Metric constraints in choosing coordinates

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  • #26
PeterDonis
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either these metric components are degrees of freedom, that are allowed to vary (constrained by equations of motion), or they are not. At least some are not. I thought you were telling me none of them are.
No, I was saying that they are degrees of freedom that don't tell you about the geometry, at least not considered by themselves. You can make them change arbitrarily without changing the underlying geometry, by changing coordinates.

what is the relationship between all these degrees of freedom, and your choice of coordinates.
That's what my analysis a while back, with the 150 total degrees of freedom of which 130 are "used up" by choosing coordinates, was trying to get at. But I agree there is more that could be said there.

I mean, if you know the mass distribution globally, i.e. in the Schwarzschild metric you know the mass of the source.
If by "globally" you mean "at every event in the spacetime", then I think I agree that that's enough information to determine the Weyl tensor. I'll have to consider some more to be sure, though.

I'll defer comment on the rest of your post until I've digested your questions some more.
 
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PeterDonis
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How does choosing coordinates cause a loss of degrees of freedom.
It's not a "loss" of degrees of freedom; it's just that not all of the degrees of freedom available tell you about the physics; some of them only tell you about which coordinates you chose.

How many degrees of freedom does it cost?
I haven't found the passage I remember from MTW, but I have found something that I think is closely related: Exercise 13.3 on pp. 314-315 of my edition. The exercise is about showing properties of a local Lorentz frame; specifically, proving that in a local Lorentz frame about some event E, the following will be true:

(1) The ten metric coefficients at event E will be those of Minkowski spacetime; i.e., (-1, 1, 1, 1) along the diagonal, with all other coefficients zero.

(2) The first derivatives of the metric coefficients at event E will all be zero.

(3) *Not* all of the second derivatives will be zero; there will be twenty second derivatives that will be nonzero, and will describe the spacetime curvature in the neighborhood of event E.

The solution goes like this (it's given as a "hint", but it basically walks you through the solution). Suppose we have an expression for the metric [itex]g_{\mu \nu}[/itex] at event E in some arbitrary coordinates, such that there are no symmetries whatsoever; i.e., as far as we can tell in the arbitrary coordinates, the metric and its first and second derivatives are completely unconstrained. Consider a coordinate transformation that will take us from the arbitrary coordinates [itex]x^{\mu}[/itex] to the coordinates [itex]x'^{\alpha}[/itex] of a local Lorentz frame at event E. We expand this transformation in powers of [itex]x^{\mu}[/itex]:

[tex]x'^{\alpha} = M^{\alpha}{}_{\mu} x^{\mu} + \frac{1}{2} N^{\alpha}{}_{\mu \nu} x^{\mu} x^{\nu} + \frac{1}{6} P^{\alpha}{}_{\mu \nu \rho} x^{\mu} x^{\nu} x^{\rho} + {} ...[/tex]

We then look at how we can choose the coefficients [itex]M^{\alpha}{}_{\mu}[/itex], [itex]N^{\alpha}{}_{\mu \nu}[/itex], [itex]P^{\alpha}{}_{\mu \nu \rho}[/itex] in order to make the metric [itex]g'_{\alpha \beta}[/itex] in the local Lorentz frame, and its derivatives, meet the above conditions.

First, there are 16 coefficients [itex]M^{\alpha}{}_{\mu}[/itex] that we can choose. This lets us set the ten metric coefficients [itex]g'_{\alpha \beta}[/itex] to the desired values, with 6 coefficients left over. However, the term "local Lorentz frame at event E" does not refer to a single frame; it refers to a 6-parameter group of frames, corresponding to the 6 Lorentz transformation parameters (as I noted a number of posts ago, we have 3 parameters to pick out the "time direction" of the frame--which we can think of as 3 velocity components relative to some chosen basis--and 3 parameters to pick the orientation of the spatial axes). So to pin down one particular local Lorentz frame, we need to be able to choose the rest of the 6 M coefficients appropriately. This uses up all of the M coefficients.

Then we have 40 coefficients [itex]N^{\alpha}{}_{\mu \nu}[/itex] (there are only 40, not 64, because we must have [itex]N^{\alpha}{}_{\mu \nu} = N^{\alpha}{}_{\nu \mu}[/itex] by symmetry). We can choose these appropriately to set all 40 first derivatives of the metric coefficients to zero. This uses up all of the N coefficients.

