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Naïve reasoning: consider two satellites A and B that are in almost identical but opposite direction orbits, just not colliding. When A meets B, B is going past at a good speed, hence its clock is running slow relative to A. Half an orbit later they meet again, and again B is going past at a good speed, its clock running slow relative to A. But since the situation is symmetrical no clock difference can have accumulated. B’s clock is running slow relative to A in at least two points in the orbit, but it doesn’t accumulate a clock difference relative to A.

How is this explained?


Addendum:

Some answers have already been posted, and it seems they all can benefit from a common description of how clock ticks are communicated from satellite B to satellite A.

To avoid doppler effects and all that stuff, the satellites are assumed to have circular orbits outside the equator, and they communicate optically via a huge relay mirror placed on a some thousand km high pole, at the geographic north pole, like this signalling from a point X on B, to A:

enter image description here

The nice thing about this scheme is that the distance from B to A along the signal path is constant, so there's a constant communications delay: simple!

In order to make sure that special relativity can be considered as a valid approximation for when the satellites pass by each other (this has to do with clock skew in the reference frames), B's clock ticking is communicated not only via the above constant length path, but also directly from the single point X on B's side to closest receiving point on A. A's side is chock full of really tiny densely packed receiving points. As measured on board A, after a receiving point receives a B clock tick from the effectively coinciding point X on B, there is a fixed time delay until that same clock tick is also received via the North Pole Mirror signal path.

The question can be reformulated in terms of the clock ticks that A receives from B: one line of reasoning (e.g. considering receipt at the mirror) dictates constant spacing, while another line of reasoning, using special relativity as a valid approximation when A and B meet, says that at those occasions A will see longer intervals between the received ticks.

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  • $\begingroup$ If the paths are different, why are you surprised that the ticks are not synchronized? Did you try sketching a space-time graphic similar to the one I linked in a comment below? Maybe your question would benefit being rephrased more precisely. $\endgroup$ – user130529 Dec 27 '16 at 18:19
  • $\begingroup$ @claudechuber: It's an incorrect understanding that this question presupposes one conclusion. I wouldn't have asked it if I had a ready conclusion. I think you're wrong that the question could be more precise, but by adding the description of a measurement scheme, which doesn't change the physics, I realized that it could be way more clear in the sense of not leaving so much room for hand-waiving in answers. ;-) Unfortunately, with less room for imagined unspecified effects, the current answers look less satisfying to me. :( $\endgroup$ – Cheers and hth. - Alf Dec 27 '16 at 22:50
  • $\begingroup$ yes, why not adding your scheme, and telling what is the point that you do not understand? As it is, I don't understand exactly what you do not understand (as far as you said that you perfectly understand the symmetry of Lorentz equations, which is all it seems about). $\endgroup$ – user130529 Dec 28 '16 at 8:08
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    $\begingroup$ BTW, would you have the same problem with the simplified (and meaningless) variant of the twin paradox where the twins start both flying in opposite directions, both decelerate, change direction and come back home (all this done symmetrically), where they observe ... that nothing has happened, they are still exactly the same age? $\endgroup$ – user130529 Dec 28 '16 at 8:46
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Special relativity applies to objects which are at rest in an inertial reference frame — that is, not accelerating. Objects which are moving relative to each other at constant velocity may have exactly one closest approach; afterwards they are moving away from each other for ever.

When your two satellites pass each other, there'll be a period of time when you may neglect the curvature of their orbits and analyze their clocks using special relativity. However that period of time does not extend for a half-orbit until their next interaction: if you neglected the satellites' accelerations, they would never meet again. So the symmetry must be restored if you treat the problem using general relativity.

