Should the stars of a relativistic binary system have tangential acceleration?

Question

Should the stars of a relativistic binary system have tangential acceleration?

Answer 1

Any object not travelling at a constant speed has a tangential acceleration because its speed (which is the magnitude of its tangential velocity vector) is changing.

In order for the stars in a binary system to have a constant speed they would have to be orbiting their joint centre of mass in perfect circles. This is very unlikely. Therefore we can say that the stars in any binary system (whether it is relativistic or not) almost certainly have some non-zero tangential acceleration.

Answer 2

Edit: I see what you are attempting to add. But consider this: if the system is in equilibrium, then it is in equilibrium even considering the fact that gravity is being applied over a distance shorter than diameter $d$.

There is something off with the summary and pic though. If we think about B doing the same to A (attracting it at an angle less than $\pi$), then we have a perpetual motion machine. The mechanism is that A “is pulling” in a direction and “being pulled on” in a different direction, which isn’t possible even under relativity, because A is one body.

Returning to the problem statement: In other words, there is a logical error in taking the system to be in equilibrium without relativistic effects, and then saying that there are relativistic effects so therefore it won’t actually be an equilibrium. If your logic is correct, then the equilibrium $\omega$ that the system is in.. is merely higher than what a classical estimate would give.

That would be a better question to ask. As it is now, there is this inconsistent assumption (that a binary system in equilibrium is in equilibrium with a classical $\omega$ and therefore is not in equilibrium because that classical $\omega$ wouldn’t result in equilibrium). There is no external setting of revolutionary speeds. Things collapse, things fly apart, things fly by without capturing each other, etc. And some things fall into a longterm binary system.

End of edit

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Relativity is not needed to analyze this situation. The stars will either have positive acceleration or not.

We can view this at first as a single object rotating about its center. This requires a centripetal force to hold the stars together. If the centripetal force from gravity exceeds the centrifugal force they will get closer and undergo angular acceleration.

But there is a stable speed too. If $$F = m_1 \omega^2 r_1 = G \frac{m_1 m_2}{r_{1,2}^2} $$

If the LHS is lower, the stars will get closer and conserve angular momentum, $wlog$ let $m_1=m_2 (\implies r_1=r_2)$: $$j=I \omega = 2mr^2 \omega = 2mvr =k$$

Assuming gravity is pulling them together:

$$\text{For } r_f<r_i: 2mvr=k \implies v_f>v_i$$

Classically, the angular velocity increases inversely proportionally with the square of the radius of revolution, but the tangential velocity still increases (linearly) with decreasing $r$. Relativistically, the magnitude may be different but there will still be an increase from any frame.

Edit 2: Yes the orbits will be elliptical per Kepler. Nothing changes. Still quadratic. Largely conceptual anyway. I’d change it if that was the actual question

Answer 3

Yes, because a relativistic system will emit gravitational waves, which will carry energy away from the system, and cause the orbit to decay in size and the rotational frequency to increase. There may be other effects as well (eccentricity, etc) leading to a tangential acceleration, but two isolated bodies orbiting in a perfectly stable circular orbit with asymptotically flat boundary conditions is not possible in General Relativity.