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 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.