# SR question about time dilation/space contraction

surfa808
Hi, I have a question that's been bugging me. For simplicity, assume everything is inertial.

Let's say I take off in a spaceship from earth towards a distant galaxy at .99999999C - so basically, the trip will be very fast, and let's say in one week time in my reference frame, it will appear to me that I've passed the galaxy.

But since these are symmetric inertial frames, isn't this equivalent to the galaxy traveling towards my spaceship at .99999999C? So if that's true, then in the frame of the distant galaxy, the galaxy will also pass my ship in one week's time.

But then this doesn't make sense: from the galaxy's perspective light from earth would appear to take billions of years to get there. So how is it that the spaceship can get there in less time than the light, which travels faster? Where is the error in my thinking?

Mentor
But since these are symmetric inertial frames, isn't this equivalent to the galaxy traveling towards my spaceship at .99999999C?

Yes.

So if that's true, then in the frame of the distant galaxy, the galaxy will also pass my ship in one week's time.

No, for two reasons. First, you're forgetting the relativity of simultaneity; second, you're not applying the time dilation factor correctly.

Take the second point first, because it's easier, but also because it will make the effect of the first one even more dramatic. In your spaceship's frame, the trip from Earth to the galaxy takes 1 week; but since in that frame, the galaxy is moving at .99999999C, the time elapsed for the galaxy will be 1 week times the time dilation factor for v = .99999999C, which is some very small number. In other words, if we just apply the time dilation factor, without considering anything else, we will conclude that, from your spaceship's point of view, far *less* than a week elapses for the galaxy while the ship is traveling from Earth to the galaxy!

Now let's go back to the first point. When you ask how much time elapses for a given observer between two events, you have to make sure you have correctly defined *which* events, because of the relativity of simultaneity. First let's work the numbers in the Earth-galaxy frame (we assume the Earth and the galaxy are mutually at rest). In this frame, we'll let the origin be the event at which your spaceship leaves Earth, so that event has coordinates $(t, x) = (0, 0)$; call this event O. The galaxy, at this instant of time in the Earth-galaxy frame, is at $(t, x) = (0, D)$; call this event G. The event at which your spaceship arrives at the galaxy, which we'll say is a distance D away, is at coordinates $(t, x) = (D/v, D)$, where v is the velocity of your spaceship relative to Earth (or the galaxy, it's the same thing); call this event S. (A light beam emitted from event O will arrive at the galaxy at event L, with coordinates $(t, x) = (D, D)$; so the spaceship arrives after the light beam, as expected.)

Now, what do things look like in the spaceship's frame? We'll keep the origins the same, so event O is at $(t', x') = (0, 0)$. Event S, where the spaceship arrives at the galaxy (or rather, where the galaxy "arrives" at the spaceship, in this frame) will be at $(t', x') = (D \sqrt{1 - v^2} / v, 0)$, where we can see that the elapsed time has been greatly reduced by time dilation. But what about event G? The Lorentz transformation gives $(t', x') = (- D v / \sqrt{1 - v^2}, D / \sqrt{1 - v^2})$. Note that the time t' is a very large *negative* value here, and the distance x' is a very large *positive* value. In other words, in the spaceship frame, event G occurs a very, very long time *before* event O, i.e., *before* the spaceship leaves Earth!

To see how much time elapses for the galaxy in the spaceship frame during the "trip" from Earth to the galaxy, we need to find the x' coordinate of the galaxy at $t' = 0$ (i.e., at the time the ship leaves Earth, in the spaceship frame). Call that event on the galaxy's worldline G1. The easiest way to find event G1's x' coordinate is to work backwards from event S, where we know the galaxy was at $x' = 0$. Since the galaxy is traveling in the minus x' direction at speed v, it will travel a distance $x' = v t'$ in a time $t'$. So at $t' = 0$, the galaxy will be at $x' = v t'(S)$, or v times the t' coordinate of event S, which gives for event G1 $(t', x') = (0, D \sqrt{1 - v^2})$.

Now, what are the coordinates of event G1 in the Earth-galaxy frame? The inverse Lorentz transformation (i.e., just use -v instead of v) gives for event G1 $(t, x) = (D v, D)$. In other words, in the galaxy's frame, event G1, the event that is simultaneous, in the spaceship frame, with the spaceship leaving Earth, is only a very short time before event S, where the spaceship *arrives* at the galaxy! In fact, we can see that the elapsed time, in the galaxy's frame, from event G1 to event S is $D \left( 1/v - v \right) = D \left( 1 - v^2 \right) / v$, which is, in fact, the time elapsed in the spaceship frame from event O to event S (i.e., the elapsed time for the Earth-galaxy trip in the spaceship frame), times the time dilation factor. So everything is fully consistent with the answer given above to the second point: from the spaceship's point of view, far *less* time elapses for the galaxy during the trip, than for the spaceship. But that's because the relativity of simultaneity means the "elapsed time" is being evaluated from event G1 to event S, instead of from event G to event S; the latter is a much *longer* elapsed time (and that shows up in the spaceship frame as the very large *negative* time coordinate of event G, as above).

Here's a summary of the key events and their coordinates, in the Earth-galaxy frame and the spaceship frame:

$$\begin{table} \centering \begin{tabular}{ l c c } Event G & (0, D) & (- D v / \sqrt{1 - v^2}, D / \sqrt{1 - v^2}) \\ Event O & (0, 0) & (0, 0) \\ Event G1 & (D v, D) & (0, D \sqrt{1 - v^2}) \\ Event S & (D/v, D) & (D \sqrt{1 - v^2} / v, 0) \\\ \end{tabular} \end{table}$$

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