4
$\begingroup$

Imagine an asymptotically flat spacetime with nothing but two stars at a certain distance. They fall into each other and form one big star, so their potential energy is converted to kinetic energy.

Looking at the definition of the stress-energy tensor, I can see that kinetic (and thermal) energy is part of it, potential energy is not. So the total energy content of the spacetime would increase when the stars approach each other.

Is that correct? So the merged star has a stronger "gravitational pull" on far objects than the two stars had before the merge?

Or is there some kind of potential energy that makes sure that the total energy of the system is conserved? I understand that the potential energy is frame-dependent, so it cannot directly be part of the stress-energy tensor.

$\endgroup$
2
$\begingroup$

No, the curvature outside does not increase, it can decrease though.

First lets consider a simpler case. You have a small spherical ball of neutral hydrogen at rest in an otherwise empty and infinite (asymptotically flat) spacetime that is not in equilibrium (and larger than equilibrium size).

It contracts, it gets smaller, it develops some pressure, but is a small ball, so not enough to ionize or fuse. Since it is spherically symmetric the field outside the original ball is Schwarzschild. But it is Schwarzschild between the old outside and new boundary too since you can't sew two Schwarzschild solutions with different mass parameters together on the same surface area of areal surface without their being stress energy on the surface. But the same applies again as long as it isn't radiating and none of the random pressures give an escape velocity.

You could have a spherical shell of water contract if you want the surface tension to keep it all in. The point is that infall doesn't increase the curvature outside. Only radiation heading out would do that, and that would make it weaker as the energy flows outwards.

Go back to that spherical shell of water, you have Minkowski on the inside and Schwarzschild onthe outside, they can be together because there is stress energy on the surface. That is what stress energy does, it allows vacuum solutions that couldn't normally fit together to be able to be sewed up.

So imagine two shells of water at different radii with a common center. Minkowski in the interior, Schwarzschild of mass $m$ in between the two shells and Schwarzschild of mass $M$ outside the outer shell. And why is $M>m$? Because that is specifically what positive energy density does, it allows a larger mass exterior to be sewed up to a smaller mass (possibly zero mass) interior.

So as the stress energy falls towards each other, the outside vacuum spacetime has to fill itself in the only way a vacuum spacetime can (subject to the evolution equations determined by the vacuum EFE, the current metric and the current rate of change of the metric). As long as the stress energy tensor evolves in a way where it has zero covariant divergence then this is possible. So that's all you need to worry about.

But get this part clearly. The amount of energy you need to place on a shell to transition between a mass $M$ Schwarzschild exterior and a mass $m$ Schwarzschild interior is not $Mc^2-mc^2.$ In fact you need more energy to make the transition at a smaller radius than at a larger radius.

How much more energy? Guess. Exactly the amount produced by the in fall. A shell of collapsing dust produces a static Schwarzschild of a fixed mass parameter even though its energy increases as it infalls. You just need more energy to transition between two Schwarzschild metrics that transition at a smaller radius.

The parameter $M$ or $m$ you see in a metric is not the sum of the mass of some parts somewhere in the spacetime, it is just a parameter. It is a parameter $M$ chosen sot hat in the weak field limit of the metric the geodesics like the acceleration of a body in Minkowksi space about a star of mass $M$ if gravity were a force like $GMm/r^2,$ which it isn't. It is just a parameter designed to have an intuitive weak field limit correspondence to old physics. Nothing more, nothing less.

OK. So we handled the easier spherically symmetric case, which hopefully addressed your misconceptions. What about your case? Same thing, it takes more energy to sew together the exterior (beyond both stars) and the interior (between both stars) solutions, by exactly the increase energy of the two stars. It all works out. It's just that neither of the solutions is as simple as a static Schwarzschild solution.

$\endgroup$
  • $\begingroup$ Thank you. This is a great qualitative explanation. Do you know the quantitative answer too? If the mass parameter $M$ of the Schwarzschild solution is not the total mass of the stress energy at the center, then how is it calculated from the stress energy? How can I calculate the difference $Mc^2-mc^2$ in your example? $\endgroup$ – Bass Aug 28 '16 at 11:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.