating the moving frame it is Minkowsky. In the moving frame, the relativity postulate holds. So, the equation for the light sphere holds in all frames, but when calculating those framea, it does not, the calculation is Minkowsky.

You have never given any mathematical illustration of what you mean by "apply Minkowsky to stationary system" (and as I told you, the name is "Minkowski" with an i) or why this leads to "failure". As I said in post #25, this is what most physicists mean by "Minkowski geometry":

My intention was to understand what is moving coordinates and what is stationary. I made clear the word distinction.

If you apply Minkowsky to your own coordinates for you, that is failure. I made this clear. That is called frame mixing.

Do you disagree that the metric $$d\tau^2 = dt^2 - (1/c^2)dx^2 - (1/c^2)dy^2 - (1/c^2)dz^2$$ is what physicists normally mean by "Minkowski geometry", or that this equation still holds if we are dealing with the t,x,y,z coordinates of the "stationary" frame?

No, I do not. That is simply one componant of a space, it metric.

How exactly are you going to make this equation true exclusively within a stationary fame?

By the way, I notice you skipped my post #52 to reply to post #55. Hopefully you will go back to #52 and address it, particularly my request for an explanation of the precise step in my mathematical derivation that you think contains an error.

Thanks to poiint this out. I will look.

 

[chinglu1998]

As with the calculation in my previous post, $$\Delta{}t'=\gamma{}\Delta{}t\Rightarrow{}\Delta{}n'=\frac{\Delta{}n}{\gamma}$$, so in each frame clocks in motion tick slower than clocks at rest (no matter which frame is taken to be at rest).

OK, how do you explain the WIKI article.

Would you please use the experiment in the article to prove reciprocal time dilation?

 

[JesseM]

Tes, I start to think WIKI is correct. The absolute light abberation causes a longer path for the light travel for the primed framer vs the unprimed frame in an absolute sense.

Let me take the unprimed frame as stationary. This promed frame light beam still longer.

c²t'² = v²t'² + c²t²

c²t'² - v²t'² = c²t²

t'²( c² - v²) = c²t²

t' = tγ

Wow, when taking the unprimed frame as stationary with absolute light abberation, the moving clock beats faster just like WIKI said.

No, any clock in the moving (primed) frame must be running slow as measured in the stationary (unprimed) frame. I showed why this is not incompatible with the wiki's time dilation equation in post #62.

You have to define the precise meanings of t and t' in your equations. t' doesn't deal with any single "moving clock", rather it is the time in the primed frame between the event of the light being emitted from the bottom of the light clock (which is at rest in the unprimed frame and therefore moving relative to the primed frame) and the event of it hitting the top of the light clock (this event will occur at a different position in the primed frame). To measure the time of these two events in the primed frame using clocks, you would need a pair of clocks at different positions in the primed frame. As I explained in #62, each of these primed clocks is individually running slow as measured in the unprimed frame, but the unprimed frame nevertheless agrees that t' is greater than t because of the way the two primed clocks are out-of-sync in the unprimed frame.

 

[darkhorror]

I did. Are they starting at (0,0,0,0) for both?

It was a very simple question, you don't have to do any math. Just read the question. If you have a spaceship moving away from Earth at .6c. From the ship's frame of reference which clock is moving slower the one of the ship or the one on earth?

 

[grav-universe]

If you do the math on WIKI, you will find if you take the unprimed frame as rest with respect to the primed frame, then t' = t γ.

If you take the primed frame as rest with respect to the unprimed frame, then t' = t γ.

So either way, it is absolute that t' = t γ. Why is this? This light beam path for the primed frame is absolutly longer than that of the unprimed frame because of light abberation. This is not a relative issue.

From the perspective of the primed frame, we have t = t' y. From the perspective of the unprimed frame, we have t' = t y. The unprimed frame is comparing the period for the light pulse to travel between the mirrors in the primed frame to the period that the primed frame measures for the same pulse to travel to the same mirrors in the primed frame also. The primed frame, however, is comparing the period for a different pulse to travel between the mirrors in the unprimed frame to the period the unprimed frame measure for that same pulse in the unprimed frame. These are two completely different sets of circumstances.

 

[chinglu1998]

There is no single "moving clock". 25s is the time measured in the primed ('moving') frame between two events on the worldline of the clock at rest in the unprimed ('stationary') frame. But these events occur at two different positions in the primed frame, so you'd need a pair of clocks at rest and synchronized in the primed frame to assign time-coordinates to both events in a local way, after which you could figure out their difference in time-coordinates. That's how position and time coordinates are supposed to be assigned in SR, using a lattice of rulers and clocks at rest relative to one another and synchronized using the Einstein synchronization convention, as illustrated here:

Einstein also discusses the idea that coordinate times and coordinate positions should be defined using local measurements on a set of rulers and synchronized clocks in sections 1 and 2 of the 1905 paper you linked to.


