# I Time dilation (something seemingly paradoxical)

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1. Sep 28, 2016

### Joker93

Hello.
Consider the following case:
Two observers, A and B, moving relative to each other with velocity v. For B, it's A that moves (with v) and so DTb=g*DTa (where DT denotes finite time difference and g is/the Lorentz factor gamma). So, (following the same logic as in Morin's Classical Mechanics book at page 515) if say g=5/4 then if A claps once every 4 second then B hears claps every 5 seconds.
Now, for A it's B that moves(with -v) and so DTa=g*DTb. Rearranging we get DTb=1/g*DTa. So, if A claps once every 4 seconds, B hears claps every 15/5 seconds.

So, what is going on here?
I think that I have not properly understood what each DTa and DTb stands for in the above relations.
For example, if in the first relation (DTb=g*DTa) DTb stands for the time difference that he sees in the clock, then what does DTa in the same relation stand for?
Analogously, in the second relation (DTa=g*DTb) if DTa stands for the time difference that he sees in the clock, then what does DTb in the same relation stand for?
Lastly, how is the DTa's and DTb's related?

Please keep in mind that this is confusing me very much both conceptually and mathematically(how to relate the intervals), so don't take anything for granted as I am surely confused about various concepts of relativity(as this question proves).

2. Sep 28, 2016

### vanhees71

Have a look at my intro to special relativity. The Minkowski diagrams should help with understanding the issues with the kinematics of relativity. Usually it's the relativity of simultaneity which makes problems in the beginning since we are used to think in terms of Newtonian spacetime:

http://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf

3. Sep 28, 2016

### Joker93

But here I am only talking about intervals and not the actual time that their clocks show. This is ensured by the analysis with claps.

4. Sep 28, 2016

### vanhees71

Time is what (ideal) clocks show.

5. Sep 28, 2016

### Joker93

But since I am talking about time intervals why would I care if one observers' clocks are not synchronized as viewed by an other observer?

6. Sep 28, 2016

### vanhees71

In Minkowski diagrams the clocks are synchronized by definition. The tics on the world line of the observers in their frames (the time axes in the diagram) are by definition unit time steps.It's simple to read off why there is time dilation and why there is no contradiction possible when comparing the time between events with respect to different inertial observers. Everything becomes clear as soon as you understand that there is no absolute simultaneity of events. Have a look at Fig. 2 (left panel) and read the referring text in my writeup.

7. Sep 28, 2016

### Joker93

But, again, if it's the respective time intervals between claps that concerns me, why would I care if the clocks are synchronized or not?

8. Sep 28, 2016

### vanhees71

Which things should it make to compare times with non-synchronized clocks? Then the relation between these times just depend on how in some arbitrary way each observer defines "time". Physics is about objective well-defined properties of nature. That's why the bureaus of standard take great care to define the units to measure quantities very accurately!

9. Sep 28, 2016

### jartsa

@Joker93

If you use a time dilation formula, then do not calculate how many claps there are in a second, because that thing is not a time, it is something divided by time.

Calculate how many seconds there are between two consecutive claps, because that is time, some number of seconds.

"every 5 seconds" is not a time interval, it is something else, its unit is 1/s.

10. Sep 28, 2016

### Ibix

If your travellers are clapping regularly, that's a clock - just add a counter.

The quick answer is that this is only paradoxical if there is a global notion of time. There is no such thing in relativity.

A pretty close analogy is two cars driving at 30mph down straight roads that diverge at angle θ. If the driver of one car looks at the other he'll see it falling behind because its speed in his direction is only 30cos(θ). But the driver of the other car can make the same argument. Is it paradoxical that the two cars both see the other falling behind? No. They just have different notions of what "forwards" and "level with me" mean.

Similarly, the two travellers have different notions of which direction in spacetime is the future and which directions "now" lies in. Just as you have no problem with both cars seeing the other going slow, there is no problem with both seeing the other's clock run slow.

To relate this to what @vanhees71 said. Clock synchronisation is a way of defining "now", which is why you need to care about it. Minkowski diagrams are an excellent tool for visualising this and I strongly recommend learning about them.

11. Sep 28, 2016

### Staff: Mentor

If you are interested in what each actually hears then you should use the relativistic Doppler formula. Here is my favorite page on that, and it can work for sound also.

http://mathpages.com/rr/s2-04/2-04.htm

OK, so first, this equation is not the right one for determining what they hear, it is for determining the actual coordinate time between claps, as measured by a system of synchronized clocks at rest.

