Metric constraints in choosing coordinates

[NanakiXIII]

Hello all,

I've been puzzled by this problem for some time now and was wondering if anyone here could help me out. Textbooks on GR (specifically when going into gravitational waves) tend not to elucidate this. It's often taken for granted that through the gauge diffeomorphism invariance (or general covariance), you can choose a 'gauge' by choosing coordinates on your manifold and that this choice deprives your metric of four degrees of freedom.

I don't quite understand this. A metric is just a set of ten functions

g^{\mu\nu} : M \to \mathbb{R}

where M is your manifold. Choosing coordinates amounts to choosing four more functions

x^{\mu} : M \to \mathbb{R}

so that the x^{\mu} label the points on your manifold. Having labelled those points, we can work with our coordinates instead of with the abstract manifold:

g^{\mu\nu} : \{x\} \to \mathbb{R}

How does doing this, i.e. just labelling your points, constrain the components of your metric?

 

[kevinferreira]

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.

 

[WannabeNewton]

The gauge freedom of GR comes from diffeomorphisms which means that (M,g_{ab},\varphi ) and (M,\phi ^{*}g_{ab},\phi ^{*}\varphi ) represent the same physical space - time where \varphi is a matter field and \phi is a diffeomorphism. The other related concept of gauge freedom is that GR has no preferred, "god - given" coordinate system for a physical space - time.

By the way, your maps are horribly incorrect. The metric tensor's domain is certainly not the manifold itself. Coordinates are not in general maps on the entire manifold itself. I don't even know what in the world the last one is.

 

[NanakiXIII]

@kevinferreira:

Of course I agree that you cannot actually write down the numerical components until you choose coordinates to express them in, but I don't see how that gives you constraints on what those components are allowed to be. You say this symmetric bilinear form has six components? How so? A four-by-four matrix (which you can write the metric as since it is bilinear) that is also symmetric has ten components.

@WannabeNewton:

You'll have to forgive my heuristic maps. I'm aware they may not be fully rigorous, I was merely trying to illustrate my point with simplicity. The last one just represents going to a coordinate-dependent notation rather than the abstract one.

I'm aware of what you said in your reply, but it hasn't so far helped me understand this issue.

 

[WannabeNewton]

Another way to look at it is that the Einstein field equations R_{\mu \nu }-\frac{1}{2}g_{\mu \nu }R = 8\pi GT_{\mu \nu } are really a set of 10 independent equations but the Bianchi identity \triangledown ^{a}G_{ab} = 0 is a constraint that reduces the number to 6.

 

[NanakiXIII]

Indeed, this is another explanation I've come across, but it only adds to my difficulties. Does enforcing the Bianchi identities choose your coordinates for you? It doesn't seem to. If it does, how can we still choose e.g. harmonic coordinates on top of that?

 

[WannabeNewton]

Harmonic coordinates can always be chosen locally and this is more a consequence of the poincare lemma. I have to head to class now but I'll try to answer after if someone else hasn't already.

 

[PeterDonis]

Another way to look at it is that the Einstein field equations R_{\mu \nu }-\frac{1}{2}g_{\mu \nu }R = 8\pi GT_{\mu \nu } are really a set of 10 independent equations but the Bianchi identity \triangledown ^{a}G_{ab} = 0 is a constraint that reduces the number to 6.

This doesn't look quite right. First, the EFE and the Bianchi identities involve curvature components, which are second derivatives of the metric; they don't involve the metric itself. Second, the Bianchi identities don't reduce the number of independent components of the curvature tensor; they just provide four alternate equations that can be used in place of four of the EFE components. Often the Bianchi identities are easier to work with.

 

[WannabeNewton]

Uh all I said in that sentence was that the Bianci identity reduces the number of independent equations. See e.g. page 112 of Carroll's notes.

 

[NanakiXIII]

@WannabeNewton:

You seem to have hit on a lapse in what I know; I was under the impression harmonic coordinate conditions were imposed globally. At least, I've not come across a text mentioning this caveat when using them.

@PeterDonis:

I was indeed under the impression that the Bianchi identity was just a property of the curvature in GR, rather than a constraint.

 

[WannabeNewton]

What you do is at each point find an open subset on which harmonic coordinates can be imposed and construct an atlas out of that. It's not the same thing as having a single global chart.

 

[NanakiXIII]

But even if only locally, they're still a constraint.

 

[PeterDonis]

Uh all I said in that sentence was that the Bianci identity reduces the number of independent equations. See e.g. page 112 of Carroll's notes.

Yes, I see what he says there. MTW says something similar, but they also say that the Bianchi identities are constraints on the source, i.e., on the stress-energy tensor (they just express local energy-momentum conservation). So there are still ten independent equations; it's just that four of them do not constrain the metric, they constrain the source. That means the metric has four (of ten) degrees of freedom that are "unphysical" (that's the word Carroll uses), because they correspond to our freedom to choose coordinates, not to any physical degree of freedom in spacetime.