Then we have 80 coefficients [itex]P^{\alpha}{}_{\mu \nu \rho}[/itex] (again, there are 80 instead of 256 because of symmetry among the three lower indexes). We can choose these appropriately to set 80 of the 100 second derivatives of the metric coefficients to zero. But that leaves 20 second derivatives that can't be changed by this process; these are the 20 degrees of freedom that tell us about spacetime curvature at event E.

I should note that this analysis is a bit different than the one I gave in an earlier post, so I was mis-remembering somewhat. In particular, the first part, with the 16 M coefficients, is different than what I said before. I apologize for the confusion.

If all the information about curvature is actually encoded in the second derivatives of the metric, then how can we possibly use the metric itself to describe gravitational waves?
By choosing coordinates that let the metric coefficients themselves tell you things of interest about the waves. However, your question also brings up another point that's worth a bit of discussion.

Your question implies that describing curvature using second derivatives of the metric, and describing gravitational waves using the metric itself, are somehow [STRIKE]different[/STRIKE] [edit: mutually exclusive] alternatives. They're not. They're just different possible descriptions of the same thing. Which description we use will depend on what we're trying to do.

If we choose local Lorentz coordinates, then the metric coefficients don't tell us anything about the geometry: after all, they're just the standard Minkowski metric coefficients, and if we took them at face value, we would think spacetime was completely flat! Also, in a local Lorentz frame, the first derivatives of the metric coefficients are all zero: there is no local "acceleration due to gravity" in such a frame. (Another way of putting this is that freely falling objects move in straight lines in a local Lorentz frame, as opposed to, for example, a frame fixed to the surface of the Earth, in which freely falling objects accelerate downward.) So the first derivatives don't tell us anything about the geometry either. We have to look at the second derivatives, and in particular the 20 that we can't set to zero by choosing our coordinates appropriately, to tell us about the geometry.

However, for many purposes it's more convenient *not* to choose local Lorentz coordinates, but instead to choose some other coordinates in which the metric coefficients, or their first derivatives, may *not* take their standard Minkowski values. If we do this, then some of the information about the geometry *will* be contained in the metric or its first derivatives. For example, if we choose coordinates fixed to the Earth's surface, then some of the first derivatives of the metric will be nonzero. This is why freely falling objects in such a frame appear to accelerate downward instead of moving in straight lines.

In the case of gravitational waves, I believe the standard harmonic coordinates in the transverse-traceless gauge are similarly set up to make the effects of the wave "visible" in the metric itself (or its first derivatives--I haven't reviewed all the details). But that's a matter of convenience, to make analysis easier. It doesn't change anything fundamental about the number of degrees of freedom it takes to fully describe the geometry; it just puts some of them in different places in the math.
 
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I haven't found the passage I remember from MTW, but I have found something that I think is closely related: Exercise 13.3 on pp. 314-315 of my edition. The exercise is about showing properties of a local Lorentz frame; specifically, proving that in a local Lorentz frame about some event E, the following will be true:

(1) The ten metric coefficients at event E will be those of Minkowski spacetime; i.e., (-1, 1, 1, 1) along the diagonal, with all other coefficients zero.

(2) The first derivatives of the metric coefficients at event E will all be zero.

(3) *Not* all of the second derivatives will be zero; there will be twenty second derivatives that will be nonzero, and will describe the spacetime curvature in the neighborhood of event E.

The solution goes like this (it's given as a "hint", but it basically walks you through the solution). Suppose we have an expression for the metric [itex]g_{\mu \nu}[/itex] at event E in some arbitrary coordinates, such that there are no symmetries whatsoever; i.e., as far as we can tell in the arbitrary coordinates, the metric and its first and second derivatives are completely unconstrained. Consider a coordinate transformation that will take us from the arbitrary coordinates [itex]x^{\mu}[/itex] to the coordinates [itex]x'^{\alpha}[/itex] of a local Lorentz frame at event E. We expand this transformation in powers of [itex]x^{\mu}[/itex]:

[tex]x'^{\alpha} = M^{\alpha}{}_{\mu} x^{\mu} + \frac{1}{2} N^{\alpha}{}_{\mu \nu} x^{\mu} x^{\nu} + \frac{1}{6} P^{\alpha}{}_{\mu \nu \rho} x^{\mu} x^{\nu} x^{\rho} + {} ...[/tex]

We then look at how we can choose the coefficients [itex]M^{\alpha}{}_{\mu}[/itex], [itex]N^{\alpha}{}_{\mu \nu}[/itex], [itex]P^{\alpha}{}_{\mu \nu \rho}[/itex] in order to make the metric [itex]g'_{\alpha \beta}[/itex] in the local Lorentz frame, and its derivatives, meet the above conditions.