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  • $\begingroup$ Re “there'll be a period of time when you may neglect the curvature of their orbits and analyze their clocks using special relativity”, yes that's what I thought originally. But the measurement scheme now described in the question, after you answered, really sharpens the apparent contradiction that the question is about. The local measurements of clock ticks when B passes by A, logically need to be the same rate, as measured in A, as the clock ticks communicated via the fixed length signal path. And the latter needs to be N ticks per orbit, exactly the same as A. How is this explained? $\endgroup$ – Cheers and hth. - Alf Dec 27 '16 at 23:07
  • $\begingroup$ Now that you've involved a signal relay out of the orbital plane, your entire experiment takes place in the Schwartzchild/Kerr metric around the Earth and you can't use special relativity at all. An explanation is beyond my GR skill, sorry. $\endgroup$ – rob Dec 28 '16 at 1:20
  • $\begingroup$ As I see it nothing can change by just adding measurements, which I did. This isn't a quantum mechanics problem where measurements can affect the experiment. So whatever analysis you did, if you did, if your answer was correct then it is still correct with the measurements added, but if it was wrong, then it is still be wrong but perhaps now more easy to recognize as wrong. $\endgroup$ – Cheers and hth. - Alf Dec 28 '16 at 1:28
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    $\begingroup$ You're right, nothing has changed: the Earth's gravitational field is an integral part of your thought experiment, so you can't use special relativity. Remove gravity, let the satellites travel in straight lines, and the "paradox" goes away. $\endgroup$ – rob Dec 28 '16 at 7:00
  • $\begingroup$ Uhm, please correct my understanding of your response here. I hear you as saying (1) this answer is incorrect because SR does not apply as an approximation for the satellite encounters in this case, since gravity is involved, and (2) that argument applies also to satellites orbiting other bodies, such as the Sun (which we're orbiting), or the Milky Way galaxy (which we're orbiting), which means essentially, SR is out as an approximation for encounters of any objects in free fall. It feels lonely without SR. $\endgroup$ – Cheers and hth. - Alf Dec 28 '16 at 18:09
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This is a variation on the well-known (apparent) twin paradox. The general description of these kind of problems is a s follows: You have worldlines $C_1$ and $C_2$ which meet at events A and B. Both travelers reset their clocks at A and you want to know which clock is running late (or fast) at B, where they meet again. In order to solve such a problem you "simply" calculate the proper time of each traveler along her own worldline and compare them.

$$\Delta\tau = \int_{C_1} \, d\tau- \int_{C_2} \, d\tau$$

In general these line integrals are calculated using the metric tensor (thus implicitly taking into account spacetime curvature, if present)

$$\int_C \, d\tau = \int_C \sqrt{-g_{\mu\nu} \; dx^\mu \; dx^\nu}$$

where the metric has signature $(-+++)$.

In your case - ignoring the spinning of the planet - the situation is symmetric, so the proper time along $C_1$ and $C_2$ is the same.

If the planet is spinnng, then the proper time on the prograde and retrograde orbit is in general different.

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  • $\begingroup$ I wouldn't say it is a variation of the twin paradox, unless you modify the latter to the simplified and meaningless variant where the twins start both flying in opposite directions, both decelerate, change direction and come back home (all this done symmetrically), where they observe ... that nothing has happened, they are still exactly the same age. $\endgroup$ – user130529 Dec 28 '16 at 8:15
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    $\begingroup$ @claudechuber that's exactly why I call it a variation of the twin paradox! In the normal twin paradox you have a twin in free fall (inertial movement), while the other is moving on an accelerated world line (since it's a loop), in flat space. Here you have 2 twins both in free fall (following geodesics), but the space is curved so the geodesics meet again after some time (half way thru the orbit). We use the same techniques (which I have shown and also mentioned in the link I posted) to solve the problem: compute the proper time along each geodesics. $\endgroup$ – magma Dec 28 '16 at 11:35
  • $\begingroup$ Yes, but this variation is trivial, in the sense it looses all the "interest" inherent to the original variant. $\endgroup$ – user130529 Dec 28 '16 at 14:28
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At any given instant, $A$'s clock is running slow in $B$'s (instantaneous inertial) frame.