I don't think event that is the whole story, though, because of the fact that I mentioned before, that the EFE and the Bianchi identities involve the curvature tensor, i.e., second derivatives of the metric. There's a passage in MTW that talks about this, but I haven't been able to find it.

 

[WannabeNewton]

I found a section in d'Inverno's GR text that might be of use but to be honest it just seems to be singing the same song again. Here's the relevant page: http://s15.postimage.org/jyd2d3gm2/d_inverno_ref.jpg

I tried to find something in my copy of Wald while in class but I couldn't (yeah I take Wald with me to class go ahead make fun of me >.<). I don't own MTW so if you do find the relevant section Peter I'd love to know thanks because MTW usually gives better physical insights into these constraints than other texts.

 

[PeterDonis]

There's a passage in MTW that talks about this, but I haven't been able to find it.

I still haven't been able to find it, but I'll try to walk through what I remember of the passage I was thinking of.

Pick any event in a spacetime and try to describe the physics of gravity at that event. How do we do that? Suppose we start by constructing a coordinate chart centered on our chosen event. How will our chosen chart describe the physics of gravity? More precisely, how will we distinguish the degrees of freedom that describe the actual physics, from the degrees of freedom that merely describe our choice of coordinates?

(1) First of all, we have four coordinates, and we can scale each of them however we like. That's four degrees of freedom.

(2) Second, we have the freedom to choose the "directions" in which each of the coordinate axes point. This corresponds to the six-parameter group of local Lorentz transformations: basically we're choosing which timelike geodesic through our chosen event is the "time axis" of our local coordinate chart, and how the spatial axes are oriented. So that's six more degrees of freedom.

You'll notice that we now have ten degrees of freedom, corresponding to the ten metric coefficients. In other words, the metric coefficients by themselves don't tell us *anything* about physics; the actual physics must be contained somewhere else. (More on this below.)

(3) Now we look at the first derivatives of the metric coefficients. There are forty of these (four derivative operators times ten metric coefficients). However, it turns out that we can always set things up so all forty first derivatives are zero at the event we choose. This is because of the equivalence principle: the first derivatives of the metric correspond to the local "acceleration due to gravity", i.e., to a coordinate acceleration of freely falling objects, and the equivalence principle says that we can always make that vanish by a suitable choice of coordinates; we can always make the "acceleration due to gravity" vanish locally.

Note that our freedom to set the first derivatives equal to zero is *independent* of our freedom to choose the scaling and orientation of the coordinate axes! Setting the first derivatives to zero amounts to saying we are working in a local Lorentz frame; but that still leaves a ten-parameter group of possible local Lorentz frames--six parameters for a local Lorentz transformation, plus four parameters for scaling the coordinates. So there's no "overlap" in degrees of freedom thus far; in other words, so far we have found *no* degrees of freedom that describe any actual physics of gravity, only degrees of freedom that describe our coordinate choice.

(4) Now we come to the second derivatives of the metric coefficients. There are one hundred of these (ten metric coefficients times ten second derivative operators--there are only ten because partial derivatives commute). The Riemann curvature tensor is built out of combinations of these second derivatives; but due to various identities, including the Bianchi identities (the full ones, not the contracted ones that are satisfied by the Einstein tensor), there are eighty constraints on the second derivatives, so there are only twenty degrees of freedom here, corresponding to the twenty independent components of the Riemann tensor.

(This is where my memory gets a bit hazy; I'm not sure if the full Bianchi identities provide all eighty constraints or if there are other identities involved as well. There are various symmetries that the Riemann tensor must satisfy due to the commutation of partial derivatives, but IIRC these only reduce the number of components from 256, the number of components of a fourth-rank tensor in 4 dimensions, to 100, the total number of independent second derivatives.)

It is these twenty degrees of freedom that describe the physics of gravity at our chosen event, i.e., they describe the properties of gravity (tidal gravity) that are independent of our choice of coordinates. The Einstein tensor, with ten independent components, describes ten of these twenty degrees of freedom; these are the ones that are constrained by the source, the stress-energy tensor. (Sometimes the Ricci tensor is used instead of the Einstein tensor; they are basically equivalent for our purposes here.) The other ten degrees of freedom are usually dealt with using the Weyl tensor, which also has ten independent components; these degrees of freedom describe how tidal gravity varies from event to event independently of whether any source (stress-energy) is present.

I should note that all of the above may be the answer to a different question than the one the OP was asking. The OP's question was about counting independent equations, as WannabeNewton pointed out, whereas the above counts independent components, or "degrees of freedom" (if that's the right term).

