Why Can't Flat Spacetime Connection Be Used in GR Covariant Derivative?

[dEdt]

In flat spacetime, there exists a natural connection that allows you to compare two vectors at different events. Now, in GR the spacetime in the neighborhood of an event is flat. It therefore seems possible to define the connection between two events close to one another to simply be the connection you'd get in flat spacetime. Why can't this connection be used in the formulation of the covariant derivative, rather than the Levi-Civita connection?

 

[Mentz114]

I don't understand what you're asking, but the local Minkowski basis has to be parallel transported along a curve - which is where the connection comes in.

 

[WannabeNewton]

Now, in GR the spacetime in the neighborhood of an event is flat.

No it isn't. Where did you see that?

I'm afraid the rest of your post makes little to no sense mathematically.

 

[pervect]

In flat spacetime, there exists a natural connection that allows you to compare two vectors at different events. Now, in GR the spacetime in the neighborhood of an event is flat. It therefore seems possible to define the connection between two events close to one another to simply be the connection you'd get in flat spacetime. Why can't this connection be used in the formulation of the covariant derivative, rather than the Levi-Civita connection?

I think you may be taking local flatness a bit too seriously. I'm not sure I positively understand your question, but I'll try to answer what I think you are asking.

Local flatness does not mean that the Riemann curvature tensor literally vanishes, it just means that you can ignore its effects.

For instance, if you look at the geodesic deviation between two parallel geodesics , it is proportional by the geodesic equation to the separation between geodesics. If the geodesics are close enough that the relative acceleration between geodesics due to the geodesic equation is negligible, you have local flatness.

NOte that the vanishing of the Riemann tensor is equivalent to saying that $\nabla_a \nabla_b = \nabla_b \nabla_a$. In flat space, the natural connection is also the Leva Civita connection, but the above commutator vanishes. In curved spaces, it doesn't, and we have a nonzero Riemann.

 

[dEdt]

Obviously I was lazy when I worded my question. If you'll indulge me I will give it another shot.

At any event it's always possible to set up a coordinate system such that the metric (at that event) is diagonal and all the partial derivatives of the metric components vanish (again at that event). I've always interpreted this statement to mean that the region of spacetime around an event is approximately flat, and that this approximation gets better as you take a smaller and smaller region. To use an analogy, a small region of a sphere's surface can be approximated as a flat plane, and this approximation becomes exact in the limit that the size of this region goes to zero.

Let's continue using the sphere for simplicity. If we have a vector field on the sphere, it's obviously a non-trivial task to compare vectors at two separate points because the sphere is curved ie there's no immediately natural map from $T_p$ to $T_{p'}$. Now suppose we wish to construct a derivative operator to act on this vector field. Ideally we'd like to find the difference between the vectors at two nearby points, but this isn't possible without some map between $T_p$ and $T_{p'}$. BUT, if $p$ and $p'$ are "close" to one another, then the region of space containing both points is approximately flat. Further, this approximation becomes exact as $p'$ gets ever closer to $p$, which is exactly what happens as we compute a derivative. Given that there's a natural map between the tangent space at two points when the space is flat, why do we have to go through all the trouble of introducing the Levi-Civita connection when we calculate covariant derivatives? Why can't we just use the "flat space" connection?

 

[PeterDonis]

why do we have to go through all the trouble of introducing the Levi-Civita connection when we calculate covariant derivatives? Why can't we just use the "flat space" connection?

You are incorrectly assuming that these are two different and distinct choices. They're not. The "flat space" connection *is* the Levi-Civita connection--when the space is flat and you are using Cartesian coordinates.

If you are working entirely within a local inertial frame (i.e., within a small enough region of spacetime that all effects of curvature are negligible), and you are using a Minkowski chart in that frame (i.e, the metric in your chart is $\eta_{ab}$), then all of the connection coefficients in the chart you are using are zero, and covariant derivatives are just partial derivatives. But that's not because the "flat space" connection is somehow different from the Levi-Civita connection: it's just because the Levi-Civita connection reduces to the "flat space" connection in this particular case.

