WIKI and Time Dilation

So where is the third frame to apply the velocity add equations?

I never took him to mean all. So, I agree with you. But, I specifically focused on the Euclidian measurements of the stationary frame and that part is not refutable.

Even if you choose to label one frame as "stationary" and another as "moving", it is still true in the moving frame that the light postulate holds true in the coordinates of that frame (light has a coordinate velocity of c in all directions in the moving frame), and that in the coordinates of that frame the position of the light at any specific time T' satisfies x'² + y'² + z'² = c² T'². So you still haven't given any non-confused explanation of what you meant in post #23 when you said "When you take a frame as stationary, you assume the Euclidian Geometry.
When you assume another frame is the moving frame, you apply the Minkowsky Geometry.
This is a vital distinction such that if the distinction is not clear, complete and total breakdown of SR occurs."[/QUOTE

In my post and statement "I assume the Euclidian Geometry" in the stationary system. Do you prove otherwise?

This is not Minkowsky. If you attempy to apply Minkowsky to stationary system, you get failure.

And can you give a numerical example of "apply the Minkowski geometry"? Normally I would say that Minkowski geometry is defined in terms of the metric, which says that the proper time between two events with coordinate separations [tex]\Delta t, \Delta x, \Delta y, \Delta z[/tex] is given by the formula [tex]\sqrt{\Delta t^2 - (1/c^2)*(\Delta x^2) - (1/c^2)*(\Delta y^2) - (1/c^2)*(\Delta z^2)}[/tex]. But this is true in all frames, even if we choose to label one of the frames as "stationary" like Einstein does. So I have no idea why you think Minkowski geometry doesn't apply in what you call the "stationary frame".

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.

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"?

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":

Do you disagree that the metric [tex]d\tau^2 = dt^2 - (1/c^2)dx^2 - (1/c^2)dy^2 - (1/c^2)dz^2[/tex] 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?

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.

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

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.

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

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.

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.

It was a very simple question, you don't have to do any math. Just read the question. If you have a space ship 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 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.

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

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

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.

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.

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.

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.

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?

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.

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.

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.

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?

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.

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