 

[NanakiXIII]

Thanks for digging that out of your memory, PeterDonis. It brings back some complications that I forgot about. However, it glances over my original problem like most textbooks do.

(2) Second, we have the freedom to choose the "directions" in which each of the coordinate axes point. This corresponds to the six-parameter group of local Lorentz transformations: basically we're choosing which timelike geodesic through our chosen event is the "time axis" of our local coordinate chart, and how the spatial axes are oriented. So that's six more degrees of freedom.

You'll notice that we now have ten degrees of freedom, corresponding to the ten metric coefficients.

My question is about this bit. Why do these degrees of freedom have anything to do with the metric components?

Also, if, indeed, none of the metric components are physical, why do we use two of them to describe gravitational waves? I was under the impression that choosing coordinates entailed four degrees of freedom.

 

[PeterDonis]

Why do these degrees of freedom have anything to do with the metric components?

Because the form of the metric depends on the coordinate scaling and orientation. For a simple example, consider the metric of a flat Euclidean plane in Cartesian vs. polar coordinates:

ds^2 = dx^2 + dy^2

ds^2 = dr^2 + r^2 d \theta^2

Both of these metrics describe the same underlying geometry, but the coefficients are different because the coordinates are different. And we could come up with even more metrics describing the same geometry by varying the coordinates in other ways. (For example, we could change the scaling of x vs. y in the Cartesian version.)

Also, if, indeed, none of the metric components are physical, why do we use two of them to describe gravitational waves?

The metric components don't describe the physics in a way that's independent of coordinates. It just so happens that we can find coordinates for describing gravitational waves that give the metric coefficients a good physical interpretation; but that interpretation still depends on using those specific coordinates. If you change coordinates, the metric coefficients will change and will no longer have any good physical interpretation. But the description in terms of the Riemann curvature tensor will always have a direct physical interpretation.

I was under the impression that choosing coordinates entailed four degrees of freedom.

People often say that because there are four coordinates, so it seems like you just have to pick four functions, for four degrees of freedom. But you can't pick all four independently; there are conditions that the functions have to jointly satisfy.

Moreover, when you're trying to count degrees of freedom, what matters isn't how many functions you have; what matters is how many different ways you can pick them. That's what my post was trying to count.

 

[NanakiXIII]

Because the form of the metric depends on the coordinate scaling and orientation. For a simple example, consider the metric of a flat Euclidean plane in Cartesian vs. polar coordinates:

ds^2 = dx^2 + dy^2

ds^2 = dr^2 + r^2 d \theta^2

Both of these metrics describe the same underlying geometry, but the coefficients are different because the coordinates are different. And we could come up with even more metrics describing the same geometry by varying the coordinates in other ways. (For example, we could change the scaling of x vs. y in the Cartesian version.)

Yes, I understand this point, but the analogy doesn't seem satisfactory. We're dealing with dynamical metrics. In the above case, the degrees of freedom are zero in either coordinate frame. However, if you do not choose coordinates, nothing changes about the geometry; you can't express the metric numerically, but it doesn't suddenly obtain degrees of freedom because the geometry is still static.

The metric components don't describe the physics in a way that's independent of coordinates. It just so happens that we can find coordinates for describing gravitational waves that give the metric coefficients a good physical interpretation; but that interpretation still depends on using those specific coordinates. If you change coordinates, the metric coefficients will change and will no longer have any good physical interpretation. But the description in terms of the Riemann curvature tensor will always have a direct physical interpretation.

Fair enough, this is one of the answers I was looking for. Presumably, this is why harmonic coordinates are chosen. But it still seems sketchy, because harmonic coordinates are prescribed using four equations, not ten. So, does it not fix all coordinates? Does it fix something else entirely? And, if you constrain the metric down to two components (where do the other four constraints come from?), your ten-dof choice of coordinates must constrain two other things besides the metric. What things, then? Components of the cuvature?

People often say that because there are four coordinates, so it seems like you just have to pick four functions, for four degrees of freedom. But you can't pick all four independently; there are conditions that the functions have to jointly satisfy.

Moreover, when you're trying to count degrees of freedom, what matters isn't how many functions you have; what matters is how many different ways you can pick them. That's what my post was trying to count.

What exactly do you mean by "ways to pick them"? I interpret counting degrees of freedom as counting the number of functions that are allowed to vary independently.

 

[PeterDonis]

In the above case, the degrees of freedom are zero in either coordinate frame. However, if you do not choose coordinates, nothing changes about the geometry; you can't express the metric numerically, but it doesn't suddenly obtain degrees of freedom because the geometry is still static.