As always, "moving" and "stationary" are arbitrary labels. Changing the labels does not magically change any facts about the amount of coordinate time between two specific events in different frames. Instead of playing these silly word-games, why don't you go through my analysis using the Lorentz transformation and figure out specifically where you think I made an error? Do you disagree that if a clock is at rest at x=0 in the unprimed frame, then the coordinates of it showing a time of 0 would be x=0,t=0 while the coordinates of it showing a time of 20 would be x=0,t=20? Do you disagree that when we transform x=0,t=0 into the primed frame we get t'=0, and when we transform x=0,t=20 into the primed frame we get t'=25? Please address these quantitative specifics instead of retreating into vague word-games.

Your conclusions are correct. Do you understand them?

Since LT matrix invertible, then t' = 25 and t = 20 regardless of which frame is at rest.

So if unprimed at rest, t' > t.

If primed frame at rest, then t' > t.

Have you figured this out yet?

 

[chinglu1998]

It was a very simple question, you don't have to do any math. Just read the question. If you have a spaceship moving away from Earth at .6c. From the ship's frame of reference which clock is moving slower the one of the ship or the one on earth?

I am discussing this same subject with Jesse right now.

 

[JesseM]

I meant when calculating the moving frame it is Minkowsky. In the moving frame, the relativity postulate holds. So, the equation for the light sphere holds in all frames, but when calculating those framea, it does not, the calculation is Minkowsky.

I don't understand what "calculation" you refer to. If you mean we must use the Lorentz transformation to go from stationary coordinates to moving coordinates I agree, but it is equally true that we must use the Lorentz transformation to go from moving coordinates to stationary coordinates. So again I don't see why you think "Minkowski" applies to the moving frame but not to the stationary frame.

If you apply Minkowsky to your own coordinates for you, that is failure. I made this clear. That is called frame mixing.

I don't know what you mean by "apply Minkowski to your own coordinates for you", but whatever it means, presumably it would be equally a mistake for the stationary observer to "apply Minkowski to his own coordinates for himself" and a mistake for the moving observer to "apply Minkowski to his own coordinates for himself"? As always the situation is completely symmetrical, there's no sense in which it is correct to "apply Minkowski" in the moving frame but not to do something equivalent in the stationary frame.

How exactly are you going to make this equation true exclusively within a stationary fame?

Huh? I never said anything about it being exclusive to the stationary frame, I said 'this equation still holds if we are dealing with the t,x,y,z coordinates of the "stationary" frame?'--that "still" was to indicate that although it's true this Minkowski equation holds in the moving frame, it also holds in the stationary frame. It holds in all inertial frames, so again I don't see how it makes sense to say that Minkowski geometry should only be used in the moving frame and Euclidean geometry should only be used in the stationary frame. For any equation (Euclidean or Minkowski) which applies to a situation in one frame, you can find an equivalent situation in any other frame where the same equation applies.

 

[chinglu1998]

From the perspective of the primed frame, we have t = t' y. From the perspective of the unprimed frame, we have t' = t y. The unprimed frame is comparing the period for the light pulse to travel between the mirrors in the primed frame to the period that the primed frame measures for the same pulse to travel to the same mirrors in the primed frame also. The primed frame, however, is comparing the period for a different pulse to travel between the mirrors in the unprimed frame to the period the unprimed frame measure for that same pulse in the unprimed frame. These are two completely different sets of circumstances.

How do you make this hsppen with the WIKI article give the fact light abberation is absolute?

That means, the clock at rest with the light source will measure t=d/c and the frame that views that clock as moving will also conclude it calculates t=d/c and it will view a light beam at an angle with a longer distance to travel. Light abberation is not reciprocal. It is only true for observers moving relative to the light source.

 

[chinglu1998]

I don't understand what "calculation" you refer to. If you mean we must use the Lorentz transformation to go from stationary coordinates to moving coordinates I agree, but it is equally true that we must use the Lorentz transformation to go from moving coordinates to stationary coordinates. So again I don't see why you think "Minkowski" applies to the moving frame but not to the stationary frame.

Oh, you are not familiar with topological spaces. You have a space and then a mapping to another space. That mapped space if not isomorphic is not the same as the original space.

So, in my mind, I see the two as two different objects.

There is the stationary space which has Euclidian properties.
There is the Minkowsky space which has odd properties. The light sphere in Minkowsky is off center and not spherical.

It is my view these are different. If you have your view fine. This is simply my view.

I don't know what you mean by "apply Minkowski to your own coordinates for you", but whatever it means, presumably it would be equally a mistake for the stationary observer to "apply Minkowski to his own coordinates for himself" and a mistake for the moving observer to "apply Minkowski to his own coordinates for himself"? As always the situation is completely symmetrical, there's no sense in which it is correct to "apply Minkowski" in the moving frame but not to do something equivalent in the stationary frame.

