# Time dilation and reference frame in an orbiting context [closed]

I have looked for other answers, but most confused me more than clarified and did not provide specific calculations. So I am still struggling with this simple question.

Consider two persons, $P_a$ and $P_b$.

From $P_a$'s reference, $P_b$ is orbiting around him at speed $v$. But from $P_b$'s reference it is the other way around, and $P_a$ is orbiting around him at speed $v$

As I understand, from $P_a$'s reference point, $P_b$'s clock is ticking slower, more specifically, $t_b = t_a \sqrt{1 - v^2/c^2}$.

However, from $P_b$'s reference it is the other way around, and he measures $t_a = t_b \sqrt{1 - v^2/c^2}$.

Assume that both clocks started at 0.

Now here is my question:

After X years from $P_a$'s perspective, they decide that they will be both at the same speed.

There are two ways to do it. Either $P_a$ will accelerate to reach $P_b$ or $P_b$ will accelerate to reach $P_a$.

Both of them expect to find that the time has passed slower for the other one. What will happen when they meet? Does it make a difference who reaches the other person? Please, provide specific calculations for the answer!

• John Doe: "From $P_a$'s reference, $P_b$ is orbiting around him at speed $v$." -- Is it therefore understood that $P_a$ is a member of an inertial system (whose members are all at rest to each other, and who determine fixed distance relations between each other, allowing them to actually determine the value of $P_b$'s speed with respect to $P_a$ and all members of this inertial system)? "But from $P_b$'s reference it is the other way around" -- If $P_a$ is a member of an inertial system then $P_b$ cannot be likewise supposed to be a member of an inertial system while orbiting $P_a$. – user12262 Oct 17 '15 at 16:07
• @user12262 from $P_a$'s perspective, $P_b$'s is moving inertially, and from $P_b$'s perspective, $P_a$ is moving inertially. – John Doe Oct 17 '15 at 16:08
• John Doe: "From $P_a$'s perspective, $P_b$'s is moving inertially, and from $P_b$'s perspective, $P_a$ is moving inertially." -- Being a member of an inertial system is not a matter of someone else's perspective, but to be determined intrinsically, properly. And by the definition how to do that, and how to determine who is "orbiting" it follows that the setup description is inconsistent. p.s. The first version of my first comment happened to be formatted so unfortunately that I cannot even delete it anymore. Would someone please take care of that. – user12262 Oct 17 '15 at 16:16
• @user12262 A is moving around B from B's perspective. And B is moving around A from A's perspective. They can't tell which one is the "inertial system" of reference. For the sake of the argument, both may be accelerating we just don't know. – John Doe Oct 17 '15 at 16:20
• John Doe: "A is moving around B from B's perspective. And B is moving around A from A's perspective." -- Oh, like Calcutta is moving around the Easter Islands "from the Easter Islands's perspective", and vice versa? "They can't tell which one is the "inertial system" of reference." -- It seems that you are yourself not quite sure how to tell that; which eventually makes your question problematic, as it stands. Because one way to say that A is (as good as) a member of an inertial system is to quantify B's motion with respect to A (and those at rest wrt. A) by a speed value $v$. – user12262 Oct 18 '15 at 5:49

From As point of view (which may or may not be an inertial frame), it looks like B is going in a circle, thus A concludes that either 1) they (A) is not moving inertially or else 2) they (A) is moving inertially and hence the laws of physics apply and so B must be accelerating and hence feeling a nonzero force. In which case B is not moving inertially (since B is accelerating).

From Bs point of view(which may or may not be an inertial frame), it looks like A is going in a circle, thus B concludes that either 1) they (B) is not moving inertially or else 2) they (B) is moving inertially and hence the laws of physics apply and so A must be accelerating and hence feeling a nonzero force. In which case A is not moving inertially (since A is accelerating).

So we conclude that at most one is moving inertially.

If neither is moving inertially we don't have enough information.

If one is moving inertially, the laws of physics hold in that frame and hence the other one has its clock run slower.

Done.

• Timaeus: "[...] So we conclude that at most one is moving inertially." -- What do you suggest is meant by "speed $v$" which is prescribed in the OP? Is "speed of $B$ with respect to $A$" defined if $A$ is not a member of an inertial system? – user12262 Oct 17 '15 at 17:08
• @user12262 Yes. You can describe kinematics in a non inertial frame. If you assign positions and times to events you can compute velocities and hence speeds. Just don't expect Newton's laws to hold. But that is dynamics not kinematics. – Timaeus Oct 17 '15 at 17:11
• Timaeus: "can describe kinematics in a non inertial frame." -- Sure: determining who met whom, in which order; and who saw whose signal indications in coincidence, or in which order. From this, geometric relations between the participants may be derived. The notion of quasi-distance is helpful ... "If you assign positions and times to events" -- You mean: coordinates?? "you can compute velocities and hence speeds." -- coordinate velocities, or coordinate speeds. But surely, the OP doesn't know better, either. – user12262 Oct 17 '15 at 17:35
• Timaeus: "I think I was clear and addressed the case where one is inertial and that there must be an asymmetry in that case." -- Of course the word "inertial" doesn't appear in the OP; but "speed $v$" does. "You are welcome to write your own answers or ask your own questions" -- Right. I'm hesitatant to put an answer to this OP because the notion of "inertial system", and hence of "speed $v$" etc., is by itself so difficult. But yes: I plan to ask about "(the, or a, coordinate-free notion of) speed wrt. members of a non-IS" ... tomorrow. – user12262 Oct 17 '15 at 19:09