1
$\begingroup$

Consider 2 point masses $A,B$ a distance $d$ away from each other without velocity or rotation spin. Is the center of mass in general relativity equal to the center of mass in newtonian gravity?

In newtonian gravity the center of mass is where the particles collide, if they collide. Is the center of mass in general relativity also the place where the particles collide, if they collide? Or maybe the center of mass in general relativity is the place where the particles collide if they had no initial velocity or rotation spin?

$\endgroup$
  • $\begingroup$ The center of mass is a mathematical construct. Maybe you wanted to ask about center of gravity. $\endgroup$ – Torsten Hĕrculĕ Cärlemän Jan 7 '14 at 12:32
  • $\begingroup$ I don't like your definition of COM as 'where the particles collide , if they collide'. And what if they don't collide? The actual definition of COM is a point with $$\mathbf{r} = \frac{\sum \mathbf{r}_i m_i } {\sum m_i}$$ $\endgroup$ – user23660 Jan 7 '14 at 12:46
  • $\begingroup$ vector valued integrals are subtle in general relativity (do you put the basis vectors inside or outside of the integral?), and the answer to this question will depend on how you generalize the notion of center of mass. $\endgroup$ – Jerry Schirmer Jan 7 '14 at 13:07
  • $\begingroup$ Yes user23660 I know that is the definition of COM ... in newtonian-gravity ! I ask about general relativity in the OP and the relation to collision. $\endgroup$ – mick Jan 7 '14 at 21:24
1
$\begingroup$

There is generally no center of mass in general relativity.

The notion of center of mass becomes useful in Newtonian (and special relativistic) mechanics since because of simplifications we obtain by using it as a reference point. For instance, we could decouple the motion of COM from the relative motion of various parts of the system.

But in general relativity there are generally no conservation of momentum. Moreover the notion of a point particle also has some problems.

In certain special cases (such as for an isolated system with an asymptotically flat metric) in general relativity we could still define the notion of center of mass. For instance in the paper:

Huisken, G., & Yau, S. T. (1996). Definition of center of mass for isolated physical systems and unique foliations by stable spheres with constant mean curvature. Inventiones mathematicae, 124(1), 281-311. doi:10.1007/s002220050054, (online pdf).

the COM is defined not as a 'inner' point of a spacetime, but rather through foliation of asymptotic exterior, that is by considering the behavior of metric in the almost flat region. Such definition is essentially Newtonian.

In general spacetime such expansion of the metric is impossible, so the notion of COM does not exist. Its nonexistence is illustrated by the process of relativistic swimming in a curved spacetime. This effect, suggested by J. Wisdom in the paper:

Wisdom, J. (2003). Swimming in spacetime: Motion by cyclic changes in body shape. Science, 299(5614), 1865-1869. (free pdf).

essentially allows motion of an initially at rest system through internal deformations. (Like Baron Münchhausen pulling himself by his hair).

$\endgroup$
  • $\begingroup$ There are no point particles in GR !? That sounds so " unmathematical " and " unpractical " ! weird. $\endgroup$ – mick Jan 7 '14 at 21:28
  • $\begingroup$ DO the masses $A,B$ collide at the same position as in newtonian ? $\endgroup$ – mick Jan 7 '14 at 22:12
  • $\begingroup$ The main objects of GR are 'fields' and not particles. If a point object has a finite mass, there will be a horizon around it. $\endgroup$ – user23660 Jan 8 '14 at 6:07
  • $\begingroup$ And no, the masses won't collide in the same positions as in Newtonian case. The spaces in which these collisions occur are simply different. $\endgroup$ – user23660 Jan 8 '14 at 6:22
  • $\begingroup$ Isn't the metric used to measure distances, and when combined with mass don't you get a center of mass? $\endgroup$ – ja72 Feb 9 '16 at 18:06
5
$\begingroup$

There is a general definition of a center of mass in General Relativity, which was proposed by Dixon (W.G. Dixon, Il Nuovo Cimento 34, 317 (1964)) and whose existence and uniqueness were proved by Beiglböck (W. Beiglbock, Commun. Math. Phys. 5, 106 (1967)). These are very old papers, so that people have been working since long ago on this topic.

In Newtonian mechanics, the center of mass of a collection of particles is an average of positions of particles weighted by their masses. In GR, everything is a bit more complicated, but it is still possible to define a center of mass. To begin with, if you keep weighting by mass, then you cannot take into account photons, so that one has to weight by something else. Additionally, you have to sum positions weighted by something, however, in a general spacetime there is no notion of "summing" positions. The solution is that you weight by energies and that the positions you sum over are actually vectors that point in the direction of the particles (the exponential map of these vectors from the position of an observer takes you to the particles' positions).

The formal definition is a bit intricate, but I will mention it very briefly. In the first place one weights with energies instead of masses. In the second place, one defines a center of momentum in the following way. Take an arbitrary observer specified by its position $x$ and its 4-velocity $u$. Extend all geodesics orthogonal to $u$ beginning at $x$. This will be the simultaneity hypersurface $\Sigma$ associated to the observer. If the particles are sufficiently close to $x$, then there is a unique geodesic joining $x$ with the intersection of the particles' world lines with $\Sigma$. The total momentum of the collection of particles will be the sum of each particle's momentum parallel transported along such geodesic. If the total momentum you compute this way is parallel to $u$, then you are in the center of momentum frame at $x$. If you repeat this procedure over all spacetime points, finding the center of momentum frame gives you a vector field. Now, in order to find the center of mass, choose a point $x$ and the velocity $v$ of the center of momentum there. You average over the vector position of each particle $i$, which is the vector such that its exponential map from $x$ takes you to the intersection point of the particle's world line with $\Sigma$, weighted by the energy $-g_{ab}v^a \tilde{p}^b_i$ computed in the center of momentum frame, where $\tilde{p}^b_i$ is the parallel transported momentum of the particle $i$. Note that this sum is well defined because all momenta have been parallel transported to the tangent space at $x$, so that its sum is a vector defined at $x$. If this vector is zero, then $x$ is at the center of mass.

Beiglböck has shown in his paper that such center of mass exists, is continuous and timelike, given very reasonable assumptions, in particular, conservation of momentum and finite extent of the particles configuration.

$\endgroup$
  • 1
    $\begingroup$ Note that the other answer is stating that "But in general relativity there are generally no conservation of momentum" . Assuming momentum conservation answers the question for a sub phase space . $\endgroup$ – anna v Feb 9 '16 at 7:19
  • 1
    $\begingroup$ Ok, but that statement has nothing to do with GR nor with Newtonian mechanics, but with the composite system one is talking about. In Newtonian mechanics conservation of momentum might be violated too, of course, when you neglect some interactions. The condition required by the center of mass by Dixon and Beiglböck is that the composite object be described by an energy-momentum tensor satisfying the standard condition $\nabla_a T^{ab} = 0$. $\endgroup$ – Pedro Aguilar Feb 9 '16 at 17:41
  • $\begingroup$ What about a vacuum solution, would it make sense the center of gravity? $\endgroup$ – jinawee Oct 1 '18 at 11:42

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.