First, there are 16 coefficients [itex]M^{\alpha}{}_{\mu}[/itex] that we can choose. This lets us set the ten metric coefficients [itex]g'_{\alpha \beta}[/itex] to the desired values, with 6 coefficients left over. However, the term "local Lorentz frame at event E" does not refer to a single frame; it refers to a 6-parameter group of frames, corresponding to the 6 Lorentz transformation parameters (as I noted a number of posts ago, we have 3 parameters to pick out the "time direction" of the frame--which we can think of as 3 velocity components relative to some chosen basis--and 3 parameters to pick the orientation of the spatial axes). So to pin down one particular local Lorentz frame, we need to be able to choose the rest of the 6 M coefficients appropriately. This uses up all of the M coefficients.

Then we have 40 coefficients [itex]N^{\alpha}{}_{\mu \nu}[/itex] (there are only 40, not 64, because we must have [itex]N^{\alpha}{}_{\mu \nu} = N^{\alpha}{}_{\nu \mu}[/itex] by symmetry). We can choose these appropriately to set all 40 first derivatives of the metric coefficients to zero. This uses up all of the N coefficients.

Then we have 80 coefficients [itex]P^{\alpha}{}_{\mu \nu \rho}[/itex] (again, there are 80 instead of 256 because of symmetry among the three lower indexes). We can choose these appropriately to set 80 of the 100 second derivatives of the metric coefficients to zero. But that leaves 20 second derivatives that can't be changed by this process; these are the 20 degrees of freedom that tell us about spacetime curvature at event E.

I should note that this analysis is a bit different than the one I gave in an earlier post, so I was mis-remembering somewhat. In particular, the first part, with the 16 M coefficients, is different than what I said before. I apologize for the confusion.
This is an interesting way of looking at it, but I'm not sure about the expansion (as I said, I don't have MTW.) I'm assuming it's just a Taylor series, in which case I'm wondering where the constant term went and where you're finding these derivatives of the metric. To clarify my point, let me expand explicitly:

For a coordinate transformation [itex] x^{\mu} \to y^{\mu} [/itex], we can expand:

[tex]
y^{\mu}(x) = y^{\mu}(x_0) + \frac{1}{2} y^{\mu}{}_{,\alpha} x^{\alpha} + \frac{1}{6} y^{\mu}{}_{,\alpha\beta} x^{\alpha} x^{\beta} + \ldots
[/tex]

You're taking derivatives of your coordinate functions, not of your metric, in this case. Am I misunderstanding the intention?

By choosing coordinates that let the metric coefficients themselves tell you things of interest about the waves. However, your question also brings up another point that's worth a bit of discussion.

Your question implies that describing curvature using second derivatives of the metric, and describing gravitational waves using the metric itself, are somehow [STRIKE]different[/STRIKE] [edit: mutually exclusive] alternatives. They're not. They're just different possible descriptions of the same thing. Which description we use will depend on what we're trying to do.

If we choose local Lorentz coordinates, then the metric coefficients don't tell us anything about the geometry: after all, they're just the standard Minkowski metric coefficients, and if we took them at face value, we would think spacetime was completely flat! Also, in a local Lorentz frame, the first derivatives of the metric coefficients are all zero: there is no local "acceleration due to gravity" in such a frame. (Another way of putting this is that freely falling objects move in straight lines in a local Lorentz frame, as opposed to, for example, a frame fixed to the surface of the Earth, in which freely falling objects accelerate downward.) So the first derivatives don't tell us anything about the geometry either. We have to look at the second derivatives, and in particular the 20 that we can't set to zero by choosing our coordinates appropriately, to tell us about the geometry.

However, for many purposes it's more convenient *not* to choose local Lorentz coordinates, but instead to choose some other coordinates in which the metric coefficients, or their first derivatives, may *not* take their standard Minkowski values. If we do this, then some of the information about the geometry *will* be contained in the metric or its first derivatives. For example, if we choose coordinates fixed to the Earth's surface, then some of the first derivatives of the metric will be nonzero. This is why freely falling objects in such a frame appear to accelerate downward instead of moving in straight lines.

