A Dirac's integral for the energy-momentum of the gravitational field

[Kostik]

TL;DR Summary

Dirac integrates his pseudo-tensor for the energy-momentum of the gravitational field over a sufficiently large 4-volume to demonstrate (subject to some conditions) that total energy-momentum is conserved. But his expression has an integral of the mixed tensor $T_\mu^\nu$ that cannot be converted to the usual contravariant $T^{\mu\nu}$ without passing the metric tensor through the integral, which cannot be done in curved space.

See Dirac's brief treatment of the energy-momentum pseudo-tensor in the attached picture. Dirac is presumably integrating eq. (31.2) over the 4D "hypercylinder" defined by $T_1 \le x^0 \le T_2$ and $\mathbf{|x|} \le R$, where $R$ is sufficiently large to include all the matter-energy fields in the system. Then
\begin{align}
0 &= \int_V \left[ ({t_\mu}^\nu + T_\mu^\nu)\sqrt{-g}\, \right]_{,\nu} d^4 x = \int_{\partial V} ({t_\mu}^\nu + T_\mu^\nu)\sqrt{-g} \, dS_\nu \nonumber\\
&= \left( \int_{\mathbf{|x|} \le R, \, x^0=T_2} - \int_{\mathbf{|x|} \le R, \, x^0=T_1} \right) ({t_\mu}^0 + T_\mu^0)\sqrt{-g}\, dx^1 dx^2 dx^3 + \int_{\mathbf{|x|} = R, \, T_1 \le x^0 \le T_2} {t_\mu}^\nu \sqrt{-g} \, dS_\nu \nonumber\\
\end{align} Subject to the assumption that the integrals converge and that the flux integral vanishes as $R\rightarrow\infty$, Dirac deduces that
$$\lim_{R\rightarrow\infty} \int_{\mathbf{|x|} \le R} ({t_\mu}^\nu + T_\mu^\nu )\sqrt{-g} \, dx^1 dx^2 dx^3$$ is conserved, and gives the total energy-momentum of the system.

The problem I have is that he's got the mixed tensor $T_\mu^0$ and mixed pseudo-tensor $t_\mu^0$. In flat spacetime (with rectilinear coordinates), the (inverse) metric $g^{\mu\nu}$ is constant and can be moved through the integrals. But in curved spacetime, $T^{\mu 0}$ is the energy-momentum, not $T_\mu^0$, and $$T^{\mu 0} = g^{\mu\alpha} T_\alpha^0 \,\, .$$ Likewise,
$$t^{\mu\nu} + T^{\mu\nu} = g^{\mu\alpha} ({t_\alpha}^\nu + T_\alpha^\nu) \,\, .$$ If $g^{\mu\alpha}(x)$ varies throughout space, then it cannot pass through the integrals. So how does Dirac justify saying "We thus have definite expressions for the total energy and momentum, which are conserved"?

 

[PeterDonis]

Dirac is presumably integrating eq. (31.2) over the 4D "hypercylinder"

Where are you getting that from? All I see in his (31.4) is integrating over a 3-volume. A conserved "total energy" over a 4-volume wouldn't make sense anyway; the whole point of "conserved" is "doesn't change with time", which means time can't be part of the integral you do to get the "conserved" quantity.

 

[Kostik]

Where are you getting that from? All I see in his (31.4) is integrating over a 3-volume. A conserved "total energy" over a 4-volume wouldn't make sense anyway; the whole point of "conserved" is "doesn't change with time", which means time can't be part of the integral you do to get the "conserved" quantity.

I think it's pretty clear from his comment "The equation (31.2) then shows that the integral (31.4) taken at one time $x^0=a$ equals its value at another time $x^0=b$." Please look at my post, you will see the same 3D integral appear.

 

[PeterDonis]

I think it's pretty clear from his comment "The equation (31.2) then shows that the integral (31.4) taken at one time $x^0=a$ equals its value at another time $x^0=b$."

What's clear to me from that comment is that the integral (31.4) is taken over a 3-volume at a particular time, not over a 4-D hypercylinder. Otherwise it would make no sense to say that the value of the integral (31.4) at two different times is the same.

 

[Kostik]

What's clear to me from that comment is that the integral (31.4) is taken over a 3-volume at a particular time, not over a 4-D hypercylinder.

You are confusing the 3D integral of the pseudo tensor with $\nu = 0$ with the 4D integral of (31.2).

 

[PeterDonis]

Please look at my post, you will see the same 3D integral appear.