We agree.

Huh? I never said anything about it being exclusive to the stationary frame, I said 'this equation still holds if we are dealing with the t,x,y,z coordinates of the "stationary" frame?'--that "still" was to indicate that although it's true this Minkowski equation holds in the moving frame, it also holds in the stationary frame. It holds in all inertial frames, so again I don't see how it makes sense to say that Minkowski geometry should only be used in the moving frame and Euclidean geometry should only be used in the stationary frame. For any equation (Euclidean or Minkowski) which applies to a situation in one frame, you can find an equivalent situation in any other frame where the same equation applies.

Good, can you apply the Minkowski metric only to the stationary frame and none other?

 

[JesseM]

Your conclusions are correct. Do you understand them?

Since LT matrix invertible, then t' = 25 and t = 20 regardless of which frame is at rest.

I still don't understand the significance of calling one frame "at rest" other than it just being a totally arbitrary label like "frame A" or "the friendly frame". Switching such arbitrary labels has no mathematical significance whatsoever, so if you switch the labels but change nothing else about the problem the mathematical analysis will be exactly the same, you don't have to "invert" the LT matrix or any other such mathematical change in your calculations.

So if unprimed at rest, t' > t.

If primed frame at rest, then t' > t.

Have you figured this out yet?

Since I think "at rest" is a totally arbitrary verbal label of course I agree that changing the label won't change any mathematical fact like t' > t. But this is assuming that aside from changing the labels we are leaving the physical facts of the problem unchanged--specifically, we are still dealing with a problem where we have two events on the worldline of a clock in the unprimed frame and want to know the time between these events in both frames. If on the other hand we had two events on the worldline of a clock in the primed frame, then in this case t > t' (and again this would be true regardless of which frame we called 'stationary'). As I said back in post #25, all that matters is which frame the clock we're analyzing is in (i.e. the clock whose worldline the two events are on), once you know that the time dilation equation always has the following form:

(time interval in frame of observer who sees clock in motion) = (time elapsed on clock)*gamma

Do you agree that this always works, regardless of what frame the clock is in or which frame we choose to label "stationary"?

 

[grav-universe]

How do you make this hsppen with the WIKI article give the fact light abberation is absolute?

That means, the clock at rest with the light source will measure t=d/c and the frame that views that clock as moving will also conclude it calculates t=d/c and it will view a light beam at an angle with a longer distance to travel. Light abberation is not reciprocal. It is only true for observers moving relative to the light source.

They are two different light pulses. Light pulse 1 bounces back and forth between the mirrors in the unprimed frame, so t1 = 2 d / c. A clock in the primed frame measures t1' = (2 d / c) / sqrt(1 - (v / c)^2) for light pulse 1 to travel between the mirrors of the unprimed frame, so t1' = t1 y. Vice versely, a light pulse 2 that bounces between the mirrors in the primed frame will give t2 = t2' y. Each sees the other's clock time dilating.

 

[JesseM]

Oh, you are not familiar with topological spaces. You have a space and then a mapping to another space. That mapped space if not isomorphic is not the same as the original space.

A change in coordinates cannot take you into a different "topological space" if the geometry (given by the metric) is unchanged.

There is the stationary space which has Euclidian properties.
There is the Minkowsky space which has odd properties. The light sphere in Minkowsky is off center and not spherical.

I have only seen physicists refer to Minkowski spacetime, not "Minkowski space" in the sense of a 3D space containing only a light sphere rather than a full 4D light cone. Is this a term you have made up? Can you give any precise mathematical definition of what you mean by "Minkowski space"?

Huh? I never said anything about it being exclusive to the stationary frame, I said 'this equation still holds if we are dealing with the t,x,y,z coordinates of the "stationary" frame?'--that "still" was to indicate that although it's true this Minkowski equation holds in the moving frame, it also holds in the stationary frame. It holds in all inertial frames, so again I don't see how it makes sense to say that Minkowski geometry should only be used in the moving frame and Euclidean geometry should only be used in the stationary frame. For any equation (Euclidean or Minkowski) which applies to a situation in one frame, you can find an equivalent situation in any other frame where the same equation applies.

Good, can you apply the Minkowski metric only to the stationary frame and none other?

Uh, did you read anything I wrote above? I was specifically pointing out that I never claimed "you apply the Minkowski metric only to the stationary frame and none other", this is a strawman that is the exact opposite of what I have been saying. What part of "it's true this Minkowski equation holds in the moving frame" and "it holds in all inertial frames" didn't you understand? It was you who seemed to claim that the moving frame is Minkowski but that the stationary frame is not Minkowski, while I am saying that all frames are on equal footing as far as all equations are concerned, and "stationary" vs. "moving" has no significance except as an arbitrary verbal label.

