# Metric constraints in choosing coordinates

1. Mar 12, 2013

### 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?

2. Mar 12, 2013

### 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.

3. Mar 12, 2013

### 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.

4. Mar 12, 2013

### 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.

5. Mar 12, 2013

### 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.

6. Mar 12, 2013

### 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?

7. Mar 12, 2013

### 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.

8. Mar 12, 2013

### Staff: Mentor

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.

Last edited: Mar 12, 2013
9. Mar 12, 2013

### 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.

10. Mar 12, 2013

### 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.

11. Mar 12, 2013

### 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.

12. Mar 12, 2013

### NanakiXIII

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

13. Mar 12, 2013

### Staff: Mentor

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.

14. Mar 12, 2013

### 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.

15. Mar 12, 2013

### Staff: Mentor

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).

16. Mar 12, 2013

### 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.

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.

17. Mar 12, 2013

### Staff: Mentor

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.)

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.

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.

18. Mar 12, 2013

### NanakiXIII

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.

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?

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.

19. Mar 12, 2013

### Staff: Mentor

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.

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.

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.

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.

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).

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

20. Mar 13, 2013

### 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.

21. Mar 13, 2013

### Staff: Mentor

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?

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.

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.

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.

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.

22. Mar 13, 2013

### Staff: Mentor

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.

23. Mar 13, 2013

### NanakiXIII

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

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

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)

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?

24. Mar 13, 2013

### Staff: Mentor

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.

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.

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.

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.

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".

Last edited: Mar 13, 2013
25. Mar 13, 2013

### NanakiXIII

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.

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.

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.

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?