How GR Resolves the Conservation of Momentum and Energy

In summary: In that case, I agree with what you're saying: active diffeomorphims don't change the actual geometry of the spacetime, they just transform which coordinate chart you use to describe it.
  • #106
stevendaryl said:
I'm a little on shakier ground when it comes to diffeomorphisms. The passive case is clear enough--it's just a coordinate change, and the equations are the same for any coordinate system. The active case is the one I'm a little fuzzy on. Roughly speaking, it works like this:

Let m be a (continuous, invertible, differentiable, blah, blah, blah) function from ℝ4 to ℝ4. Let X[itex]\alpha[/itex] be some coordinate system (function from spacetime points [itex]P[/itex] to ℝ4). Let S be some solution to the full equations of motion. Then if the theory is invariant under active diffeomorphisms, then we can get another solution S' in the following round-about way:

Let X'[itex]\alpha[/itex] be the coordinate system obtained from X[itex]\alpha[/itex] by applying the transformation m. Then pick S' so that the description of S' in terms of the original coordinate system X[itex]\alpha[/itex] looks the same as the description of S in terms of the new coordinate system X'[itex]\alpha[/itex].

But S' is related to S by (an obvious) coordinate transformation. So you haven't generated a new solution at all; you've generated the same solution, in different coordinates.

Solutions are only distinct up to isometries. If you want to generate a truly different solution, you will have to consider diffeomorphisms which are not isometries; i.e., you will have to forget about coordinate transformations, either "active" or "passive", and do something entirely different. See my earlier post about changing g by an infinitesimal perturbation.

Remember there is no invariant notion of what it means to "hold coordinates fixed while changing the manifold under them", unless you have a suitable mathematical definition you'd like to explain. Coordinates are just labels for points on the manifold; they have no sense of location other than that!
 
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  • #107
Ben Niehoff said:
Remember there is no invariant notion of what it means to "hold coordinates fixed while changing the manifold under them", unless you have a suitable mathematical definition you'd like to explain. Coordinates are just labels for points on the manifold; they have no sense of location other than that!

I'd like to expand on this, because there was some confusion about it earlier in the thread.

There is no such thing as "writing the Kerr solution in Schwarzschild coordinates", or anything similar. There is no invariant notion of what "Schwarzschild coordinates" are, except on the Schwarzschild geometry itself, where they can be given precise definitions:

t is chosen such that [itex]\partial_t[/itex] is the timelike Killing vector (this is only possible because such a Killing vector exists).

[itex]\theta[/itex] and [itex]\phi[/itex] are chosen as spherical coordinates on the spherical shells orthogonal to [itex]\partial_t[/itex] (this is only possible because of spherical symmetry; i.e., three Killing vectors that generate SO(3)).

r is defined such that [itex]\partial_r[/itex] is orthogonal both to [itex]\partial_t[/itex] and to the spherical shells, and such that the area of each spherical shell is [itex]4 \pi r^2[/itex] (again this is only possible because of spherical symmetry).

These definitions only make sense on geometries that have the appropriate Killing vectors (so, for example, the Kerr geometry will not work). And seeing the precise definitions, it hardly seems appropriate to call these "Schwarzschild" coordinates at all.

Note: This is typically how coordinate systems are defined when looking for new solutions to GR. First one carefully finds (or imposes) constraints on symmetry or other algebraic properties the manifold must have; then one can define coordinates consistent with these symmetries. Finally one writes down a metric ansatz in these coordinates with some arbitrary functions chosen to be consistent with the symmetries, and then one crosses one's fingers and hopes that Einstein's equations will reduce to something solvable.
 
  • #108
Ben Niehoff said:
There is no such thing as "writing the Kerr solution in Schwarzschild coordinates", or anything similar.

Yes, this is a good point. I think I was one of the ones who said something like this upthread: what I meant to say was that one can think of some 4-tuple (t, r, theta, phi) as labeling some point in Schwarzschild spacetime, *and* as labeling some point in Kerr spacetime. It seems to me that at least some of the talk about "active diffeomorphisms" implies drawing some kind of connection between the point (t, r, theta, phi) in one spacetime and the point in the other spacetime labeled with the same 4-tuple (t, r, theta, phi); as if somehow an active diffeomorphism "holds the points the same" (in terms of keeping the labeling the same), but changes the metric "underneath them". I'm not sure that this has any actual physical interpretation, though.
 
