Confusion in General Relativity

[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.