0
$\begingroup$

I've seen this question How does Einstein field equations interact with geodesic equation?, but it doesn't make any sense to me. If spacetime is a Lorentzian manifold, then surely one thing general relativity tells us is what the possible manifolds are when gravity is the only "force". And in that context, the field equations themselves don't restrict the manifold at all -- any manifold has an Einstein tensor $G_{\mu\nu}$ that represents a possible matter distribution in that it is "conserved" (zero divergence).

So the form of the manifold is restricted only by the geodesic equation. How so? Well, here's my thought process:

First of all, we can say that $G(\vec u, \vec u)$, for a timelike unit vector $\vec u$, is the mass density flowing along $\vec u$. And all we require is that that mass follows a geodesic. So, if we take the geodesic along $\vec u$, the same mass should remain on it the whole time. But since the mass can spread out spatially over time, we actually have to consider a tight family of initially parallel geodesics, and say that the density integrated over their spatial cross-section (volume) $V$ is what remains constant. So we would basically get

$$\nabla_{\mu} (G_{\mu\mu} \cdot V_{geodesic}) = 0.$$

Now, the variation of the cross-sectional volume along a tight family of initially parallel geodesics is exactly what the Ricci curvature describes. Except that it is the second derivative of that volume, because the first derivative is identically zero, because they are initially parallel. But we wouldn't want the $\nabla_{\mu}$ to be ${\nabla_{\mu}}^2$, because that would allow the total mass to change linearly. So clearly something is off with the above equation. Or maybe I'm barking up the wrong tree, but the conceptual statement makes sense to me. Can anyone clarify this, and/or point me to an online reference that explains conceptually how the two equations are combined?

$\endgroup$
1
  • $\begingroup$ "then surely one thing general relativity tells us is what the possible manifolds are when gravity is the only "force"." I'm not sure I agree with this statement -- the matter Lagrangian is included in the terms whose variation gives $T_{ab}$, and these terms certainly include forces in them. This is certainly a well-known procedure for Maxwell theory. $\endgroup$ – Jerry Schirmer Feb 21 at 18:27
3
$\begingroup$

I feel there is something interesting to explore here, but it is not physical to assert "any manifold ... represents a possible matter distribution". The equivalent statement in Newtonian gravity would be to assert that there are no constraints on mass, not even that it cannot be negative. But such a constraint is an important aspect of the theory as a physical theory; it cannot be dropped or ignored. In G.R. the constraints on $T^{ab}$ (going by such names as weak energy condition, strong energy condition and so on) are part of the story of general relativity considered as a physical theory, as opposed to merely some mathematical tools to deal with manifolds. But it has never been quite fully settled which of these constraints one should insist upon. It is an open area of research at the boundary of quantum theory and general relativity.

A widely discussed aspect of G.R. is that the geodesic equation, with its interpretation as an equation of motion for particles, follows from the field equation (see e.g. MTW book). This is a bit like the way you can deduce the Lorentz force equation from the Maxwell equations plus energy conservation. So there are connections between geodesic equation, field equation and conservation. But that's about as far as this answer is going to take it. It is perhaps more of a long comment than strictly an answer, but I hope you may find it helpful.

$\endgroup$
6
  • $\begingroup$ This is certainly helpful, I hadn't considered that kind of inherent constraint. But as to the second point, I don't get physically how geodesic motion could be deduced. Because we can contemplate a universe that is governed only by gravity, or we can contemplate a universe with electric charge. And surely they correspond to different manifolds, right? But wouldn't they both satisfy the kind of constraint that you say is implicit in the Einstein field equation? So then they would both obey geodesic motion -- except that charges don't follow geodesics... $\endgroup$ – Adam Herbst Feb 21 at 19:11
  • $\begingroup$ @AdamHerbst my comment on geod. eqn. as eqn of motion for particles refers just to freefall motion; i.e. no forces except gravity. Little bumps of spacetime manifold evolve as the field eqn tells them to. In the limit this evolution tells them to follow timelike geodesics. $\endgroup$ – Andrew Steane Feb 21 at 20:16
  • $\begingroup$ I really appreciate you taking the time to respond, but I still don't get it. What about the case of two repelling charges? Each charge has a little lump of spacetime, and these lumps deviate from geodesics, according to the Lorentz force. But how could this manifold not obey the Einstein field equation, if the only inherent constraint in the EFE is that of positive energy? You're saying EFE implies geodesic motion "when gravity is the only force" -- yet what else is there to tell us gravity is the only force, but the geodesic equation itself? $\endgroup$ – Adam Herbst Feb 21 at 20:53
  • $\begingroup$ @AdamHerbst The derivation from the field eqn applies to the case $T^{ab} = 0$ surrounding the lump in question, but if there is an e.m. field then $T^{ab} \neq 0$. $\endgroup$ – Andrew Steane Feb 21 at 20:59
  • $\begingroup$ Wow! Okay. That is really cool. Thank you so much for helping me through my ignorance! $\endgroup$ – Adam Herbst Feb 21 at 21:13

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.