  • #109
PeterDonis said:
Yes, this is a good point. I think I was one of the ones who said something like this upthread: what I meant to say was that one can think of some 4-tuple (t, r, theta, phi) as labeling some point in Schwarzschild spacetime, *and* as labeling some point in Kerr spacetime. It seems to me that at least some of the talk about "active diffeomorphisms" implies drawing some kind of connection between the point (t, r, theta, phi) in one spacetime and the point in the other spacetime labeled with the same 4-tuple (t, r, theta, phi); as if somehow an active diffeomorphism "holds the points the same" (in terms of keeping the labeling the same), but changes the metric "underneath them". I'm not sure that this has any actual physical interpretation, though.

I'm not sure how there is any meaning to this. Here's a spacetime labelled with [itex](t, r, \theta, \phi)[/itex]:

[tex]ds^2 = -d\theta^2 + dt^2 + dr^2 + d\phi^2.[/tex]
Certainly this breaks some conventions, but it doesn't break anything mathematical. And recall from my earlier post, before we can even think about imposing the usual conventions, we must have a manifold with certain symmetries. Otherwise we can't define what we mean by things like "time coordinate", "spherical coordinate", "radial coordinate", etc.
 
  • #110
Ben Niehoff said:
What you are really going after is the idea of "preferred frames". With u included, there is a preferred frame. But you must make a distinction between "having no preferred frame", and "being invariant under Lorentz/Galilean/etc. symmetry", because it is always possible to specify the preferred frame in an invariant way; it is, after all, a real, geometrical object. Being invariant under some set of coordinate transformations is a property of how the equations are written down; having a preferred frame is the result of placing some real object that doesn't participate in dynamics.
The idea of preferred frames is directly related to spontaneous symmetry breaking.
 
  • #111
Ben Niehoff said:
Come now, you must be fair. The transformation must also include

u' = u - v

just as it does in the passive case. Then the transformed solution works! There is no material distinction between "active" and "passive".

That does not seem correct to me. Intuitively, if there is a preferred rest frame (the frame in which the "friction" force is zero), then spacetime is not invariant under Galilean boosts. But you can always write the equations of motion so that they have the same form under changes of arbitrary coordinates.

Maybe it has to do with whether or not you consider the u field to be part of the initial conditions, or part of the laws of physics?

That is, when people do physics, there is usually (maybe not always) a separation between the "laws of physics" and the "solutions to those laws". In the simplest kind of particle dynamics of the Newtonian sort, you have some number of particles, and the "solution" that you're trying to solve for is a set of functions

Xnj(t)

the position of the nth particle as a function of time. The "laws" are a bunch of equations of the form

mn d2Xnj/dt2 = Fnj

where Fnj is the force on the nth particle, given by some force law.

For a given form of the force law, we can ask whether it has the property that the solutions are closed under "boosts"; that is, whether the transformation

Xnj(t) → Xnj(t) + vj t

takes you from solutions to solutions. For some force laws, it will, and for some force laws, it won't.

What you are saying, I think, is that if we also consider Fnj to be part of the solution, then we can get another solution by messing with F, as well. That's true; if you consider absolutely everything to be fair game for the transformation, then maybe there is no difference between passive and active transformations. But I think that for a given division between "laws" and "solutions", the distinction makes sense.

You might complain that I'm not allowed to touch u because it's a constant in the equation of motion. But "active" vs. "passive" is always a matter of interpretation. I am actively changing the physical situation, not just by shifting x(t), but also by changing u.

What you are really going after is the idea of "preferred frames". With u included, there is a preferred frame. But you must make a distinction between "having no preferred frame", and "being invariant under Lorentz/Galilean/etc. symmetry", because it is always possible to specify the preferred frame in an invariant way; it is, after all, a real, geometrical object.

Yes, that's what I thought the active/passive distinction was supposed to be getting at.

Being invariant under some set of coordinate transformations is a property of how the equations are written down; having a preferred frame is the result of placing some real object that doesn't participate in dynamics.

