# Twin paradox - easy question

1. Nov 15, 2013

### Nikitin

Hey! So the formula for time dilation is:
$$\gamma \Delta t_0 = \Delta t$$, where $\Delta t_0$ is the time elapsed for a traveller $S'$ who has a relative constant velocity $u$ compared to a stationary observer $S$, while $\Delta t$ is the time elapsed for the stationary observer.

However.. Since everything is relative, you could also say that it's the other way around: $S$ is "travelling" with a constant velocity $-u$ compared to the "stationary" $S'$, and thus the foruma for time elapsed is also $$\gamma \Delta t = \Delta t_0$$

Uhm, which one is correct? I assume it is the first one, provided $S'$ was accelerated from a relative speed of $0$ to $u$? Or are both correct, depending on the observer?

I'm asking because I bumped into this problem when I was figuring out the lorentz-transformations.

2. Nov 15, 2013

### HallsofIvy

If you start with each in motion relative to the other, both are correct. Each correctly sees the other as being younger. The point you are missing is that to have a "twin paradox" they cannot both continue moving at a constant velocity with respect to each other. One or the other, or both, has to stop and turn around (if both stop and turn around symmetrically, there will be no paradox- they will both see the other as the same age when they are back together again) so that they can be together in the same coordinate frame. If one accelerates, moves away from the other, who does not accelerate, then accelerates again to come back, that one will be younger.

3. Nov 15, 2013

### nitsuj

considering the other detail in your post, I'm surprised geometric symmetry isn't mentioned.

That's the term for those particular situations.

4. Nov 15, 2013

### Staff: Mentor

In any inertial reference frame a moving clock runs slower than a stationary clock. Since changing reference frames can change which clock is moving and which is stationary it also changes which is slow and which is normal.

5. Nov 15, 2013

### Nikitin

So are both formulas in the OP correct at the same time? Say S was standing still while S' was travelling in a high-speed train: both formulas apply, right?

But what I'm confused about, is that they seem to be contradicting each-other..

6. Nov 15, 2013

### Psychosmurf

Your equations are wrong. For S':

$$\Delta t = \gamma \left( {\Delta t' - v\Delta x'} \right)$$

While for S:

$$\Delta t' = \gamma \left( {\Delta t - v\Delta x} \right)$$

Where the primed coordinates are the measurements in the primed frame and the unprimed coordinates are measurements in the unprimed frame.

Also, plug in the formulas

$$\Delta x' = \gamma \left( {\Delta x - v\Delta t} \right)$$

$$\Delta x = \gamma \left( {\Delta x' - v\Delta t'} \right)$$

and you'll see that there are no contradictions.

EDIT: Oh I think I see where the confusion is. Basically, if two events occur some time apart in the S frame but at the same place, then they must be in different places in a moving S' frame. Omitting the $$\Delta x$$ 's makes it look like there's a contradiction.

Last edited: Nov 15, 2013
7. Nov 15, 2013

### Staff: Mentor

Suppose that A and B are two inertially moving clocks. In A's frame B will be running slow. In B's frame A will be running slow. The two different statements do not contradict each other. They refer to completely different comparisons. The first one compares the coordinate time for A to the proper time for B. The second one compares the coordinate time for B to the proper time for A.

8. Nov 15, 2013

### yuiop

Generally speaking the Lorentz transformations for time intervals compare the coordinate time interval ($\Delta t$) in one reference frame (S) to the coordinate time interval ($\Delta t'$) in another reference frame (S') for the same pair of events.

The proper time interval between two events $(\Delta t_0)$ is equivalent to the time measured by a single clock that is present at both events. There is only one unique reference frame where the velocity of that single clock is zero and $\Delta x = 0$ and it is only in that unique reference frame that the coordinate time interval ($\Delta t'$) is equal to the proper time interval $(\Delta t_0)$ and it is always true that $\Delta t_0 <= \Delta t$. Therefore there is no symmetrical relationship between proper time and coordinate time.

