# Conservations law or equations of motion, and other constrains

1. Jan 17, 2014

### sergiokapone

In the special relativity the conservation of energy and momentum is represented by the equation:
$\partial_{\mu}T^{\mu\nu}=0$, where $T^{\mu\nu}$ - stress-energy tensor.
In the case of perfect fluid $T^{\mu\nu}=(\rho+p/c^2)u_{\mu}u_{\nu}-pg^{\mu\nu}$ this equations leads to relativistic equations of continuity and the relativistic equations of motion for a perfect fluid. Some times this eqn $\partial_{\mu}T^{\mu\nu}=0$ call the conservation law, some times - equatin of motion. Why this duality? In non-relativistic mechanics we even know what is meant by the law of conservation, and what is meant by the equation of motion.

2. Jan 17, 2014

### WannabeNewton

In Newtonian mechanics conservation of energy refers specifically to local conservation of a scalar energy density field. $T^{\mu\nu}$ on the other hand includes much more information because it's a tensor; for example the stress-energy tensor of a perfect fluid codifies not only the energy density of the fluid but also the pressure of the fluid. The conservation equation $\partial_{\mu}T^{\mu\nu}$ is really local conservation of stress-energy-momentum so it results in a variety of dynamical equations for a given matter field. For a perfect fluid it straightforwardly leads to the relativistic continuity and Euler equations whereas for the electromagnetic field it leads to Maxwell's equations (https://www.physicsforums.com/showthread.php?t=733219). By combining $\partial_{\mu}T^{\mu\nu} = 0$ with a given foliation of Minkowski space-time into a one-parameter family of space-like hypersufaces we can also derive global conservation of angular momentum etc.

3. Jan 17, 2014

### sergiokapone

$\nabla_{a}T^{ab} = -\frac{3}{2}F_{ac}\nabla^{[a}F^{bc]} + F_{c}{}{}^{b}\nabla_{a}F^{ac} =0$

I do not quite understand how you have proved that these terms
$\nabla^{[a}F^{bc]}=0$
$\nabla_{a}F^{ac}=0$
is equal to zero?
(Maxwell's equations in vacuum)

4. Jan 17, 2014

### WannabeNewton

The proof is in section 20.6 of MTW.

5. Jan 17, 2014

### sergiokapone

Ok, I try to understand. Tnx.

6. Jan 17, 2014

### sergiokapone

And another strange thing. Eqn $\partial_{\mu}T^{\mu\nu}=0$ is the only four equations. But number of Maxwell's equations in 3D is 12!

7. Jan 17, 2014

### dextercioby

Add the equations $\partial^{[\mu}F^{\nu\rho]} = 0$ which, as explained by MTW, are a byproduct of the final result, and you'll have all of them.

8. Jan 17, 2014

### WannabeNewton

$\partial_{\mu}T^{\mu\nu}=0$ directly implies $\partial_{\mu}F^{\mu\nu} = 0$ so we still end up with 4 equations. $\partial_{[\gamma}F_{\mu\nu]} = 0$ is trivially implied by $F_{\mu\nu} = 2\partial_{[\mu}A_{\nu]}$ where $A_{\mu}$ is the 4-potential so it's not really a consequence of $\partial_{\mu}T^{\mu\nu}=0$.

*dexter beat me to it! :)

9. Jan 17, 2014

### Staff: Mentor

In non-relativistic solid and fluid mechanics, the continuity equation is interchangeably called the conservation of mass equation, and the equations of motion are interchangeably called the conservation of momentum equation. So they are called both continuity equations and conservation equations. In relativity, the divergence of the stress energy tensor (or whatever they call it) is the relativistic continuity equation, and it automatically combines the continuity equation (conservation of mass) with the equation of motion (conservation of momentum) into a single equation.

Chet

10. Jan 18, 2014

### sergiokapone

Ok, thanks for all.
I have another quastion about integral conservation law. In MTW Sec. 20.2 (Formulaes 20.9) said
Using first formula of 20.9, and following what has been said, for Schwarzschild black hole I get the $P^{0}=mc^2$, what is reasonable, where $m$ - is mass of BH. But if I decline to use instructions "Minkowskian coordinates" (mean Cartesian) and using flat spherical coordinates, I get negative infinity (calculations was made in Maple 12).
Why condition to using of Minkowskian coordinates are necessary to obtain the correct result?
Is intact here the condition of general covariance?

