Variation of CMFR mass density ##\rho## along a world line

[Kostik]

TL;DR Summary

Dirac shows that, along a world line, $\frac{d\rho}{ds} = -\rho {v^\mu}_{;\mu}.$ He states "this is a condition that fixes how $\rho$ varies along the world line of an element of matter."

Dirac ("GTR" p. 47) makes an interesting observation immediately after obtaining Einstein's field equations with the simple energy-momentum tensor $T^{\mu\nu}=\rho v^\mu v^\nu$. ($v^\mu$ is the four-velocity.)

First, the conservation of matter $\left( \rho v^\mu \sqrt{-g} \right)_{,\mu}=0$ implies $\left( \rho v^\mu \right)_{;\mu}=0$. Hence $$\rho_{;\mu}v^\mu + \rho {v^\mu}_{;\mu}=0. \quad(*)$$ On a geodesic, Dirac previously showed that ${v^\mu}_{;\nu}=0$ so the second term in $(*)$ vanishes, hence $\rho_{;\mu}=\rho_{,\mu}=0$. This doesn't seem surprising: a small element of matter in free-fall shouldn't experience a change in its CMFR density $=\rho$.

However, Dirac takes $(*)$ one step further. We have $$\frac{d\rho}{ds} = \frac{\partial\rho}{\partial x^\mu}v^\mu = -\rho {v^\mu}_{;\mu}.$$ He states "this is a condition that fixes how $\rho$ varies along the world line of an element of matter."

This is intriguing. Does this mean that along a world line other than a geodesic, the CMFR density $\rho$ is changing by a factor of ${v^\mu}_{;\mu}$? Why?

I suppose that if a small element of matter is not moving on a geodesic then it must be subject to some non-gravitational force, but the factor ${v^\mu}_{;\mu}$ doesn't contain any information about such a force. Can someone shed a little light here?

 

[renormalize]

I suppose that if a small element of matter is not moving on a geodesic then it must be subject to some non-gravitational force, but the factor ${v^\mu}_{;\mu}$ doesn't contain any information about such a force. Can someone shed a little light here?

Per Dirac's General Theory of Relativity eq.(25.4), the energy-momentum tensor of a dust (i.e., pressure-less) material of mass-density $\rho$ and 4-velocity $v^{\mu}$ is $T_{M}^{\mu\nu}=\rho v^{\mu}v^{\nu}$. Suppose now that this material is not in freefall, but is rather acted-upon by an external force-field having energy-momentum tensor $T_{X}^{\mu\nu}$. The divergence of the sum of these two tensors must vanish:$$0=\left(T_{X}^{\mu\nu}+T_{M}^{\mu\nu}\right){}_{;\nu}=T_{X}^{\mu\nu}{}_{;\nu}+\left(\rho v^{\mu}v^{\nu}\right)_{;\nu}=T_{X}^{\mu\nu}{}_{;\nu}+v^{\mu}\left(\rho v^{\nu}\right)_{;\nu}+\rho v^{\nu}v^{\mu}{}_{;\nu}=T_{X}^{\mu\nu}{}_{;\nu}+\rho v^{\nu}v^{\mu}{}_{;\nu}\tag{1}$$where eq.(25.3) has been used to set $\left(\rho v^{\nu}\right)_{;\nu}=0\tag{1}$. The next-to-last term in (1) is by definition the negative of the 4-force-density $f_{X}^{\mu}$ exerted by the external field on the material:$$T_{X}^{\mu\nu}{}_{;\nu}\equiv-f_{X}^{\mu}\tag{2}$$whereas the last term in (1) is:$$\rho v^{\nu}v^{\mu}{}_{;\nu}=\rho\left(v^{\nu}\frac{\partial v^{\mu}}{\partial x^{\nu}}+\Gamma_{\sigma\nu}^{\mu}v^{\nu}v^{\sigma}\right)=\rho\frac{Dv^{\mu}}{ds}\equiv\rho a^{\mu}\tag{3}$$where $a^\mu$ is the proper 4-acceleration experienced by the material. Combining (1), (2) and (3) gives:$$f_{X}^{\mu}=\rho a^{\mu}\tag{4}$$which is just the relativistic version of Newton's second-law, defining the acceleration of the the material caused by an arbitrary external force, like an imposed electric field.

 

[Kostik]

Is there a way to express $d\rho/ds$ in terms of $f^\mu$ in some explicit way?

 

[PeterDonis]

if a small element of matter is not moving on a geodesic then it must be subject to some non-gravitational force, but the factor ${v^\mu}_{;\mu}$ doesn't contain any information about such a force.

That's because the force doesn't directly affect $\rho$. As @renormalize showed, the force directly affects $v^\mu$. In other words, it affects the worldlines of the matter, which are described by the 4-velocity field $v^\mu$.

