Confusion in General Relativity

[PeterDonis]

The distinction I was referring to was between Riemannian and pseudo-riemannian geometry.

I don't see any real distinction between the two with regard to the equations I wrote down; both of them are valid regardless of the metric signature.

You can see this explained here: https://www.physicsforums.com/threads/einstein-hilbert-action-origin.838487/ in posts #15 and #17

The point being made in those posts is that, if we look at the Einstein-Hilbert action for gravity alone, it can only be linear in second derivatives of the metric (i.e., in $R$, the Ricci scalar); otherwise we would get a third or higher order PDE for the field equation for gravity.

But we haven't been looking at the field equation for gravity; we've been looking at the field equation for electromagnetism, which is derived by varying the Lagrangian with respect to the vector potential $A_{\mu}$, not the metric. Terms of higher order in $R$ aren't functions of the vector potential, so, from the standpoint of Maxwell's Equations, if we're going to include any curvature terms, we should include them all, regardless of order.

We could, however, look at how adding those terms to the Lagrangian would affect the Einstein Field Equation; what we will find is that even the $\alpha$ term would add extra unwanted terms to that equation. The usual EFE is derived from the action (leaving out irrelevant factors in the terms and including the EM Lagrangian--we assume there is no other matter present):

$$
S = \int d^4 x \sqrt{-g} \left( R - \frac{1}{4} F_{\mu \nu} F^{\mu \nu} \right)
$$

Varying this with respect to the metric gives the usual EFE with the EM stress-energy tensor on the RHS (which I won't bother writing out explicitly since its form won't matter here):

$$
R_{\mu \nu} - \frac{1}{2} g_{\mu \nu} R = T_{\mu \nu}
$$

But now suppose we include the extra $\alpha$ term in the Lagrangian, so that varying with respect to the vector potential puts the extra $\alpha$ term in Maxwell's Equations:

$$
S = \int d^4 x \sqrt{-g} \left( R - \frac{1}{4} F_{\mu \nu} F^{\mu \nu} - \frac{1}{4} F_{\mu \nu} F^{\mu \nu} \alpha R \right)
$$

Now, when we vary this with respect to the metric, we get:

$$
\left( R_{\mu \nu} - \frac{1}{2} g_{\mu \nu} R \right) \left( 1 - \alpha \frac{1}{4} F_{\mu \nu} F^{\mu \nu} \right) = T_{\mu \nu} \left( 1 + \alpha R \right)
$$

This mess is clearly not a clean field equation for the metric (i.e., an equation where the LHS has no matter terms and the RHS has no curvature terms. (And if we try to make it into one by algebraically moving terms and factors around, what we end up with will not be a clean second-order PDE anyway, since it will have to divide the Einstein tensor by the factor $\left( 1 + \alpha R \right)$ on the LHS.)

In other words, you can't add any curvature terms to the field equations for matter (such as Maxwell's Equations) without also adding unwanted terms to the field equation for gravity (as above). That's a powerful argument in favor of the minimal coupling principle.

 

[loislane]

I don't see any real distinction between the two with regard to the equations I wrote down; both of them are valid regardless of the metric signature.

They both use the formulation of EM in pseudo-riemannian space, so how would you know? Are you aware of any formulation of the electromagnetic field tensor in Riemannian geometry to compare?

The point being made in those posts is that, if we look at the Einstein-Hilbert action for gravity alone, it can only be linear in second derivatives of the metric (i.e., in $R$, the Ricci scalar); otherwise we would get a third or higher order PDE for the field equation for gravity.

But we haven't been looking at the field equation for gravity; we've been looking at the field equation for electromagnetism, [...]

Well, this thread is about GR and the EP and so is the reference the equations are taken from, not about electromagnetism. The last equation is an example of what one equation of EM in the Lorentzian space formulation would look like in the curved spacetime of GR if the minimal coupling was not used. I agree they should have used the equation you wrote in your post instead if the point was about electromagnetism, but Carroll's point was about GR and the EP and I guess that is why he used 4.31 instead of your version. And that's why the posts I linked that are about GR and the EP in relation to the absence of higher derivatives are pertinent

 

[Dale]

The disagrement here was about teaching a "useful guideline" as if was a fundamental principle

So you dislike that the Equivalence Principle is called a principle? OK. But that is what it is called. Weren't you earlier complaining that other people were focusing on semantics?

Do you have a substantive objection or question about GR , or just want to complain about the "window dressings"?

