Energy in GR

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  • #101
stevendaryl
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Wrt to the passive vs active diffeomorphism issue I can't see how it can have any physical consequence, it is just a way to allude to the fact that every diffeomorphism is bijective and therefore every active (passive)transformation that is a diffeomorphism has its inverse passive (active) transformation
I think that the distinction between active and passive is perhaps clearer when you use a simpler transformation than a full diffeomorphism. So let's take invariance under Galilean transformations:

x' = x - vt
t' = t

There is an active and a passive interpretation of this transformation. The passive interpretation is about the form of the equations of motion: If you start with the equations expressed in terms of (x,t), and then you rewrite the equations in terms of (x',t'), then the equations have the same form. For example, if the equations of motion are:

m d2x/dt2 = F

then since d2x/dt2 = d2x'/dt'2, the equations have the same form in terms of (x',t'), provided that the force F transforms as:

F' = F.

On the other hand, the active interpretation is about solutions to the equations of motion: If x = f(t) is a solution to the equations of motion, then x = f(t) - vt is also a solution to the equations of motion.

Notice the distinction: In the passive case, we are considering the same situation expressed in different coordinates. In the active case, we are consider a different situation expressed in the same coordinates. Newton's equations, with a velocity-independent force, are invariant under both types of transformations.

But now let's spoil the invariance by throwing in a friction force proportional to the velocity. We can still write the equations in a way that is invariant under passive transformations:

m d2x/dt2 = F - k (dx/dt - u)

This is still invariant in form under a Galilean transform, provided that we transform things as follows:

x' = x - vt
t' = t
F' = F
u' = u - v

So things are still invariant under passive transforms. But what about active transforms?

If x = f(t) is a solution to the equations of motion, is x = f(t) - vt also a solution to the equations of motion? Clearly not.
 
  • #102
haushofer
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So is it then right to say that the fact that because the Einstein/geodesic equation AND their solutions in GR are invariant under general coordinate transformations, makes some people say that GR is invariant under both passive AND active general coordinate transformations?
 
  • #103
stevendaryl
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So is it then right to say that the fact that because the Einstein/geodesic equation AND their solutions in GR are invariant under general coordinate transformations, makes some people say that GR is invariant under both passive AND active general coordinate transformations?
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].

So a very simple example is the following: we have two charged point masses A and B, each of mass M, and in the coordinate system X[itex]\alpha[/itex] they are a distance L apart in the x-direction (coordinate X[itex]1[/itex]), with a nonzero relative velocity in the y-direction (coordinate X[itex]2[/itex]). Otherwise, we have an empty universe (with some specified boundary conditions at infinity, such as being asymptotically flat). So we use GR, plus electrodynamics in curved spacetime, to compute the trajectories of those two masses.

Now, consider the simple coordinate transformation x → λx, where λ is just some constant.

Under a passive transformation, this is just a rescaling of the x-coordinate, just like changing from feet to inches. Clearly, the physics is not changed by such a rescaling.

Under an active transformation, this is the same as if you physically changed the location of the two masses. This is a drastically different situation; it can make the difference between the case where the two masses are gravitational bound and the case where they are not. The claim made for General Relativity is that if you make the corresponding changes to the metric, and to the electric fields, and to trajectories, etc., then the new situation will be a solution, as well. That's a much stronger claim than invariance under passive transformations.

Where I'm a little fuzzy is exactly what does it mean to say that we change the distance between the masses and change everything else appropriately.
 
  • #104
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You raise some good points.

I think that the distinction between active and passive is perhaps clearer when you use a simpler transformation than a full diffeomorphism. So let's take invariance under Galilean transformations:

x' = x - vt
t' = t

There is an active and a passive interpretation of this transformation. The passive interpretation is about the form of the equations of motion: If you start with the equations expressed in terms of (x,t), and then you rewrite the equations in terms of (x',t'), then the equations have the same form. For example, if the equations of motion are:

m d2x/dt2 = F

then since d2x/dt2 = d2x'/dt'2, the equations have the same form in terms of (x',t'), provided that the force F transforms as:

F' = F.

