WIKI and Time Dilation: The Possible Error in Relative Velocity

[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.

 

[JesseM]

What does this have to do with light aberration?

Did you even read the article?

Yes, I did. I am not asking about the fact that light aberration is seen in the frame moving relative to the clock, I am asking about what light aberration has to do with your argument in the post I was responding to:

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.

You said "let's follow through with light aberration", but none of the subsequent sentences seemed to have anything to do with light aberration.

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.

You continually use vague ill-defined terminology, I don't know what it even means to say a frame "elapses less time", the time dilation equation deals with the time between a pair of events on the worldline of a physical clock. If we pick two events on the worldline of a clock, and t is the time between them in the clock's frame while t' is the time between those same events in the frame of an observer who is moving relative to the clock (though we are free to label the observer as "stationary" and the clock as "moving" rather than vice versa), then the equation will always be t' = t γ. As I said in my previous post, as long as those conditions are met this equation is correct, it's completely irrelevant which frame you label as "stationary" and which you label as "moving".

 

[chinglu1998]

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.

Einstein used the label "stationary" 62 times in his paper.
http://www.fourmilab.ch/etexts/einstein/specrel/www/

Are you calling Einstein wrong. Let's get it out now!

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, I can go through a numerical example to show you're wrong.

I agree with all above. Where is your reciprocal time dilation as required by SR?

 

[JesseM]

Einstein used the label "stationary" 62 times in his paper.
http://www.fourmilab.ch/etexts/einstein/specrel/www/

Are you calling Einstein wrong. Let's get it out now!

Do you have zero reading comprehension? I said 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. Einstein of course calls one frame "stationary", but this is "understood to be an arbitrary label for the sake of convenience", that's why he says 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.'' Calling the system "stationary" is purely to distinguish it verbally, it has no physical relevance and doesn't affect any aspect of his mathematical derivation.

I agree with all above.

So you agree it makes no difference to the equation which frame we call "stationary"? And you agree the wiki article was correct to write t' = t γ since their scenario matched the one I described?

Where is your reciprocal time dilation as required by SR?

The reciprocity lies in the fact that if we picked a different clock which was in the primed frame rather than the unprimed frame, and picked two events on the worldline of that clock, then the time between these events in each frame would be given by t = t' γ. Again, it's all about which frame is the clock's frame, which is why I said it's best to think of the time dilation equation as (time interval in frame of observer who sees clock in motion) = (time elapsed on clock)*gamma

 

[Dale]

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/

In what way is my statement in conflict with Einsteins?

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

Yes, you do. The first postulate requires that the metric which applies to any inertial frame must apply to all of them. If it applies to one frame then it applies to all. The light cone is the set of all points:
ds² = -c²dt² + dx² + dy² + dz²
such that ds²=0
The same in one inertial frame as in all others. You cannot pick one metric for one inertial frame and another metric for another inertial frame without violating the first postualte.

 

[yuiop]

I agree with all above. Where is your reciprocal time dilation as required by SR?

Read post #90 for an example of reciprocal time dilation.

 

[darkhorror]

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.




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.

I think I see your problem in reasoning and understanding of time dilation. I'll ask a few questions see if you agree or disagree and hopefully we can find exactly what the problem your having is. I made a post earlier that had most of the questions but i'll try to simplify.

You have the Earth and a ship. The ship is moving at a constant .6c away from the earth.

1. Is the ship "stationary" in the ship's frame of reference?

2. If in the Earth's frame of reference the ship's clock is ticking slower does this mean that in the ships frame of reference that the Earth's clock is ticking faster?

3. In which frame of reference is the Earth's clock ticking slower than the ship's clock?

4. Does the clock on the ship always tick slower than the clock on the Earth no matter which frame of reference you chose?

 

[darkhorror]

I think I might have figured out what is being said. How does this sound?

In the wiki article the observer is in the primed frame. So it doesn't matter what you label the frames as if it be "moving" or "stationary". To the observer which ever frame he is in he will always be at rest. He doesn't care if you call him "moving" or "stationary".

 

[grav-universe]

Okay, I think I finally figured out a way to clarify that part of the Wiki article with the light clock that fits in well with it and the diagrams, by adding a simple paragraph to the end. How does this sound?