To see this another way, suppose we are using polar coordinates instead of Cartesian coordinates on a plane. Then the connection coefficients are *not* all zero, even though the space we are in (the plane) is flat. So when you compute covariant derivatives, you can't just equate them to partial derivatives; you have to take the connection coefficients into account or you will get the wrong answers. In other words, even in a flat space, the correct connection is really the Levi-Civita connection. The standard Minkowski coordinates in flat spacetime just obscure this fact by happening to have connection coefficients that are all identically zero.

 

[dEdt]

Thank you PeterDonis. Final question: is the Levi-Civita connection the only connection that reduces to the flat-space connection, or is this a property of all well-defined connections?

 

[WannabeNewton]

Thank you PeterDonis. Final question: is the Levi-Civita connection the only connection that reduces to the flat-space connection, or is this a property of all well-defined connections?

It works for any symmetric linear connection $\nabla$ meaning $T(X,Y) = \nabla_X Y - \nabla_Y X - [X,Y] = 0$ must hold. $T$ is called the torsion of $\nabla$.

 

[haael]

If someone is curious, nonzero torsion and its physical consequences are analyzed by Einstein-Cartan theory.

 

[dEdt]

WannabeNewton

I'm having trouble seeing how there could be more than one connection which reduces to the "flat space connection" in a locally inertial coordinate system. After all, if you have a locally inertial coordinate system with origin at event $p$, and you assert that the covariant derivative of a vector field $A^\alpha$ at $p$ is just $\partial_\alpha A^\beta (p)$ in this coordinate system, then it seems like you've uniquely specified the covariant derivative, and all that's left is to make the equation manifestly covariant. Am I wrong?

 

[PeterDonis]

is the Levi-Civita connection the only connection that reduces to the flat-space connection, or is this a property of all well-defined connections?

I'm not sure this question makes sense as you state it. "Reducing to the flat-space connection" just means covariant derivative = partial derivative when all the connection coefficients are zero. That's a tautology. Once again, the "flat space connection" is not some different connection that the Levi-Civita connection "reduces" to. It's just what you get when all the connection coefficients are zero.

 

[pervect]

WannabeNewton

I'm having trouble seeing how there could be more than one connection which reduces to the "flat space connection" in a locally inertial coordinate system. After all, if you have a locally inertial coordinate system with origin at event $p$, and you assert that the covariant derivative of a vector field $A^\alpha$ at $p$ is just $\partial_\alpha A^\beta (p)$ in this coordinate system, then it seems like you've uniquely specified the covariant derivative, and all that's left is to make the equation manifestly covariant. Am I wrong?

I believe one can define a connection in flat space (I'm thinking of a 2d euclidean space, not space-time, because it's simpler) that has torsion, which would make it different from the torsion free Levi-Civita connection in flat space that you are calling "the flat space connection".

But I don't really do much with torsion, so I can't provide an example. I can say that the absence of torsion can be regarded as requiring that geodesic parallograms close to third order (see Penrose, "Road to Reality", pg 307, if you happen to have it). And that this is related to an interesting geometric construct for parallel transport called Schild's ladder. But unfortunately these remarks are all about the absence of torsion, and what we want to do to answer your question is exhibit a connection that has it.

 

[dEdt]

I'm not sure this question makes sense as you state it. "Reducing to the flat-space connection" just means covariant derivative = partial derivative when all the connection coefficients are zero. That's a tautology. Once again, the "flat space connection" is not some different connection that the Levi-Civita connection "reduces" to. It's just what you get when all the connection coefficients are zero.

Here's (hopefully) a more precise way of wording my question: if we assert that the connection coefficients must vanish in a locally inertial coordinate system, do we arrive at a unique connection?