In the case of gravitational waves, I believe the standard harmonic coordinates in the transverse-traceless gauge are similarly set up to make the effects of the wave "visible" in the metric itself (or its first derivatives--I haven't reviewed all the details). But that's a matter of convenience, to make analysis easier. It doesn't change anything fundamental about the number of degrees of freedom it takes to fully describe the geometry; it just puts some of them in different places in the math.
This raises another question, though. Your explanation above, using the expansion of the coordinate transformation, doesn't seem to leave room for "putting things elsewhere in the math." Your [itex]M[/itex] constrain the metric components, [itex]N[/itex] the first derivatives and [itex]P[/itex] the second derivatives. For the metric componens to vary, then, I would say you would have to somehow make [itex]M[/itex] and [itex]N[/itex] constrain the second derivatives. Either that, or going to harmonic coordinates in the TT gauge does not uniquely determine your coordinates, which also seems problematic to me.
 
  • #29
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NanakiXIII said:
How does choosing coordinates cause a loss of degrees of freedom.

It's not a "loss" of degrees of freedom; it's just that not all of the degrees of freedom available tell you about the physics; some of them only tell you about which coordinates you chose.
I remember an old thread about degrees of freedom of the Riemann tensor, and a sort of agreement was reached that to go from the 20 dof that are left after the usual symmetries of the tensor are taken into account, to 6 dof (in a local frame) we used the fact that by metric compatibility we can always choose an orthonormal frame, that is, we have, and I quote directly from Ben Niehoff's post in that thread: "an invertible linear transformation on each tangent space, chosen continuously on some open patch U to an orthonormal frame, and in such a frame the Riemann tensor must take a special form which has only (n)(n-1)/2 degrees of freedom.... [So]Given a metric-compatible connection, R(X,Y) at any point P is an element of the group that preserves a metric sphere in the tangent space at P (where a sphere is defined as the locus of points of distance 1 from the origin, according to the inner product given by the metric tensor). The group that preserves a unit sphere in R^n is precisely SO(n), which has dimension n(n-1)/2."
 
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I remember an old thread about degrees of freedom of the Riemann tensor, and a sort of agreement was reached that to go from the 20 dof that are left after the usual symmetries of the tensor are taken into account, to 6 dof (in a local frame) we used the fact that by metric compatibility we can always choose an orthonormal frame, that is, we have, and I quote directly from Ben Niehoff's post in that thread: "an invertible linear transformation on each tangent space, chosen continuously on some open patch U to an orthonormal frame, and in such a frame the Riemann tensor must take a special form which has only (n)(n-1)/2 degrees of freedom.... [So]Given a metric-compatible connection, R(X,Y) at any point P is an element of the group that preserves a metric sphere in the tangent space at P (where a sphere is defined as the locus of points of distance 1 from the origin, according to the inner product given by the metric tensor). The group that preserves a unit sphere in R^n is precisely SO(n), which has dimension n(n-1)/2."
I'm afraid I don't understand this. Why does it force the Riemann tensor to lie in SO(n)? Could you elaborate?
 
  • #31
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I'm afraid I don't understand this. Why does it force the Riemann tensor to lie in SO(n)? Could you elaborate?
It forces the 20 dof of the general Riemann tensor to just 6 when we consider the special situation of an orthonormal frame in a patch of our manifold. That is if we restrict ourselves to the tangent bundle's associated frame bundle.
You need certain acquaintance with differential geometry to fully understand this, and I'm no real expert so here's the post from the above mentioned thread so you can read it for yourself.


https://www.physicsforums.com/showpost.php?p=3511530&postcount=63
 
  • #32
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So basically and since in yor OP you only talk about the metric, we have that by symmetry and general covariance, the metric has 6 dof and we know the metric completely determines the Riemann tensor so the doubt in the other thread was how to reconcile that with the fact that the Riemann tensor once we have included all the symmetries has 20 dof, and the orthonormal frame explanation was a way to bridge that disparity.

I see originally you were concerned with the reduction of the dof of the metric from 6 to 2 for the special case of the GW in linearized gravity, and maybe this is too far from that.
 
  • #33
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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?

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.
 