 

[yuiop]

If you do the math on WIKI, you will find if you take the unprimed frame as rest with respect to the primed frame, then t' = t γ.

If you take the primed frame as rest with respect to the unprimed frame, then t' = t γ.

So either way, it is absolute that t' = t γ. Why is this? This light beam path for the primed frame is absolutly longer than that of the unprimed frame because of light abberation. This is not a relative issue.

If you take the unprimed frame as at rest with respect to the primed frame, then t' = t.

If you take the primed frame as at rest with respect to the unprimed frame, then t' = t.

The v in the gamma factor is the velocity of the two frames relative to each other. If the two frames are at rest with respect to each other then v=0 and gamma=1 and t'=t.

Under SR, a moving clock is supposed to beat slower. What about the WIKI article?

From the equation:

$$t' = t \gamma$$

the elapsed time t' is greater than t for any none zero v. Agree?

t is the time measured by a single clock (A) at rest in frame S. Let us say t= 10 seconds. In frame S' clock A appears to be moving at 0.8c which is the velocity of frame S' relative to S. Let us say we have two clocks (B' and C') that are at rest in frame S' and that clock A is initially adjacent to clock B' and ends up adjacent to clock C'. The time measured by the synchronised clocks (B' and C') at rest in S' is 16.666 seconds. So the time measured in frame S' by the moving clock (A) is less than the time measured by clocks B' and C' that at rest in frame S'. This is in agreement with what you say should happen. Note that the time that elapses on clock A is the same (10 seconds) according to observers in frame S AND frame S'. The time interval of 16.666 seconds measured in S' is measured by two spatially separated clocks and is NOT the time measured by clock A according to any observer. Clock A is only running slow in a relative sense compared to the coordinate time difference measured by separate clocks.

 

[IsometricPion]

Would you please use the experiment in the article to prove reciprocal time dilation?

The upper picture is of the light clock in its rest frame and is used to define the time interval between ticks of the clock in its rest frame. The second picture is of the same light clock in a frame in which it is moving and is used to derive the time interval between ticks in that frame. Since the velocity of the light clock is perpendicular to that of the path the light takes in the light clock, this path length is invariant (under the transformation from the unprimed frame to the primed frame). Then follow a few lines of algebra culminating in a relation between the interval between ticks in the primed and unprimed frames (this algebra only assumes that space is euclidean, the velocity is perpendicular to the light path, and that the primed position is the time-integral of the primed velocity, all of which are valid in SR). If one measures the number of ticks (N) of the unprimed clock in the primed frame during a time period $$\alpha{}$$ as measured by N' ticks of an equivalent clock in the primed frame, $$\Delta{}t'=\gamma{}\Delta{}t\Rightarrow{}\frac{\alpha}{\Delta{}t'}=\frac{\alpha{}}{\gamma{}\Delta{}t}\Rightarrow{}N=\frac{\alpha}{\gamma{}\Delta{}t}\,\,\,\,N'=\frac{\alpha}{\Delta{}t}\Rightarrow{}N=\frac{N'}{\gamma}$$
Thus, the unprimed clock ticks fewer times than an equivalent clock at rest in the primed frame for an arbitrary time period measured in the primed frame. So, a clock that ticks once per second in its rest frame would be measured to tick 4 times when traveling at 0.6c relative to a frame in which an equivalent clock at rest would tick 5 times.

 

[Dale]

when calculating the moving frame ... In the moving frame ... what is moving coordinates and what is stationary ... within a stationary fame

which frame is at rest ... if unprimed at rest ... If primed frame at rest

You still seem to not understand that velocities are relative. None of these statements have any meaning because velocity is a relative quantity and you have not specified what the velocity is relative to in any of these! Each of these needs to be re-written to specify "at rest wrt ___", "stationary wrt ___", "moving wrt ___", etc.

VELOCITIES ARE RELATIVE

 

[Dale]

There is the stationary space which has Euclidian properties.

I think you mean Galilean rather than Euclidean, but this is incorrect. All frames use the Minkowski metric to determine spacetime intervals, regardless of whether or not they are moving wrt some given frame or object.

There is the Minkowsky space which has odd properties. The light sphere in Minkowsky is off center and not spherical.

I think you mean the light cone. The light cone is a right cone in all frames.

Good, can you apply the Minkowski metric only to the stationary frame and none other?

It applies to all frames.

 

[darkhorror]

To chinglu1998
Trying to get an example that might make sense to see what they are talking about.

If we have two frames of reference the stationary frame and the moving frame. moving frame is moving away from stationary frame at .6c.

In the "moving frame" frame of reference the "stationary frame" is moving at .6c, and the "moving frame" is at rest.