 

[PeterDonis]

They both use the formulation of EM in pseudo-riemannian space, so how would you know?

Are we talking about mathematics, or about physics? If we're talking about physics, then the objections you're raising don't seem relevant, since they're mathematical, not physical. You can't argue that a particular term "should" be in a physical equation because "differential geometry says so". What belongs in a physical equation depends on the physics.

If we're talking about mathematics, then my statement means just what it says: you can't tell, just by looking at either of the equations under discussion, what metric signature they apply to. They are valid equations, mathematically, in a Riemannian metric just as they are in a pseudo-Riemannian metric. Of course they say different things physically if the metric is Riemannian vs. pseudo-Riemannian, but that's a question of physics, not differential geometry.

the posts I linked that are about GR and the EP in relation to the absence of higher derivatives are pertinent

Do you have any comment on the rest of my post, where I argue that if those arguments are pertinent, they mean that the $\alpha$ term also doesn't belong?

 

[loislane]

If we're talking about mathematics, then my statement means just what it says: you can't tell, just by looking at either of the equations under discussion, what metric signature they apply to. They are valid equations, mathematically, in a Riemannian metric just as they are in a pseudo-Riemannian metric. Of course they say different things physically if the metric is Riemannian vs. pseudo-Riemannian, but that's a question of physics, not differential geometry.

I guess I'm missing something here, or you are missing my point. What I'm saying is that the electromagnetic field tensor in those equations is formulated in Lorentzian space, again is there a formulation of the electromagnetic tensor that doesn't have indefinite signature, i.e. that uses a Riemannian metric?

Do you have any comment on the rest of my post, where I argue that if those arguments are pertinent, they mean that the $\alpha$ term also doesn't belong?

No. It doesn't sem obviously wrong to me but my knowledge of physics is limited.

 

[martinbn]

I am a bit confused what the discussion is about? I always thought that the equivalence principle is quite clear and any non-physicist would understand it (that includes mathematicians). There are a couple of threads discussing it and it seems that there are different opinions, but I am not sure what they are. Loislane, can you state what your position is as clearly as possible? Now it seems to me that you are arguing for the sake of arguing and I am sure I am wrong.

 

[PeterDonis]

What I'm saying is that the electromagnetic field tensor in those equations is formulated in Lorentzian space

Physically, yes, we only use the version of those equations that is formulated with a Lorentzian metric signature. But that's a physical restriction, not a mathematical restriction. Mathematically, the equations themselves are perfectly valid equations with a Riemannian metric.

In other words, it isn't differential geometry that tells us to use a Lorentzian metric signature when we apply the equations; it's physics, picking a particular formulation out of all the possible ones that are valid mathematically.

 

[loislane]

Physically, yes, we only use the version of those equations that is formulated with a Lorentzian metric signature. But that's a physical restriction, not a mathematical restriction. Mathematically, the equations themselves are perfectly valid equations with a Riemannian metric.

Let's consider 4.24, how could it be valid in the context of Riemann geometry with definite positive metric? Let's say $F^{\mu\nu}$ is a general second order tensor field(meaning not necessarily euclidean), it's divergence would possibly include corrections in the form of Christoffel coefficients, what do you do with the nonvanishing connection coefficients? You are assuming they are zero as if it could only be a Euclidean tensor.

 

[PeterDonis]

Let's consider 4.24, how could it be valid in the context of Riemann geometry with definite positive metric?

Um, because you can take the covariant divergence of a tensor field, and validly equate it to some vector field, just as easily in Riemannian geometry as in pseudo-Riemannian geometry? I don't understand what the problem is. Remember we are talking math, not physics, in this particular case; we're not putting any physical interpretation on $F^{\mu \nu}$ or $J^{\nu}$, we're just taking the covariant divergence of the first and equating it to the second. Mathematically, that can obviously be done in Riemannian geometry.

If you want a concrete example, suppose $F^{\mu \nu}$ and $J^{\nu}$ are tensor and vector fields on a 2-sphere, a curved Riemannian manifold. Are you saying we can't take the covariant divergence of $F$, using the 2-sphere's metric, and equate it to $J$?

Let's say $F^{\mu\nu}$ is a general second order tensor field(meaning not necessarily euclidean)

What is a "Euclidean" tensor field? I understand what a Euclidean metric is, but we're not assuming the metric (in the Riemannian case you are talking about) is Euclidean. Tensor fields other than the metric aren't Euclidean or non-Euclidean (or, in the pseudo-Riemannian case, Minkowskian or non-Minkowskian). They're just tensor fields on the manifold, which has whatever metric we are assuming for the particular problem under discussion.

it's divergence would possibly include corrections in the form of Christoffel coefficients, what do you do with the nonvanishing connection coefficients?