The geometry is only "static" because we chose that particular geometry. But the question is, of all the "degrees of freedom" that we can see, which ones actually specify the geometry, and which ones can vary without changing the geometry? In the case of the Euclidean plane, there are two coordinate degrees of freedom, but we can vary them without varying the geometry at all. (And this would be true of *any* geometry, not just a flat Euclidean plane. An irregular two-dimensional blob, with curvature varying from point to point, doesn't have any more "degrees of freedom" in its metric; it's just a different particular geometry, and could still be described by lots of different coordinates.)

So the coordinates by themselves don't tell us about the geometry. The coordinates in combination with the metric do. But even then you can't necessarily just "read off" the geometry from the metric coefficients. The real information about the geometry, the information that is independent of coordinates, is in the curvature tensor.

harmonic coordinates are prescribed using four equations, not ten.

I think it's a bit misleading to think of coordinates as prescribed using four equations, or four functions. A coordinate chart is a one-to-one mapping of 4-tuples of real numbers to points in a manifold, which has to satisfy certain conditions. It's not four separate, independent functions or equations.

So, does it not fix all coordinates? Does it fix something else entirely?

If you mean, what's fixed by the geometry, as opposed to by the choice of coordinates, it's the curvature tensor. See above and further comments below.

And, if you constrain the metric down to two components (where do the other four constraints come from?)

For a gravitational wave, you are choosing coordinates to make the metric look simple, given some assumptions about the wave (that it's of small amplitude and that you're far enough from the source that it can be treated as a plane wave). The "constraints" on the form of the metric come from the choice of coordinates.

your ten-dof choice of coordinates must constrain two other things besides the metric.

No. Let me summarize briefly how it goes.

The degrees of freedom are: 10 metric coefficients; 40 first derivatives of metric coefficients; 100 second derivatives of metric coefficients.

The choice of coordinates "uses up" the following degrees of freedom: 10 in the metric coefficients (4 choices of scaling--1 per coordinate--plus 6 choices of "spacetime direction"--the six Lorentz transformation parameters that tell which worldline is "at rest" in our coordinates at a given event and how the spatial axes are oriented at that event); 40 in the first derivatives (to enforce the equivalence principle--we have to be able to set all 40 first derivatives equal to zero in a local Lorentz frame); 80 in the second derivatives (the Bianchi identities, and possibly other geometric constraints--this is what I need to check when I can find the right passage in MTW). So of 150 total "degrees of freedom", 130 of them are taken up in the choice of coordinates, leaving 20 to tell us about the geometry (the 20 independent components of the Riemann tensor).

What exactly do you mean by "ways to pick them"? I interpret counting degrees of freedom as counting the number of functions that are allowed to vary independently.

I didn't express myself very well with that phrase. Hopefully the above will help clarify things.

 

[NanakiXIII]

Thanks, Peter. While everything you say sounds reasonable enough, there are a few issues preventing me from coming to an actual understanding of what you are saying.

1. I don't know and am unable to find the why and how of fixing these degrees of freedom through a choice of coordinates. Your statement that 130 degrees of freedom are fixed by this choice may be completely accurate, but I don't understand why.

2. This enumeration of reasons to lose degrees of freedom is rather ecclectic; though perhaps this perception is due to my lack of understanding as described under point 1. It is not clear to me, besides the point that these reasons actually throw out the degrees of freedom you say they do, that there are not more constraints that get rid of even more.

3. Different authors take completely different approaches to this and I'm unable to reconcile the different ways of counting. For example, Wald on page 266 does something completely different. He constructs GR in an initial value formulation and uses completely different reasoning to throw out degrees of freedom. He ends up with 2, not 20.

I've also seen a more hand-waving approach in linearized gravity where (gauge) symmetries of the equations are simply used as arguments to throw out "8 of 10" degrees of freedom in the metric. First, harmonic coordinates are chosen, which throw out 4, then additional symmetry is noted in the equations and four more are lost (I think by going to the transverse traceless gauge), leaving the two polarizations of gravitational waves. I think Carroll does something along these lines as well.

All these different approaches make the issue highly confusing.

 

[PeterDonis]

1. I don't know and am unable to find the why and how of fixing these degrees of freedom through a choice of coordinates.

...

This enumeration of reasons to lose degrees of freedom is rather ecclectic; though perhaps this perception is due to my lack of understanding as described under point 1.

Yes, it is a somewhat eclectic list. I'm not sure there's any way around that. The only way I have found to make sense of it is to just take each item in turn and understand how it works. Are there particular ones you're wondering about?

It is not clear to me, besides the point that these reasons actually throw out the degrees of freedom you say they do, that there are not more constraints that get rid of even more.

That's why I would like to find the passage I've been referring to in MTW. IIRC they go into more detail about how we know the list of constraints is complete.