Then in the "stationary frame" frame of reference the "moving frame" is moving at .6c, and the "stationary frame" is at rest.

This also means that which ever frame you chose the other frame's clock is going to be ticking slower.

So if you ask which clock is moving slower you have to ask in which frame of reference. Otherwise the question is meaningless.

So let's say you have two clocks t and t'. Clock t is in the "stationary frame", t' is in the "moving frame"

Now if I put the observer in the "moving frame", to this observer clock t is ticking slower than t'.

This is basicly what they are talking about on wiki

From the frame of reference of a moving observer traveling at the speed v (diagram at lower right), the light pulse traces out a longer, angled path.

 

[yuiop]

Let me try again with a simple example and maybe you can let me know which parts you disagree with.

Anne is at rest in frame S. Anne considers herself stationary.
Bob is at rest in frame S'. Bob considers himself stationary.
According to Bob, Anne is moving at 0.6c in the x direction.
According to Anne, Bob is moving at 0.6c in the -x direction.

The fact that Bob thinks Anne's velocity is +0.6c and Anne thinks Bob's velocity is -0.6c is not important because the velocity is squared in the gamma factor and the sign of the velocity is "lost".
The gamma factor in this case is $$\gamma = 1/\sqrt{1-(\pm 0.6)^2} = 1.25$$ from either point of view.

From Bob's point of view:

Anne passes Bob at time zero on both their clocks. (Event 1)
Bob has a brother Bob2 who is at rest with Bob in frame S'.
Bob and Bob2's clocks are synchronised.
Anne passes Bob2 when 25 seconds are showing on Bob2's clock. (Event 2)
When Anne passes Bob2, 20 seconds have elapsed on her clock.
Since Bob and Bob2's clcoks are synchronised, Bob concludes that 25 seconds elapses on his clock between events 1 and 2, when 20 seconds elapses on Anne's clock between those same two events.
Therefore according to Bob:

$$t_{Bob} = \gamma ( t_{Anne})$$

From Anne's point of view:

Bob is moving and passes Anne at time zero on both their clocks. (Event 1)
Anne has a sister Anne2 who is at rest with Anne in frame S.
Anne and Anne2's clocks are synchronised.
Bob passes Anne2 when 25 seconds are showing on Anne2's clock. (Event 3)
When Bob passes Anne2, 20 seconds have elapsed on his clock.
Since Anne and Anne2's clocks are synchronised, Anne concludes that 25 seconds elapses on her clock between events 1 and 3, when 20 seconds elapses on Bob's clock between those same two events.
Therefore according to Anne:

$$t_{Anne} = \gamma ( t_{Bob})$$

==============================

This appears to be a contradiction to the earlier conclusion that according to Bob:

$$t_{Bob} = \gamma ( t_{Anne})$$

but there is a slight "decepton" going on here.

In Anne's rest frame, the time $t_{Anne}$ in the equation $t_{Anne} = \gamma ( t_{Bob} )$ is really the difference between $t_{Anne}$ and $t_{Anne2}$'s clocks while $t_{Bob}$ is the time measured by a single clock.
In Bob's rest frame, the time [itex]t_{Bob}[/tex] in the equation $t_{Bob} = \gamma ( t_{Anne})$ is really the difference between $t_{Bob}$ and $t_{Bob2}$'s clocks while $t_{Anne}$ is the time measured by a single clock.

In other words, the $t_{Bob}$ in $t_{Bob} = \gamma ( t_{Anne})$ is not the same as the $t_{Bob}$ in $t_{Anne} = \gamma ( t_{Bob})$. $t_{Bob}$ in the first equation is measured by two spatially separated clocks (coordinate time) and $t_{Bob}$ in the second equation is measured by a single clock (proper time). $t_{Bob}$ in the first equation is the difference between events 1 and 2 while $t_{Bob}$ in the second equation is the difference between events 1 and 3. This is the "deception" that is hidden when using the primed notation that gives the appearance of a reciprocal time dilation relationship.

Note that the time on the right hand side of both equations is measured by a single clock that is present at both the start and finish events. This is the proper time and I have mentioned it several times before, but I repeat it again, because it is an important concept to understand. All observers agree on the proper time of a clock between two events.

 

[espen180]

I present here an alternative derivation of the result found in the wiki article. Hopefully the fact that different approaches lead to the same result will convince you that the result is not erroneous.

The Minkowski metric, which really is fundamental to SR, gives a very short derivation of the formula you are critiquing. This metric illustrates a very fundamental aspect of Relativity, namely the invariance of the spacetime element $$ds$$ between frames. We have that

$$ds^2=c^2dt^2-dx^2-dy^2-dz^2$$ (Note that I am using a convention different from, but physically equivalent to, the one used by JesseM. (This can also be interpreted as "The norms of all vectors (ct,x,y,z) in Minkowski spacetime are equal.")