I've convinced myself that the answer should be 'yes', but I can't prove it. Here's my attempt at such a proof:

Let $A$ be a vector field, and suppose in a locally inertial coordinate system at event $p$ the components of $A$ are $A^{\alpha'}$. I define the covariant derivative of $A$ at $p$, denoted by $\nabla A(p)$, to be a (1,1) tensor such that in this locally inertial coordinate system its components take the form $A^{\alpha'}{}_{;\beta'}(p)=\partial_{\beta'}A^{\alpha'}(p)$ ($A^{\alpha'}{}_{;\beta'}$ being the components of $\nabla A$, obviously). The only thing left to do is to find the components of $\nabla A$ in a generic coordinate system, which can be done by applying the transformation law for tensors: $A^\alpha{}_{;\beta}=\frac{\partial x^\alpha}{\partial x^{\alpha'}} \frac{\partial x^{\beta'}}{\partial x^\beta}\partial_{\beta'}A^{\alpha'}$. After some algebra, I found that
A^\alpha{}_{;\beta}=\partial_\beta A^\alpha + \left(-\frac{\partial x^{\alpha'}}{\partial x^\gamma}\partial_\beta \frac{\partial x^\alpha}{\partial x^{\alpha'}} \right) A^\gamma.

If we define $\Gamma^\alpha_{\beta\gamma}=-\frac{\partial x^{\alpha'}}{\partial x^\gamma}\partial_\beta \frac{\partial x^\alpha}{\partial x^{\alpha'}}$, then it seems like we've found the unique connection that reduces to regular partial derivatives in locally inertial coordinate systems.

My only problem with this proof is that I've been unable to get a formula for $\Gamma^\alpha_{\beta\gamma}$ in terms of the metric and its derivatives, which would allow me to show that we just get the regular Christoffel symbols. I haven't even been able to prove that this connection is torsion free! So I'm doubting my "proof".

 

[PeterDonis]

The only thing left to do is to find the components of $\nabla A$ in a generic coordinate system, which can be done by applying the transformation law for tensors

No, it can't, because the connection coefficients aren't tensors, not even the standard Levi-Civita ones. They obviously can't be tensors, because you can make them vanish at a particular event by switching to a local inertial coordinate chart, even if they are non-vanishing at the same event in a generic chart. You can't do that with tensors.

I've been unable to get a formula for $\Gamma^{\alpha}{}_{\beta \gamma}$ in terms of the metric and its derivatives

Every relativity textbook I'm aware of has them. Also see Wikipedia:

http://en.wikipedia.org/wiki/Christoffel_symbols

 

[dEdt]

No, it can't, because the connection coefficients aren't tensors, not even the standard Levi-Civita ones. They obviously can't be tensors, because you can make them vanish at a particular event by switching to a local inertial coordinate chart, even if they are non-vanishing at the same event in a generic chart. You can't do that with tensors.

That doesn't stop $\nabla A$ from being a tensor.

Every relativity textbook I'm aware of has them.

You misunderstand. I'm defining $\Gamma^\alpha_{\beta\gamma}$ to be $-\frac{\partial x^{\alpha'}}{\partial x^\gamma}\partial_\beta \frac{\partial x^\alpha}{\partial x^{\alpha'}}$, and trying to show that you can recuperate the regular formula for the Christoffel symbols from this equation. Ie, I'm trying to show that
-\frac{\partial x^{\alpha'}} {\partial x^\gamma}\partial_\beta \frac{\partial x^\alpha}{\partial x^{\alpha'}}= \frac{1}{2} g^{\alpha \delta} \left( \partial_\beta g_{\gamma \delta} +\partial_\gamma g_{\delta \beta}- \partial_\delta g_{\beta \gamma}\right).

 

[PeterDonis]

I'm defining $\Gamma^\alpha_{\beta\gamma}$ to be $-\frac{\partial x^{\alpha'}}{\partial x^\gamma}\partial_\beta \frac{\partial x^\alpha}{\partial x^{\alpha'}}$, and trying to show that you can recuperate the regular formula for the Christoffel symbols from this equation.

