Metric constraints in choosing coordinates

[TrickyDicky]

It's an interesting idea. But does this orthonormality property also imply the frame is orthonormal in the conventional sense under the metric? If so, how?

Not sure what you mean here, what conventional sense? All Riemannian manifolds can be given orthonormal frames locally and they are obviously orthonormal wrt the metric.

It seems strange to me because I would think that the choice of frame should not actually matter; parallel transport will conserve the length of the vector no matter what frame you choose, is my intuition.

Well the fact that spacetime manifolds are dynamic introduces some complications in that intuition.
In GR an orthonormal frame always exists locally, that is in the neighbourhood of any point of the manifold; but the existence of an orthonormal frame globally (wich one would need to guarantee that the choice of local frame doesn't matter) would require some global topology information which is not available in GR, that is only concerned with the local topology.
IOW there is no unique way of extending the choice of frame at an event to other points, and so a choice must be made, unlike in SR.

 

[NanakiXIII]

Sort of. One note, though; I got things backwards as far as which coordinates are which. The $x'^{\alpha}$ coordinates are the ones for the arbitrary coordinate system, and the $x^{\mu}$ coordinates are those of the local Lorentz frame. The equation I wrote down with the M, N, and P coefficients gives the coordinates in the arbitrary frame in terms of the coordinates in the local Lorentz frame.



The coefficients M, N, P will appear when you take derivatives of $x'^{\alpha}$ with respect to $x^{\mu}$, yes. But they also will appear in the transformation of the metric, because the transformation matrix for the metric contains the derivatives of $x'^{\alpha}$ with respect to $x^{\mu}$:

$$g_{\mu \nu} = \frac{\partial x'^{\alpha}}{\partial x^{\mu}} \frac{\partial x'^{\beta}}{\partial x^{\nu}} g'_{\alpha \beta}$$

Similarly, we find that the transformation of first derivatives of the metric contains second derivatives of $x'^{\alpha}$ with respect to $x^{\mu}$, and the transformation of second derivatives of the metric contains third derivatives of $x'^{\alpha}$ with respect to $x^{\mu}$.

They may appear, but I don't see how they constrain anything. If $g'_{\alpha \beta}$ has degrees of freedom, then so does

$$g_{\mu \nu} = \frac{\partial x'^{\alpha}}{\partial x^{\mu}} \frac{\partial x'^{\beta}}{\partial x^{\nu}} g'_{\alpha \beta}$$

even if you fix the coordinate derivatives you you contract with.

I'm not sure how you're coming to that conclusion. From the above, it should be clear that the M coefficients can only affect the transformation of the metric coefficients themselves, because the M coefficients only appear in terms like $\partial x'^{\alpha} / \partial x^{\mu}$; they don't appear in any higher derivatives. So the M coefficients can't affect the transformation of the first or second derivatives of the metric. Similarly, the N coefficients can't affect the transformation of second derivatives of the metric; only the P coefficients can.

That is my point exactly. If the M coefficients can do nothing but constrain the metric, and they do indeed constrain the metric, how can you suddenly choose coordinates in which some of the degrees of freedom are in the metric, rather than in the second derivatives, like for example in linearized gravity?

Not sure what you mean here, what conventional sense? All Riemannian manifolds can be given orthonormal frames locally and they are obviously orthonormal wrt the metric.

Well the fact that spacetime manifolds are dynamic introduces some complications in that intuition.
In GR an orthonormal frame always exists locally, that is in the neighbourhood of any point of the manifold; but the existence of an orthonormal frame globally (wich one would need to guarantee that the choice of local frame doesn't matter) would require some global topology information which is not available in GR, that is only concerned with the local topology.
IOW there is no unique way of extending the choice of frame at an event to other points, and so a choice must be made, unlike in SR.

Let me rephrase. I mean, does it imply that the frame is orthonormal in the sense that all its basis vectors are unit vectors all mutually orthogonal? Because I am not sure I see the connection between this property and the property that the Riemann tensor is in SO(n).

 

[NanakiXIII]

I've understood the argument in the link better; but I am still not sure how the two ways of counting are equivalent. I did have an idea about how they are different, though, but I am not sure if it is correct and I'm hoping some one can help me out:

When you're talking about the metric and diffeomorphism invariance parameterized by four functions, you're thinking of specifying the metric globally (or at least in some extended part of the manifold) and therefore you immediately fix all the derivatives as well.