For our purposes we may assume $$dy=dz=0$$ and let the $$x$$-axes of all frames point in the direction the clock is moving. We will now derive from this the relationship between the relevant frames.

1) The rest frame of the observer. This frame is stationary relative to the observer (Note that ALL measurements MUST have an observer).

2) The rest frame of the clock, moving with velocity $$v$$ with respect respect to the observer.

The question we want to answer is: As the clock moves past the observer (we assume it does), at what rate does the observer measure the passing clock to tick relative to his own wristwatch, assuming they tick at the same rate when stationary with respect to one another? (This problem statement was a mouthful, be sure you got everything).

Of course, the clock is at rest with respect to its own rest frame, so $$ds^2=c^2d\tau^2$$, where $$d\tau$$ is an infinitesimal interval of time. By convention, we call this the proper time of the clock along its trajectory.

However, the clock is moving with constant velocity $$v=\frac{dx}{dt}$$ in the observer's rest frame, and by the invariance of the spacetime element we therefore have

$$c^2d\tau^2=c^2dt^2-dx^2$$

$$\left(\frac{d\tau}{dt}\right)^2=1-\frac{v^2}{c^2}$$

Giving us the time dilation formula

$$\frac{dt}{d\tau}=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=\gamma$$

or

$$dt=\gamma d\tau$$

And I restate my convention that $$dt$$ is an infinitesimal time interval in the observer's rest frame, where the clock is moving, and $$d\tau$$ is an infinitesimal time interval in the clock's rest frame.

 

[chinglu1998]

I still don't understand the significance of calling one frame "at rest" other than it just being a totally arbitrary label like "frame A" or "the friendly frame". Switching such arbitrary labels has no mathematical significance whatsoever, so if you switch the labels but change nothing else about the problem the mathematical analysis will be exactly the same, you don't have to "invert" the LT matrix or any other such mathematical change in your calculations.

Since I think "at rest" is a totally arbitrary verbal label of course I agree that changing the label won't change any mathematical fact like t' > t. But this is assuming that aside from changing the labels we are leaving the physical facts of the problem unchanged--specifically, we are still dealing with a problem where we have two events on the worldline of a clock in the unprimed frame and want to know the time between these events in both frames. If on the other hand we had two events on the worldline of a clock in the primed frame, then in this case t > t' (and again this would be true regardless of which frame we called 'stationary'). As I said back in post #25, all that matters is which frame the clock we're analyzing is in (i.e. the clock whose worldline the two events are on), once you know that the time dilation equation always has the following form:

(time interval in frame of observer who sees clock in motion) = (time elapsed on clock)*gamma

Do you agree that this always works, regardless of what frame the clock is in or which frame we choose to label "stationary"?

The WIKI article states clearly that ∆t' = t γ. This does not agree with your assertion yest you agreed the WIKI article is correct. How do you do this?

 

[JesseM]

The WIKI article states clearly that ∆t' = t γ. This does not agree with your assertion yest you agreed the WIKI article is correct. How do you do this?

Of course it agrees with my assertion, why do you think otherwise? I said that the time dilation formula always has the following form:

(time interval in frame of observer who sees clock in motion) = (time elapsed on clock)*gamma

In the wiki article ∆t' is the time in the frame of the observer who is moving relative to the clock, and t is the time elapsed on the clock itself. Do you disagree?

 

[chinglu1998]

I think you mean Galilean rather than Euclidean, but this is incorrect. All frames use the Minkowski metric to determine spacetime intervals, regardless of whether or not they are moving wrt some given frame or object.

How do you square your statement with Einstein's?

Let us take a system of co-ordinates in which the equations of Newtonian mechanics hold good.2 In order to render our presentation more precise and to distinguish this system of co-ordinates verbally from others which will be introduced hereafter, we call it the ``stationary system.''

http://www.fourmilab.ch/etexts/einstein/specrel/www/

And no, you do not apply the metric to the stationary system. That is the light postulate defined as c²t² = x² + y² + z².