The same can happen in electromagnetism if I consider a situation with a fixed, infinite current-carrying wire along the z axis. There is now a preferred frame, but I can still write everything in a Lorentz-invariant way.

Like I said, it seems that it has to do with what you consider to be part of the solution, versus part of the laws.

Another way that I heard people make similar distinctions is to ask whether there are fundamental, non-dynamic scalar fields, vector fields or tensor fields. In Newtonian physics, there is a fundamental non-dynamic scalar field, the universal time. It's a scalar field in the sense that it is a real number associated with each point in spacetime. It's non-dynamic in that it can affect things, but nothing can affect it.

In Special Relativity, the metric tensor gαβ is a fundamental non-dynamic scalar field.

General Relativity has no fundamental non-dynamic fields.
 
  • #112
Ben Niehoff said:
But S' is related to S by (an obvious) coordinate transformation. So you haven't generated a new solution at all; you've generated the same solution, in different coordinates.

I don't see that. Surely the physical situation in which two point-masses are 1 million miles apart is different from the physical situation in which they are 1 mile apart, if we keep the same dynamics. In the one case, the masses may fly off to infinity, while in the other case, the masses may form a bound system (an orbit). It's certainly true that if you move the masses from 1 million miles away to 1 mile away, and correspondingly weaken the gravitational constant G, then we will have done nothing.

Solutions are only distinct up to isometries. If you want to generate a truly different solution, you will have to consider diffeomorphisms which are not isometries; i.e., you will have to forget about coordinate transformations, either "active" or "passive", and do something entirely different. See my earlier post about changing g by an infinitesimal perturbation.

Remember there is no invariant notion of what it means to "hold coordinates fixed while changing the manifold under them", unless you have a suitable mathematical definition you'd like to explain. Coordinates are just labels for points on the manifold; they have no sense of location other than that!

I think you would agree that a universe with two point masses that fly by each other and escape to infinity is a different situation than a universe with two point masses that orbit each other. So what is the difference?
 
  • #113
stevendaryl said:
I don't see that. Surely the physical situation in which two point-masses are 1 million miles apart is different from the physical situation in which they are 1 mile apart, if we keep the same dynamics. In the one case, the masses may fly off to infinity, while in the other case, the masses may form a bound system (an orbit).

Agreed those are two completely different physical situations. However, you haven't shown that you can transform one situation into the other via the process you've described. In fact, it is impossible to do so, since your process is equivalent to a coordinate transformation.

Statements like "Two masses are 1 million miles apart" are coordinate-invariant. Distance is measured by rulers laid out along geodesics (if more than one geodesic connects the two masses, take the infimum). There is no change of coordinates that can change measured distances; that's the whole point of "coordinate invariance".

(To be more precise in scattering problems, you can say "impact parameter", "closest approach", or something like that...in any case you will have a precisely-defined invariant quantity.)
 
  • #114
PeterDonis said:
Yes, this is a good point. I think I was one of the ones who said something like this upthread: what I meant to say was that one can think of some 4-tuple (t, r, theta, phi) as labeling some point in Schwarzschild spacetime, *and* as labeling some point in Kerr spacetime.

That doesn't make sense. You cannot first label points, and then consider the metric in GR. That's the moral of the hole-argument.
 
  • #115
stevendaryl said:
In Newtonian physics, there is a fundamental non-dynamic scalar field, the universal time. It's a scalar field in the sense that it is a real number associated with each point in spacetime. It's non-dynamic in that it can affect things, but nothing can affect it.

In Newtonian physics you have two metrics which are kept invariant under the Galilei group. One is a temporal metric, which is related to the absolute time function you mention. The other metric is simply the constant spatial metric. If you multiply one metric with the other, you get zero; the metric structure is degenerate.

This should be compared to the Minkowski metric.
 
  • #116
haushofer said:
That doesn't make sense. You cannot first label points, and then consider the metric in GR. That's the moral of the hole-argument.

I agree that it doesn't really make sense physically, for the reasons Ben gave in recent posts.
 
  • #117
Ben Niehoff said:
Agreed those are two completely different physical situations. However, you haven't shown that you can transform one situation into the other via the process you've described. In fact, it is impossible to do so, since your process is equivalent to a coordinate transformation.