The coordinate time interval measured by S is greater than the time interval that elapses on a clock that is at rest in S' such that $\Delta t >= \Delta t'$ and the coordinate time interval measured by S' is greater than the time interval measured by a clock that is rest in in S such that $\Delta t' >= \Delta t$. In this case a symmetrical relationship exists, but it is important to be aware that in this case, we are not comparing the intervals between the same pair of events.

You need to be clear in your mind about the difference between proper and coordinate time intervals. Sometimes, some texts imply that the primed frame refers to the frame that measures proper time and length, but that is not always true and you need to be aware of that.

Yes, the first one ($\gamma \Delta t_0 = \Delta t$) is correct (and the second one is not correct) if by $\Delta t_0$ you mean proper time.

Last edited: Nov 15, 2013
9. Nov 15, 2013

### Staff: Mentor

These formulas are not entirely correct; the sign of the $v$ term has to be different for the two "directions" of formulas (primed to unprimed, vs. unprimed to primed). The usual convention is that the sign of the $v$ term is negative when going from unprimed to primed, i.e., if the primed frame is moving at $v$ relative to the unprimed frame, then

$$\Delta t' = \gamma \left( {\Delta t - v\Delta x} \right)$$

$$\Delta x' = \gamma \left( {\Delta x - v\Delta t} \right)$$

But then the primed to unprimed formulas must be

$$\Delta t = \gamma \left( {\Delta t' + v\Delta x'} \right)$$

$$\Delta x = \gamma \left( {\Delta x' + v\Delta t'} \right)$$

Note the plus signs on the $v$ terms. To see why, consider what the motion of the spatial origin of the primed frame, $x' = 0$, looks like in the unprimed frame.

10. Nov 15, 2013

### ghwellsjr

Just to avoid confusion, you should make it clear that pair of events are for an inertially moving object or clock (like DaleSpam said).

11. Nov 15, 2013

### ghwellsjr

Let's do an example where the twins start off colocated and then one of them departs with a speed of 0.6c. At this speed, gamma is 1.25. I'm not going to try to interpret your formulas, instead, I'm going to advise you to stick with the Lorentz Transformations and to draw some spacetime diagrams which I will do here for you.

The first one shows the stationary observer in blue and the traveler in red. The dots represent one-second intervals of time for each twin. The question is, what do they actually observe? To show this, I draw in thin red lines to show the images of the traveler's clock ticking off one-second intervals of time:

As you can see, the stationary observer sees his own clock going twice as fast as the traveler's clock even though the traveler's clock takes 1.25 seconds of Coordinate Time to progress through each second of his Proper Time.

Now if we draw in what the traveler sees of the stationary clock, he sees his clock going twice the rate of the stationary clock but the Coordinate Times are still for the stationary observer:

However, we can transform to the frame in which the traveler is at rest while he is traveling:

And now the Coordinate Times indicate the gamma effect for the traveler.

But this raises the question of how does each twin determine the gamma effect for the other twin? The answer is by using radar signals in which each twin sends a signal at the speed of light to the other twin which reflects off of him along with the time on the other twin's clock and when he sees these, he can assume that it took the same time for the radar signal to get to the other twin as it did for the signals to get back and from that he can calculate when the other twin's clock was at a particular time relative to his own clock. In other words, he averages the time he sent the radar signal with the time he received the reflection.

Here is a diagram to show the process for the stationary twin:

If you average 1 and 4, you get 2.5 as the time his twin's clock was at 2 and if you average 2 and 8, you get 5 as the time his twin's clock was at 4. This allows him to construct the same diagram that we are looking at so that he can determine the value of gamma.

Similarly, the traveling twin can do the same thing and we can show it in the stationary observer's rest frame:

But we can also transform to the traveler's rest frame:

And now we see the Coordinate Times reflecting what the traveler calculated.

Does this all makes sense and clarify all you issues?

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12. Nov 15, 2013

### Nikitin

ahh, I see. So they apply in two completely separate systems? OK I'm starting to understand.

I appreciate all the answers given here (especially the very thorough one by ghwellsjr)!

thanks :)

Last edited: Nov 15, 2013