Dirac's Conservation of Matter: A Closer Look

[Kostik]

TL;DR Summary

Dirac's comments on conservation of matter from the Einstein field equations

In Dirac's "General Theory of Relativity", at the end of Ch. 25 (p. 47), right after deriving the full Einstein equation $R^{\mu\nu} - \frac{1}{2}g^{\mu\nu}R = -8\pi\rho v^\mu v^\nu = -8\pi T^{\mu\nu}$, he makes a reference to the conservation of mass (Eq. 25.3):
$$0 = (\rho v^\mu)_{:\mu} = \rho_{:\mu}v^\mu + \rho v^\mu_{\,\,:\mu}.$$
(Here $v^\mu \equiv dx^\mu/ds$.) This can be deduced from the Einstein field equation by application of the Bianci relation, but that's not important here. He then writes (this is just the chain rule, since $\rho$ is a scalar, hence $\rho_{:\mu}=\rho_{,\mu}$):
$$\frac{d\rho}{ds} = \frac{\partial\rho}{\partial x^\mu} v^\mu = -\rho v^\mu_{\,\,:\mu}.$$
So far, so good. What I fail to understand is Dirac's comment that follows:

"This is a condition that fixes how $\rho$ varies along the world line of an element of matter. It allows $\rho$ to vary arbitrarily from the world line of one element to that of a neighboring element. Thus we may take $\rho$ to vanish except for a packet of world lines forming a tube in space-time. Such a packet would compose a particle of matter of a finite size. Outside the particle we have $\rho=0$, and Einstein's field equation for empty space hold."

Can someone interpret this? The matter density $\rho(x^\mu)$ is whatever it is; I don't see how we are free to change this. It is true that the previous equation means that you can look at an infinitesimal particle of matter, and as its proper time advances by $ds$ the previous equation gives a condition for the change $d\rho$ in matter density. However, I don't follow the comment "It allows $\rho$ to vary arbitrarily from the world line of one element to that of a neighboring element. Thus we may take $\rho$ to vanish except for a packet of world lines forming a tube in space-time." I'm just lost here.

Appreciate if someone could clarify this.

 

[PeterDonis]

right after deriving the full Einstein equation $R^{\mu\nu} - \frac{1}{2}g^{\mu\nu}R = -8\pi\rho v^\mu v^\nu = -8\pi T^{\mu\nu}$

Note that the $\rho v^\mu v^\nu$ part is not the general Einstein equation; it is only valid under a particular very specific assumption about the stress-energy present: that it is matter with zero pressure (often called "dust"). It might help to read everything Dirac says with that in mind.

Can someone interpret this?

It's very simple: the conservation equation only relates quantities along a single worldline. Which means, as Dirac says, that it tells you nothing about the relationship between those quantities on different worldlines.

The matter density $\rho(x^\mu)$ is whatever it is; I don't see how we are free to change this.

The conservation equation is not an equation for $\rho(x^\mu)$ everywhere at once. It is only an equation for $\rho$ along a single worldline. So you cannot infer anything about $\rho(x^\mu)$ in general from the conservation equation. That is Dirac's point.

Of course in any particular solution you will have to know $\rho(x^\mu)$ in general. But Dirac is not talking about a single particular solution here. He is talking about the entire general class of solutions that satisfies the conservation equation he derived.

 

[ergospherical]

Can someone interpret this? The matter density $\rho(x^\mu)$ is whatever it is; I don't see how we are free to change this.

I think he's just showing how you could construct the worldtube representing a single blob of matter by choosing the density in the way he described.

 

[Kostik]

I think he's just showing how you could construct the worldtube representing a single blob of matter by choosing the density in the way he described.

But the mass density/flow is taken as a given; I don't see how one can "choose" it.

 

[PeterDonis]

the mass density/flow is taken as a given

Where?

 

[Kostik]

Where?

Chap. 25, p. 45, where he says "$\rho v^0 \sqrt{}$ is the density and $\rho v^m \sqrt{}$ is the flow." (Dirac uses $\sqrt{}$ as short-hand for $\sqrt{-g}$.)

 

[PeterDonis]

Chap. 25, p. 45, where he says "$\rho v^0 \sqrt{}$ is the density and $\rho v^m \sqrt{}$ is the flow." (Dirac uses $\sqrt{}$ as short-hand for $\sqrt{-g}$.)

Those are just definitions of symbols. Where does he specify a functional form for $\rho$? Nowhere, as far as I can see.

 

[vanhees71]

Looking at the book, what Dirac did was to note that both the conservation of mass and the motion of dust particles on geodesics follows from Einstein's field equation alone due to the Bianchi identity ${G^{\mu \nu}}_{;\mu}=0$, because from this it follows
$${T^{\mu \nu}}_{;\mu}=(\rho u^{\mu} u^{\nu})_{; \mu} =0. \qquad (1)$$
To prove this we note that
$$u_{\nu} u^{\nu}=1 \; \Rightarrow \; u_{\nu} {u^{\nu}}_{;\mu}=0. \qquad (2)$$
Now we can write (1) as
$$(\rho u^{\mu})_{;\mu} u^{\nu} + \rho u^{\mu} {u^{\nu}}_{;\mu}=0. \qquad (3)$$
contracting this with $u_{\nu}$ and using (2) gives
$$(\rho u^{\mu})_{;\mu}=0, \qquad (4)$$
i.e., (local) conservation of mass (since $\rho$ is the proper mass density of the dust). Using this in (3) yields
$$u^{\mu} {u^{\nu}}_{;\mu}=0,$$
which is the geodesic equation for the dust particles.

As you see, no specific solution for $\rho$, $u^{\mu}$, and $g_{\mu \nu}$ is needed. What's shown however is that to solve the Einstein field equation for dust needs a full solution of the dynamical problem as a whole, i.e., if you find a solution of the Einstein field equation, you've also solve the continuity equation of mass and the geodesic equation for dust.

An example is the FLRW solution, where you use symmetry arguments to get the metric. These symmetries then determine that the "matter content" (including also radiation of course) is necessarily an ideal fluid. To close the equations of motion you need an "equation of state". For dust it's simply $P=0$ (where $P$ is the pressure), which describes massive "non-relativistic matter" in a sense.