# Whose time dilates?

1. Oct 8, 2015

### Heatherfield

This question has probably been asked millions of times and me coming back here will be nagging, but the threads I read on here did not seem satisfactory and I hope someone can offer me a quick but personal explanation. My problem is rather classic and asks why one of the twin paradox' twins ages and one doesn't. While I came across this problem reading Brian Greene's Elegant Universe, I have previous experience with Special Relativity with some (very minor) maths background in the calculations concerned.

As proven per thought experiment (and later, of course, per real-life experiment), if one inertial reference frame moves against another, the reference frame "standing still" (I know this frame doesn't exist but for the sake of explaining, bear with me) will experience the other reference frame having a slower "time" (as light's speed is constant but needs more time to move distances in the frame). Thus, if a friend sprinted away from me and then came back, we could say he was a very tiny fraction of a millisecond younger than me. Furthermore, if one twin makes a spaceship journey through the cosmos, the twin will come back to see its other half much older.

What I don't get is how this observation/thought experiment combines with the theory that all reference frames are equal. Thus, one would not be entirely wrong in saying I float away from my sprinting friend, or the planet flies away from the twin's rocket. Other threads argue that in both reference frames, the other ages slower. As per the physics of SR, I would agree with this.

What I don't understand, is how a twin coming back to its other half, or my friend sprinting back towards me could still be decisively younger than me. Doesn't this imply that there is an absolute state of lower velocity?

2. Oct 8, 2015

### Orodruin

Staff Emeritus
No. The travelling twin is changing rest frames and you are not taking relativity of simultaneity into account.

3. Oct 8, 2015

### Staff: Mentor

All inertial frames are equivalent.
The sprinter's frame and the frame of the travelling twin are non inertial.

4. Oct 8, 2015

### Heatherfield

Thank you much for the eloquent response, glad I didn't have to battle through a long story like my own :-)

So would I be correct in explaining the "solution" to my "problem" as such:
- In the Twin Paradox, both twins see the other age more slowly
- Due to the relativity of simultaneity, both "truths" are true
- Once the travelling twin comes back and joins the frame of the non-travelling twin, the "truth" of that frame becomes true, thus the travelling twin is younger?

5. Oct 8, 2015

6. Oct 8, 2015

### Heatherfield

That really put me back into my place :') Thanks for the very clear link. The thing causing my confusion was a passage in Greene's book where he writes about two astronauts passing each other who wouldn't know who ages faster. It does make me wonder (I tried to model this myself, but SR was too long ago to be certain about implementing the gamma factors correctly) about the following:

Say the two astronauts Alice and Bob pass each other and at the moment of passing both set their clocks to zero. Subsequently, after ten seconds or so, Bob sends a light beam in the direction of Alice. At the instant moment Alice this beam of light she sends a radio signal (which is electromagnetic -> travels at light speed) to Bob stating the time on her clock. Would Bob always receive the same radio message back?

Greene does speak of a modified version of this problem but he solves it by hiding most of the Maths from the reader which makes it a really vague part of the book. I really want to understand him before continuing.

7. Oct 8, 2015

### Staff: Mentor

By "at the instant moment Alice this beam of light" do you mean that Alice sends her message at the moment that Bob's light signal reaches her? I've also made an addition in bold above.

Assuming this.... and just for definiteness, let's take Alice and Bob's relative speed to be .6c
- Using Bob's notion of time and position, they pass each other and synchronize clocks at t=0,x=0.
- Bob sends his light signal at time t=10,x=0 (x is zero because Bob considers himself to be at rest while Alice is moving). Alice's position in spacetime is t=10,x=6 (because she's been moving at .6c for 10 seconds).
- Bob's flash of light reaches Alice at t=25,x=15 (after the light had 15 seconds to cover a distance of 15 light-seconds, which is where Alice is after 25 seconds at .6c).

Alice's clock read 20 seconds when the light beam reached her (calculated from the Lorentz transformation which converts Bob's notion of time and position into Alice's, and vice versa). So the message Bob receives from Alice is "My clock reads 20 when the light signal reached me". That message reaches Bob at t=40,x=0 (because the message was sent at time t=25 and spent 15 seconds in flight).

The experiment is symmetrical. If Alice also sends a light signal to Bob when her clock reads 10 seconds, Bob will receive it when his clock reads 20.

8. Oct 8, 2015

### Heatherfield

Thank you! I don't have a pen/paper/calculator at the moment but I am assuming this all still holds when Bob is moving AND sends the light signal?

Edit: google's calculator is better than expected and shows this to be true :-)

Last edited: Oct 8, 2015
9. Oct 8, 2015

### Staff: Mentor

What do you mean, "when Bob is moving"? He IS moving.

