Who is Moving Faster: Planet or Probe?

[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.

 

[weirdoguy]

We will ignore the acceleration period.

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

 

[calinvass]

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]

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.

 

[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?

 

[weirdoguy]

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.

 

[calinvass]

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

But special relativity cannot use acceleration.

 

[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 :)

 

[Nugatory]

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.

 

[Nugatory]

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.

 

[Nugatory]

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.

 

[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 ?

 

[Orodruin]

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.

 

[pixel]

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]

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.

 

[calinvass]

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.

 

[calinvass]

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.

 

[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.

 

[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]

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]

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.

 

[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.

 

[Nugatory]

...the larger the equivalent gravitational potential, hence the larger the difference between the clocks at the end of the experiment?

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

 

[Orodruin]

GR uses SR to create a 4d spacetime.

No it doesn't. GR is a generalisation of the space-time concept to curved space-times including a prescription for how the curvature relates to the energy momentum tensor.

 

[Ibix]

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.

In that case, OP is using "imaginary frame" to mean some frame in which the corner of your lab is moving with some four-velocity. I don't think that "real" or "imaginary" are helpful labels, to be honest (my last post notwithstanding). You can define them directly in terms of actual physical things (Einstein's rods and clocks, for example) or in terms of derived quantities (e.g. the imaginary rods and clocks of a hypothetical observer in inertial motion). They're both either real or not depending on your definition of "real".

 

[A.T.]

A reference frame is something real.

Depends on what you mean by "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.

You can use a material object to define infinitely many reference frames, but that doesn't make all those reference frames material objects.

 

[pervect]

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.

It would be best (not that people conform strictly tho what it's best) to study SR first, and GR second. So the answer to the GR part of your question can and should be put off.

Let's do a variant on your problem to illustrate where there is a potential for confusion. Let's have two identical spaceships in your universe, and no planet at all. We can either have them both at rest at some distance away, and have one of the spaceships accelerate towards the other, or we can have the two spaceships initially in relative motion, whichever seems simpler to you.

In either case, we start considering the time issue when both spaceships are at the same location in space. They compare their watches, and synchronize them. Then time passes, and the two spaceships separate.

The two spaceships cannot compare their watches directly. They have to do it through an indirect process, for instance they might watch the other spaceship through a large telescope and read a large clock, or they might exchange radio signals with encoded timestamps. The visual or radio readings obtained in this manner mean that the observed readings are always out-of-date. Mathematical computations based on some model have to be done to figure out what the clock is reading "now".

The tricky part of the problem is that each spaceship has a different concept of "now". This is frequently misunderstood, and it's hard to even get the initial idea across, as words are very slippery things, and the idea is so strange that people tend to reject the favored interpretation of what is meant by this in termis of some other unintended interpretation. Unfortunately, it's not clear how to fix this problem - you'd think there would be some set of words so precise that they'd prevent misunderstandings, but words that precise tend not to be read and fully understood.

The best way that I know of for conveying the details is to go through a rather elaborate description of exactly what signals are exchanged, and how, in order to compare the clock readings. I'll try that approach, though it's a bit long.

The following is a typical example. Let T be the time at which the two spaceships are co-located. At T+1hr, according to spaceship #1, spaceship #1 sends out a signal which encodes the time, that it is now 1 hour after T. This signal is received at spaceship #2 at T + 2hr, according to spaceship 2's clock. Spaceship #2 immediately sends a signal back to spaceship #1, encoding the fact that it was sent out at T+2hr. This signal is received by spaceship #1 at T+4hr.

You may notice a pattern here, if not I'll point it out. If either spaceship sends a signal out at time X, it is received by the other at time 2x. The factor of 2 here is for ease of exposition - the general rule is that if a spacehip sends out a signal at time X, it is received by the other spaceship at time k*X, where k is some constant that depends on the relative speed. If you'd like more detail on this general approach or justification of it, I'd suggest Bondi's very old book, "Relativity and Common Sense". The general approach is usually called k-calculus, though it only involves algebra, not calculus.

Here is how spaceship #1 interprets these facts. It sent out the signal at T+1hr, and received it at T+4hr, so the signal took 3 hours to get to spaceship #2 and come back. The speed of light is constant, spaceship #1 concludes that at T+2.5 hr, spaceship #2 was 1.5 light-hours away.

Spaceship #1 also concludes that at T+2.5 hr, spaceship #2's clock read 2.0 hours, rather than 2.5 hours, so it must have been slow.

Spaceship #2 can make exactly the same observations, and come to the same conclusion. So at this point , there is no answer to the question of which clock is running faster.

If spaceship #2 turns around and accelerates so that it rejoins spaceship #1, spaceship 2's clock will read the lower time when the reunite. If spaceship #1 turns around and accelerates so as to catch spaceship #2, spaceship #1's clock will read the lower when they re-unite.

The fundamental idea here is that each spaceship has a different concept of "now". This is called the relativity of simultaneity. There is a lot written about this - understanding this is one of the tricky points about understanding SR.

For one paper on the topic (aimed, however, at teachers rather than students), see http://cds.cern.ch/record/571967/files/0207081.pdf, "The Challenge of overcoming deeply held student beliefs about the relativity of simultaneity".

 

[calinvass]

I was confused because I thought that if we use the reference frame in which the planet is at rest and the probe at a velocity the clock on board will tick slower. But if we use the frames the other way round, the clock on the planet will slow down. I thought it didn't make sense. To answer the topic question, we don't need the third frame of reference. We define a frame in which both the planet and the probe are initially at rest. Then we know that the probe will be moving in this reference frame.
If we go to the probe reference frame, the planet will appear as leaving the probe, and clock on board we can suppose is runs slower. But then the probe will need to do something to come back and it will need to catch the planet. This way in the probe reference frame the probe will not be at rest anymore but moving faster than the planet previously moved in the probe reference frame. So we get the same result. We can ignore the acceleration.

 

[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).

Yes, you were right. I was using multiple frames attributed to objects that are always at rest in those frames of reference, except for the clocks and frames moving relative to each other. That generated confusion and contradictory ideas.

 

[GeorgeDishman]

If we go to the probe reference frame, the planet will appear as leaving the probe, and clock on board we can suppose is runs slower. But then the probe will need to do something to come back and it will need to catch the planet. This way in the probe reference frame the probe will not be at rest anymore but moving faster than the planet previously moved in the probe reference frame. So we get the same result. We can ignore the acceleration.

It's a good idea when starting with SR to think primarily of inertial reference frames. You can start with a frame in which the probe is initially at rest and the planet is moving, the planet-based clock therefore ticks at a lower rate in that frame. To catch up though, the probe must then accelerate and move faster than the planet which means that although both clocks are now "ticking slowly" in the chosen frame, the probe's clock is affected to a greater degree. When the probe catches up with the planet, it therefore shows a lesser total elapsed time, by the same amount as would be calculated in the planet's rest frame of course.

 

[calinvass]

There is a different version of a thought experiment without using acceleration by introducing a third object, sincronizing clocks, like in a relay race. Basically, one probe passes by, synchronizes its clock with the planet, then meets another probe, in the opposite direction,same speed, synchronizes clocks again. The third probe, passes by the planet again and they read their clocks.

 

[GeorgeDishman]

That's right, and if you consider the picture from the first probe, the second probe is moving much faster than the planet in order to catch up. It's the same as what I said but avoids the need for an engine burn.

 

[Mister T]

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.

In order for the two clocks to end up at the same place at least one of them will have had to change direction. If only one of the two clocks changed its direction of motion (in your example it's the one on the probe) then that's the one for which a smaller amount of time will have elapsed.