Second, you don't rearrange this equation as you have done. In that equation DTa is the proper time between two events as measured by a single (moving) clock at both events. DTb is the coordinate time measured by a pair of synchronized clocks (at rest) one at each event. You cannot rearrange them like that because they represent physically distinct quantities.

12. Sep 28, 2016

### stevendaryl

Staff Emeritus
First, you don't really mean "hear", because sound takes time to propagate, so A can only hear B's clap a while after B claps. So what does it really mean, operationally, that A considers B time-dilated, and B considers A time-dilated? Well, operationally, if B is moving away from A, A can't know what time B claps unless A has a way of figuring out the time that a distant event took place. If A is right next to B when B claps, then A can just look at her watch to know what time it happened. But if B is far away from A when he claps, then to figure out what time B clapped, A can do one of the following things:
1. Beforehand, A can set up a system of synchronized clocks, and when B claps, A can look through a telescope to see what time it is on the clock right beside B when he claps.
2. Or, A can note the time that she sees B clap, and compute what time he must have clapped by taking into account the travel time for light.
3. Or, maybe A has some other bright idea for figuring it out.
Anyway, considering the first method, A has to be able to synchronize distant clocks. Relativity shows us that if A thinks of the clocks as synchronized, then B will think of them as out-of-synch, and vice-versa. So, A and B disagree about both whose clocks are running slower, and about whose clocks are correctly synchronized. This resolves the apparent paradox of mutual time dilation:

According to A:
1. A has two clocks that are rest relative to A, and they always show the same time.
2. A's clocks run at the normal rate.
3. B claps for the first time just as he is passing the first clock, which shows time t=0.
4. B claps for the second time just as he is passing the second clock, showing time t=5.
So A concludes that B claps once every 5 seconds.

According to B:
1. A's clocks are not synchronized; the first clock is 1.8 seconds behind the second clock.
2. A's clocks run slow by a factor of $\frac{4}{5}$.
3. B claps for the first time as he is passing the first clock, which shows time t=0. The second clock shows time t=1.8.
4. 4 seconds later, B claps for the second time as he is passing the second clock. A's clocks have advanced only 3.2 seconds, so the time on the second clock is t=5 seconds (1.8 + 3.2).
So both A and B agree that A's first clap is when the first clock shows t=0, and both agree that A's second clap is when the second clock shows t=5, but they disagree about whether those clocks are synchronized, and whether they are running at the normal rate, or not.

The second method of determining the time of a distant event, by taking into account of the time for light to travel, will also lead to disagreements about synchronization. The time required for light to travel from B to A depends on whether A is stationary or moving away from B. A thinks of herself as stationary, so she will use one method to calculate the travel time. B thinks of A as moving, so he will use a different method to calculate the travel time. Specifically:

According to A:
1. B's first clap takes place as he passes A, and she can see by her watch that the time is $t=0$.
2. A sees B's second clap when her watch shows 8 seconds. She computes that it took place at time $t=5$, and the light took 3 seconds to travel back to A.
According to B:
1. B's first clap takes place as he passes A, when her watch showed the time $t=0$.
2. B's second clap takes place at time t=4, but it takes an additional 6 seconds for the light to reach A. So the light reaches A at time $t=10$. But since A's watch is time-dilated, it only shows $8$ seconds have passed (Because $\frac{4}{5} \cdot 10 = 8$).
So both agree that the first clap reaches A when her watch shows $t=0$, and that the second clap reaches A when her watch shows $t=8$, but they disagree about the travel time for light, and they disagree about whether A's watch is time-dilated.

13. Sep 28, 2016

### Mister T

Because the two observers are in relative motion. If A claps his hands twice, those two events occur at the same place in A's frame. They do not occur at the same place in B's frame, B observes them happening at different locations. For B to determine the time interval that elapses between them he will need to know the clock reading at each event, so he will need two clocks. One at each location. And he will need to make sure they're synchronized with each other, because otherwise the difference in their readings will be a meaningless number instead of the number that's equal to the time that elapses between the events.

And here's the rub. If B synchronizes those clocks, and A watches him do it, A will point out an error. And it's that error that's responsible for the time dilation symmetry that prompted your original query.

14. Oct 4, 2016