# I Who is moving faster?

1. Oct 18, 2016

### calinvass

Suppose we have a single planet in the entire universe and we send a probe through space. There is a clock on the probe and one on the planet. The probe will fly at a constant speed relative to the planet then it comes back. We will ignore the acceleration period. Which clock will show a lower value ?

My understanding is that it is impossible to tell until the probe gets back, using special relativity , but there will be a difference.

Last edited: Oct 18, 2016
2. Oct 18, 2016

### weirdoguy

You can't ignore the accelereation period! Thats the whole point.

3. Oct 18, 2016

### calinvass

Sorry, i forgot to tell, that we can only use special relativity only for this example. I have modified my post.

4. Oct 18, 2016

### Staff: Mentor

The one on the probe. You should read and try to understand this FAQ before asking any followup questions.
We can calculate exactly what both clocks read throughout the journey. What we cannot do is answer the question "which clock is really behind the other?"; that question is ambiguous as long as the two clocks are not at the same place at the same time, which only happens at thevbeginning and end of thevexperiment.

Last edited: Oct 18, 2016
5. Oct 18, 2016

### calinvass

Actually, we can consider planet frame at rest and the probe is also at rest before leaving. So, special relativity can also be used. Is that correct?

6. Oct 18, 2016

### weirdoguy

That doesn't change anything, still you can't ignore the acceleration.

7. Oct 18, 2016

### calinvass

But special relativity cannot use acceleration.

8. Oct 18, 2016

### weirdoguy

That is not true (a common misconception), you can have accelerated motion in SR. Actually you HAVE TO consider acceleration, because without it the probe can't come back to the planet.

EDIT: So as Nugatory below said, you don't have to, but it's the easiest way. Sorry for confusion :)

Last edited: Oct 18, 2016
9. Oct 18, 2016

### Staff: Mentor

That is common misunderstanding, but the acceleration is in fact irrelevant. One way to see this is to consider that the accelerations are the same no matter how far the probe travels, but the difference between the clocks at the end of the experiment is not. There are also variants of the twin paradox in which there is no acceleration.

What's really going on: the two clocks follow different paths through spacetime; the different paths have different lengths and because they're paths in spacetime not space a different amount of time passes on the different paths. Acceleration only comes into the picture because the easiest way of setting the probe on a different path than the planet is to accelerate the probe.

10. Oct 18, 2016

### Staff: Mentor

Special relativity handles acceleration just fine - it's gravity that it doesn't work with.

Google for "Rindler coordinates" and "momentarily comoving frame" for examples of how SR works with accelerations.

11. Oct 18, 2016

### Staff: Mentor

Special relativity can be used, but not because "we can consider planet frame at rest" - those words don't make sense because a frame is not something that can be moving or at rest.

We can use special relativity because there is no significant gravity involved.

12. Oct 18, 2016

### calinvass

The increase in mass of the probe will also affect its clock by increased gravity in and around the probe according to GR. Is this the same amount we get from SR or it is an additional value ?

13. Oct 18, 2016

### Orodruin

Staff Emeritus
Another common misconception. This is not how gravity works in SR. In fact, SR does not include gravity. The concept of relativistic mass is another remnant from how relativity was historically considered.

14. Oct 18, 2016

### pixel

I'm familiar with those variants, so I accept your statement that acceleration is irrelevant. But can't your argument for proving this - that the accelerations are the same no matter how far the probe travels, but the difference between the clocks at the end of the experiment is not - be countered by use of the equivalence principle and the fact that the further away the probe is when it accelerates, the larger the equivalent gravitational potential, hence the larger the difference between the clocks at the end of the experiment?

15. Oct 18, 2016

### jbriggs444

You seem to be contemplating a "twins paradox" where there is a uniform gravity field in which the Earth is freely falling and which the probe is continuously resisting with its rocket motors or, equivalently, that there is no gravity and the probe is continuously accelerating throughout the scenario. In this scenario, the required acceleration depends on how far the probe goes. For a fixed launch speed, the required acceleration for a longer trip is smaller.

