Classical fields in GR

1. May 7, 2013

TrickyDicky

I've seen different objects described as the field in GR: the metric, the connection (and then the metric is seen as the "potential" by analogy with EM field theory), both...
Could someone comment on what object should most reasonably be considered the field of the GR theory? Could for example the Riemannian tensor be considered the field of GR?

Thanks.

2. May 7, 2013

Mentz114

Since the Lagrangian and the action are written in terms of $R\sqrt{-|g|}d^4x$ I don't see how it can be anything but.

3. May 7, 2013

pervect

Staff Emeritus
They are all reasonable. I think the connection is closest to the Newtonian view of gravity "as a force". If you set the majority of the Christoffel symbols to zero (or just assume they are small), leaving only three of the symbols, the $\Gamma^{space}{}_{{time}\,{time}}$ terms as having values significantly different from zero, you'll have something that looks and transforms in a force-like manner (but only if you're careful about what sort of transformations you allow).

To my mind this isn't very fundamental and the analogy breaks down fairly quickly, but it's helpful when talking to laypeople who think of gravity as "being a force". They are familiar with gravity being a force from Newtonian theory, and are generally don't have the background (or willingness) to adjust their thinking.

I don't find it particularly helpful to actually mention the Christoffel symbols by name :-(, but I keep it in mind as representing the "force of gravity".

I personally favor the Riemann as the most fundamental, but really all the choices are logical. When going with the Riemann approach, I often view the Riemann as being "tidal gravity". (It's really only the Electrogravitic parts of the tensor that can be described as tidal gravity, but those are usually the ones of concern).

I've used to "push" the tidal gravity idea to laypersons more (ala MTW), but I feel like the idea didn't actually get across to many, so I've mostly given up on that approach. It's not a big change - just change forces to force gradients - but it seems to confuse them more than it helps.

4. May 7, 2013

WannabeNewton

There are two standard ways of arriving at the vacuum Einstein equations $R_{ab} = 0$ (which are of course the field equations of GR in vacuum) by applying a variational principle to a dynamical field related to space-time. The first of these is also arguably the most famous: we take $g^{ab}$ as our field variable and define the Lagrangian density $\mathcal{L}_{G} = \sqrt{-g}R$ so that our action is $S[g^{ab}] = \int \mathcal{L}_{G}\epsilon$ where $\epsilon$ is the volume element (recall that integration of Riemann or Lebesgue integrable functions over manifolds is done via a volume n-form); this is of course the Hilbert action. Varying this action yields $R_{ab} = 0$.

The other usual way is to view $\nabla_{a}$ as being an independent field variable from $g^{ab}$ (that is, $R_{ab}$ is a function of only $\nabla_{a}$) and define the Palatini action $S[\nabla_{a},g^{ab}] = \int \sqrt{-g}R_{ab}g^{ab}\epsilon$. Interestingly, variation of the Palatini action not only yields the vacuum Einstein equations but also the metric compatibility $\nabla_{c}g^{ab} = 0$. The calculations involving in the variation of the Palatini action are not as trivial as the ones involved in the variation of the Hilbert action (although they are more elegant) so if you are interested, I can show the calculations in a subsequent post.

Both of these viewpoints are valid since they both yield the vacuum Einstein equations.

EDIT: By the way, I'm assuming that when you ask "what should be considered the field of GR", you do indeed mean what field variable can yield the vacuum equations of GR through a variational principle. Perhaps as a comparison, if I have a Klein-Gordon field propagating on curved space-time, I can take my field variable to be the scalar field $\varphi$ and define an action $S[\varphi] = \int \mathcal{L}_{K}\epsilon$ where $\mathcal{L}_{K} = -\frac{1}{2}\sqrt{-g}(\nabla^{a}\varphi\nabla_{a}\varphi + m^{2}\varphi)$. Of course as far as calculations go, showing that the variation of this action yields the Klein-Gordon equation on curved space-time is much more simple than the calculations involved in the variation of the two actions written above for GR.

Last edited: May 7, 2013
5. May 7, 2013

TrickyDicky

Thank you for the quick replies.
I was thinking that since in Riemannian geometry both the affine connection and the curvature tensor can be derived from the metric tensor the three are valid, that is they amount to the metric field, IOW the geometry is the field, right? At least I think this is the conventional view on this.
I was interested also in stressing the difference between this geometric field approach to physical theories and the more frequently used in the rest of field theories, either Quantum or classical, where there's the manifold on one side and the field as a function of the manifold on the other.

WN, I have access to the calculations from the usual textbooks on GR, thanks.

6. May 7, 2013

WannabeNewton

The term "the geometry is the field" can be quite ambiguous since when we speak of a field variable for a theory we have in mind a concrete function of space-time. As I said above, you can view the metric tensor as the field variable and look at the associated derivative operator as well as curvature quantities defined in terms of the derivative operator as being derived from the metric or you can with equal validity view the connection as an independent field variable from the metric tensor and still get the field equations of GR so I don't see any reason to say one view is any more "fundamental" than another.

