# 3 inertial frames compared in STR

Suppose an event is observed in 3 inertial frames K, K' and K''. The coordinates in K are $(x,t)$ in K' are $(x',t')$ in K'' are $(x'',t'')$. The K' and K'' coordinates are then Lorentz-transformed to K-coordinates: $(x',t') \rightarrow (x_1, t_1)$ and $(x'',t'') \rightarrow (x_2, t_2)$. Now there are 2 sets of coordinates in K for the same event and they are proably different.

$x_1 = \gamma_1(x' + v_1t') \neq \gamma_2(x'' + v_2t'') = x_2$?

What is the correct interpretation?

Note that it is not allowed to assume that e.g. $x'$ and $x''$ are transformations of $x, t$.

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No, $(x_1, t_1)=(x_2, t_2)=(x,t)$. –  Raskolnikov Sep 4 '12 at 13:00
Is it really so, then why, zee edit? –  Gerard Sep 4 '12 at 20:40
Why don't you check instead of saying they are "probably" different. They aren't different, and if you check you will see this, and save people time. –  Ron Maimon Sep 5 '12 at 1:30

I think I understand what you're asking: if I've misunderstood you ignore the rest of this answer.

Suppose we're in the frame K and we observe the event at $(x, t)$. To find out what this looks like to the observer in the K' frame we just apply the Lorentz transform to get:

$$x' = \gamma (x - vt)$$

$$t' = \gamma (t - \frac{vx}{c^2})$$

But now let's do the reverse as you suggest. Now we're in frame K' and we want to know what the event looks like in frame K: I'll call this $(x_1, t_1)$ as you did. We use the Lorentz transform just as before, but of course this time the velocity is $-v$ because it's in the opposite direction so:

$$x_1 = \gamma (x' + vt')$$

If you substitute for x' and t' you get the expression:

$$x_1 = \frac{x - vt + vt - \frac{v^2x}{c^2}}{1 - \frac{v^2}{c^2}} = x\frac{1 - \frac{v^2}{c^2}}{1 - \frac{v^2}{c^2}} = x$$

So if you Lorentz transform from K to K' then Lorentz transform from K' to K you get back to the same point, which is of course hardly surprising. I won't show this is true for $t$ as well because it's a bit messy - I'll leave this as an exercise for the reader.

So to go back to your question. Going from $(x, t)$ to $(x', t')$ and back to $(x_1, t_1)$ will take you back to the same point i.e. $(x, t)$ = $(x_1, t_1)$ and likewise for the K'' frame.

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Yes your calculation is of course correct: the Lorentz transformations form a group, so it is not surprising. Your interpretation differs from my question, in the respect that you assume x' is transformed from x. I take x' and x'' as starting point and I assume that the transformation of x' to x1 is not equal to the transformation of x'' to x2. –  Gerard Sep 4 '12 at 20:21
I guess I'm missing something because I don't understand what you're asking. If it's the same point we're talking about it doesn't matter how and which order you switch between frames, in any one frame you'll always end up with the same result. How could it be otherwise? Can you give a concrete example of where you end up with different representations of the same point? –  John Rennie Sep 5 '12 at 5:59
No, I cannot. The Lorentz group of course is 100% consistent. What is bothering me is that it is somehow strange that observed x', x'' result in the same x after transformation, you wouldn't expect that looking at the formulas sec. –  Gerard Sep 6 '12 at 8:59

This statement:

$x_1 = \gamma_1(x' + v_1t') \neq \gamma_2(x'' + v_2t'') = x_2$

Note that it is not allowed to assume that e.g. $x'$ and $x''$ are transformations of $x, t$.

is a direct contradiction of this one:

Suppose an event is observed in 3 inertial frames K, K' and K''. The coordinates in K are $(x,t)$ in K' are $(x',t')$ in K'' are $(x'',t'')$

If a single event is observed in these three inertial frames with coordinates as you've specified, then practically by definition, $(x_1,t_1) = (x_2,t_2) = (x,t)$.

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I can see your point. –  Gerard Sep 6 '12 at 9:00

They are relative to each other this is where the special theory of relativity comes in, if you think about it. If they were all observing something like a spacecraft flying at the speed of sound, it would be relative to the different coordinates: $K' \to K_1$ would be seeing something totally different from what $K''\to K_2$ would see.

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