17
$\begingroup$

Assume a body with some mass moving in a straight line in empty space without any objects around. Now according to general relativity, it curves space-time around it. Is it possible in any way that the curved space time affect the objects motion?

$\endgroup$
2
  • 1
    $\begingroup$ Are your asking if there is frame-dragging resistance. Like air resistance? There can't be. That would violate conservation of momentum in some reference frames $\endgroup$
    – Jim
    Sep 12 at 14:50
  • 1
    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/3436/2451 and links therein. $\endgroup$
    – Qmechanic
    Sep 12 at 21:35

4 Answers 4

20
$\begingroup$

Answering the title question: yes, but not in the particular example you provided.

This is an example of a self-interaction, or self-force, and it is, in a well-defined sense, a higher-order correction to geodesic motion. There is a 2009 preprint by R. M. Wald (arXiv: 0907.0412 [gr-qc]) describing the general ideas behind these effects. There is also a 2011 Living Reviews in Relativity (doi: 10.12942/lrr-2011-7) discussing the motion of point-particles in GR. Wald's Advanced Classical Electromagnetism discusses the analogous problem for Electromagnetism in Chap. 10. I refer you to those references (and references therein) for further detail. In here, I'll keep to some conceptual notions.

As pointed out by rob, the particular situation you prescribed is symmetric, so by taking a look at what happens in the particle's reference frame leads you to the conclusion that it can't be "pulled" to any direction. You'll get a Schwarzschild solution (notice that you can't really take a massive point-like particle in GR).

In other situations, however, you can get emission of gravitational radiation by accelerated masses, in a fashion similar to what happens with Electromagnetism (Do accelerating masses generate gravitational waves?). As a consequence, you have a "reaction force" on the body. This means that the geodesic equation (for gravity) or the Lorentz force (for electromagnetism) does not take into account all of the effects involved in the motion of a point-particle and one has some more trouble describing the motion correctly, meaning in particular that the differential equations for the position of the point-particle as a function of time get correction terms, making them deviate from the "usual" solutions.

$\endgroup$
1
  • 3
    $\begingroup$ Note the self-force has a conservative component that is not directly linked to the emission of gravitational waves. (Although, in a Feynman diagram sense the conservative self-force is related to processes involving the emission and reabsorbtion of gravitational waves. $\endgroup$
    – TimRias
    Sep 12 at 20:28
11
$\begingroup$

The answer to this question depends on the nature of the object and its motion. The standard Kerr solution to Einstein's equation, which describes a spinning black hole, can be expressed in the form (Kerr-Schild coordinates):

$$g_{ab} = \eta_{ab} + C \ell_{a}\ell_{b}$$

where $\eta$ is ${\rm diag}\left(-1,1,1,1\right)$, which we will henceforth call the "Minkowski metric" and $\ell$ is a null vector relative to both $\eta$ and $g$, and $C$ is a constant. There is a particular form that $\ell$ and $C$ have to take, but for the scope of this question it's not important. What is important is that since you have reduced the metric to a minkowski metric and a null vector on a minkowski background, you can freely Lorentz transform it, and if the base metric describes a non-moving spinning and charged black hole, the boosted metric describes a moving one. But all of this is just coordinate transforms, and no physics has happened. This means that, from the perspective of back-reaction on the background, we still have a timelike Killing vector, and the spacetime of a black hole moving at constant velocity is still static, and no back-reaction happens.

But, if you go and extend the solution to describe a system where a piece of matter has a physical extent, or where the black hole is accelerating, the situation changes. In that case, it requires a lot of fine-tuning to get it so that the higher time derivatives of the quadrupole moment of the matter distribution (or the radiation-related Weyl scalars, for the case of vaccum solutions) are non-zero. This means that the moving object will emit gravitational waves, and those waves will cost the moving object some energy, and it will feel a force.

$\endgroup$
4
  • 2
    $\begingroup$ Very interesting. Is this related to the Unruh effect, where an accelerating observer sees different radiation than an inertial observer? $\endgroup$
    – rob
    Sep 12 at 14:58
  • 1
    $\begingroup$ @rob: you could probably make a connection here (though I usually think about the Unruh effect being a quantum thing making an analogy between teh horizon in Schwarzchild and the horizon in an accelerated frame, and then applying Hawking's radiation argument), but this is a much simpler thing than the unruh effect -- it is just pure classical GR, and the radiation I'm talking about is classical gravitational radiation, not Hawking/Unruh radiation. $\endgroup$ Sep 12 at 15:11
  • 2
    $\begingroup$ @rob It might be worth recalling that the Hawking/Unruh radiation relates to any quantum field, while the effect mentioned by Jerry is the emission of gravitational waves. However, there is indeed a relation between radiation by accelerated charges and the Unruh effect (doi: 10.1103/PhysRevD.45.R3308) and it makes sense to me that a similar effect generalizes to other fields. $\endgroup$ Sep 12 at 18:08
  • 1
    $\begingroup$ Quoting from the paper I mentioned, "The agreement between the results (26) and (31) indicates that the ordinary QED bremsstrahlung can be interpreted by an observer coaccelerated with the charge as the emission (absorption) of zero-energy Rindler photons into (from) the Rindler thermal bath corresponding to the Minkowski vacuum." $\endgroup$ Sep 12 at 18:10
8
$\begingroup$

No. General relativity in flat spacetime reduces to special relativity. If an isolated (massive) object moves with constant momentum in flat spacetime, a reference frame exists in which that object is at rest. Any observable event, like a collapse to a black hole or a radioactive decay, will be agreed upon by observers in all reference frames.

$\endgroup$
2
  • 4
    $\begingroup$ But the spacetime is not flat; the object has mass. The reduction to SR fails except locally. GR is itself invariant under coordinate transforms. You can find a coordinate system where the object is at rest and therefore conclude that there can be no difference between a moving mass and one at rest, but I don't believe you can justify this via SR. $\endgroup$
    – HTNW
    Sep 12 at 14:42
  • 1
    $\begingroup$ Correct. But in terms of pedagogy, I think it’s fine to leave that clarification as a comment. $\endgroup$
    – rob
    Sep 12 at 14:54
2
$\begingroup$

Assume a body with some mass moving in a straight line in empty space without any objects around.

A single material object in spacetime generated by it does not move in space but in time. Without other objects around the notion of relative movement in the space makes no sense.

$\endgroup$

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.