# Who is moving faster?

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## Main Question or Discussion Point

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.

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

You can't ignore the accelereation period! Thats the whole point.
Sorry, i forgot to tell, that we can only use special relativity only for this example. I have modified my post.

Nugatory
Mentor
Which clock will show a lower value ?
The one on the probe. You should read and try to understand this FAQ before asking any followup questions.
My understanding is that it is impossible to tell until the probe gets back, but there will be a difference.
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.

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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?

Sorry, i forgot to tell, that we can only use special relativity only for this example.
That doesn't change anything, still you can't ignore the acceleration.

That doesn't change anything, still you can't ignore the acceleration.
But special relativity cannot use acceleration.

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 :)

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

m4r35n357, calinvass and weirdoguy
Nugatory
Mentor
But special relativity cannot use acceleration.
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.

calinvass
Nugatory
Mentor
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?
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.

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 ?

Orodruin
Staff Emeritus
Homework Helper
Gold Member
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 ?
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.

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.
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?

jbriggs444
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2019 Award
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?
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.

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.
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.

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.
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.

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.

robphy
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).

calinvass
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).
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.

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.

vanhees71
Gold Member
2019 Award
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).

Nugatory
Mentor
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.
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.

Dale and SiennaTheGr8
vanhees71