# Mass conservation charged dust

1. Aug 10, 2013

### WannabeNewton

Hi guys! This is related to a recent thread but since that thread became cluttered, I figured it would be more coherent to just ask the question here. Say we have a congruence of charged dust particles in some space-time with tangent field $\xi^a$. The energy-momentum of the charged dust is given by $T^{\text{charges}}_{ab} = \rho \xi_a \xi_b$ where as usual $\rho$ is the mass density as measured by observers comoving with the dust. The charge density $\sigma$ is also as measured by comoving observers hence the 4-current $j^a = \sigma \xi^a$ because in a frame field comoving with the dust the 4-current will only have a time-like component (which will be the charge density). However there is an additional energy-momentum from the electromagnetic field carried by the charged dust particles and this is given by $T^{\text{EM}}_{ab} = F_{ac}F_{b}{}{}^{c} - \frac{1}{4}g_{ab}F^{cd}F_{cd}$.

Now $\nabla^a T^{\text{charges}}_{ab}, \nabla^aT^{\text{EM}}_{ab}\neq 0$ by themselves. I've seen authors prove that $\nabla^a T_{ab} = \nabla^a( T^{\text{charges}}_{ab}+ T^{\text{EM}}_{ab}) = 0$ by assuming that (1) the individual charged fluid elements satisfy the Lorentz force law, which can be written as $\rho \xi^b\nabla_b \xi^a = \sigma F^{ab}\xi_b$ for the congruence itself and that (2) the dust satisfy conservation of mass current $\nabla_a (\rho \xi^a) = 0$.

Usually, if we have charge free dust (so that $T_{ab} = \rho \xi_a \xi_b$ is the only energy-momentum source), one first assumes that $\nabla ^a T_{ab} = 0$ and then derives the mass current conservation for the dust from this. Here, in the presence of both the energy-momentum of the dust and the energy-momentum of the electromagnetic fields they carry, these authors are trying to show that $\nabla^a T_{ab} = 0$ for the total energy-momentum $T_{ab}$, and to do this they first assume that mass current conservation holds. Why can we assume it holds, before even showing that $\nabla^a T_{ab} = 0$?

EDIT: Ok nevermind, I overlooked a very simple thing. Well that's that :)!

Last edited: Aug 10, 2013
2. Aug 10, 2013

### Staff: Mentor

What was the simple thing you overlooked?

3. Aug 10, 2013

### ProfDawgstein

Would be nice to know :)

I would say that mass is conserved by arguing that:
-there is nothing which changes the mass
-the electric field just effects their movement
-even if they clump together, mass should still be the same?
(not sure about any binding energies though)

There should be a nice way to derive this though...

Or you just use global conservation and then "calculate backwards" to get mass conservation?

4. Aug 10, 2013

### WannabeNewton

Well at least I hope I overlooked a simple thing! Geroch talks about the mass current conservation here: http://postimg.org/image/brlpqy61z/ and http://postimg.org/image/8w8mk321z/

He assumes the dust is non-interacting (which is not what we have above wherein the dust is interacting) but his proof of mass current conservation doesn't use $\nabla^a T_{ab} =0$ (in fact he proves that later by first showing that mass current is conserved for the dust); he just seems to assume that the dust particles themselves are conserved.

5. Aug 10, 2013

### ProfDawgstein

Where is that from?
Looks very old.

