# Continuity equation (charge vs matter) in SR

1. Feb 27, 2014

### center o bass

If we consider a perfect relativistic fluid it has energy momentum tensor

$$T^{\mu \nu} = (\rho + p) U^\mu U^\nu + p\eta^{\mu \nu}$$

where $U^\mu$ is the four-velocity field of the fluid. $\partial_\mu T^{\mu \nu} = 0$ then
implies the relativistic continuity equation

$$\partial_\mu(\rho U^\mu) + p \partial_\mu U^\mu = 0$$

which reduces to the ordinary continuity equation for matter

$$\partial_t \rho + \nabla \cdot (\rho \vec v) = 0$$

in the non-relativistic limit $v << c$ and $p << \rho$. Charge obeys an identical equation in the same limit, but does it has a relativistic analogue? If not, why?

2. Feb 27, 2014

### WannabeNewton

Sure. Just consider a charged fluid.

3. Feb 27, 2014

### TrickyDicky

Sure, instead of the SET you use the EM SET in free space(electrovacuum). Look up the Einstein-Maxwell equations.

4. Feb 27, 2014

### center o bass

What does SET stand for? Could you provide a reference where I could find the relevant
equation? I could not find it when looking up the Einstein-Maxwell equations.

5. Feb 27, 2014

### TrickyDicky

SET= Stress-energy tensor
EM SET= Electromagnetic Stress-energy tensor, this is the SET of a charged fluid.
It is in Wikipedia, i think they even treat the continuity equation at the end of the article on the EM SET.

6. Feb 27, 2014

### center o bass

Alright. I found two equations on the wikipedia article on the EM SET. But these are for the electromagnetic field itself. However the equation

$$\partial_t u_{em} + \nabla \cdot \vec{p}_{em} + \vec J \cdot \vec E = 0$$

where $u_{em}, \vec p_{em}$ is the electromagnetic field energy/momentum density. This does very much look like the one for a perfect fluid when $\vec E = 0$.

7. Feb 27, 2014

### center o bass

You suggest just switching the energy density $\rho$ with the charge density and use exactly the same tensor?

8. Feb 27, 2014

### Staff: Mentor

There is a charge-current density 4-vector in relativity, which obeys a similar continuity equation to the stress-energy tensor. The 4-vector is usually written $J^{\mu}$, where $J^0 = \rho_c$ is the charge density and $J^k = \left( j_x, j_y, j_z \right)$ is the current density. The continuity equation is just $\partial_{\mu} J^{\mu} = 0$, which when written out in components becomes

$$\frac{\partial \rho_c}{\partial_t} + \nabla \cdot \vec{j} = 0$$

If we think of the charge as being carried by a fluid with ordinary velocity $\vec{v}$, then $\vec{j} = \rho_c \vec{v}$ and the above equation becomes

$$\frac{\partial \rho_c}{\partial_t} + \nabla \cdot \left( \rho \vec{v} \right) = 0$$

9. Feb 27, 2014

### center o bass

Indeed, but the relativistic generalization of the matter continuity equation also contains a pressure term. Is there any reason why a charged relativistic fluid should not contain pressure term?

10. Feb 27, 2014

### pervect

Staff Emeritus
If you look at, for instance, http://web.mit.edu/edbert/GR/gr2b.pdf, you'll see that the stress energy tensor is the tensor product of the number-flux 4-vector and the energy-momentum 4-vecctor. This yields the pressure terms.

But for the charge continuity equation, you only need to take the tensor product of the charge (a scalar) and the number-flux 4-vector. Hence, you have a conserved 4-vector and no pressure terms. The components of the 4-vector are charge density and current.

For an informal diagram/description of the number-flux 4-vector and the charge continuity equation, take a look at the "flow of stuff" diagrams in https://www.physicsforums.com/showthr...04#post4632145 [Broken]. And ignore the typo's (I switch midstream from meters to inches, for instance).

Once you know how worldline density transforms, you get the charge/current density by considering that each worldline carries charge, the energy-momentum 4-vector by considering that each worldline carries energy-momentum (a 4-vector).

Last edited by a moderator: May 6, 2017
11. Feb 27, 2014

### Staff: Mentor

There will in general be one, but it won't be contained in the charge-current density 4-vector. There are really two separate sets of equations: one describing the behavior of the charge-current and the EM fields, the other describing the behavior of stress-energy. The latter set of equations is what will contain the pressure term.

As far as how the presence of charged matter affects the latter set of equations, the electromagnetic stress-energy tensor on the Wikipedia page can help with that; but you should realize that this SET does not have the form of the perfect fluid, because the "pressure" it describes (which is stress-energy contained in the EM fields) is not isotropic; it is different in different spatial directions. So things won't look as simple as they do for a perfect fluid.

12. Feb 27, 2014

### TrickyDicky

Unless he is considering specific solutions like say isotropic radiation of a spherical EM wave, but yes in general the SET won't be a perfect fluid.

13. Feb 27, 2014

### Staff: Mentor

The proper term for what I think you're describing here is "null dust"--a perfect fluid composed of massless particles moving on null worldlines, whose pressure is one-third of its energy density.

A "spherical EM wave" is not physically possible: the lowest-order EM radiation is dipole, not monopole, and the SET associated with any kind of EM radiation will not be of the perfect fluid form. But null dust is a good approximation of a sort of spherically symmetric "average" of null radiation going in all directions equally. The Vaidya metric is the corresponding line element for this case.

14. Feb 27, 2014

### TrickyDicky

Right.
Well, I guess the spherical wave example is not the best example here. But saying they are physically imposible seems to me a bit of a strong statement. One should discuss first the possibility of point sources, certainly in practice they are accepted in astrophysics for distant stars light, pulsar radio sources... but I guess they are just an idealization.

Quoting Wikipedia:"a point source is a singularity from which flux or flow is emanating. Although singularities such as this do not exist in the observable universe"
Unless one decides on physical impossibility judging only from what is directly observable, most of modern physics including elementary point particles-charges, and BH's point masses would go out the window.

Last edited: Feb 27, 2014
15. Feb 27, 2014

### Staff: Mentor

Yes, it is, but it's a simple consequence of the fact that the EM field is spin-1, plus conservation laws. You can construct approximate models of spherically symmetric EM radiation, but they will always be approximate; there are no spherically symmetric exact solutions.

Exactly. They're not exact models; they're just approximations that work well enough for the purpose intended.

Those are idealizations as well; when taken literally they have obvious problems, such as the infinite self-field of a point particle. But as approximate models (in this case to an underlying physics that we don't fully understand right now, perhaps something like string theory at the Planck scale) they work fine for what we use them for.

16. Feb 27, 2014

### TrickyDicky

Agree.

Ironically all the important theories, CM, CED, QED, GR.. have those singularities built in their theoretical structure not just as idealizations or approximations but as a necessary part for them to be mathematically coherent.