Um, the same thing you do with them in a curved pseudo-Riemannian metric? Are you saying that Christoffel symbols somehow don't work in Riemannian geometry?

You are assuming they are zero as if it could only be a Euclidean tensor.

I'm assuming no such thing. The equation under discussion has a covariant divergence in it, not an ordinary divergence; the whole point of using $\nabla$ instead of $\partial$ is to emphasize that we are in a curved manifold so you have to include the connection coefficients when you take derivatives.

 

[martinbn]

Let's consider 4.24, how could it be valid in the context of Riemann geometry with definite positive metric? Let's say $F^{\mu\nu}$ is a general second order tensor field(meaning not necessarily euclidean), it's divergence would possibly include corrections in the form of Christoffel coefficients, what do you do with the nonvanishing connection coefficients? You are assuming they are zero as if it could only be a Euclidean tensor.

This makes no sense! For tensors you only need a manifold, so you can have a tensor $F^{\mu\nu}$ (antisymetric or not) on a given manifold. If on top of that you have a connection, say from a Riemannian metric, then you can differentiate the tensor.

 

[loislane]

If you want a concrete example, suppose $F^{\mu \nu}$ and $J^{\nu}$ are tensor and vector fields on a 2-sphere, a curved Riemannian manifold. Are you saying we can't take the covariant divergence of $F$, using the 2-sphere's metric, and equate it to $J$?
Um, the same thing you do with them in a curved pseudo-Riemannian metric? Are you saying that Christoffel symbols somehow don't work in Riemannian geometry?

No, I was asking you where are the Christoffel symbols in 4.24.

I'm assuming no such thing. The equation under discussion has a covariant divergence in it, not an ordinary divergence; the whole point of using $\nabla$ instead of $\partial$ is to emphasize that we are in a curved manifold so you have to include the connection coefficients when you take derivatives.

Exactly, I gues what you are saying is that they are implicit in the equation in vector field in the rhs.
I think we are basically talking at cross purposes here, so I'm leaving the thread, (apparently there are some attempts at trolling too so it is better to leave it alone).

 

[PeterDonis]

I was asking you where are the Christoffel symbols in 4.24.

Um, by definition, the $\nabla$ operator includes them. That's why I said the whole point of writing $\nabla$ instead of $\partial$ in a curved manifold is to ensure that the Christoffel symbols are taken into account.

I gues what you are saying is that they are implicit in the equation in vector field in the rhs.

No, they are implicit in the $\nabla$ operator on the LHS.

 

[loislane]

So in your own words, since you agreed with Caroll's quote, why doesn't the EP deserve to be treated as a fundamental principle?

And subsequently why should the practical consequence of something that is not a fundamental principle, i.e. the minimal coupling, should be taught as something exact in GR?

 

[PeterDonis]

why doesn't the EP deserve to be treated as a fundamental principle?

Because GR is not a theory of everything, and on quantum field theoretical grounds, we expect the additional terms that GR does not include, because of the minimal coupling principle, to be present, but suppressed by inverse powers of something like the Planck mass, which makes them completely unobservable in any experimental regime we are likely to be able to reach any time soon (or even not so soon). So including them in our models would just clutter up the models to no purpose, since our calculations would crank out a bunch of additional terms that are harder to deal with mathematically and predict effects that are way too small to measure.

why should the practical consequence of something that is not a fundamental principle, i.e. the minimal coupling, should be taught as something exact in GR?

Because, as I said before, if you're going to teach GR, you teach GR. In GR, the minimal coupling principle is exact, so that's how we teach it. Just as, when we teach Newtonian physics, we treat the principle of conservation of mass as exact, without cluttering students' heads with all the complications brought in by mass-energy equivalence in relativity. Or, when we teach special relativity, we treat flat Minkowski spacetime as exact, without cluttering students' heads with all the complications brought in by the fact that there is no such thing as true flat Minkowski spacetime in the real world, since there is nonzero stress-energy present.

In other words, physics is not taught as the theory of everything, with all its complications, all in one lump. It is taught in stages, and at each stage things are taught as exact which turn out to be not quite exact in later stages.