I just checked Cheng again, he does only show it for non-interacting dust.
Lets hope the next book does better

6. Aug 10, 2013

### vanhees71

I'm not an GR expert, but as I understand the model from the manuscript, the author considers "non-relativistic particles", neglecting pressure and collisions but including the long-range forces in terms of gravitational and electromagnetic fields (the gravitational fields of course in the sense of GR as the space-time "metric"). In continuum language that means he considers a mean-field (Vlasov) approximation. First of all this implies that the velocity of "fluid elements" is independent of the choice of the Landau or the Eckart description, i.e., $u^{\mu}$ is the flow four-velocity as defined from the energy-momentum tensor or the electric four-current. The equations of motion are then given by the conservation of energy and momentum, i.e., the covariant divergence of the total energy-momentum tensor (consisting of the matter contribution $\rho u^{\mu} u^{\nu}$, the em. field $F_{\mu \rho} {F_{\nu}}^{\rho} - \frac{g_{\mu \nu}}{4} F_{\mu \nu} F^{\mu \nu}$), the Maxwell equations (with the current $j^{\mu}=\sigma u^{\mu}$, and the Einstein equations for the gravitational field. This set of equations should imply the conservation of mass, i.e., in GR $(\rho u^{\mu})_{;\mu}=0$.

7. Aug 10, 2013

### ProfDawgstein

If I understand you correctly, you mean

$\nabla^a T_{ab} = \nabla^a( T^{\text{charges}}_{ab}+ T^{\text{EM}}_{ab}) = 0$

and then "derive" that

$(\rho u^{\mu})_{;\mu}=0$

or

$\nabla_{a} (\rho u^a) = 0$

which is 'mass conservation'

8. Aug 13, 2013

### WannabeNewton

Hey ProfDawgstein. I'm still not fully grasping Geroch's argument because in my mind there are some physical subtleties surrounding his proof (the one I linked above) that he doesn't explain. However I think I can give somewhat of an intuitive argument. Choose any dust particle in the congruence $\xi^a$ and consider an observer comoving with the chosen dust particle. Take a volume $V$ carried along the worldline of the dust particle with $V$ small enough so that the mass density $\rho$ is essentially uniform in the space within the volume i.e. $\rho = \frac{mN}{V}$ where $N$ is the number of dust particles contained in the volume and $m$ is the rest mass of each dust particle contained in the volume; here 'contained' means both the interior and surface of the volume.

As the proper time $\tau$ on the comoving observer's clock passes, $V$ will be increasing or decreasing because the dust particles on the surface of $V$ (and the ones in the interior as well) will be expanding away or contracting towards the chosen dust particle. However the total mass contained in $V$ must be constant because the rest mass of each dust particle isn't changing with $\tau$ and neither is the number of dust particles contained in $V$ so $\partial_{\tau}(N m) = 0 \Rightarrow \partial_{\tau}(V\rho) = 0$. Consequently, computing in the comoving coordinates setup by the comoving observer, we have $\nabla_{\mu}(\rho \xi^{\mu}) = \partial_{\tau}\rho + \rho\nabla_{\mu}\xi^{\mu} = \partial_{\tau}\rho + \frac{\rho}{V}\xi^{\mu}\nabla_{\mu}V = \frac{1}{V}(V\partial_{\tau}\rho + \rho\partial_{\tau}V) = 0$ where one can show that the expansion $\nabla_{\mu}\xi^{\mu}$ is given by $\nabla_{\mu}\xi^{\mu} = \frac{1}{V}\xi^{\mu}\nabla_{\mu}V$.

9. Aug 14, 2013

### ProfDawgstein

I just read the part in Inverno's book.

He does something like this

$\rho_0$ : mass density in comoving frame

$T^{ab} = \rho_0 u^a u^b$

$\nabla_b T^{ab} = 0$

$\nabla_b [ \rho_0 u^a u^b ] = 0$

using trick : $[ (\rho_0 u^b) u^a ]$ and leibniz rule

$u^a \nabla_b (\rho_0 u^b) + \rho_0 u^b (\nabla_b u^a) = 0$ (13.10)

contracting with $u_a$ and $u_a u^a = 1$

-> $u_a (\nabla_b u^a) = 0$ 2nd term vanishes and 1st term remains

$\nabla_b (\rho_0 u^b) = 0$

which leads to (divide by $\rho_0$ and use 13.10)

$u^b \nabla_b u^a = 0$

Now he says that $u^a$ is tangential to the geodesic, which means that the particles
move on a geodesic.