He is moving at .6c relative to Alice (that's how I did the calculation that said that the situation is symmetrical so that if Alice were to send a flash of light at her ten-second point, she'd get a reply from Bob that said his clock read 20 seconds), he is moving at several miles per second relative to an observer on Mars who is watching this experiment with great interest, he is moving at .9999c relative to someone living on a rogue planet hurtling through our solar system at a high speed relative to the sun. All of these observers will agree about the time on Alice's clock at the moment that the light signal reaches Alice.

You need to be equally careful about saying that something is "not moving". Any time you find yourself thinking "A is not moving and B is", you can just as easily say that it is B who not moving while A is moving in the other direction.

10. Oct 9, 2015

### Heatherfield

I made my calculations like this. If my reasoning somewhere violates a postulate of relativity please tell, because I seem not to be the most brilliant conformer to relativistic principles (such as, from time to time still thinking there is an 'absolute truth').

When Bob and Alice pass each other in space, they have no way of knowing who is moving (which implies there is an absolute truth to the one of them moving, which seems to absolutely destroy my reasoning right away, but in the small chance my words actually make sense, bear with me). Therefore, the calculations for:
- Bob sending the light beam, under the assumption Alice is moving away from him, thus Alice being younger;
- Bob sending the light beam, under the assumption he is moving away from Alice, thus Alice being older;
Must provide equal results (which I calculated, and which happened to give equal results).

Above, I am using relativity's postulates, but yet I'm also implying there is a truth about one of them moving, which, as Nugatory explained is completely false. Does my above statement still hold?

11. Oct 9, 2015

### A.T.

This is self contradictory. The 2nd postulate says there is no absolute truth about who is moving. Note that this Galilean Relativity was introduced centuries before Einstein:

https://en.wikipedia.org/wiki/Galilean_invariance

12. Oct 9, 2015

### vanhees71

The twin paradox is the one that is most easy to resolve of all the socalled paradoxes in special relativity. A massive body's world line in Minkowski space is timelike, and the time an ideal clock moving with the body is its proper time. Using Minkowski coordinates of an arbitrary inertial frame of reference this proper time is given by
$$\tau=\int_{t_1}^{t_2} \mathrm{d} t \sqrt{1-\beta^2}, \quad \beta=\frac{1}{c} \left | \frac{\mathrm{d} \vec{x}}{\mathrm{d} t} \right|.$$
This means the proper time meausres a Minkowski length of a time-like curve in Minkowski space, and the proper time depends on the specific shape of this world line.

That's analogous to ordinary Euclidean geometry. If you travel from one place A to another place B the travel distance depends on the route you choose.

13. Oct 9, 2015

### Staff: Mentor

You may have been victimized by one of the more common misstatements of time dilation; you'll hear time dilation described as "a moving clock runs slower than one at rest" or some such. That's not right (although you'll hear it even from experts who know what they're talking about - that just shows that you should be cautious about informal explanations not backed up with math no matter who they're from).

A more precise statement would be "a clock that is moving relative to you runs slow compared to a clock that is at rest relative to you". Bob's clock is at rest relative to him while Alice's is not so he finds her clock to be slow while his clock correctly measures the passage of time that he experiences. Alice's clock is at rest relative to her while Bob's is not, so she comes to the exact opposite conclusion.

In this formulation, the notion of which clock is "really" moving disappears - there are just clocks that are moving relative to you and ones that aren't.

Last edited: Oct 9, 2015
14. Oct 9, 2015

### vanhees71

Just as an aside: Here, some authors claim that there are no ideal clocks. I'm not sure, whether I buy it, given the overwhelming accuracy with which time dilation has been verified with unstable particles/nuclei in storage rings, but I also cannot find any obvious flaw in the argument. Perhaps it's fun to discuss this paper (perhaps in a separate thread):

Krzysztof Lorek, Jorma Louko, Andrzej Dragan, Ideal clocks - a convenient fiction, Class. Quantum Grav. 32, 175003 (2015)
http://iopscience.iop.org/article/10.1088/0264-9381/32/17/175003/meta
http://arxiv.org/abs/1503.01025v2

15. Oct 9, 2015

### Staff: Mentor

To make it even more precise, we could say "runs slow relative to you", to make it clear that the definition of "running slow" is also relative.

16. Oct 14, 2015

### 15characters

17. Oct 15, 2015

### 15characters

In fact I think for a beginner the invariant interval is the best place to start, because you can take any two events and convert them to any other "frame" and through doing this in excel you can test scenarios and improve your understanding.

Here's my excel file - the boxes C-E rows 1 - 35 and G to J rows 3 to 11 are useful for playing around. Unfortunately I can't attach it but you can make one yourself link the cells and try different scenerios.

For example if you know the time and distance measurements for a specific observer, you can calculate the invariant interval and then use that to work out the time and distance measurements for another observer moving relative to the first.