Different authors take completely different approaches to this and I'm unable to reconcile the different ways of counting. For example, Wald on page 266 does something completely different. He constructs GR in an initial value formulation and uses completely different reasoning to throw out degrees of freedom. He ends up with 2, not 20.

He's counting something different. He's counting (if I'm thinking of the correct passage, to do with gravitational waves) the degrees of freedom in an idealized plane gravitational wave. That is a very constrained system. The 20 degrees of freedom refers to a generic curvature tensor, which has to be able to describe *any* gravitating system, not just an idealized plane gravitational wave.

If you're only interested in the degrees of freedom of a plane gravitational wave, then most of what I and others have said in this thread is irrelevant. In the OP you mentioned gravitational waves, but then you talked about choosing coordinates and the degrees of freedom in the metric in general terms, not specifically in reference to gravitational waves. So it really depends on what you're interested in.

I've also seen a more hand-waving approach in linearized gravity where (gauge) symmetries of the equations are simply used as arguments to throw out "8 of 10" degrees of freedom in the metric. First, harmonic coordinates are chosen, which throw out 4, then additional symmetry is noted in the equations and four more are lost (I think by going to the transverse traceless gauge), leaving the two polarizations of gravitational waves.

Yes, for gravitational waves specifically, this is one way of thinking about it that a number of authors discuss. (There's a discussion of it in MTW; I'll look that up when I get a chance.) But I'm not sure this analysis is complete, because it only talks about the degrees of freedom in the metric; it doesn't talk about the first and second derivatives. The first derivatives are all set to 0 by working in a local inertial frame (which is implied by choosing harmonic coordinates and working in the TT gauge); but that still leaves the second derivatives. So I'm not sure that I've seen a good discussion of how this analysis, which is specific to plane gravitational waves, is reconciled with the general analysis I gave in my previous post.

All these different approaches make the issue highly confusing.

I think part of the confusion is that different authors are talking about different questions, and they don't always try to fit all the different questions into a single scheme that can cover all of them. For a textbook that may be a necessary compromise; I don't know if anyone has tried to cover this more rigorously in a paper in the literature.

 

[PeterDonis]

I think part of the confusion is that different authors are talking about different questions

To expand on this a bit more, even if we're talking about general questions (as opposed to a particular solution such as a plane gravitational wave), there are at least two different ones that can be asked:

(1) How many independent degrees of freedom does it take to completely describe a spacetime geometry? I believe the answer is 20, based on my earlier analysis; that's the number of independent components of the curvature tensor.

(2) How many independent equations do you have to solve to solve the Einstein Field Equation? I believe the answer to this can be either 6 or 10; it's 6 if you only want a description of the metric, but it's 10 if you also want a description of the source.

The reason for the difference between these two answers is that the Einstein Field Equation only involves the Ricci tensor, not the full Riemann tensor. The Ricci tensor (or the Einstein tensor) has only 10 independent components, not 20; the other 10 independent components of the Riemann tensor are often referred to as the Weyl tensor. The Einstein Field Equation doesn't tell you the Weyl tensor; more precisely, for a given Ricci tensor (or Einstein tensor), there can be *multiple* solutions of the Einstein Field Equation that describe different geometries, and therefore have different Weyl tensors.

For example, take the simplest case of a vacuum spacetime, with an Einstein tensor of zero. Flat Minkowski spacetime, Schwarzschild spacetime, and Kerr spacetime are all solutions of the EFE with this Einstein tensor, and they are different geometries with different Weyl tensors. So just specifying that the 10 components of the Einstein tensor are zero isn't enough to fully constrain the geometry. You need to also specify the 10 components of the Weyl tensor, for 20 degrees of freedom total.

This point appears to be glossed over in a lot of discussions, such as the one in Carroll's lecture notes, where it is said that specifying the 10 independent components of the metric is all that's needed, and therefore solving the EFE's 10 independent equations is "sufficient". In a sense this is true: all of the tensors we've been talking about, Riemann, Ricci, Einstein, Weyl, are derived from the metric, so if the metric is fully specified, so are all these tensors. But because of the freedom to choose coordinates, a metric is not "fully specified" just by writing down a line element in some coordinate chart. If you look at actual cases, you find that there are always additional constraints imposed; they just aren't emphasized, so the question of how many additional degrees of freedom it takes to specify them, over and above the 10 independent metric components or the 10 independent equations in the EFE, is not addressed.