  • #34
PeterDonis
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I'm assuming it's just a Taylor series
Sort of. One note, though; I got things backwards as far as which coordinates are which. The [itex]x'^{\alpha}[/itex] coordinates are the ones for the arbitrary coordinate system, and the [itex]x^{\mu}[/itex] 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.

You're taking derivatives of your coordinate functions, not of your metric
The coefficients M, N, P will appear when you take derivatives of [itex]x'^{\alpha}[/itex] with respect to [itex]x^{\mu}[/itex], yes. But they also will appear in the transformation of the metric, because the transformation matrix for the metric contains the derivatives of [itex]x'^{\alpha}[/itex] with respect to [itex]x^{\mu}[/itex]:

[tex]g_{\mu \nu} = \frac{\partial x'^{\alpha}}{\partial x^{\mu}} \frac{\partial x'^{\beta}}{\partial x^{\nu}} g'_{\alpha \beta}[/tex]

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

Your [itex]M[/itex] constrain the metric components, [itex]N[/itex] the first derivatives and [itex]P[/itex] the second derivatives. For the metric components to vary, then, I would say you would have to somehow make [itex]M[/itex] and [itex]N[/itex] constrain the second derivatives.
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 [itex]\partial x'^{\alpha} / \partial x^{\mu}[/itex]; 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.
 
  • #35
stevendaryl
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The way I see it, when you write ##g_{\mu\nu}## you've already chosen your coordinate basis. The metric is defined as being a symmetric bilinear form on the tangent bundle of your manifold (thus having only 6 components by definition). Then at a point p ##g_p(X,Y)## is a real number where ##X,Y## are vectors in ##T_pM##.
If you choose a basis ##(E_{\mu})## at p of ##T_pM##, you have then ##g_{\mu\nu}:=g(E_{\mu},E_{\nu})## and this naturally defines a local coordinate basis for your manifold by ##E_{\mu}=\dfrac{\partial}{\partial x^{\mu}}##. So, if you have ##g_{\mu\nu}## you have a vector basis, and hence a coordinate system.
The way that many physicists use tensor indices is to indicate
  1. The type of tensor.
  2. The way tensors are contracted with each other.

The specific coordinate system is arbitrary. So the fact that someone writes [itex]g_{\mu \nu}[/itex] doesn't mean that they have already chosen a coordinate system.
 
  • #36
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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 wich 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.
 
  • #37
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Sort of. One note, though; I got things backwards as far as which coordinates are which. The [itex]x'^{\alpha}[/itex] coordinates are the ones for the arbitrary coordinate system, and the [itex]x^{\mu}[/itex] 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 [itex]x'^{\alpha}[/itex] with respect to [itex]x^{\mu}[/itex], yes. But they also will appear in the transformation of the metric, because the transformation matrix for the metric contains the derivatives of [itex]x'^{\alpha}[/itex] with respect to [itex]x^{\mu}[/itex]:

[tex]g_{\mu \nu} = \frac{\partial x'^{\alpha}}{\partial x^{\mu}} \frac{\partial x'^{\beta}}{\partial x^{\nu}} g'_{\alpha \beta}[/tex]

Similarly, we find that the transformation of first derivatives of the metric contains second derivatives of [itex]x'^{\alpha}[/itex] with respect to [itex]x^{\mu}[/itex], and the transformation of second derivatives of the metric contains third derivatives of [itex]x'^{\alpha}[/itex] with respect to [itex]x^{\mu}[/itex].
They may appear, but I don't see how they constrain anything. If [itex]g'_{\alpha \beta}[/itex] has degrees of freedom, then so does

[tex]g_{\mu \nu} = \frac{\partial x'^{\alpha}}{\partial x^{\mu}} \frac{\partial x'^{\beta}}{\partial x^{\nu}} g'_{\alpha \beta}[/tex]

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 [itex]\partial x'^{\alpha} / \partial x^{\mu}[/itex]; 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 wich 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).
 
  • #38
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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.
 
  • #39
PeterDonis
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If [itex]g'_{\alpha \beta}[/itex] has degrees of freedom, then so does

[tex]g_{\mu \nu} = \frac{\partial x'^{\alpha}}{\partial x^{\mu}} \frac{\partial x'^{\beta}}{\partial x^{\nu}} g'_{\alpha \beta}[/tex]

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 [itex]g_{\mu \nu}[/itex] are fixed to the Minkowski values. That means the degrees of freedom in [itex]g'_{\alpha \beta}[/itex] 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.
 
