# What do relativistic observers "see"?

In many popular descriptions, in particular those related to the so-called "twin paradox" (which is not really a paradox), one finds descriptions about "what a particular observer sees". This description is misleading. A more accurate description would be "what a particular observer thinks if he believes that he is, just now, at absolute rest".

## How accurate language avoids the twin paradox

The difference is essential. One can see this if one looks at the twin "paradox". So, we assume two twins, one of them travelling to Sirius and back and the other remaining at home.

If described in the "he sees" language, it looks like this: The twin at Earth "sees" that the clock of the travelling twin goes slower. Instead, the travelling twin "sees" that the clock of the twin at Earth goes slower. A paradox, indeed.

Let's translate into the correct language. The twin at Earth, thinking that he is at rest all the time, concludes from his observations that the clock of the travelling twin goes slower. The second sentence already cannot be translated in such a simple way - however stupid, the travelling twin will not think that he is at rest all the time of the travel.

He may think that he is at rest on the trip to Sirius. Then he concludes that during the trip to Sirus his clock was faster than that of the twin at Earth. But then he will not think that he is at rest travelling home. Instead, he will think that travelling back to Earth he will have a larger speed than the twin at home. So, during the travel home his own clock would be slower than that of the twin remaining at Earth. What will be larger - the loss of time on the trip to Sirius or the gain of time travelling home? One would have to look at the formulas. But, very roughly, once the speed on the trip home is greater, the relativistic effect will be greater, so that gain will overcompensate the initial loss and he will remain younger.

And the same holds, only slightly different, if he believes that he is at rest during the time of travelling home. Then he was not at rest during the travel to Sirius, and his speed was during the first part was larger than the speed of the twin at Earth. So, we had the travel home to Earth where clock of the twin at Earth is slower, and we have the first part where his own clock is slower. The result remains unclear until one looks at the formulas. But, again, one roughly estimate that, given that the velocity was greater, the relativistic effect was greater too and overcompensating the loss during the second part, so that he will remain younger.

Thus, a translation into accurate language automatically removes anything paradoxical from the "twin paradox".

## Why "the observer sees this" is wrong

To see that the use of the "the observer sees this" is misleading the consideration of the twin paradox would be sufficient. But one can see this immediately too. Namely, one has to consider the question what motivates the Einstein synchronization procedure.

The Einstein synchronization procedure is, of course, the key of the definition of the inertial frame "of the observer". So, to find out what is contemporary to himself at another location B, A measures the time a signal needs to travel from A to B and back. At time $$t^A_0$$ he sends the signal to B, at time $$t^B$$ it arrives at B and is send back, and at time $$t^A_1$$ the signal is back at A. Then, Einstein synchronization prescribes that the event $$(B,t^B)$$ is contemporary to $$(A,\frac12(t^A_0+t^A_1))$$.

But is there a justification for this choice? Why not, say, $$(A,\frac13 t^A_0+ \frac23 t^A_1)$$? Of course, $$(A,\frac12(t^A_0+t^A_1))$$ looks a little bit simpler, but it is obvious that this value could be true only for one particular velocity. If $$(A,\frac12(t^A_0+t^A_1))$$ would be correct for me, it would be incorrect for somebody moving relative to me, say, in direction toward B. Assume he is at A at the time $$\frac12(t^A_0+t^A_1)$$, and has send his signal so that it is mirrored at B at the same event $$(B,t^B)$$. Let's see what happens for him. When I have send the signal to B, he was farther afield from B, thus, for his signal arriving at the same time he had to send it earlier, $$t^{A'}_0 < t^A_0$$. When the light comes back from the mirror, he is closer to B, thus, he will see the signal earlier, $$t^{A'}_1 < t^A_1$$. That means, $$(\frac12(t^{A'}_0+t^{A'}_1) < \frac12(t^A_0+t^A_1)$$, his result for Einstein synchronization differs from my result.

If we look at this from the point of view of the Lorentz ether, there is at least clear what can be used to justify this choice: The Einstein synchronization $$\frac12(t^A_0+t^A_1)$$ would be correct for me if I'm at rest relative to the ether.

In the Minkowski space, the situation is less clear. Here all the inertial frames are on completely equal foot, none of them is correct. This suggests an advantage of simplicity for the inertial frame where he is at rest. But this works only if he moves inertially himself. If he changes his velocity, there is no longer such a preference. The frame where he is at rest now is a different one than the one where he is at rest after some time.