On the other hand, the active interpretation is about solutions to the equations of motion: If x = f(t) is a solution to the equations of motion, then x = f(t) - vt is also a solution to the equations of motion.

Notice the distinction: In the passive case, we are considering the same situation expressed in different coordinates. In the active case, we are consider a different situation expressed in the same coordinates. Newton's equations, with a velocity-independent force, are invariant under both types of transformations.

But now let's spoil the invariance by throwing in a friction force proportional to the velocity. We can still write the equations in a way that is invariant under passive transformations:

m d2x/dt2 = F - k (dx/dt - u)

This is still invariant in form under a Galilean transform, provided that we transform things as follows:

x' = x - vt
t' = t
F' = F
u' = u - v

So things are still invariant under passive transforms. But what about active transforms?

If x = f(t) is a solution to the equations of motion, is x = f(t) - vt also a solution to the equations of motion? Clearly not.
Thanks, this helps me see where the physical problem with the passive/active distinction comes from. Certainly this problem shows up in GR with any solution of the EFE that is not static. Since the mathematical definition of diffeomorphism includes both types of transformation this problem would make many GR solutions simply not invariant wrt diffeomorphisms (in the GR sense that preserves the metric), even if the equations are.
It also reminds me of the situation in Hamiltonian mechanics with the difference between symplectic and contact manifolds.
Would this be linked to the spontaneous symmetry breaking concept mentioned above?

So is it then right to say that the fact that because the Einstein/geodesic equation AND their solutions in GR are invariant under general coordinate transformations, makes some people say that GR is invariant under both passive AND active general coordinate transformations?
The problem is that mathematically is not very rigorous to separate passive from active transformations, it makes little sense.
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].

So a very simple example is the following: we have two charged point masses A and B, each of mass M, and in the coordinate system X[itex]\alpha[/itex] they are a distance L apart in the x-direction (coordinate X[itex]1[/itex]), with a nonzero relative velocity in the y-direction (coordinate X[itex]2[/itex]). Otherwise, we have an empty universe (with some specified boundary conditions at infinity, such as being asymptotically flat). So we use GR, plus electrodynamics in curved spacetime, to compute the trajectories of those two masses.

Now, consider the simple coordinate transformation x → λx, where λ is just some constant.

Under a passive transformation, this is just a rescaling of the x-coordinate, just like changing from feet to inches. Clearly, the physics is not changed by such a rescaling.

Under an active transformation, this is the same as if you physically changed the location of the two masses. This is a drastically different situation; it can make the difference between the case where the two masses are gravitational bound and the case where they are not. The claim made for General Relativity is that if you make the corresponding changes to the metric, and to the electric fields, and to trajectories, etc., then the new situation will be a solution, as well. That's a much stronger claim than invariance under passive transformations.

Where I'm a little fuzzy is exactly what does it mean to say that we change the distance between the masses and change everything else appropriately.
Do you agree that mathematically is not possible to separate active from passive transformations, because basically they are dual notions and it is the same operation only conventionally the object motion wrt the observer is the reference POV in the active case and the observer motion wrt the object is the reference in the passive one?
Once again the physical interpretation might be an instance of spontaneous symmetry breaking IMO.
 
  • #105
Ben Niehoff
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But now let's spoil the invariance by throwing in a friction force proportional to the velocity. We can still write the equations in a way that is invariant under passive transformations:

m d2x/dt2 = F - k (dx/dt - u)

This is still invariant in form under a Galilean transform, provided that we transform things as follows:

x' = x - vt
t' = t
F' = F
u' = u - v

So things are still invariant under passive transforms. But what about active transforms?

If x = f(t) is a solution to the equations of motion, is x = f(t) - vt also a solution to the equations of motion? Clearly not.
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".

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. 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.
 
  • #106
Ben Niehoff
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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!
 
  • #107
Ben Niehoff
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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
PeterDonis
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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
Ben Niehoff
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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
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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
stevendaryl
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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
stevendaryl
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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
Ben Niehoff
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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
haushofer
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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
haushofer
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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
PeterDonis
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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
stevendaryl
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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
Ben Niehoff
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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
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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 wont 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
Haelfix
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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
Ben Niehoff
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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 wont 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
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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 wont 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
Haelfix
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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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