That's not obvious. Let me try to be slightly more abstract, and then we can see if the abstraction applies to GR, and if not, where it breaks down. The following is really my thinking aloud, rather than a specific question that I want a response to (although a response is certainly welcome).

Suppose I have a set [itex]O[/itex] of objects of some sort. I have a set [itex]D[/itex] of descriptions of objects. I have a "description function" m of type [itex]O[/itex] → [itex]D[/itex]. Let's suppose that this mapping is actually a bijection: there is one object for each description, and one description for each object.

Now, suppose I have a different description function m'. So m' is a different way of describing the same objects O.

Then it is possible to put m and m' together to get a map
[itex]f[/itex] of type [itex]O[/itex] → [itex]O[/itex] as follows:

[itex]f(x) = m'^{-1}(m(x))[/itex]

Now, the question is: under what circumstances will [itex]f[/itex] be the identity function?

The relevance to GR is that O is the set of solutions to the equations of motion (Einstein's field equations, plus Maxwell's equations, plus whatever else). m and m' are two different coordinate descriptions of those solutions.

As I write, I think I see what is going on: Although there can be multiple coordinate descriptions for the same solution, so there can be many functions of type
[itex]O[/itex] → [itex]D[/itex], the inverse is unique: there is only one
function of type [itex]D[/itex] → [itex]O[/itex]. So [itex]m'^{-1} = m^{-1}[/itex].

I guess this conclusion is true by definition, or it can be made true by definition: We can define a solution to be an equivalence class of descriptions, where the equivalence relation is that one description is transformable into the other by a coordinate transformation.
 
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  • #118
Steven,

Technically speaking, in GR (and differential geometry in general), [itex]D \rightarrow O[/itex] is not unique; it is only locally unique, because the metric tensor is a local object. There may be global issues to consider.

For example, an infinite cylinder has the same metric tensor as R^2, but they are not the same topological manifold. Likewise, RP^2 has the same metric tensor as S^2, and a catenoid has the same metric tensor as a helicoid:

http://en.wikipedia.org/wiki/Catenoid

The point is, there are global topological properties that are not constrained by the metric tensor (there are also global topological properties that are constrained by it, however).

But all this is tangential to the issue to transforming between solutions. To go between a cylinder and a plane, one doesn't need to transform any coordinates at all; one just makes the identification [itex]y \sim y + 2\pi[/itex].
 
  • #119
Ben Niehoff said:
There is a lot of confusion in this thread, and it seems like I caused some of it, so I'll jump in.

Metric-preserving diffeomorphisms, which are maps [itex]\varphi : (M, g) \rightarrow (N, h)[/itex], smooth with smooth inverse, such that [itex]g = \varphi^* h[/itex] (i.e. the metric on M is the pullback along [itex]\varphi[/itex] of the metric on N). Metric-preserving diffeomorphisms do care about size and shape, and are fully equivalent to coordinate transformations.

Finally, there is the question of what kinds of transformations leave Einstein's equations invariant? First, look at vacuum solutions with cosmological constant:

[tex]R_{\mu\nu} - \frac12 R g_{\mu\nu} + \Lambda g_{\mu\nu} = 0.[/tex]
Under conformal rescaling by a constant, [itex]g \mapsto a^2 g[/itex], we have [itex]R_{\mu\nu} \mapsto R_{\mu\nu}[/itex] and [itex]R \mapsto R/a^2[/itex], and hence the equation is invariant if we also assume [itex]\Lambda \mapsto \Lambda/a^2[/itex]. What about more general conformal rescaling? From here:

http://en.wikipedia.org/wiki/Ricci_curvature#Behavior_under_conformal_rescaling

one has for [itex]g \mapsto e^{2 \varphi} g[/itex]

[tex]\tilde{\operatorname{Ric}}=\operatorname{Ric}+(2-n)[ \nabla d\varphi-d\varphi\otimes d\varphi]+[\Delta \varphi -(n-2)\|d\varphi\|^2],[/tex]
which indicates that Einstein's equation cannot be invariant under general such transformations. Therefore it is clear that general diffeomorphisms are not a symmetry of Einstein's equations!

So I agree with most of your post, except the last part. I am a little uneasy with the terminology. Following Wald and Nakahara:

A conformal isometry is a diffeomorphism:
Psi: M --> M such that (Psi*G)uv = Omega^2 Guv provided omega is everwhere real and positive. The case Omega = 1 is just a regular isometry.