By contrast, Nugatory seems to be contemplating a probe that coasts out, briefly fires its motors and then coasts back. In this scenario, the required acceleration is independent of how far the probe goes. For a fixed launch speed, the same turnaround maneuver will always work, no matter how far away it is performed.

Edit: Though you do not explicitly say it, you must contemplate a gravitational field that covers the entire extent of the trip because you speak of its potential.

16. Oct 18, 2016

### calinvass

I though we need to use SR to calculate the time difference and then wondering whether to use GR to add an extra value by adding gravity effect of its own mass. But, GR uses the spacetime concept which is already based on SR. Only SR is limited to certain problems. I was mixing gravity with SR hoping for a more accurate result, when if fact gravity doesn't even exist in this context, but the spacetime geometry makes it feel as it were a force. GR uses SR to create a 4d spacetime.

17. Oct 18, 2016

### calinvass

What I meant was, if we define a third frame of reference, before the journey, the planet frame and probe frame do not move in respect of this third one. But when after the probe leaves, its frame will be moving relative to the third frame, but the planet frame will not. That is how I understand how you can tell which time will slow down.

18. Oct 18, 2016

### calinvass

The meaning of the question of this topic was actually, if there is an imaginary third reference frame, how do you know which frame is moving relative to the third frame, the planet's or the probe frame, since you can't establish the third frame physically. The answer is, if you don't do anything to the planet, it's reference frame will not move relative the the imaginary one. But, obviously you do something to the probe.

19. Oct 18, 2016

### Ibix

Neither time really slows down. Time as experienced by you is a measure of the length of your path through spacetime, and the planet and probe took different routes with different lengths. That's all there is to it. I think you are just confusing yourself thinking about frames, especially as you are implicitly using non-inertial frames, and talking about "imaginary frames" (which makes no sense - all frames are imaginary, when all's said and done).

20. Oct 18, 2016

### calinvass

Frames can be attributed to real objects even though the frames don't exist as entities. By imaginary frame I mean a frame that has no real object associated with that you can see but you can imagine there is an object out there.

21. Oct 18, 2016

### Ibix

Then the answer to your question is trivial: if you choose a frame in which the planet is moving then it is moving in that frame.

22. Oct 18, 2016

### vanhees71

An ideal clock is showing its proper time. Given it's world line $x^{\mu}(\lambda)$ the proper time between two points is
$$\tau=\int_{\lambda_1}^{\lambda_2} \mathrm{d} \lambda \sqrt{\eta_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}}.$$
Now you can choose, e.g., $\lambda$ as the coordinate time of any reference frame you like, and then you can directly compare $\tau$ with $t=\lambda_1-\lambda_2$.

It is also clear that you can have non-inertial frames in special relativity (as is the case in Newtonian physics).

23. Oct 18, 2016

### Staff: Mentor

A frame is a convention for assigning position and time coordinates to points in spacetime.

They aren't attached to or attributed to objects, although it's easy to form this misunderstanding when you hear things like "my frame", "your frame", "the spaceship frame". We use the sloppy and confusing term "the frame of <something>" informally because the more accurate "using a frame in which the spatial coordinates of <something> are not changing with respect to proper time along its worldline" is just too clumsy to keep on repeating.

24. Oct 18, 2016

### vanhees71

As usual at this point I disagree. A reference frame is something real. You can take the corner of your lab and the clock on the wall as a reference frame as the most simple example. Physics is after all an empirical science and deals with the quantified description of real things you can measure with the adequate measurement devices. That can be a simple yard stick to measure distances up to a ultraprecise interferometer like LIGO to measure tiny distortions of its arms by hitting gravitational waves, but it's a material things that defines reference frames.

25. Oct 18, 2016

### Staff: Mentor

No, because there is no equivalent gravitational potential - we have a non-constant acceleration and that can't be modeled as a gravitational potential.