I know. Now please look at my post, where I asked about a specific statement you made about an integral over a 4-D hypercylinder. That's the statement I'm questioning.

 

[PeterDonis]

You are confusing the 3D integral of the pseudo tensor with $\nu = 0$ with the 4D integral of (31.2).

(31.2) is not an integral. It's a differential equation.

 

[Kostik]

(31.2) is not an integral. It's a differential equation.

You integrate both sides of (31.2).

 

[PeterDonis]

You integrate both sides of (31.2).

Where does Dirac do that over a 4-D hypercylinder? Nowhere that I can see.

 

[Kostik]

Where does Dirac do that? Nowhere that I can see.

It is implied in the sentence I quoted. I did the work in my post above. Dirac doesn't fill in every step. Please just kindly read the original post and all should be clear.

 

[PeterDonis]

Dirac doesn't fill in every step.

That's certainly true of Dirac, not just here but in general. And reading the text following (31.4), his assumptions (a) and (b) do correspond to the assumptions you describe in your OP. I agree that the equations in your OP that derive Dirac's (31.4) look valid if those assumptions (which, as Dirac points out, are often not satisfied in actual cases) are true.

 

[Kostik]

@PeterDonis I would appreciate your comments if any on how to deal with the index positions, to get a contravariant quantity which can justify Dirac's interpretation of the integral as being "the total energy and momentum".

 

[PeterDonis]

I would appreciate your comments if any on how to deal with the index positions

I was just about to hit "Post reply" when I saw this. See below.

in curved spacetime, $T^{\mu 0}$ is the energy-momentum

This might be one of those cases where Dirac's habit of implicitly picking a particular class of coordinate charts instead of writing everything in invariant form makes things difficult. Let me try to rewrite things in invariant form.

First, there is no such thing as "the energy-momentum" without some kind of qualification. The qualification Dirac is implicitly using is that the coordinate chart he's implicitly picking is adapted to a particular class of observers, whose worldlines are the "time" gridlines of the chart. Thus their 4-velocities are parallel to the coordinate basis vector $\partial_0$, so we can write them as $u^\mu = (k, 0, 0, 0)$, where $k$ is a coefficient that can vary from event to event. The reason $k$ has to vary is that $u^a$ has to be a unit vector, so we must have $g_{\mu \nu} u^\mu u^\nu = 1$ (using the timelike convention that Dirac appears to prefer), which means $k = 1 / \sqrt{g_{00}}$.

Now, what is the (non-gravitational) energy-momentum density measured by those observers? In covector form, it will be $E_\mu = T_{\mu \nu} u^\nu$; that is the most "basic" form, not involving the metric tensor explicitly. Note that this has to be the most "basic" form, because the energy density measured by the observers with 4-velocity $u^\mu$ is $E = T_{\mu \nu} u^\mu u^\nu$. In other words, the (0, 2) tensor $T_{\mu \nu}$ is the most "basic" form of the stress-energy tensor, because it contracts with two vectors (without requiring the metric tensor to be included explicitly in the contraction) to form scalars that give the energy density, momentum flux, pressure, shear stress, etc. that are actually measured by particular observers (and along particular spatial directions which are embodied by spacelike unit vectors carried along in a tetrad by the observers).

So the statement by you that is quoted above is not correct as you state it. The covector $E_\mu$ that I wrote down above is the form of "energy momentum" adapted to particular observers that doesn't involve the metric tensor explicitly in its invariant expression. But in our chosen coordinate chart, that covector becomes $E_\mu = T_{\mu 0} / \sqrt{g_{00}}$. Note that a metric coefficient now appears there--but it comes from the 4-velocity $u^\mu$ and its normalization, not from $T_{\mu \nu}$. In Dirac's notation, which uses $Y$ instead of $T$ for the non-gravitational stress-energy, this would be $E_\mu = Y_{\mu 0} / \sqrt{g_{00}}$.

Note that this expression already has the $\mu$ index as a lower index, which is where we want it for Dirac's (31.4). We just need to raise the $0$ index; we would do that by multiplying by $g^{00}$. I think that by taking into account the particular choice of coordinate chart that we're using, we can end up with $Y_\mu{}^0 \sqrt{-g}$, which is what we need. (I realize this part is hand-waving; I would need to take some more time to work through the details of an invariant version to verify that it simplifies the way we want.)

The point is that all of this would be done before doing any integral; we just need to justify using $Y_\mu{}^0 \sqrt{-g}$ as our expression for the "density of non-gravitational energy momentum seen by observers at rest in our chosen coordinate chart". Once we've justified that, we just use that as our integrand and there's no need to pass the metric or inverse metric through the integral, everything inside the integral is already the way we want it.