A conformal transformation (or Weyl rescaling):
Guv' = Omega^2 Guv

is NOT in general a diffeomorphism! If you don't take the pullback, then it won't be invariant.

Anyway, the equations of GR (the EH action) is a scalar under linearized (orientation preserving) diffeomorphisms in the usual way that I am sure all of you have seen before (it takes a chapter to show it). It is in that sense that we say that there is general covariance.
 
  • #120
Ben Niehoff said:
By the way, here is how to identify what transformations preserve Einstein's equations.

[tex]\Box h_{\mu\nu} + 2 R_\mu{}^\rho{}_\nu{}^\sigma h_{\rho\sigma} = 0.[/tex]
Any solutions to this equation give infinitesimal perturbations that preserve Einstein's equation. However, the gauge condition on h does not fully gauge-fix, so some of the solutions will just be infinitesimal coordinate transformations; you have to throw those ones out.

Anything left will give you a non-trivial invariance of GR. These will be local diffeomorphisms which are not isometries.

Correct! There is also, as you mentioned a sublety with so called large diffeomorphisms (diffeormorphisms which are not continously connected to the identity), as well as with so called boundary diffeomorphisms (which change the asymptotic structure in some way). The linearization procedure obscures these facts, and they have to be added by hand.
 
  • #121
Haelfix said:
A conformal isometry is a diffeomorphism:
Psi: M --> M such that (Psi*G)uv = Omega^2 Guv provided omega is everwhere real and positive. The case Omega = 1 is just a regular isometry.

Can you give an example? I'm having a hard time imagining such a map. Is it possible to have a continuous family of such maps?

A conformal transformation (or Weyl rescaling):
Guv' = Omega^2 Guv

is NOT in general a diffeomorphism! If you don't take the pullback, then it won't be invariant.

But I'm looking for things which are not invariant. Is a round 2-sphere of radius A diffeomorphic to a round 2-sphere of radius B, or not?

I say it is. In patches, [itex]\varphi: (\theta, \phi) \mapsto (\theta, \phi)[/itex], which is clearly differentiable, and poses no problems with the transition functions. But [itex]g_A[/itex] is not the pullback of [itex]g_B[/itex] along [itex]\varphi[/itex].

One does have [itex]\varphi^*(g_B) = (b^2/a^2) \, g_A[/itex], is that what you mean above? In this case, [itex]\varphi : A \rightarrow B[/itex], not [itex]\varphi : A \rightarrow A[/itex].
 
  • #122
Haelfix said:
So I agree with most of your post, except the last part. I am a little uneasy with the terminology. Following Wald and Nakahara:

A conformal isometry is a diffeomorphism:
Psi: M --> M such that (Psi*G)uv = Omega^2 Guv provided omega is everwhere real and positive. The case Omega = 1 is just a regular isometry.

A conformal transformation (or Weyl rescaling):
Guv' = Omega^2 Guv

is NOT in general a diffeomorphism! If you don't take the pullback, then it won't be invariant.
You seem to be contradicting yourself here.
A Weyl rescaling is a conformal transformation of the metric, and all conformal transformations are diffeomorphisms (they are defined as the subgroup of diffeomorphisms that preserve the metric up to a scale, the conformal factor).
 
  • #123
TrickyDicky said:
You seem to be contradicting yourself here.
A Weyl rescaling is a conformal transformation of the metric, and all conformal transformations are diffeomorphisms (they are defined as the subgroup of diffeomorphisms that preserve the metric up to a scale, the conformal factor).

Real quick, b/c I have to go. Yes. I did just change conventions from a few posts back, but that's b/c the definition of a conformal transformation differs between the texts I'm consulting and I just switched to Wald's convention. (Nakahara calls the former definition a conformal transformation, Wald calls the latter a conformal transformation). What's important is that they are distinct mathematical concepts. (scroll through a few pages in Wald as well)

http://books.google.com/books?id=9S...A#v=onepage&q=wald conformal isometry&f=false

http://books.google.com/books?id=cH...QHf78HeAw&ved=0CDcQ6AEwAA#v=onepage&q&f=false

(See the example from the latter).
 
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