This is the time dilation found only by the frame of reference that is considered stationary as it observes the light pulse which is bouncing back and forth in the moving frame. To clarify this further, let us now place a mechanical clock at mirror A in the moving frame. All frames must agree upon the readings of the mechanical clock as the light pulse departs mirror A in the moving frame and then arrives back at mirror A again, since those two readings of the mechanical clock directly coincide in the same place as the two events of the light pulse departing and arriving, therefore all frames observe the difference between these two readings in the moving frame to be Δt as that frame measures itself. According to our frame of reference, however, the same light pulse travels a longer path in the time Δt', whereby the time dilation that the stationary frame of reference observes of the moving frame's clock, as found earlier, is the ratio of the time that the stationary frame observes passing upon the moving frame's clock as compared to the time that passes upon the stationary frame's own clock.

 

[espen180]

OP, maybe the article

http://en.wikipedia.org/wiki/Introduction_to_special_relativity

is more suited to you. I suggest reading it closely before continuing the discussion, as you are demonstrating a misunderstanding of very fundamental topics in SR, and progression to more advanced topics will be impossible unless this is cleared up.

Okay, I think I finally figured out a way to clarify that part of the Wiki article with the light clock that fits in well with it and the diagrams, by adding a simple paragraph to the end. How does this sound?

It may just be me, but I don't think that paragraph, or any such clarification is needed.

 

[yuiop]

This is the time dilation found only by the frame of reference that is considered stationary as it observes the light pulse which is bouncing back and forth in the moving frame. To clarify this further, let us now place a mechanical clock at mirror A in the moving frame. All frames must agree upon the readings of the mechanical clock as the light pulse departs mirror A in the moving frame and then arrives back at mirror A again, since those two readings of the mechanical clock directly coincide in the same place as the two events of the light pulse departing and arriving, therefore all frames observe the difference between these two readings in the moving frame to be Δt as that frame measures itself. According to our frame of reference, however, the same light pulse travels a longer path in the time Δt', whereby the time dilation that the stationary frame of reference observes of the moving frame's clock, as found earlier, is the ratio of the time that the stationary frame observes passing upon the moving frame's clock as compared to the time that passes upon the stationary frame's own clock.

You really should stop talking about the "stationary frame" as it just confuses the heck out everyone here and you too.

The observer at rest in frame S considers himself to be at rest in his own reference frame and likewise the observer in frame S' considers himself to be at rest in his own reference frame. Frames S and S' are moving with a velocity magnitude of v relative to each other. There is no clear definition of whether S or S' is the stationary frame so as soon as you talk about the "stationary frame" no one knows what you are talking about and it just leads to confusion as it has in your case.

It is better to say something like the frame (S) in which the light clock is at rest, or the frame (S') in which the light clock has relative motion -v.

It *might* be OK to use phrases like "the stationary frame" as shorthand for "the reference frame in which the light clock is at rest" as long you clearly define the expression (as Einstein did) but using an identifier such as "frame S" is just a good a shorthand. Your expression "the frame of reference that is considered stationary as it observes the light pulse which is bouncing back and forth in the moving frame" gives the reader no clue as to whether "the stationary frame" is the frame in which the light clock is at rest or not.

 

[grav-universe]

You really should stop talking about the "stationary frame" as it just confuses the heck out everyone here and you too.

The observer at rest in frame S considers himself to be at rest in his own reference frame and likewise the observer in frame S' considers himself to be at rest in his own reference frame. Frames S and S' are moving with a velocity magnitude of v relative to each other. There is no clear definition of whether S or S' is the stationary frame so as soon as you talk about the "stationary frame" no one knows what you are talking about and it just leads to confusion as it has in your case.

It is better to say something like the frame (S) in which the light clock is at rest, or the frame (S') in which the light clock has relative motion v.

Right, okay. Maybe just the observing frame rather than the stationary frame, then? The paragraph I added has already been cut in the article, but I still want to clarify it. I also considered adding a second paragraph which shows the same time dilation can be found from either frame, but I think it still requires some rewording. Perhaps you can help with that. So far it reads

"If we were to now reverse the roles of the observing and moving frames, making the original frame of the light clock the observing frame while considering the other frame to be moving, while also placing another light clock in the other frame with a second light pulse bouncing back and forth between two mirrors in the same way as in the original frame, then the first frame will measure the same time dilation of the second frame's clock as well, found in the same way as before from the other frame, demonstrating that each frame measures the same time dilation of the other."