 

[chinglu1998]

The upper picture is of the light clock in its rest frame and is used to define the time interval between ticks of the clock in its rest frame. The second picture is of the same light clock in a frame in which it is moving and is used to derive the time interval between ticks in that frame. Since the velocity of the light clock is perpendicular to that of the path the light takes in the light clock, this path length is invariant (under the transformation from the unprimed frame to the primed frame). Then follow a few lines of algebra culminating in a relation between the interval between ticks in the primed and unprimed frames (this algebra only assumes that space is euclidean, the velocity is perpendicular to the light path, and that the primed position is the time-integral of the primed velocity, all of which are valid in SR). If one measures the number of ticks (N) of the unprimed clock in the primed frame during a time period $$\alpha{}$$ as measured by N' ticks of an equivalent clock in the primed frame, $$\Delta{}t'=\gamma{}\Delta{}t\Rightarrow{}\frac{\alpha}{\Delta{}t'}=\frac{\alpha{}}{\gamma{}\Delta{}t}\Rightarrow{}N=\frac{\alpha}{\gamma{}\Delta{}t}\,\,\,\,N'=\frac{\alpha}{\Delta{}t}\Rightarrow{}N=\frac{N'}{\gamma}$$
Thus, the unprimed clock ticks fewer times than an equivalent clock at rest in the primed frame for an arbitrary time period measured in the primed frame. So, a clock that ticks once per second in its rest frame would be measured to tick 4 times when traveling at 0.6c relative to a frame in which an equivalent clock at rest would tick 5 times.

Fine, let us assume the WIKI experiment.

Let us further assume the unprimed frame has the light source like WIKI .

Now, let's conduct the experiment from the view of the unprimed frame.

It's view is the light will satisfy t = y/c for some y on the axis.

Now, in the primed frame, it will see light aberration, hence the light path is longer in the view of the primed frame.

Apply the Pythagorean Theorem and you calculate t' = t γ just as WIKI said, but this is not time dilation since the unprimed frame is at rest.

Now assume the primed frame is at rest. Same thing, light aberration will force that frame to conclude t' = t γ.

So, it does not matter which frame is stationary, t' = t γ.

 

[chinglu1998]

Everyone in this thread is making the same error.

Apply light aberration when the primed frame is stationary or the unprimed frame is, it does not matter.

It will always be the case if the light source is at rest with the unprimed frame, t' = t γ

 

[yuiop]

Everyone in this thread is making the same error.

Apply light aberration when the primed frame is stationary or the unprimed frame is, it does not matter.

I think you are making the mistake of not actually reading or trying to understand anything the other posters are saying. For example Dalespam has tried to tell you several times it is meaningless to state velocity without reference to another object or frame and yet you are still saying things like "primed frame is stationary".

The "primed frame is stationary" with respect to what ?

Secondly, I asked if disagreed with anything in the long example I gave in #90 and you have not responded.

It will always be the case if the light source is at rest with the unprimed frame, t' = t γ

If by the light source you mean the light clock is at rest in the unprimed frame so the light clock measures a time of t between "bounces" then yes, the time interval measured in the primed frame (in which the light clock appears to be moving) will be t' = t γ, exactly as claimed by Wikipedia. Is that the end of the matter?

 

[JesseM]

It will always be the case if the light source is at rest with the unprimed frame, t' = t γ

Um, that's exactly what I have been saying. Read again:

(time interval in frame of observer who sees clock in motion) = (time elapsed on clock)*gamma

If we label the clock frame as unprimed, then (time elapsed on clock) = t and (time interval in frame of observer who sees clock in motion) = t'. And of course this is what the wiki article says as well.

 

[chinglu1998]

I think you are making the mistake of not actually reading or trying to understand anything the other posters are saying. For example Dalespam has tried to tell you several times it is meaningless to state velocity without reference to another object or frame and yet you are still saying things like "primed frame is stationary".

The "primed frame is stationary" with respect to what ?

If by the light source you mean the light clock is at rest in the unprimed frame so the light clock measures a time of t between "bounces" then yes, the time interval measured in the primed frame (in which the light clock appears to be moving) will be t' = t γ, exactly as claimed by Wikipedia. Is that the end of the matter?

Maybe you are correct.

Let's follow through with light aberration and see if you are.

primed frame (in which the light clock appears to be moving) will be t' = t γ

You are wrong here. It should be t' = t/γ or t' γ = t.

It is amazing how many refute relativity.

 

[chinglu1998]

Um, that's exactly what I have been saying. Read again:

(time interval in frame of observer who sees clock in motion) = (time elapsed on clock)*gamma

If we label the clock frame as unprimed, then (time elapsed on clock) = t and (time interval in frame of observer who sees clock in motion) = t'. And of course this is what the wiki article says as well.

(time interval in frame of observer who sees clock in motion) = t'

So, WIKI says t' = γt and you say the unprimed frame is at rest and sees the primed frame as moving, you claim if unprimed is at rest t' = γt is your argument? Can you explain how this is time dilation?

 

[JesseM]

(time interval in frame of observer who sees clock in motion) = t'

So, WIKI says t' = γt and you say the unprimed frame is at rest

I didn't say anything about either frame being "at rest", this is your own ill-defined terminology which everyone and their mother is telling you to stop using because it obviously causes you endless confusion. I just said the unprimed frame was the frame of the clock (the clock is at rest relative to the unprimed frame), and the primed frame was the frame of the observer (the observer is at rest relative to the primed frame). In this case, if we pick two events on the worldline of the clock, the times between these events in each frame will be related by t' = γt. This will be true regardless of which frame you choose to label "at rest" and which you choose to label "moving", as long as the conditions I mention above are met. Do you disagree with this or not? Yes or no?

