# 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.