 

[espen180]

Right, okay. Maybe just the observing frame rather than the stationary frame, then? The paragraph I added has already been cut in the article, but I still want to clarify it. I also considered adding a second paragraph which shows the same time dilation can be found from either frame, but I think it still requires some rewording. Perhaps you can help with that. So far it reads

"If we were to now reverse the roles of the observing and moving frames, making the original frame of the light clock the observing frame while considering the other frame to be moving, while also placing another light clock in the other frame with a second light pulse bouncing back and forth between two mirrors in the same way as in the original frame, then the first frame will measure the same time dilation of the second frame's clock as well, found in the same way as before from the other frame, demonstrating that each frame measures the same time dilation of the other."

This is unneeded, because the following paragraph, titled "Time dilation due to relative velocity symmetric between observers" covers this.

 

[yuiop]

I also considered adding a second paragraph which shows the same time dilation can be found from either frame, but I think it still requires some rewording. Perhaps you can help with that.

Thanks, but I am probably one of the worst people to do this task as I use a very informal language myself. There are plenty of others skilled in the formal language here such as Jesse or Dalespam who would be much better suited to the task, but from what I can tell the Wikipedia article on time dilation is pretty accurate and well explained and does not need much, if anything adding to it.

 

[grav-universe]

This is unneeded, because the following paragraph, titled "Time dilation due to relative velocity symmetric between observers" covers this.

That isn't clear either. Okay, maybe just a quick mention on that, adding it to the overall paragraph. How about this?

"This is the time dilation found only by the frame of reference that observes the light pulse bouncing back and forth in the moving frame. To clarify this further, let us now place a mechanical clock at mirror A in the moving frame. All frames must agree upon the readings of the mechanical clock as the light pulse departs mirror A in the moving frame and then arrives back at mirror A again, since those two readings of the mechanical clock directly coincide in the same place as the two events of the light pulse departing and arriving, therefore all frames observe the difference between these two readings in the moving frame to be Δt as that frame measures itself. According to the observing frame, however, the same light pulse travels a longer path in the time Δt', whereby the time dilation that the observing frame observes of the moving frame's clock, as found earlier, is the ratio of the time that the observing frame views passing upon the moving frame's clock as compared to the time that passes upon the observing frame's own clock. In addition, if we were to now place another light clock in the second frame with a second light pulse bouncing back and forth between two mirrors in that frame in the same way while reversing the roles of the observing and moving frames, then the first frame will now measure the same time dilation of the second frame's clock as found as before from the other frame, demonstrating that each frame measures the same time dilation of the other."

 

[yuiop]

That isn't clear either. Okay, maybe just a quick mention on that, adding it to the overall paragraph. How about this?

"This is the time dilation found only by the frame of reference that observes the light pulse bouncing back and forth in the moving frame. To clarify this further, let us now place a mechanical clock at mirror A in the moving frame. All frames must agree upon the readings of the mechanical clock as the light pulse departs mirror A in the moving frame and then arrives back at mirror A again, since those two readings of the mechanical clock directly coincide in the same place as the two events of the light pulse departing and arriving, therefore all frames observe the difference between these two readings in the moving frame to be Δt as that frame measures itself. According to the observing frame, however, the same light pulse travels a longer path in the time Δt', whereby the time dilation that the observing frame observes of the moving frame's clock, as found earlier, is the ratio of the time that the observing frame views passing upon the moving frame's clock as compared to the time that passes upon the observing frame's own clock. In addition, if we were to now place another light clock in the second frame with a second light pulse bouncing back and forth between two mirrors in that frame in the same way while reversing the roles of the observing and moving frames, then the first frame will now measure the same time dilation of the second frame's clock as found as before from the other frame, demonstrating that each frame measures the same time dilation of the other."

"The moving frame" is as poorly defined as the "the stationary frame". An observer in frame S considers frame S' to be moving relative to his frame and the observer in frame S' considers frame S to be moving relative to his own frame. Which is the "the moving frame"? Frame S or frame S'? If someone reads your paragraph very carefully they can deduce that by "the moving frame" you mean the frame in which the light clock is at rest, but it is hard work because it is not made clear early on.