Is there a way to express $d\rho/ds$ in terms of $f^\mu$ in some explicit way?

The "Newton's Second Law" equation should make it clear that there is not: that equation relates $f^\mu$ to $\rho$, not $d \rho / ds$.

 

[Ibix]

That's because the force doesn't directly affect $\rho$. As @renormalize showed, the force directly affects $v^\mu$. In other words, it affects the worldlines of the matter, which are described by the 4-velocity field $v^\mu$.

So we should read $\rho$ like a flux density in EM, one that measures the density of cosmological dust worldlines piercing a spacelike hypersurface?

 

[PeterDonis]

So we should read $\rho$ like a flux density in EM, one that measures the density of cosmological dust worldlines piercing a spacelike hypersurface?

Sort of. It's the energy density measured locally by an observer who is comoving with the fluid. But you can, if you squint a little, think of it as something like "flux of timelike stuff in the time direction across a local now surface".

 

[Kostik]

I interpret $d\rho/ds=-\rho {v^\mu}_{ ;\mu}$ as the density being inversely proportional to the divergence of the world lines, which makes intuitive sense. Why the world lines diverge — must be because of the action of the external force on $\rho$.

We have $$f^\mu = \rho v^\sigma {v^\mu}_{;\sigma}$$ and $$d\rho/ds = -\rho {v^\mu}_{;\mu}$$ but there seems to be no more direct way of linking $f^\mu$ with the divergence ${v^\mu}_{;\mu}$.

 

[PeterDonis]

I interpret $d\rho/ds=-\rho {v^\mu}_{ ;\mu}$ as the density being inversely proportional to the divergence of the world lines

No, it's the change in density that is inversely proportional to the divergence of the worldlines. That's what the equation says. The proportionality constant is the density at the point where you are evaluating the equation.

Or, if you want to look at it another way, rewrite the equation as $(1 / \rho) d \rho / ds = d ( \ln \rho ) / ds = - v^\mu_{; \mu}$. So the change in the logarithm of the density is inversely proportional to the divergence of the worldlines, with a proportionality constant of $1$.

Why the world lines diverge — must be because of the action of the external force on $\rho$.

No, it's because of the action of the external force on $v^\mu$, since $v^\mu$ is what describes the worldlines. Which, again, is what the Second Law equation says.

 

[Kostik]

No, it's the change in density that is inversely proportional to the divergence of the worldlines. That's what the equation says. The proportionality constant is the density at the point where you are evaluating the equation.

Or, if you want to look at it another way, rewrite the equation as $(1 / \rho) d \rho / ds = d ( \ln \rho ) / ds = - v^\mu_{; \mu}$. So the change in the logarithm of the density is inversely proportional to the divergence of the worldlines, with a proportionality constant of $1$.


No, it's because of the action of the external force on $v^\mu$, since $v^\mu$ is what describes the worldlines. Which, again, is what the Second Law equation says.

The force acts on $\rho$ which is what alters $v^\mu$.

 

[PeterDonis]

The force acts on $\rho$

What actual math are you basing this on?

 

[Kostik]

What actual math are you basing this on?

What does force act on?

 

[PeterDonis]

What does force act on?

As has already been said, the Newton's Second Law equation says it acts on $v^\mu$; that's the quantity whose rate of change appears in the force equation. The density $\rho$ is a proportionality constant in that equation; it tells you how much force (or more precisely force density) you need to produce a given amount of proper acceleration (rate of change of $v^\mu$).

 

[Kostik]

If $\rho$ is charged matter, the external force (an electric field) acts on $\rho$. The divergence in $v$ is the result.

In $f=ma$, $f$ acts on $m$; $a$ is the result.

 

[PeterDonis]

If $\rho$ is charged matter, the external force (an electric field) acts on $\rho$.

What actual math says so?

The divergence in $v$ is the result.

What actual math says so?

In $f=ma$, $f$ acts on $m$; $a$ is the result.

You seem to have a problem reading equations. $f = ma$ says that the force is equal to mass times acceleration. Acceleration is the rate of change of velocity. Mass is not the rate of change of anything. So if you're going to say that $f$ "acts on" anything, since "acts on" means "causes a change in", it can only be velocity, because that's the only thing whose rate of change appears in the equation.

 

[Kostik]

Semantics. By "acts on" I am thinking of the object the external force physically acts on, not the change in its motion.

 

[PeterDonis]

Semantics.

No, physics. See below.

By "acts on" I am thinking of the object the external force physically acts on, not the change in its motion.

In other words, you are ignoring the actual physics in favor of vague informal handwaving. Which, apart from anything else, is why you are having trouble grasping what Dirac was actually saying in what you quoted in your OP of this thread, as well as what @renormalize told you in post #2.

 

[PeterDonis]

Thread closed for moderation.

 

[PeterDonis]

After moderator review, the thread will remain closed.