There's a big difference between talking about a field variable for GR and a field variable for a physical theory on curved space-time because the latter involves a field propagating on curved space-time whereas the former involves a dynamical field that describes space-time itself.

7. May 7, 2013

WannabeNewton

By the way, in the linearized approximation you can view the perturbation $h_{ab}$ as a symmetric tensor field propagating on a background flat Minkowski space-time $(\mathbb{R}^{4},\eta_{ab})$, in which case the vacuum field equations come out of varying the action of the Lagrangian $\mathcal{L}_{G} = \frac{1}{2}[(\partial_{\nu}h)(\partial_{\mu}h^{\mu\nu}) - (\partial_{\mu}h^{\rho\sigma})(\partial_{\rho}h^{\mu}{}{}_{\sigma}) + \frac{1}{2}\eta^{\mu\nu}(\partial_{\mu}h^{\rho\sigma})(\partial_{\nu}h_{\rho\sigma}) - \frac{1}{2}\eta^{\mu\nu}(\partial_{\mu}h)(\partial_{\nu}h)]$ with respect to $h_{ab}$. This is of course akin to other classical fields that propagate on a background flat Minkowski space-time.

8. May 8, 2013

TrickyDicky

I don't think it is ambiguous if one thinks about GR as Einstein presented it in 1915, or at least what he aimed to present, wich was a geometric theory. It can be argued that during the last hundred years physicists have gotten used to think of GR in a different way, influenced at first by the varaitional aproach Hilbert initiated and Einstein quickly adopted, and later by the success of quantum field theories.
Leaving historical issues aside, yes I understand that nowadays when speaking about a field variable for a theory, a specific function for the manifold at hand is understood, I was just highlighting the peculiarity of GR. It is usually described as a classical field theory, but its particularity sets it apart from the rest of the classical field theories.

Well, to an extent it is a matter of taste what one considers "fundamental" right?
In this case maybe I'm missing something but just from the Fundamental theorem of Riemannian geometry it seems natural that the Palatini variation gives that the connection must be the unique Christoffel metric connection, I mean the connection is only a nominally independent variable from the metric. Note that if instead of the Einstein-Hilbert Lagrangian we use a different one with other fields like say spinor fields, we no longer obtain the the unique connection without torsion.
Right, so I guess you consider ambiguous to call the field describing the space-time in GR a geometric field because of the local nature of the gravitational field or something like that?

9. May 8, 2013

WannabeNewton

Well mathematically one view is not more fundamental than the other but beyond that sure personal taste can be involved I guess. As for your second remark, note that if we take the non-variational / non Palatini approach, there might not seem to be a clear physical justification from first principles for why in general relativity we want to use the derivative operator guaranteed to us by the fundamental theorem of Riemannian geometry but with the Palatini approach, if we use the same Lagrangian used in the Hilbert action but now view the derivative operator as an independent field variable, it comes out naturally from variation of the associated action that the derivative operator must be metric compatible. I find that quite remarkable since, IMO, there is no a priori reason to expect that a Lagrangian describing gravitation (in the sense of general relativity) would end up giving, via a variational principle, metric compatibility which we usually take for granted in general relativity in the non-variational approach thanks to the aforementioned theorem.

Nono I'm not objecting to calling it a "geometric field" because either way (Hilbert or Palatini action) we are talking about field variables that do pertain to structures related to the space-time manifold itself. What I'm saying is that the phrase "the field variable is the geometry" can be ambiguous because there are many different geometrical objects on a given smooth manifold.

10. May 8, 2013

TrickyDicky

Hmm... but isn't precisely this theorem of Riemannian geometry together with the fact that in GR we are dealing with a non-linear (curved) metric field demanding the use of a covariant derivative, what fully justifies physically the use of this specific nonlinear operator in GR?
Well as I said above, I find it a bit less remarkable taking into consideration the two elements I mentioned: pseudo-Riemannian geometry wich is an essential assumption in GR, and curvature of the metric tensor field variable (and this information is contained in its lagrangian) which I also think it's considered a basic assumption in GR.
But it's not at all an assumption for general field theories field variables so I also think that from that point of view the Palatini variation is rightly considered remarkable.

Ok.

BTW, you changed equations in your signature, what are the new ones?

11. May 8, 2013

WannabeNewton

Well we need some connection when talking about curved space-times in order to have a differentiation of tensor fields from tangent space to tangent space but aside from the usual arguments saying we want a connection that preserves the metric (inner product) during parallel transport, I have never seen a purely physical argument that naturally associates the metric compatibility condition with the general relativistic notion of gravitation if you go the non-variational route whereas the Palatini action does provide that; this is what I meant (I apologize if I wasn't clear before).