When you're talking about the metric and its first and second derivatives as independent, you're talking about the values these things take at a single point. Technically, you would need to specify all the derivatives, not just the first and second ones, for this to be equivalent to knowing the metric globally, but higher order derivatives do not appear in GR, so you do not expect degrees of freedom there.

I feel like I've hit the solution to my conceptual problem, at least in reconciling these two ways of looking at it, but if anything is erroneous about this, please let me know.

 

[PeterDonis]

If $g'_{\alpha \beta}$ has degrees of freedom, then so does

$$g_{\mu \nu} = \frac{\partial x'^{\alpha}}{\partial x^{\mu}} \frac{\partial x'^{\beta}}{\partial x^{\nu}} g'_{\alpha \beta}$$

even if you fix the coordinate derivatives you contract with.

How do you figure that? The whole point is to set the coordinate derivatives so that all ten components of $g_{\mu \nu}$ are fixed to the Minkowski values. That means the degrees of freedom in $g'_{\alpha \beta}$ are "cancelled", so to speak, by the freedom to choose the derivatives.

If the M coefficients can do nothing but constrain the metric, and they do indeed constrain the metric, how can you suddenly choose coordinates in which some of the degrees of freedom are in the metric, rather than in the second derivatives, like for example in linearized gravity?

Because in the latter case, the M coefficients don't fully constrain the metric. The M coefficients are different for different transformations.

In fact, for a general coordinate transformation, as opposed to one that is specifically designed to set the metric coefficients to their Minkowski values, I'm not sure the expansion in terms of the M, N, P coefficients is valid. MTW only discuss it in the context of a transformation to a local Lorentz frame, not as a general transformation.

 

[PeterDonis]

When you're talking about the metric and diffeomorphism invariance parameterized by four functions, you're thinking of specifying the metric globally (or at least in some extended part of the manifold) and therefore you immediately fix all the derivatives as well.

When you're talking about the metric and its first and second derivatives as independent, you're talking about the values these things take at a single point. Technically, you would need to specify all the derivatives, not just the first and second ones, for this to be equivalent to knowing the metric globally, but higher order derivatives do not appear in GR, so you do not expect degrees of freedom there.

This looks reasonable to me.

 

[NanakiXIII]

How do you figure that? The whole point is to set the coordinate derivatives so that all ten components of $g_{\mu \nu}$ are fixed to the Minkowski values. That means the degrees of freedom in $g'_{\alpha \beta}$ are "cancelled", so to speak, by the freedom to choose the derivatives.

I see your point now. I think I was again confusing global knowledge of the metric with knowledge in just one point.

I think I've grasped this bit now and I'd like to thank you all for your elaborate explanations.

Of course, I'm not satisfied yet! The next thing I want to know is this: in linearized gravity, you get to throw out four more degrees of freedom. Again, I've seen the handwaving "you can do this (add a derivative term if I remember correctly), so we have more gauge freedom" explanation, but what exactly is the symmetry and why does it appear only in linearized gravity?

 

[WannabeNewton]

Are you taking about the $\partial ^{b}h _{ab} = 0$ condition? It comes out of the fact that two different perturbations propagating on the background minkowski space - time, $h_{ab},h'_{ab}$ represent the same physical space - time if there exists a vector field $\xi ^{a}$ such that $h'_{ab} = h_{ab} - \mathcal{L}_{\xi }\eta _{ab}$. Using this you can derive the condition $\partial ^{b}h _{ab} = 0$.

 

[NanakiXIII]

Are you taking about the $\partial ^{b}h _{ab} = 0$ condition? It comes out of the fact that two different perturbations propagating on the background minkowski space - time, $h_{ab},h'_{ab}$ represent the same physical space - time if there exists a vector field $\xi ^{a}$ such that $h'_{ab} = h_{ab} - \mathcal{L}_{\xi }\eta _{ab}$. Using this you can derive the condition $\partial ^{b}h _{ab} = 0$.

That is just choosing harmonic coordinates, right? It reduces the degrees of freedom to six. But you can impose another set of conditions that leave only two polarizations in GR.