Not 100% satisfying though :|

I kind of had the same argument.

Last edited: Aug 14, 2013
10. Aug 14, 2013

### vanhees71

But the dust particles only move on geodesics if you neglect the electromagnetic interaction! I thought you want to describe the motion under consideration of the electromagnetic forces in the mean-field approximation (Vlasov equation).

11. Aug 14, 2013

### ProfDawgstein

Yes, this was about the non-interacting case.

I would still like to know the case with interacting particles...

12. Aug 14, 2013

### WannabeNewton

I've already mentioned the case with interacting particles in post #1 of this thread but what does that even matter? We were talking about the conservation of mass current not the equations of motion of the dust. For charged dust, the equations of motion are just $\rho u^b \nabla_b u^a = \sigma F^{ab}u_b$.

13. Aug 14, 2013

### WannabeNewton

Well yeah because d'Inverno is assuming that $\nabla_a T^{ab} = 0$ holds beforehand. We can't do that here because we're trying to show that this is true (for the total energy-momentum) by first justifying mass-current conservation such as in the manner done by Geroch above.

14. Aug 14, 2013

### ProfDawgstein

We are talking about the case with $T^{ab}_{field}$ and $T^{ab}_{dust}$ (dust is charged), right?

Yes. But somehow it has to be true globally, otherwise the whole thing would be a mess...
It should be derivable though.

If $\nabla_a T^{ab} \neq 0$, there would be a source somewhere, but there is not.
But since the system is closed, there should be no sources (overall), so everything is self-contained.
Thus the whole system should obey $\nabla_a T^{ab} = 0$.
Quite hard to imagine something coming out of nowhere.
Isn't energy conservation one of the foundations, which simply has to be true?

[just a thought]

15. Aug 14, 2013

### WannabeNewton

Yes and the aforementioned equations of motion will be important in showing that $\nabla_a T^{ab} = 0$ but it won't be relevant for the justification of $\nabla_a (\rho \xi^a) = 0$ itself.

Yes it is usually taken as a basic assumption (it also follows directly from the Lagrangian formulation) but here the authors are trying to show that when dust particles are interacting with an electromagnetic field, the total energy momentum is conserved by first utilizing mass-current conservation of the dust.

Basically it all just comes back to conservation of particles, which says that $\nabla_{a}(n \xi^a) = 0$ where $n$ is the number density of the fluid particles as measured by comoving observers ($n\xi^a$ is usually called the number flux 4-vector). This can be assumed safely if the fluid particles are taken to be Baryons. $\rho = mn$ where $m$ is the rest mass of each fluid particle so $\nabla_{a}(\rho \xi^a) = m\nabla_{a}(n \xi^a) = 0$.

16. Aug 14, 2013

### ProfDawgstein

It seems that Zee does not assume mass conservation when showing $\nabla_a T^{ab} = 0$.
He uses actions and variations. (VI.4. p383-385)
Actually I can't see anything about mass conservation there.

I guess they all just assume that mass current is conserved by argument...

17. Aug 14, 2013

### WannabeNewton

Yes as I stated it follows directly from the matter-field Lagrangian (I don't have Zee with me so I can't see your reference but it's ok) but this isn't relevant to the argument you saw in Cheng and elsewhere. They are not using the variational principle, they are simply using mass-current conservation and the equations of motion (all for charged dust that is). It is easy to see in flat space-time that $\partial_{a}(n \xi^a) = 0$. Using the equivalence principle and minimal coupling we could lift this up to curved space-time and write $\nabla_{a}(n \xi^a) = 0$ from which we get $\nabla_a (\rho \xi^a) = 0$ or you can try to justify it directly for curved space-time. Using $\nabla_a (\rho \xi^a) = 0$ and $\rho \xi^b \nabla_b \xi^a = \sigma F^{ab}\xi_b$ we get $\nabla_a T^{ab} = 0$ where $T^{ab}$ is the sum of the EM energy-momentum and the dust energy-momentum.

