# Inertial motion in GR

1. Dec 31, 2013

### TrickyDicky

Does inertial motion (understood in the SR sense) exist in a curved spacetime?

2. Dec 31, 2013

### PAllen

Trick question? Inertial motion, yes (for a test body). Extended inertial frames, no. Perhaps you are asking about and exact treatment for body of any size? Then, of course, no due to gravitational radiation and self gravity.

3. Dec 31, 2013

### TrickyDicky

Ok, let's center it on test bodies just for now. Aren't they just useful idealizations that allow very good approximations? Or are you including in the test body concept elementary particles together with the assumption that they are point-particles?
In the first case do you agree that geodesics in a curved manifold are by definition non-inertial, and otherwise how would you define their difference with the straight line of flat space? Remember that in any Riemannian manifold one can always reference the geodesics to the locally straight lines of flat space.

4. Dec 31, 2013

### Staff: Mentor

I am not sure what you mean by this. What is "inertial motion (understood in the SR sense)"?

5. Dec 31, 2013

### PAllen

Wait, I consider geodesics by definition inertial. If you mean some definition of straight 'as if the space weren't curved', then I have no idea how to define that. I don't think it could be defined, except at infinity in an asymptotically flat spacetime.

6. Dec 31, 2013

### TrickyDicky

You don't need to go to infinity, locally you have flat spacetime and inertial motion by the equivalence principle and the definition of manifold.

7. Dec 31, 2013

### PAllen

But that defines a geodesic. A geodesic is locally straight in exactly the sense in which a pseudo-riemannian manifold is locally Minkowski.

8. Jan 1, 2014

### TrickyDicky

Exactly. And I'm inquiring whether this reasoning makes sense: that if we identify local straightness as local inertial motion for test bodies, we should consider its motion when not looking at it locally as non-inertial, that is what tells us that we are dealing with curved spacetime as opposed to flat one where we cannot draw that distinction in the path of test bodies not subjected to any force.

9. Jan 1, 2014

### PAllen

What we call geodesics doesn't change anything about them. It seems you are proposing to call geodesic motion 'non-inertial' simply reversing the definition everyone else uses to distinguish these motions from all other motions in the spacetime which everyone calls non-inertial. No, I see only downsides to such an - we have non-inertial motion and what ? even more non-inertial for the other paths?

10. Jan 1, 2014

### Staff: Mentor

That reasoning seems to presuppose the existence of some non-local definition of "straightness" which is distinct from "geodesic". I don't know of any such definition.

11. Jan 1, 2014

### vanhees71

I think that the (weak) equivalence principle in General Relativity can be stated as that the inertial motion of a test body (to avoid the word "particle", which always involves quantum theory, and quantum theory in curved spacetime becomes an awfully tricky issue; even the definition of what's meant by a "free particle" in curved spacetime is not easy!) is described as a geodesic world line in a pseudo-Riemannian fourdimensional manifold, the space-time of general relativity, whose pseudo-metric (fundamental tensor) is determined by the energy-momentum tensor of matter and radiation according to the Einstein equation. The fundamental tensor is in this sense representing the gravitational field in General Relativity.

This implies that there are always local Galileian reference frames, where locally the local natural laws obey the corresponding special-relativistic laws. If a true gravitational field is present, of course, one cannot find a global Galileian reference frame. That's only the case if the space-time is the special relativistic Minkowski spacetime which is a solution of the Einstein equation for an empty universe (i.e., at vanishing energy-momentum momentum tensor).

12. Jan 1, 2014

### TrickyDicky

I see your point and how my question could be misinterpreted as a semantics issue, but certainly I'm not interested in renaming geodesics, only the physics interests me.
You implicitly seem to agree that motion along geodesics in GR is non-inertial but would rather keep calling it inertial so to avoid confusiĆ³n with motion in the presence of non-gravitational interactions wich is fine. But note that by that logic there is also the risk of confusiĆ³n with the inertial motion as understood in Minkowkian SR when the geodesics are only locally equivalents. Probably in both cases the context should prevent misunderstandings.
So again, conceptually, is it safe to say that in GR curved spacetime there is no such thing as inertial motion except as idealized approximation?

13. Jan 1, 2014

### WannabeNewton

No it doesn't but you're looking at this backwardly from how you should be looking at it.

Let $(M,g)$ be an arbitrary time-oriented space-time with an associated Levi-Civita connection $\nabla$ and consider a worldline $\gamma:I\subseteq \mathbb{R}\rightarrow M$. Let $e_{0}$ denote the unit future-directed tangent vector to $\gamma$ at $\tau \in I$. Using $e_{0}$ construct an orthonormal basis $\{e_{\alpha}\}$ for $T_{\gamma(\tau)}$ where orthonormal means $g_{\gamma(\tau)}(e_{\alpha},e_{\beta}) = \eta_{\alpha\beta}$; $\{e_{\alpha}\}$ is called a local Lorentz frame. Then $\gamma$ constitutes "inertial motion" $\forall \tau \in I$ if and only if $\nabla_{e_0}e_0 = 0$ along $\gamma$ and the local Lorentz frame $\{e_{\alpha}\}$ is an "inertial frame" $\forall \tau \in I$ if and only if $\nabla_{e_0}e_{\alpha} = 0$.

Therefore the concepts of "inertial motion" and "inertial frame" are inherently local.

It is only in the special case of $M = \mathbb{R}^{4}$ and $g = \eta$ that the concepts of "inertial motion" and "inertial frame" become global.

14. Jan 1, 2014

### TrickyDicky

I didn't intend it so if it appears that way I must have expressed myself poorly.

15. Jan 1, 2014

### PAllen

No, I don't. Calling some motion non-inertial implies there is some definition of a different motion that is inertial. Instead, I believe geodesic motion is inertial, and no other motion could possibly be singled out as inertial to give meaning to 'non-inertial'.

16. Jan 1, 2014

### PAllen

So it seems we have a slight disagreement. I agree with what you say on inertial frames, but consider that geodesic motion is inertial; it certainly meets your criterion at each point in its history, and is the only such motion that does.

17. Jan 1, 2014

### WannabeNewton

Yes so $\gamma$ constitutes "inertial motion" iff $\nabla_{e_0}e_0 = 0$ iff $\gamma$ is a geodesic so I don't think there's any disagreement between us. What I'm trying to convey is that the concept of "inertial motion" is inherently local in the sense that at any given event on the worldline, we can find a coordinate system in which the image of the worldline at that event is a straight line. Only in the special case of Minkowski space-time can we extend this straight line to every event on the worldline so as to get a straight line globally. The same goes for the space-time extent of the "inertial frame". It seems to me TrickyDicky is defining "inertial motion" starting from the latter and finding that it only holds locally in the more general context of curved space-times and hence sees an ostensible discrepancy.

18. Jan 1, 2014

### TrickyDicky

Thanks, WN. This clears it up for me for the moment.

BTW Happy new year everyone.