 

[WannabeNewton]

Let me start by first asking: what is your definition of harmonic coordinates? If $(M,g_{ab})$ is a space - time and $p\in M$, we can construct a chart $(U,\varphi )$, where $U$ is a neighborhood of $p$, such that the associated coordinates $(x^0,...,x^3)$ all satisfy $\triangledown ^{a}\triangledown _{a}x^{\mu } = 0$. We say such coordinates are harmonic coordinates.

In what way are you relating harmonic coordinates to finding a vector field $\xi^{a}$ that solves $\partial ^{b}\partial _{b}\xi _{a} = -\partial ^{b}\bar{h_{ab}}$ and making a gauge transformation $h_{ab} \rightarrow h_{ab} + 2\partial _{(a}\xi _{b)}$ so that in thia new gauge, $\partial ^{b}\bar{h'_{ab}} = 0$. This is of course similar to the Lorenz gauge from EM.

 

[atyy]

(yeah I take Wald with me to class go ahead make fun of me >.<)

Forest management? :tongue2:

 

[WannabeNewton]

Forest management? :tongue2:

Lol the only word in that entire website I could recognize was Wald :p

 

[NanakiXIII]

Let me start by first asking: what is your definition of harmonic coordinates? If $(M,g_{ab})$ is a space - time and $p\in M$, we can construct a chart $(U,\varphi )$, where $U$ is a neighborhood of $p$, such that the associated coordinates $(x^0,...,x^3)$ all satisfy $\triangledown ^{a}\triangledown _{a}x^{\mu } = 0$. We say such coordinates are harmonic coordinates.

In what way are you relating harmonic coordinates to finding a vector field $\xi^{a}$ that solves $\partial ^{b}\partial _{b}\xi _{a} = -\partial ^{b}\bar{h_{ab}}$ and making a gauge transformation $h_{ab} \rightarrow h_{ab} + 2\partial _{(a}\xi _{b)}$ so that in thia new gauge, $\partial ^{b}\bar{h'_{ab}} = 0$. This is of course similar to the Lorenz gauge from EM.

I was under the impression these two things are supposed to be equivalent, though I do not know how. At least, the terms are used interchangeably, e.g. in http://www.physics.uoguelph.ca/poisson/research/postN.pdf, section 1.3 on p. 4.

 

[WannabeNewton]

Is the gothic thing the same as $\sqrt{-g}g^{\mu \nu }$? If so then yes I agree that the harmonic coordinate condition leads to the statement that $\frac{1}{\sqrt{-g}}\partial _{\nu }(\sqrt{-g}g^{\mu \nu }) = 0$ by a simple matter of computation but I am still a bit uneasy in terms of looking at the gauge fixing as a coordinate condition in the converse. Very nice notes though, thanks for the link.

 

[NanakiXIII]

Yes, that's what the gothic g stands for. I was under the impression that this coordinate condition is used to choose a "diffeomorphism gauge", i.e. it chooses coordinates. As I said, I don't know exactly how this constraint corresponds to actual harmonic coordinates, it's another thing I'm trying to figure out.

 

[NanakiXIII]

I looked at the initial value formulation of GR, which I hadn't seen before, and it seems to give some more physical insight into the choice of harmonic coordinates. I'm looking at Carroll's treatment of the subject in Ch. 4 of his notes, but one thing still puzzles me.

You choose coordinates by choosing harmonic conditions. However, it seems to me this obviously doesn't fix your coordinates uniquely, since they are second-order differential equations for the coordinates, not direct definitions. Therefore, you still have the freedom to perform any transformation

$$x^\mu \to x^\mu + \delta^\mu; \square \delta^\mu = 0$$

i.e. you need to specify exactly which harmonic functions you're using.

Am I right in thinking that at this point, you still have only got rid of four degrees of freedom, despite having eight equations? The $\delta^\mu$ do not separately fix anything, right? They just choose a specific solution out of the ones allowed by the harmonic coordinate conditions.

Now, what puzzles me is that Carroll says that choosing these $\delta^\mu$ specifies coordinates on an initial spacelike hypersurface, but there are four functions, don't they specify coordinates on all of space-time?