At any given instant, $B$'s frame is not the same as it was an instant ago, and therefore $B$ changes his mind from one instant to the next both about how long it's been since the clocks were synchronized and about how fast $B's$ own clock (as well as $A$'s own clock) was running at various times in the past.

By the time $A$ and $B$ come back together, $B$ says:

There's $A$. His clock is running slow at the moment. It has also run slow by various factors at various times since we last met. My own clock has also run slow by various factors at various times since we last met. As a result, both of our clocks are ``incorrect'' in the sense that the time that (according to my current frame) has passed since we synchronized to zero is different from the time currently showing on both of our clocks.

You can, of course, quantify this, reconstructing (from $B$'s point of view at the instant of the reunion) exactly how slow each clock was running at each point in the past, and verifying that the total slowdown on one clock is equal to the total slowdown on the other. But of course, you already know by simple symmetry considerations how this is going to turn out.

Edited to add: (No new ideas in this addendum, just a bit more mathematical detail)--- Take the radius of the earth to be $1$, and suppose both satellites travel at speed $v$ with respect to an earthbound observer I will call Jack.

Then according to Jack: At time $t$, satellite $A$ is over the point $(\cos(vt),\sin(vt))$ while satellite $B$ is over the point $(\cos(vt),-\sin(vt))$. They synchronize their clocks to $0$ at time $0$, when they are both over the point $(0,1)$.

When the satellites pass each other again (according to Jack) at time $\pi/v$ and location $(0,-1)$. According to either satellite, whose velocity with respect to Jack is $(v,0)$, this event takes place at time

$$T_0={\pi\over v\sqrt{1-v^2}}$$

(That is, this expression is, according to either satellite, the time interval between their first crossing and their second crossing.)

But at the event of the second crossing, the time shown on either satellite's clock is the length of the path $t\mapsto (t,\cos(vt),\sin(vt))$, which is

$$T_1=\int_{0}^{\pi/v} (1-v^2)dt={\pi(1-v^2)\over v}=(1-v^2)^{3/2}T_0 < T_0$$

Thus, at the moment of their second passing, each satellite says

We synchronized our clocks $T_0$ minutes ago, but now our clocks both show time $T_1$, which is less than $T_0$. That's because both of our clocks have been running slow, by different amounts at different times. At this particular moment, my own clock is keeping perfect time, but his is running slow by a factor of $(1-v^2)/(1+v^2)$.
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  • $\begingroup$ I'm sorry, I'd forgotten about this question. In a commentary exchange I promised to post the conceptually simple solution. Happily some mod has apparently deleted or moved that comment! :) Anyway, the key to resolving this, for me, is to look not at "running slow" or "running fast", which are invalid simplifications in this context, but rather at clock skew, progressively larger difference in time as the distance gets greater, in two inertial systems with some relative speed. I like to visualize the two system's time-denoted axes moving by each other. Again, sorry for forgetting about it. $\endgroup$ – Cheers and hth. - Alf May 30 '17 at 23:43
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It is a standard observation in special relativity that for A and B having constant velocity relatively to each other (and no acceleration), both A and B experiment the same thing and see the other's clock running slower: the situation is perfectly symmetric. Here, in your case, the situation is similar, the symmetry is perfect and thus each time A and B cross each other, their clocks will be perfectly synchronized.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind Dec 28 '16 at 4:16
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If we peel this back to the original heart it comes down to this: Take 2 spacecraft orbiting a single mass in opposite circular directions and both experiencing zero g.

There appears to be a general consensus that as the ships pass each other they will both see the others' clock passing time slower than their own clock. Let them synchronise their clocks in the instant they pass. Clearly, that being the case, a moment later both ships would observe the others' clock to be running very slightly behind their own. Yet there appears to be a general consensus that when the ships next pass they will each observe their clocks to be synchronised once again. All statements allowing for Doppler effects of course.

