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I've read the other questions here on orbiting bodies but don't know how to solve the following problem, so any input would be appreciated.

If Alice and Bob are launched into orbit from their home world (same location and time of launch) but are put into perfectly opposite orbits, for this thought expirment, it is my understanding that they are both in free fall and can be considered in inertial frames of reference.

If their orbits are the same altitude, and same speed (according to an earth viewer) but in perfectly opposite directions, they should both view eachother's time as passing more slowly than their own.

From my understanding can treat eachother, mathematically the same as two observers traveling past eachother in a straight line in non-curved spacetime. Their gravitational field will be the same, since their altitude is, and won't effect how they percieve eachother's time.

please correct me if my setup is wrong up to this point.

Now, Alice and Bob should still be able to communicate via radio as they pass eachother (twice per rotation).

Here's the communication problem I'm stumped with.

From Bob's perspective, he sees Alice as talking slowly back to him, he's able to talk a great deal more in the same amount of time than she is. Ten years go by for him, but only 8 for her. Slow Alice falls in love with slick and fast talking Bob, who falls in love with younger and beautiful Alice.

From Alice perspective, bob is slow talking, and she's the fast one. He ages slower, and she begins to view him as immature. They don't fall in love, in fact she hates him.

From an earth viewer's (easedropping) perspective, they are the same speed, and a third scenario plays out where they treat eachother professionally.

Only one of these three realities can be real.

How can this be reconciled, is it that orbiting bodies in opposite direction can't communicate in this way, or that they actually perceive eachother as being in the same inertial frame while going opposite directions..

Or is there another way to reconcile?

Please go easy on me, I'm just trying to comprehend what I'm missing that I couldn't find in the other questions on here.

EDIT: I've just read the criteria for homework and check-my-work type questions, and have re-read my question. I'm not sure how this classifies as these, as there are absolutely no specific computations. This question only asks about the underlying physics to a thought experiment.

"Homework-like questions and check-my-work questions are considered off-topic here, particularly when asking about specific computations instead of underlying physics concepts"

EDIT: For the sake of clarity, when I say (i.e.) Bob "sees" Alice, I mean what Bob derives, calculates, or inferrs from his frame of reference. This is meant to remove the additional doppler complication while attempting to not fundamentally change the essence of the question. Let me know if I need to rephrase this edit.

Mr. Green
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  • They are not in a inertial frame, during one turn there velocity is opposite only once, the same at another point, and in different directions in other point. just sketch both there velocities during one turn starting in opposite directions at the same point. – trula Sep 09 '23 at 18:45
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    I would never use the word "see" in relation to "time dilation". To get anywhere near the meaning of "see" you need at least to use the Doppler Effect. So, as they approach they both see each other's clocks going faster (yes, as through a telescope!), and as they recede, they both see each others clocks going slower. Too many respondents on this site use the word "see" where it simply does not make sense. IMO time dilation is harmful to learning, and "see" should only be used for light that actually enters someone's eyes. https://en.wikipedia.org/wiki/Relativistic_Doppler_effect – m4r35n357 Sep 09 '23 at 19:27
  • M4r35n357... even with the doppler effect, won't Bob "see" a younger and younger Alice as he passes her with each consecutive orbit? – Mr. Green Sep 09 '23 at 19:38
  • No, they will see each other moving/breathing more quickly, in colours tinged towards the blue end of the spectrum, i.e aging "faster", until they pass, then aging more slowly and red-shifted. In any case you are seeking to reconcile a nonexistent "contradiction". – m4r35n357 Sep 09 '23 at 19:49
  • Due to the doppler effect, right? But they'll still see and interact with a younger Alice and Bob in the moment of passing, right? And after Bob has experienced 10 years of orbiting, he will still see Alice as having experienced less total time (8 years in my example) even if we take doppler into account, right? Or is that wrong? – Mr. Green Sep 09 '23 at 21:02
  • Could you elaborate in an answer of your own how this contradiction is nonexistent? I'm really interested in hearing your complete version of the explanation. – Mr. Green Sep 09 '23 at 21:06
  • You have defined no assymetry, so no. The faster apparent aging as they approach will be exactly cancelled by the slower aging after they pass. I say apparent, because their motion is symmetrical and free-fall, so for each of them (proper) time passes as fast as it possibly can wrt the space they are travelling in. Of course, the circular motion makes all this far more confusing than it needs to be . . . – m4r35n357 Sep 09 '23 at 21:07
  • Ah. Got it. They see doppler effect but the math is not the same as two inertial travelers passing eachother in flat spacetime. Only the classic doppler effect can be observed by Bob and Alice, they don't calculate eachothers time to be slower than their own because their orbits are symmetrical. We can think of all same altitude and same speed orbits as being in the same "speed of time" as eachother, even the ones going in opposite directions. Is this correct? Thanks for your help – Mr. Green Sep 09 '23 at 21:11
  • Not really (globally) the same, but I have only explained the bit where they pass where it is "mostly" (i.e. locally) linear. The "speed of time" (careful with that terminology!) is the maximal proper time for free fall that I mentioned above. I think you have it . . . – m4r35n357 Sep 09 '23 at 21:22
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    I agree that this is a conceptual question, so it's on-topic. – PM 2Ring Sep 10 '23 at 00:56
  • BTW, the relative speed isn't constant, it varies sinusoidally. Also, they'd have to be orbiting a black hole or neutron star to have that much SR time dilation. – PM 2Ring Sep 10 '23 at 02:27
  • PM 2Ring... It varies sinusoidally.. is that because of elliptical orbits or because their orbits aren't geodesic (perceived as a straight line in space time)? I put that in parenthesis because I "think" I'm using that term correctly :) – Mr. Green Sep 10 '23 at 17:42
  • Freefall motion is geodesic in GR. (However, a spacetime geodesic isn't a straight line in space. I have some info on that here: https://physics.stackexchange.com/a/587025/123208 ). But even in plain Newtonian mechanics, with your 2 ships going in opposite directions on a perfect circle, the relative speed varies (following the cosine function). – PM 2Ring Sep 11 '23 at 14:16

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Hint: There is no reason a slow-moving clock can't show the right time. Try setting your alarm clock foeward an hour, then let it run slow. It will show the right time eventually.

Now: At any given moment, Alice's clock runs slow in Bob's instantaneous frame, and vice versa.

When they pass each other, their clocks agree. In Alice's (instantaneous) frame, this is because Bob's slow-moving clock was initially set wrong (as, most likely, was her own). Both clocks have continuously changed speeds, and they happen to agree right now. Likewise with Bob and Alice reversed.

WillO
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  • Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on [meta], or in [chat]. Comments continuing discussion may be removed. – Buzz Sep 11 '23 at 15:58
  • The only reason I bring gravity into the equation is it is required to have circular orbits. If the two inertial travelers were not in circular orbits, than this "problem" doesn't exist. It's not intended to over complicate, but instead to reconcile a special case where my normal reasoning about SR seems to fail me. – Mr. Green Sep 20 '23 at 20:43
  • If I were to take a stab at what feels like the right answer, it's that orbiting bodies are in motion relative to their local barycenter, and that two same-speed-and-altitude opposite moving orbiting bodies would not see any total time dilation after each complete revolution, because they agree on their speed relative to their local barycenter. however I've read that orbiting bodies in freefall can be viewed like inertial frames traveling in flat spacetime. This seems to conflict with my intuition, and thus why I pose this question. thankyou both for your answers – Mr. Green Sep 20 '23 at 20:43