 

[loislane]

Because GR is not a theory of everything, and on quantum field theoretical grounds, we expect the additional terms that GR does not include, because of the minimal coupling principle, to be present, but suppressed by inverse powers of something like the Planck mass, which makes them completely unobservable in any experimental regime we are likely to be able to reach any time soon (or even not so soon).

You mean like GR is some sort of classical "effective field theory"?

 

[PeterDonis]

You mean like GR is some sort of classical "effective field theory"?

That is pretty much the current mainstream view, yes. The question we don't yet know the answer to is what underlying quantum theory it is the classical effective field theory of.

 

[loislane]

That is pretty much the current mainstream view, yes. The question we don't yet know the answer to is what underlying quantum theory it is the classical effective field theory of.

I see. I must say I can see why the OP is a bit confused by curved space-time being locally flat (Minkowskian). It is not easy to grasp but in fact it comes straight from the indefinite signature geometry in a curvature context. In the absence of a particular absolute invariant group, like say the Lorentz group in the Minkowski space of SR, the general symmetry of GR is reduced to its coordinate covariance(invariance under general local coordinate transformations), in other words a trivial property of any theory is raised to the category of physical symmetry(dynamical metric-gravitational fields).

 

[PeterDonis]

I can see why the OP is a bit confused by curved space-time being locally flat (Minkowskian). It is not easy to grasp but in fact it comes straight from the indefinite signature geometry in a curvature context.

It has nothing to do with the indefinite signature. A Riemannian manifold is locally Euclidean in the same way that a pseudo-Riemannian manifold of the type used in GR is locally Minkowskian.

 

[martinbn]

I see. I must say I can see why the OP is a bit confused by curved space-time being locally flat (Minkowskian). It is not easy to grasp but in fact it comes straight from the indefinite signature geometry in a curvature context.

As PeterDonis said the signature is irrelevant. To me the intuitive picture is that there are no sharp point or edges, the space-time has a tangent space at each point and in a small enough neighborhood the space-time and the tangent space are almost the same (including the metric).

In the absence of a particular absolute invariant group, like say the Lorentz group in the Minkowski space of SR, the general symmetry of GR is reduced to its coordinate covariance(invariance under general local coordinate transformations), in other words a trivial property of any theory is raised to the category of physical symmetry(dynamical metric-gravitational fields).

I don't understand this and how it is related to the above.

 

[loislane]

It has nothing to do with the indefinite signature. A Riemannian manifold is locally Euclidean in the same way that a pseudo-Riemannian manifold of the type used in GR is locally Minkowskian.

Not in the same way. That's an important misconception you should look into.
-Being locally euclidean has nothing to do with the Euclidean metrics or signature types, it is a topological or differentiable structure equivalence with ℝn locally depending on the kind of manifold(topological or smooth) one is talking about. It has nothing to do with having a Euclidean metric no matter how small a region you want to consider.
-Being locally Minkowskian obviously involves having locally a specific Minkowskian metric tensor with vanishing curvature and a specific signature.
This is basically the EP content and one could infer logically it is the reason why mathematically it must be limited to the first derivatives of the metric, otherwise it would collide with the differential geometry concept of curvature at a point.
I think that is what is behind the split between the "spacetime tells mtter how to move" and "matter tells spacetime how to curve" separate parts put forward by Wheeler and also determines the peculiar implementation of "general covariance" in GR that I refer to in my last post.

 

[martinbn]

Not in the same way. That's an important misconception you should look into.
-Being locally euclidean has nothing to do with the Euclidean metrics or signature types, it is a topological or differentiable structure equivalence with ℝn locally depending on the kind of manifold(topological or smooth) one is talking about. It has nothing to do with having a Euclidean metric no matter how small a region you want to consider.

That is true, but in the context of this discussion, it is clear what PeterDonis means.

 

[PeterDonis]

Being locally euclidean has nothing to do with the Euclidean metrics or signature types

It depends on how we interpret the word "Euclidean"--it can be a purely topological term, as you claim here, or it can be a term describing the metric. I was using it in the latter sense. The former sense is irrelevant to the discussion because we are talking about equations with covariant derivatives in them, and you can't take covariant derivatives unless you have a metric.

I think that is what is behind the split between the "spacetime tells matter how to move" and "matter tells spacetime how to curve" separate parts put forward by Wheeler and also determines the peculiar implementation of "general covariance" in GR that I refer to in my last post.

I disagree. All of these things require a metric. The purely topological sense of "Euclidean" is irrelevant.