  • #40
PeterDonis
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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.
 
  • #41
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How do you figure that? The whole point is to set the coordinate derivatives so that all ten components of [itex]g_{\mu \nu}[/itex] are fixed to the Minkowski values. That means the degrees of freedom in [itex]g'_{\alpha \beta}[/itex] 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?
 
  • #42
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Are you taking about the [itex]\partial ^{b}h _{ab} = 0[/itex] condition? It comes out of the fact that two different perturbations propagating on the background minkowski space - time, [itex]h_{ab},h'_{ab}[/itex] represent the same physical space - time if there exists a vector field [itex]\xi ^{a}[/itex] such that [itex]h'_{ab} = h_{ab} - \mathcal{L}_{\xi }\eta _{ab}[/itex]. Using this you can derive the condition [itex]\partial ^{b}h _{ab} = 0[/itex].
 
  • #43
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Are you taking about the [itex]\partial ^{b}h _{ab} = 0[/itex] condition? It comes out of the fact that two different perturbations propagating on the background minkowski space - time, [itex]h_{ab},h'_{ab}[/itex] represent the same physical space - time if there exists a vector field [itex]\xi ^{a}[/itex] such that [itex]h'_{ab} = h_{ab} - \mathcal{L}_{\xi }\eta _{ab}[/itex]. Using this you can derive the condition [itex]\partial ^{b}h _{ab} = 0[/itex].
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.
 
  • #44
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Let me start by first asking: what is your definition of harmonic coordinates? If [itex](M,g_{ab})[/itex] is a space - time and [itex]p\in M[/itex], we can construct a chart [itex](U,\varphi )[/itex], where [itex]U[/itex] is a neighborhood of [itex]p[/itex], such that the associated coordinates [itex](x^0,...,x^3)[/itex] all satisfy [itex]\triangledown ^{a}\triangledown _{a}x^{\mu } = 0[/itex]. We say such coordinates are harmonic coordinates.

In what way are you relating harmonic coordinates to finding a vector field [itex]\xi^{a}[/itex] that solves [itex]\partial ^{b}\partial _{b}\xi _{a} = -\partial ^{b}\bar{h_{ab}}[/itex] and making a gauge transformation [itex]h_{ab} \rightarrow h_{ab} + 2\partial _{(a}\xi _{b)}[/itex] so that in thia new gauge, [itex]\partial ^{b}\bar{h'_{ab}} = 0[/itex]. This is of course similar to the Lorenz gauge from EM.
 
  • #47
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Let me start by first asking: what is your definition of harmonic coordinates? If [itex](M,g_{ab})[/itex] is a space - time and [itex]p\in M[/itex], we can construct a chart [itex](U,\varphi )[/itex], where [itex]U[/itex] is a neighborhood of [itex]p[/itex], such that the associated coordinates [itex](x^0,...,x^3)[/itex] all satisfy [itex]\triangledown ^{a}\triangledown _{a}x^{\mu } = 0[/itex]. We say such coordinates are harmonic coordinates.

In what way are you relating harmonic coordinates to finding a vector field [itex]\xi^{a}[/itex] that solves [itex]\partial ^{b}\partial _{b}\xi _{a} = -\partial ^{b}\bar{h_{ab}}[/itex] and making a gauge transformation [itex]h_{ab} \rightarrow h_{ab} + 2\partial _{(a}\xi _{b)}[/itex] so that in thia new gauge, [itex]\partial ^{b}\bar{h'_{ab}} = 0[/itex]. 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 Poisson's notes on the PN formalism, section 1.3 on p. 4.
 
  • #48
WannabeNewton
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Is the gothic thing the same as [itex]\sqrt{-g}g^{\mu \nu }[/itex]? If so then yes I agree that the harmonic coordinate condition leads to the statement that [itex]\frac{1}{\sqrt{-g}}\partial _{\nu }(\sqrt{-g}g^{\mu \nu }) = 0[/itex] 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.
 
  • #49
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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.
 
  • #50
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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

[tex]x^\mu \to x^\mu + \delta^\mu; \square \delta^\mu = 0[/tex]

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 [itex]\delta^\mu[/itex] 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 [itex]\delta^\mu[/itex] specifies coordinates on an initial spacelike hypersurface, but there are four functions, don't they specify coordinates on all of space-time?
 

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