 

[Kostik]

@PeterDonis That's probably a little over my head. Yes, I agree it makes sense to deal with $({t_\mu}^0 + T_\mu^0)\sqrt{-g}\,$ prior to integrating, since there's no way to pass the metric through the integral.

I guess I have to wonder, if $T^{\mu 0}$ is the energy and momentum density (obviously coordinate/observer dependent), what exactly is $T_\mu^0$? Even in flat space, if you choose non-rectilinear coordinates (for example, polar coordinates), the metric will not be constant, so $T_\mu^0 = g_{\mu\alpha}T^{\alpha 0}$ is a very different "thing" than the energy and momentum density.

Dirac must be right in saying that "We thus have [in the integral (31.4)] definite expressions for the total energy and momentum", because the integral (subject to the various provisos) IS conserved, and -- except for the index positions -- it looks an awful lot like the energy-momentum tensor.

I somehow suspect the issue I have raised isn't really that deep. After all, there are other energy-momentum pseudo-tensors, like the Landau-Lifshitz pseudo-tensor $t^{\mu\nu}_{LL}$ which satisfy the same equation (31.2) but with all contravariant indices, so the entire question I am asking simply doesn't arise.

 

[PeterDonis]

if $T^{\mu 0}$ is the energy and momentum density

It isn't. I explained what is in my previous post.

Dirac must be right in saying that "We thus have [in the integral (31.4)] definite expressions for the total energy and momentum", because the integral (subject to the various provisos) IS conserved, and -- except for the index positions -- it looks an awful lot like the energy-momentum tensor.

It's a pseudotensor, not a tensor (because $t$ is a pseudotensor, not a tensor), so it only describes "total energy and momentum" in a coordinate-dependent sense. Which, according to most GR textbooks, means it isn't physically meaningful. Dirac tries to extract a limited physical meaning for it; whether that viewpoint is valid is a matter on which physicists in the field differ. But in any case, since the integral is only valid when his conditions (a) and (b) are true, and as he admits, they aren't true in most practical cases, the usefulness of what he's doing is definitely limited.

I somehow suspect the issue I have raised isn't really that deep.

That depends on how seriously you take what I said in my previous post about what actually is the energy and momentum density. Some physicists pay more attention than others to actual index positions and whether you need to bring in the metric or its inverse to get an expression whose indexes are where you want them. Dirac is probably near the "don't really care" end of that spectrum.

 

[Kostik]

@PeterDonis I'm still struggling with this. Perhaps I should have phrased the question a more direct way:

(1) If $({M_\mu}^\nu)_{,\nu}=0$, does it follow that $(M^{\mu\nu})_{,\nu}=0 \,\,?$

(2) If $$\int {M_\mu}^0 \, dx^1 dx^2 dx^3$$ is conserved, does it follow that $$\int M^{\mu 0} \, dx^1 dx^2 dx^3$$ is conserved?

(No need to assume that ${M_\mu}^\nu$ is a tensor.) The answer to (1) is clearly no. If the answer to (2) is also no, which I suspect, then I question Dirac's statement "We thus have [in the integral (31.4)] definite expressions for the total energy and momentum".

This whole issue goes away by using the L-L pseudotensor, which satisfies $$\left[ -g(t^{\mu\nu}_{LL} + T^{\mu\nu}) \right]_{,\nu}=0 \,\, ,$$ and is symmetric. I wonder why Dirac did not base his chapters 32-33 on this pseudotensor, since it was discovered and published in the first edition of The Classical Theory of Fields in 1951, long before Dirac's book in 1975.

 

[PeterDonis]

(1) If $({M_\mu}^\nu)_{,\nu}=0$, does it follow that $(M^{\mu\nu})_{,\nu}=0 \,\,?$

Well, let's see.

$$
M^{\mu \nu} = g^{\mu \alpha} M_\alpha{}^\nu
$$

Therefore,

$$
M^{\mu \nu}_{,\nu} = \left( g^{\mu \alpha} M_\alpha{}^\nu \right)_{,\nu}
$$

which gives

$$
M^{\mu \nu}_{,\nu} = g^{\mu \alpha} \left( M_\alpha{}^\nu{}_{,\nu} \right) + M_\alpha{}^\nu \left( g^{\mu \alpha}{}_{,\nu} \right)
$$
The first term vanishes by hypothesis, but the second doesn't. So I think the answer to your question is no. And you agree with this later in your post.