Go back to my example above, of vacuum solutions of the EFE. Two such solutions, as I said, are flat Minkowski spacetime and Schwarzschild spacetime. Observe that we can write *both* of these solutions using the *same* coordinates and the *same* line element:

ds^2 = - \left( 1 - \frac{2M}{r} \right) dt^2 + 1 / \left(1 - 2M / r \right) dr^2 + r^2 \left( d\theta^2 + sin^2 \theta d\phi^2 \right)

This is the general line elment for Schwarzschild spacetime; flat Minkowski spacetime is just the limiting case where M = 0. So just choosing the coordinate chart and writing down the line element doesn't fully specify the geometry. At a minimum, we need to also specify M; but even that's not all. We also had to specify that the spacetime was spherically symmetric and asymptotically flat. I haven't seen an accounting of how many additional degrees of freedom that takes, but it certainly takes some.

Once again, in a textbook it might be unavoidable to gloss over these technical issues; but it would be nice if someone had gone through them in detail in a paper somewhere in the literature. I haven't seen one, though.

 

[NanakiXIII]

Yes, it is a somewhat eclectic list. I'm not sure there's any way around that. The only way I have found to make sense of it is to just take each item in turn and understand how it works. Are there particular ones you're wondering about?

Mostly still how the choice of coordinates specifies the metric. I don't understand this intuitively, nor technically.

That's why I would like to find the passage I've been referring to in MTW. IIRC they go into more detail about how we know the list of constraints is complete.

I don't have MTW myself but I'll have a look when I'm near a copy.

He's counting something different. He's counting (if I'm thinking of the correct passage, to do with gravitational waves) the degrees of freedom in an idealized plane gravitational wave. That is a very constrained system. The 20 degrees of freedom refers to a generic curvature tensor, which has to be able to describe *any* gravitating system, not just an idealized plane gravitational wave.

If you're only interested in the degrees of freedom of a plane gravitational wave, then most of what I and others have said in this thread is irrelevant. In the OP you mentioned gravitational waves, but then you talked about choosing coordinates and the degrees of freedom in the metric in general terms, not specifically in reference to gravitational waves. So it really depends on what you're interested in.

In this particular paragraph he seems to be talking about an initial value formulation for GR in general. He gives us the same routine I originally started with: diffeomorphism invariance leads to four nonphysical gauge degrees of freedom that drop out of the metric. So, you choose a gauge, e.g. the harmonic gauge, using four constraints. (p. 260)

Yes, for gravitational waves specifically, this is one way of thinking about it that a number of authors discuss. (There's a discussion of it in MTW; I'll look that up when I get a chance.) But I'm not sure this analysis is complete, because it only talks about the degrees of freedom in the metric; it doesn't talk about the first and second derivatives. The first derivatives are all set to 0 by working in a local inertial frame (which is implied by choosing harmonic coordinates and working in the TT gauge); but that still leaves the second derivatives. So I'm not sure that I've seen a good discussion of how this analysis, which is specific to plane gravitational waves, is reconciled with the general analysis I gave in my previous post.

Indeed, it feels handwaving and incomplete, which is where my troubles started. Perhaps it's useful to state my problem more clearly. I was wondering what the symmetry was that allowed us to reduce the metric to two components for gravitational wave descriptions (at least in linearized gravity). I found that, while GR doesn't have any inherent symmetries in the conventional sense, the fact that you raise the metric to a dynamical field means that the theory is generally diffeomorphism invariant. This gets rid of four degrees of freedom, leaving six.

I was under the impression that of the remaining six degrees of freedom, losing four more is something to do with the particular structure of GR; other theories of gravitation (which I'm by no means familiar with in detail) retain up to six polarizations. I was looking for the reason for this.

Because of all the confusion in the texts, I didn't find a solution, only more worries. Then I started to doubt about the validity of the coordinate "gauge" explanation, for losing four degrees of freedom, itself, hence my original question (which I still do not think answered.)

While I am interested in this question in the context of gravitational waves mostly, and first of all I want to understand the linearized case, I am interested in the more general issues as well, because I want to understand them in the context of the post-Newtonian formalism. I don't know that the considerations from linearized gravity carry over.

In response to your second post:

This is very interesting, I had never realized this. However, presumably, when working with a known matter distribution, you don't need to worry about this, since these Weyl components become fixed. There aren't any equations of motion for these degrees of freedom, right?

 

[PeterDonis]

Mostly still how the choice of coordinates specifies the metric. I don't understand this intuitively, nor technically.

It's not that the choice of coordinates specifies the metric; it's that you can't write down the metric as a set of functions of the coordinates (which is what the 10 metric coefficients are) without specifying the coordinates.

In this particular paragraph he seems to be talking about an initial value formulation for GR in general.

Yes, you're right, I've seen that in other references as well. But this is still a somewhat different problem than specifying a geometry as a single mathematical object; you're separating the problem into two parts, specifying the initial conditions and evolving the solution forward in time. Again, there should be some way of reconciling these two viewpoints with respect to the total number of degrees of freedom, but I haven't seen such a discussion in any of the references I've read.

