4
$\begingroup$

Could we simply change coordinates of the Schwarzschild metric in order to obtain the metric of a moving massive particle? Which would those coordinates be? Rindler coordinates? Maybe there is a change of coordinates describing an arbitrary accelerating movement.

I'm looking for a Lienard-Wiechert type of solution.

$\endgroup$

1 Answer 1

1
$\begingroup$

What is the gravitational field of an accelerated particle?

General relativity is a theory with constraints and those constraints imply conservation of energy and momentum. So a gravitational field of a particle that is spontaneously accelerating without any external influence would contradict general relativity, so it is meaningless to ask about it.

Compare this with the situation in electromagnetism: Maxwell equations say nothing about energy and momentum of charges, so it is reasonable to ask about EM field of a charge moving along with arbitrary acceleration. On the other hand, electromagnetism does have its own constraints and those constraints imply conservation of electric charge, so it would be meaningless to ask about e.g. EM field of an isolated charge varying with time.

Could we simply change coordinates of the Schwarzschild metric in order to obtain the metric of a moving massive particle?

Change of coordinates does not change spacetime and it would not make a particle accelerate. But suitably chosen coordinate system can make it easier to interpret spacetime from the viewpoint of specific observer (or a set of observers). For example, a change to coordinate system where the particle is moving with constant velocity gives the so-called boosted Schwarzschild solution, see this answer for details.

Rindler coordinates?

Schwarzschild metric viewed from the reference frame of an observer with asymptotically constant acceleration could be seen as a particle plunging into the Rindler horizon. Note, that while the particle is not accelerating, such a description can provide insight into the physics of black hole mergers in the extreme mass ratio limit, see e.g. this or this paper.

Now, if we do provide the reason for particle's acceleration, we would have to include it as an additional source for the gravitational field (forces gravitate), and so, instead of a single solution for an accelerating particle, we would have multiple solutions depending on the physics behind particle's acceleration.

When acceleration is constant we then can have an isometry generated by a Killing vector field that approaches generator of the boost Lorentz transformation in the asymptotic region of Minkowski background spacetime, such spacetimes are called boost symmetric. If, additionally, there is axial symmetry with the axis along the direction of acceleration then there are two independent Killing vector fields and spacetime is called boost-rotation symmetric. Einstein equations are often fully integrable in this case and there are powerful solution generating techniques, see this paper for a review. Here are some solutions of this class:

  • C-metric (see this) describing particle (or a black hole) being accelerates by a conical singularity (cosmic string). Actually, there are two black holes, accelerating in the opposite directions, but they are not in causal contact with each other, the second black hole is behind acceleration horizon of the first and vice versa.

  • Bondi dipole: a self-accelerating system of positive and negative mass (see this). Negative mass repels the positive which, in turn, attracts the negative and so the system accelerates as a whole. The presence of negative mass arguably makes the solution rather unphysical.

  • Ernst metric, massive charged particle/black hole is accelerated by electric field (or particle with magnetic monopole charge is accelerated by magnetic field). Away from the particle metric asymptotes to Melvin solution.

…describing an arbitrary accelerating movement. I'm looking for a Lienard-Wiechert type of solution.

There closest is probably Kinnersley's photon rocket (original paper), a solution describing mass moving with arbitrary acceleration due to recoil from anisotropic emission of photon streams (or rather null dust). Note, that despite an arbitrary motion of the mass, there is no overall gravitational radiation here: contributions from moving mass and photon streams cancel each other (see here).

$\endgroup$
8
  • $\begingroup$ Really cool answer, I appreciate it. However, wouldn't an observer that's accelerating with respect to a particle perceive that particle as being accelerated? The question then shifts to how could that observer be accelerating, but I thought that all observers were valid according to GR. $\endgroup$
    – K. Pull
    Commented Dec 10, 2023 at 17:57
  • 1
    $\begingroup$ thought that all observers were valid according to GR valid does not mean equal, in GR, like in Newtonian mechanics or SR, there are privileged inertial observers. And if the spacetime in question describes particle in some background (which means that geometric size of a particle is much smaller than the scale of background) then whether particle accelerates is a question that has an objective answer indepent of coordinate systems, observers etc. $\endgroup$
    – A.V.S.
    Commented Dec 10, 2023 at 18:42
  • $\begingroup$ I did not know that! Thanks $\endgroup$
    – K. Pull
    Commented Dec 10, 2023 at 20:41
  • $\begingroup$ I've thinking about your answer for a while. What interpretation would you give to the solution of $G^{\mu\nu} = \tfrac{m\delta^3(x-q(t))}{\sqrt{g_{\alpha\beta}\dot{q}^{\alpha}(t)\dot{q}^{\beta}(t)}}\dot{q}^{\mu}(t)\dot{q}^{\nu}(t)$? Isn't that the gravitational field of an accelerated particle. That is really what interested me in the first place $\endgroup$
    – K. Pull
    Commented Dec 11, 2023 at 1:32
  • $\begingroup$ If $G$ does not satisfy constraint $\nabla_\mu G^{\mu\nu} =0$ (that constraint is known as contracted Bianchi identity), then there could be no such solution. For $q$ this is equivalent to $q^\mu(t)$ satisfying geodesic equation. $\endgroup$
    – A.V.S.
    Commented Dec 11, 2023 at 3:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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