Note that if we were taking covariant derivatives instead of partial derivatives (i.e., semicolons instead of commas), then the answer to (the analogue to) your question would be yes, since the metric is covariantly constant.

(2) If $$\int {M_\mu}^0 \, dx^1 dx^2 dx^3$$ is conserved, does it follow that $$\int M^{\mu 0} \, dx^1 dx^2 dx^3$$ is conserved?

I would say no because raising the index under the integral sign would introduce a factor of the inverse metric that would change the value of the integral.

I question Dirac's statement "We thus have [in the integral (31.4)] definite expressions for the total energy and momentum".

That's because you keep persisting in thinking of $T^{\mu \nu}$ with two upper indexes as the density of stress-energy, even though I've told you more than once now that that's wrong, and explained why.

 

[Kostik]

Yes, I am thinking of the contravariant $T^{\mu 0}$ as the energy and momentum. If you lower one of the indices with the metric, you may well change the units, since the various components of the metric may have different units (depending upon the coordinate system). Consider for example the energy momentum tensor of a dust $T^{\mu\nu}=\rho v^\mu v^\nu$ or a perfect fluid $T^{\mu\nu} = (E+p)v^\mu v^\nu - pg^{\mu\nu}$. I don't see how you can lower an index with the metric (in curved spacetime) willy-nilly, and say that the physical interpretation remains the same.

Sorry if I have not grasped your explanations.

 

[PeterDonis]

This whole issue goes away by using the L-L pseudotensor

Only if you think $T^{\mu \nu}$ represents what I've said it doesn't represent, and explained why.

I can't remember if L-L use the same restricted class of coordinate charts that Dirac does.

and is symmetric.

$T_{\mu \nu}$ with two lower indexes, which is what I have said is the correct expression for the density of stress-energy, and explained why, is also symmetric. So there is no reason on these grounds to prefer the form with two upper indexes.

 

[PeterDonis]

Yes, I am thinking of the contravariant $T^{\mu 0}$ as the energy and momentum.

I know you are. And you have not responded at all to my argument in post #13 for why that is not correct--why $T_{\mu \nu}$, with two lower indexes, is physically the correct expression for the density of stress-energy, the expression that does not intrinsically involve any factors of the metric tensor.

 

[PeterDonis]

I don't see how you can lower an index with the metric (in curved spacetime) willy-nilly, and say that the physical interpretation remains the same.

You can't--at least, not if you're going to take the viewpoint of Dirac (and L-L, for that matter) that it's perfectly okay to write curved spacetime equations using partial derivatives instead of covariant derivatives. In other textbooks, such as MTW, that viewpoint is not taken--everything in curved spacetime is done with covariant derivatives and covariant volume elements in integrals, so the issue doesn't arise; you can always raise or lower indexes as you please and the equations remain physically the same.

But you're not taking that viewpoint here. And if you don't, you can't just assert without argument that $T^{\mu \nu}$ is the density of stress-energy. You have to give a physical reason why that form is the one that intrinsically involves no factors of the metric. I've given such a reason for $T_{\mu \nu}$, the (0, 2) tensor, in post #13. Please read it.

 

[Kostik]

$T_{\mu \nu}$ with two lower indexes, which is what I have said is the correct expression for the density of stress-energy, and explained why, is also symmetric. So there is no reason on these grounds to prefer the form with two upper indexes.

I meant the Landau-Lifshitz pseudotensor is symmetric, while the Dirac-Einstein pseudotensor is not. I wasn't referring to the energy-meomentum tensor of the matter-energy fields -- which is symmetric.

 

[PeterDonis]

I meant the Landau-Lifshitz pseudotensor is symmetric

As far as I know, the same would be true if we lowered both indexes, so again, this is not a reason to prefer the form with two upper indexes.

 

[Kostik]

As far as I know, the same would be true if we lowered both indexes, so again, this is not a reason to prefer the form with two upper indexes.

The Dirac-Einstein pseudotensor is definitely not symmetric (with the indices in any position). So this seems a good reason to prefer the LL pseudotensor.

 

[PeterDonis]

The Dirac-Einstein pseudotensor is definitely not symmetric (with the indices in any position). So this seems a good reason to prefer the LL pseudotensor.

You're missing the point. I wasn't saying there was no reason to prefer the L-L pseudotensor to the Dirac pseudotensor; I think that's a matter of opinion. I was saying that "it's symmetric" is not a reason to prefer the L-L (2, 0) pseudotensor to the L-L (0, 2) pseudotensor.