Indeed, it feels handwaving and incomplete, which is where my troubles started. Perhaps it's useful to state my problem more clearly.

...

You give a good clear statement of the issue, I think, and it matches what I've seen in various references. But I still think it leaves out the relationship between the degrees of freedom in the metric, and the degrees of freedom in the curvature tensor. Again, I don't think I've seen a good discussion of this specific issue in any reference.

when working with a known matter distribution, you don't need to worry about this, since these Weyl components become fixed.

Not quite, no. Take the vacuum example again. Minkowski spacetime and Schwarzschild spacetime are both vacuum spacetimes--no matter anywhere (meaning we know the matter distribution)--but they have different Weyl tensors.

There aren't any equations of motion for these degrees of freedom, right?

The Einstein Field Equation only gives you equations of motion involving the Ricci tensor (and the stress-energy tensor), true. But as the vacuum case illustrates, you can have spacetimes with the same Ricci tensor but different Weyl tensors, so the equations of motion derived from the EFE can't be the whole story.

Physically, the Weyl tensor is thought of as being due to "propagation" of curvature from one part of spacetime to another, on the assumption that the curvature is ultimately created via the EFE by the presence of stress-energy somewhere in the past. So, for example, in a physically realistic solution for a black hole, we don't just use the vacuum Schwarzschild spacetime by itself; we have to combine it with some non-vacuum spacetime in the past, representing the massive object that originally collapsed to form the hole. The nonzero Weyl tensor in the vacuum region outside the hole, which is what distinguishes that solution from flat Minkowski spacetime, is due to the spacetime curvature originally caused by the collapsing matter (which is due to the non-zero Ricci tensor in the non-vacuum region of the spacetime, because of the presence of stress-energy there and the EFE) propagating forward in time to the vacuum region. But whether this notion of "propagation" qualifies as an equation of motion depends on your definitions of terms; I've seen equations written down for how the Weyl tensor propagates (IIRC MTW has a discussion of this), but I haven't seen them called "equations of motion".

 

[NanakiXIII]

It's not that the choice of coordinates specifies the metric; it's that you can't write down the metric as a set of functions of the coordinates (which is what the 10 metric coefficients are) without specifying the coordinates.

This I understand quite well and isn't my problem. But 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.

This is a good clear statement of the issue, I think, and matches what I've seen in various references. But I still think it leaves out the relationship between the degrees of freedom in the metric, and the degrees of freedom in the curvature tensor. Again, I don't think I've seen a good discussion of this specific issue in any reference.

It leaves out a good many things, since I don't have a grasp on them yet. It's exactly what I'm looking to figure out: what is the relationship between all these degrees of freedom, and your choice of coordinates.

Not quite, no. Take the vacuum example again. Minkowski spacetime and Schwarzschild spacetime are both vacuum spacetimes--no matter anywhere (meaning we know the matter distribution)--but they have different Weyl tensors.

I understand; my terminology was poor. I mean, if you know the mass distribution globally, i.e. in the Schwarzschild metric you know the mass of the source.

Physically, the Weyl tensor is thought of as being due to "propagation" of curvature from one part of spacetime to another, on the assumption that the curvature is ultimately created via the EFE by the presence of stress-energy somewhere in the past. So, for example, in a physically realistic solution for a black hole, we don't just use the vacuum Schwarzschild spacetime by itself; we have to combine it with some non-vacuum spacetime in the past, representing the massive object that originally collapsed to form the hole. The nonzero Weyl tensor in the vacuum region outside the hole, which is what distinguishes that solution from flat Minkowski spacetime, is due to the spacetime curvature originally caused by the collapsing matter (which is due to the non-zero Ricci tensor in the non-vacuum region of the spacetime, because of the presence of stress-energy there and the EFE) propagating forward in time to the vacuum region. But whether this notion of "propagation" qualifies as an equation of motion depends on your definitions of terms; I've seen equations written down for how the Weyl tensor propagates (IIRC MTW has a discussion of this), but I haven't seen them called "equations of motion".

I'm not sure I understand this. It's certainly not something I've seen before. However (though it may well be interesting) with the risk of sounding rude, I'm going to try to focus the thread back on topic, because I'm sure we could digress into these things indefinitely.


My original question still stands, but perhaps I can reformulate it to be slightly more to the point:

How does choosing coordinates cause a loss of degrees of freedom. I do not understand this mechanism.

How many degrees of freedom does it cost? PeterDonis has reasonably explained why it should cost ten, if any. Most textbooks, however, including Wald, talk about losing four.

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?

 

[PeterDonis]

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.

 

[PeterDonis]

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 g_{\mu \nu} 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 x^{\mu} to the coordinates x&#039;^{\alpha} of a local Lorentz frame at event E. We expand this transformation in powers of x^{\mu}:

x&#039;^{\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} + {} ...