Then I don't understand what you mean by "a formula for $\Gamma^{\alpha}{}_{\beta \gamma}$ in terms of the metric and its derivatives". You're applying a coordinate transformation to the components of $\nabla A$; that coordinate transformation must already contain all the information about how to transform the metric and its derivatives. You're basically defining $\Gamma^{\alpha}{}_{\beta \gamma}$ as whatever is left over in the transformed components, besides partial derivatives with respect to the transformed coordinates. That information must already be inherent in whatever coordinate transformation you pick.

 

[dEdt]

You're applying a coordinate transformation to the components of $\nabla A$; that coordinate transformation must already contain all the information about how to transform the metric and its derivatives. You're basically defining $\Gamma^{\alpha}{}_{\beta \gamma}$ as whatever is left over in the transformed components, besides partial derivatives with respect to the transformed coordinates. That information must already be inherent in whatever coordinate transformation you pick.

Exactly. But I believe that it should be possible to express "whatever is left over in the transformed components" in terms of the components of the metric and their partial derivatives in that coordinate system-- at least, this should be possible if my claim that "connection coefficients = 0 in local inertial coordinate system" implies "Levi-Civita connection" is true.

 

[Markus Hanke]

I am no expert on the matter, but to my understanding, if your coordinate system is a locally inertial one, then the connection coefficients vanish by definition. This is true if you globally use a Levi-Civita connection, but it would also be true of you use any other connection, for example Weizenboeck's. The fact that any small enough free-fall frame is locally Minkowskian ( i.e. the laws of SR apply ) couldn't depend on which connection you are working with, can it ?

 

[pervect]

I'm not sure of the provenance (though it looks good), but http://www.slimy.com/~steuard/teaching/tutorials/GRtorsion.pdf

says

Theorem 3.1.1
Let $g_{ab}$ be a metric and $T^c{}_{ab}$ be a torsion. Then there exists a unique derivative operator
$\nabla_a$ with this torsion satisfying $\nabla_a g_{bc} = 0$

If this is true, then for any value of torsion we specify, there is a unique derivative operator with this torsion that is compatible with the metric.

A different value of torsion implies different connection coefficients. (The connection coefficients are written out in 3.1.30 in the above paper).

None of my GR textbooks cover the case with torsion to confirm the truth of this statement. This web article has a couple of citations - but I don't see any publication history.

 

[WannabeNewton]

None of my GR textbooks cover the case with torsion to confirm the truth of this statement. This web article has a couple of citations - but I don't see any publication history.

It's an exercise from chapter 3 of Wald. It's very easy to show.

 

[pervect]

Baez has a little tutorial on torsion that doesn't answer the original question, but is helpful in understanding what torsion is.

http://math.ucr.edu/home/baez/gr/torsion.html

I also like the section in Penrose "Road to Reality" that talks about this.

If we compare the connection with torsion to the Levi-Civita connection, a connection with nonzero torsion causes vectors to rotate when we parallel transport them. In the example Baez shows, the vectors point up, after being parallel transorted to the right, the vector is rotated by an angle that's proportional to the distance transported.

So the connection with torsion seems a bit artificial it has a different notion of "parallel" that while mathematically consistent, is different from the intuitive one.

Also, it seems reasonably clear to me now that the Christoffel symbols are only zero for an inertial observer if he chooses the Levi-Civita connection.

 

[dextercioby]

Pervect, the geometric meaning of torsion is the subject of many sources. I particularly like the one by the Japanese (mathematical) physicist Nakahara.

https://www.amazon.com/dp/0750306068/?tag=pfamazon01-20

 

[pervect]

Pervect, the geometric meaning of torsion is the subject of many sources. I particularly like the one by the Japanese (mathematical) physicist Nakahara.

https://www.amazon.com/dp/0750306068/?tag=pfamazon01-20

Thanks, I don't have the textbook in question, can you give a brief summary of the author's position?

 

[dextercioby]

He builds the torsion tensor on pure geometric grounds, from the failure of a 'reasonable' parallel transport. What I like is that the sections on the geometric meanings of curvature and torsion fit nicely in his very well written standard diff. geom. stuff which is about 60% of his book. I would consider Nakahara as the mathematical companion to Wald or Hawking/Ellis.