18. Aug 14, 2013

### ProfDawgstein

They were just too lazy to show that it's true.
The way you explained it is totally fine with me.

19. Aug 14, 2013

### WannabeNewton

Cool! If you want to see a nice explanation for why $\partial_{a}(n\xi^a) = 0$ in flat space-time (where as before $\xi^a$ is the tangent field to the fluid flow) then Schutz has a nice explanation in his text "A First Course in General Relativity" (2nd ed.) p.100. If I ever become fully satisfied with Geroch's argument after working out the possibly ostensible subtleties parading around in my head, I'll post again here. By the way since you asked, it's from Geroch's notes on GR which you can download from here: http://home.uchicago.edu/~geroch/Links_to_Notes.html [Broken]

Last edited by a moderator: May 6, 2017
20. Aug 14, 2013

### ProfDawgstein

Thanks :)
Schutz will be my next book, I really can't wait start working trough it.
It won't be the last time I will hear about fluids and EM-charges/fields.

Last edited by a moderator: May 6, 2017
21. Aug 14, 2013

### vanhees71

Note that in relativistic theory in general not particle numbers but other conserved charges are conserved. E.g., not the baryon number but the net-baryon number (=number of baryons minus number of antibaryons) is conserved. Of course for dust in the universe that's no issue, because the collision energies of dust particles are way below the threshold for pair creation of baryond :-).

22. Aug 14, 2013

### WannabeNewton

Thanks for the point vanhees! Is it possible for antibaryons to be produced without collisions amongst the baryons making up the fluid? The reason I ask is, in most of the GR texts I know, the vector field describing the worldlines of the fluid particles is taken to be non-self-intersecting i.e. the fluid particles are assumed not to collide with another.

23. Aug 14, 2013

### ProfDawgstein

What I actually meant was 'what about mass current conservation if they are charged/interacting'...

I was thinking about collisions and other things, but let's not do any QM/QFT for now.
Maybe later... :P

If there were different types of particles, like electrons & protons, it should
be possible that they interact in such a way that they combine and release energy.
Not sure if this would produce atoms, but I did not want to get this far.
Having anti-particles makes the whole thing even more interesting...

This should be 'QFT with GR background' then, right?

How do you want to produce antibaryons without any interactions?
You could produce some by relativistic collisions.
Are you talking about vacuum fluctuations?

Last edited: Aug 14, 2013
24. Aug 14, 2013

### WannabeNewton

Yeah I really can't seem to come to terms with Geroch's argument, ProfDawgstein. Right from the start he says to take a tubular space-time volume with sides parallel to the tangent field $\xi^a$ of the congruence and with faces orthogonal to $\xi^a$. The second requirement can only hold if $\xi^a$ is locally hypersurface orthogonal i.e. $\xi^{[a}\nabla^{b}\xi^{c]} = 0$ but this certainly need not be true for an arbitrary time-like congruence.

25. Aug 15, 2013

### ProfDawgstein

[I am not yet ready to imagine 4D, still a bit tired.]
But if all sides are parallel to the tangent field, then shouldn't all faces be orthogonal to the $\xi^a$?

Considering intersecting tangent field lines (flow lines), it might cause you trouble if you try to construct a volume at that intersection point (where all faces are orthogonal to $\xi^a$). Maybe you can in 4D, but I really do not know (I don't think you can).

They probably rule that out, because particles may combine and do some fancy binding/quantum/collision stuff.

In the general case the fluid / dust could also be relativistic, so it might combine and clump together somewhere, so they rule out the case where they might intersect.
Seems legitimate to me...

If you had intersecting tangent field lines and the particles hit each other with enough energy you might produce some other particles, thus needing an extra term in the energy momentum tensor (for the collision results, whatever they are).

I really do not know ALL the details (yet), so as always, just a thought...