We then look at how we can choose the coefficients M^{\alpha}{}_{\mu}, N^{\alpha}{}_{\mu \nu}, P^{\alpha}{}_{\mu \nu \rho} in order to make the metric g&#039;_{\alpha \beta} in the local Lorentz frame, and its derivatives, meet the above conditions.

First, there are 16 coefficients M^{\alpha}{}_{\mu} that we can choose. This let's us set the ten metric coefficients g&#039;_{\alpha \beta} 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 N^{\alpha}{}_{\mu \nu} (there are only 40, not 64, because we must have N^{\alpha}{}_{\mu \nu} = N^{\alpha}{}_{\nu \mu} 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 P^{\alpha}{}_{\mu \nu \rho} (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.

 

[NanakiXIII]

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 g_{\mu \nu} 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 x^{\mu} to the coordinates x&#039;^{\alpha} of a local Lorentz frame at event E. We expand this transformation in powers of x^{\mu}:

x&#039;^{\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} + {} ...

We then look at how we can choose the coefficients M^{\alpha}{}_{\mu}, N^{\alpha}{}_{\mu \nu}, P^{\alpha}{}_{\mu \nu \rho} in order to make the metric g&#039;_{\alpha \beta} in the local Lorentz frame, and its derivatives, meet the above conditions.

First, there are 16 coefficients M^{\alpha}{}_{\mu} that we can choose. This let's us set the ten metric coefficients g&#039;_{\alpha \beta} 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 N^{\alpha}{}_{\mu \nu} (there are only 40, not 64, because we must have N^{\alpha}{}_{\mu \nu} = N^{\alpha}{}_{\nu \mu} 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 P^{\alpha}{}_{\mu \nu \rho} (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 x^{\mu} \to y^{\mu}, we can expand:

<br /> 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<br />

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 M constrain the metric components, N the first derivatives and P the second derivatives. For the metric componens to vary, then, I would say you would have to somehow make M and N 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.

 

[TrickyDicky]

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."

 

[NanakiXIII]

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?

 

[TrickyDicky]

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

 

[TrickyDicky]

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.

 

[NanakiXIII]

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.

 

[PeterDonis]

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 x&#039;^{\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.

You're taking derivatives of your coordinate functions, not of your metric

The coefficients M, N, P will appear when you take derivatives of x&#039;^{\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&#039;^{\alpha} with respect to x^{\mu}:

g_{\mu \nu} = \frac{\partial x&#039;^{\alpha}}{\partial x^{\mu}} \frac{\partial x&#039;^{\beta}}{\partial x^{\nu}} g&#039;_{\alpha \beta}

Similarly, we find that the transformation of first derivatives of the metric contains second derivatives of x&#039;^{\alpha} with respect to x^{\mu}, and the transformation of second derivatives of the metric contains third derivatives of x&#039;^{\alpha} with respect to x^{\mu}.

Your M constrain the metric components, N the first derivatives and P the second derivatives. For the metric components to vary, then, I would say you would have to somehow make M and N 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 \partial x&#039;^{\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.

 

[stevendaryl]

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 g_{\mu \nu} doesn't mean that they have already chosen a coordinate system.

 

[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&#039;^{\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&#039;^{\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&#039;^{\alpha} with respect to x^{\mu}:

g_{\mu \nu} = \frac{\partial x&#039;^{\alpha}}{\partial x^{\mu}} \frac{\partial x&#039;^{\beta}}{\partial x^{\nu}} g&#039;_{\alpha \beta}

Similarly, we find that the transformation of first derivatives of the metric contains second derivatives of x&#039;^{\alpha} with respect to x^{\mu}, and the transformation of second derivatives of the metric contains third derivatives of x&#039;^{\alpha} with respect to x^{\mu}.

They may appear, but I don't see how they constrain anything. If g&#039;_{\alpha \beta} has degrees of freedom, then so does

g_{\mu \nu} = \frac{\partial x&#039;^{\alpha}}{\partial x^{\mu}} \frac{\partial x&#039;^{\beta}}{\partial x^{\nu}} g&#039;_{\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&#039;^{\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&#039;_{\alpha \beta} has degrees of freedom, then so does

g_{\mu \nu} = \frac{\partial x&#039;^{\alpha}}{\partial x^{\mu}} \frac{\partial x&#039;^{\beta}}{\partial x^{\nu}} g&#039;_{\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&#039;_{\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&#039;_{\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&#039;_{ab} represent the same physical space - time if there exists a vector field \xi ^{a} such that h&#039;_{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&#039;_{ab} represent the same physical space - time if there exists a vector field \xi ^{a} such that h&#039;_{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&#039;_{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?

 

[WannabeNewton]

Forest management?

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&#039;_{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?