It's an exercise from Wald (problem 7.5 - chapter 7 is the one about methods for solving Einstein's equations) to show the equality in my signature actually holds true. I haven't been able to solve it as of yet (and it is getting really frustrating!) so I just put it there to remind me that if I don't solve it soon I might go mad and beat my copy of Wald to the ground :tongue2:. $\xi^{a}$ is a killing vector field, $\zeta = \xi^{a} \xi_{a}$, and $\alpha$ is a scalar field such that $\nabla_{a} \alpha = \epsilon_{abcd}\xi^{b}\nabla^{c}\xi^{d}$ (which is the twist of the killing vector field). Feel free to take a crack at solving it if you want and drop me some hints or something via pm before I do go clinically insane lol. Thanks for asking, that was nice of you :)

Last edited: May 8, 2013
12. May 8, 2013

TrickyDicky

No need to, you were clear. I guess my justification is mathematical rather than physical.
Haha, I see. I very much doubt that I can help you with it but if I can find some hint I'll let you know.

13. May 8, 2013

pervect

Staff Emeritus
The idea that connections are related to forces seems to me to be implicit in Einstein's equivalence principle. When we look at the famous Einstein elevator experiment, Einstein is saying that what is in Newtonian physics is a force, can be replaced by a system in which there is no force, the accelerating elevator.

What is it in the accelerating elevator that directly replaces the force? Physicists often talk loosely about "inertial forces", but it's long been known and taught that "inertial forces are fictitous forces". The whole idea of being "in" an accelerating elevator is just a change of coordinates, from an inertial reference frame to the non-inertial reference frame of the elevator.

Forces that transform as a tensor can't possibly suddenly appear if we change coordinates. But yet, we are trying to replace something that is in Newtonian physics, a force, with these "fictitious" forces.

We need something with the right units (meters/second^2). And, this something, in order to replace force, must appear directly to the equations of motion, which would be the geodesic equations in flat space-time. And it can't transform as a tensor, or it couldn't suddenly appear in our elevator just because we changed to elevator coordinates. The obvious candidate is the Christoffel symbol. I believe Einstein knew this, though I dont have any specific references (I wish I did - it can be hard to find references for the really simple stuff, alas).

I do have to agree that I haven't seen much on the topic in modern textbook, but the "connection" between connnections and the Newtonian idea of gravity is so compelling that I have to wonder why it's omitted. My best guess is that the whole topic of forces has been discounted - if you look up "force" in the usual GR textbook, you won't find much.

One might find an occasional reference that if you have an object moving under the influence of a force (such as an electric charge in both an electric field) "that you just add it to the geodesic equation, perhaps, which rather emphasizes the close relation between Christoffel symbols and forces.

The next point is showing that the Christoffel symbols transform properly - not under general coordinate transformations, we already know that they can't. But that they transform as tensors under some _restricted_ coordinate transformations. This is (or should be) very reminiscent of the way we insist on using "inertial frames" in Newtonian mechanics - we can't use just any old coordinate system (until we get to Lagrangian mechanics), we need to use specific ones.

The mathematics involved are not terribly complex. One starts off with the observation that a Christoffel symbol had 16 components, and a force has 4, one of which is zero for a 3-force. So we need to ignore most of the Christoffel symbols.

The next question is whether the 3 we pick out, $\Gamma^x{}_{tt}, \Gamma^y{}_{tt}, \Gamma^z{}_{tt}$ transform properly, and what restrictions are required.

I worked this out the necessary conditions for this approximation to be true myself in a short post. I really wish I would have seen something like this in a textbook, I don't feel like anything I've been doing should be "breaking new ground".

14. May 9, 2013

WannabeNewton

No worries, I managed to figure it out haha. Now to get back to studying for finals xD, thanks!

15. May 9, 2013

stevendaryl

Staff Emeritus
By "connection" do you mean the Christoffel coefficients $\Gamma^\mu_{\nu \lambda}$? How does it work to consider those as fundamental, since its not a tensor?

16. May 9, 2013

WannabeNewton

Not exactly. See here for the definition of an affine connection: http://en.wikipedia.org/wiki/Affine_connection
and here in particular for the Levi-Civita connection:
http://mathworld.wolfram.com/Levi-CivitaConnection.html

In particular if $(\partial_{i})_{i}$ is a coordinate basis then $\nabla_{\partial_{i}}\partial_{j} = \Gamma ^{k}_{ij}\partial_{k}$. This is the relationship between the Levi-Civita connection and the Christoffel symbols.

Fundamental here was being used to refer to functions in the argument of the action functional whose variation gives the vacuum field equations. The Palatini action above allows one to make the action $S$ have functional dependence on $g^{ab}$ as well as $\nabla_{a}$ (recall that the curvature endomorphism is actually a function of $\nabla_{a}$ in general, not $g_{ab}$; in GR we use the levi-civita connection so we just take the curvature endomorphism as a function of the metric tensor but we drop this restriction a priori for the Palatini action). Its variation gives both the vacuum field equations and metric compatability.

17. May 9, 2013