So... if A is to observe B's clock catch up up with its own there must be a period during which A will observe B's clock to be running faster than its own. Similarly if B is to observe A's clock catch up up with its own there must be a period during which B will observe A's clock to be running faster than its own. In this scenario both ships are experiencing zero g so we're left only with the ships relative velocities - and that difference drops momentarily to zero when the 2 ships are opposite one another so that's not going to resolve the apparent paradox.

The only answer I can come up with relates to each ship's observation of the other ships' orbit.

I think they will each observe the other ship to be in an elliptical orbit. Furthermore, as they each carry out their multiple mid-orbit calculations to monitor the other ships position and "clock time when the light left the other ship" they will indeed observe the other clock to be passing time faster than their own peaking at the point where they are opposite one-another. At that point they will both observe the other to be further from the object being orbited than they themselves are.

Reasonable?

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  • $\begingroup$ "If if A is to observe B's clock catch up up with its own there must be a period during which A will observe B's clock to be running faster than its own." --- This is exactly wrong!! $A$ must in fact always see $B$'s clock running slower than his own. But $A$ is also constantly revising his opinion of his own clock's past speed. So at any given moment $M$, $A$ always says "B's clock is currently running slower than mine" --- but $A$ can also say "At the previous moment $N$, $B$'s clock was running faster than mine" --- though $A$ did not believe this at moment $N$. $\endgroup$ – WillO May 30 '17 at 21:51
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The correct answer is, both clock would tick at identical speed, since their absolute speed is identical (7.9 km/sec). Absolute speed is the one making the clocks tick slower, not relative speed. Special Relativity is wrong. It produces usable results only when you compare high-speed vs low-speed objects, where absolute and relative speeds are similar.

In your case the objects have identical orbital speed, and orbital speeds are absolute. You don't need another object to measure orbital speed, so rotational speed doesn't need relativity (it's absolute). Hence, using rotation is the easiest way to refute relativity.

This doesn't mean Lorentz transformation is wrong. It's correct, but has to be used with absolute speeds, not relative. In other words, clocks on satellites tick slower than clocks on Earth's surface.

Why? Because satellites orbit the Earth at 7.9 km/s to keep their orbit, while surface of the planet, at equator, travels at 0.46 km/s. So you have a difference of 7.4 km/s between the surface and satellite, which makes satellite clock slower.

However, satellites have the equally ticking clocks since they all travel at the same speed. This is why time difference won't accumulate. There is no time-dilation between the satellites. Their atomic clocks tick at the same rate, because satellites have the same speed.

However, relativists refuse to accept this simple truth and are ready to introduce the most complex explanations and formulas to attempt to keep the dead theory alive. It's not possible. Special Relativity is wrong, but Lorentz transformations using absolute speeds are still used to adjust GPS devices and other measurements.

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  • $\begingroup$ Uhm, I see nothing but assertions here. As stated in a comment elsewhere, in my view (arrived at after posting the question) “the key to resolving this, for me, is to look not at "running slow" or "running fast", which are invalid simplifications in this context, but rather at clock skew, progressively larger difference in time as the distance gets greater, in two inertial systems with some relative speed.” So, it's not really a big mystery to me, any more. Still, even if SR works out for this problem, it doesn't work out for e.g. a closed spherical universe. So you're right to be skeptical. $\endgroup$ – Cheers and hth. - Alf Jan 22 '18 at 11:56
  • $\begingroup$ What is this "absolute speed" you keep referring to? $\endgroup$ – JMac Jan 22 '18 at 13:25
  • $\begingroup$ Champeney and Moon experiment is perfect demonstration of correctness of said above. iopscience.iop.org/article/10.1088/0370-1328/77/2/318/meta Even inertial observers, if they choose a frame, in which they move with equal velocities, will not measure any dilation of each other clocks. mathpages.com/home/kmath587/kmath587.htm Below diagram: "Since both emitter and receiver have the speed v relative to this system of reference, there is no differential time dilation." That "I am slower than you, you are slower than me" is the masterpiece of nonsense even in relativity. $\endgroup$ – Albert Jan 22 '18 at 14:24

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