Also, you seem to be rather evasive about answering my simple, oft-repeated question about whether you agree or disagree that the time dilation equation always works like this:

(time interval in frame of observer who sees clock in motion) = (time elapsed on clock)*gamma

If you disagree, please give me a counterexample.

 

[JesseM]

Maybe you are correct.

Let's follow through with light aberration and see if you are.

primed frame (in which the light clock appears to be moving) will be t' = t γ

You are wrong here. It should be t' = t/γ or t' γ = t.

What does this have to do with light aberration? And why do you preface that equation with "primed frame"? The equation t' = t γ isn't written from the point of view of any specific frame, it relates the times in two different frames. If the clock is moving at 0.6c in the primed frame, and we look at the pair of events on the clock's worldline where the clock reads a time of 0 and where it reads a time of 20, do you disagree that t (the time between these events in the unprimed frame) is 20 while t' (the time between these events in the primed frame) is 25, satisfying t' = t*1.25?

 

[chinglu1998]

I didn't say anything about either frame being "at rest", this is your own ill-defined terminology which everyone and their mother is telling you to stop using because it obviously causes you endless confusion. I just said the unprimed frame was the frame of the clock (the clock is at rest relative to the unprimed frame), and the primed frame was the frame of the observer (the observer is at rest relative to the primed frame). In this case, if we pick two events on the worldline of the clock, the times between these events in each frame will be related by t' = γt. This will be true regardless of which frame you choose to label "at rest" and which you choose to label "moving", as long as the conditions I mention above are met. Do you disagree with this or not? Yes or no?

Can you explain how time dilation is valid if a frame is not taken stationary? I mean, you seem to think this idea is dirty. I am sorry, but you must take one frame as stationary to determine time dilation.

(time interval in frame of observer who sees clock in motion) = (time elapsed on clock)*gamma

If you disagree, please give me a counterexample.

Yes, this is how time dilation works. But, the reason I see not need to answer this question is your evasiveness to indicates a frame is stationary.

Are you ready to step up to the plate and claim the unprimed frame in WIKI is stationary and t' = γt. I want to see you declare this.

 

[chinglu1998]

What does this have to do with light aberration? And why do you preface that equation with "primed frame"? The equation t' = t γ isn't written from the point of view of any specific frame, it relates the times in two different frames. If the clock is moving at 0.6c in the primed frame, and we look at the pair of events on the clock's worldline where the clock reads a time of 0 and where it reads a time of 20, do you disagree that t (the time between these events in the unprimed frame) is 20 while t' (the time between these events in the primed frame) is 25, satisfying t' = t*1.25?

What does this have to do with light aberration?

Did you even read the article?

satisfying t' = t*1.25

Actually, you got me. When the unprimed frame is stationary, it elapses less time than the primed frame because of light aberration. Hence, the moving frame (primed frame) beats faster instead of slower as SR would demand for time dilation. You just refuted SR.

 

[JesseM]

Can you explain how time dilation is valid if a frame is not taken stationary? I mean, you seem to think this idea is dirty. I am sorry, but you must take one frame as stationary to determine time dilation.

I am sorry, but this a nonsensical notion of yours that no physicist would agree with, physicists do sometimes choose to label a frame as "stationary" but this is understood to be an arbitrary label for the sake of convenience with no relevance to any equation or physical result.

Let's say the clock is in the unprimed frame and the observer is in the primed frame. Pick two events on the clock's worldline, and say the time between them in the unprimed frame is t while the time between these same two events in the primed frame is t'. Then, if we label unprimed as "stationary" and primed as "moving" the correct time dilation equation is:

t' = t γ

But if we label the primed frame as "stationary" and unprimed as "moving", while keeping everything else the same, the correct time dilation equation is:

t' = t γ

Do you disagree? Yes or no, please. If you don't disagree, then you can see that the choice of which frame to label as "stationary" is utterly irrelevant to the equation, which is exactly the same either way. If you do disagree, please specify in which case you think the equation is wrong, and then I can go through a numerical example to show that the equation is still correct in that case.

Are you ready to step up to the plate and claim the unprimed frame in WIKI is stationary and t' = γt. I want to see you declare this.

There is no convention in physics that you must label one frame as "stationary", and in fact the wiki doesn't do so. But for the sake of argument, let's pretend the wiki did label the unprimed as "stationary". If they did, the correct equation given their description of the problem would still be t' = t γ, since as I said above it is utterly irrelevant to the equation which frame you choose to pin the arbitrary label "stationary" on.