11. Jan 18, 2014

### WannabeNewton

The ADM energy-momentum integrals are defined with respect to special Euclidean charts for which the metric tensor representation satisfies necessary convergence criteria. See Wald "General Relativity" problem 11.2. See also here: http://homepage.univie.ac.at/piotr.chrusciel/teaching/Energy/Energy.pdf (p.12 in particular)

As a bit of a digression, for stationary asymptotically flat space-times the total energy of the space-time can be most easily calculated from the Komar integral for energy. More precisely, if we have an asymptotically flat space-time (for the rigorous definition of an asymptotically flat space-time see section chapter 11 of Wald) with a time-like killing field $\xi^{\mu}$, the Komar energy is given by $E = -\frac{1}{8\pi}\int _{S}\epsilon_{\mu\nu\alpha\beta}\nabla^{\alpha}\xi^{\beta}$ where $S$ is a 2-sphere taken at spatial infinity. For Schwarzschild space-time, using spherical coordinates on $S$, a very simple calculation yields $E = M$ where $M$ is the Schwarzschild mass parameter.

12. Jan 18, 2014

### sergiokapone

Thanks for the references.

It's a pity I'm not familiar with the Komar integral's, so I now difficult to understand their physical meaning.

13. Jan 18, 2014

### WannabeNewton

Section 11.2 of Wald is an excellent place to start if you want to grasp their physical meaning. I made a thread regarding a calculation from this section a few months back and in the first paragraph of the thread summarized the motivation behind the Komar energy so if you don't have access to Wald you can take a quick scan of the thread: https://www.physicsforums.com/showthread.php?t=688422

There are also tons of references online regarding both ADM energy-momentum and Komar integrals.

An absolutely excellent reference is the following: http://relativity.livingreviews.org/open?pubNo=lrr-2004-4&amp;page=articlesu4.html [Broken] and it also addresses your previous question.

Last edited by a moderator: May 6, 2017
14. Jan 18, 2014

### sergiokapone

Clarify the issue. Condition asymptotically flat metric to calculate the mass, I understand. I just do not understand why you want to use is not arbitrary coordinates, for example, spherical ($ds^2=c^2dt^2-dr^2-r^2d\Omega^2$) or cylindrical, but only Cartesian $ds^2=c^2dt^2-dx^2-dy^2-dz^2$?

Last edited: Jan 18, 2014
15. Jan 18, 2014

### sergiokapone

Oh! I find my quastion in your reference http://homepage.univie.ac.at/piotr.chrusciel/teaching/Energy/Energy.pdf
page 6 (above formula 1.1.14 )
(my bold emphasis)
I ﬁnds a mass m in the coordinate system in which g takes the form $g_{\mu\nu}=(1+2\phi)\delta_{\mu\nu}$. Obviously it asymptotically flat and Cartesian.
But, if I try tu use spherical coordinates, $g_{\theta\theta} = r^2$, for example, increases with increasing distance, I get infinity.

Last edited: Jan 18, 2014
16. Jan 18, 2014

### WannabeNewton

It is indeed coordinate independent but the actual computation of the ADM energy, under its normal definition, requires one to use asymptotically Cartesian coordinates. What you actually did above in post #15 is rewrite the Schwarzschild metric in isotropic form (spatially conformally flat form: http://en.wikipedia.org/wiki/Schwar...c.29_formulations_of_the_Schwarzschild_metric), whose spatial part is clearly asymptotically Cartesian as $r\rightarrow \infty$. So we can easily evaluate the ADM energy for Schwarzschild space-time if we use isotropic coordinates, again because the spatial part of the metric in said coordinates is manifestly asymptotically Cartesian.

Why do we care if it's asymptotically Cartesian, regardless of the fact that the ADM energy is coordinate-independent? Well take a look at how MTW goes from (20.6) to (20.7). What key equation do they use for $g_{\mu\nu}$ in relation to the asymptotic behavior of the gravitational field of a self-gravitating isolated source?

17. Jan 18, 2014

### sergiokapone

Yes, thats interesting.

???

You mean $g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}$?

18. Jan 18, 2014

### WannabeNewton

Yes. So they already make use of an asymptotically Minkowskian representation of the metric tensor far away from the isolated source when deriving (20.7) so you can't evaluate (20.7) in a non-asymptotically Minkowskian coordinate system!

19. Jan 18, 2014

### sergiokapone

Yes, but if I will use the equation (20.6) directly (with Schwarzschild $g_{\mu\nu}$'s, or Kruskal-Szekeres), what I will get? Mass or dummy number?

20. Jan 18, 2014

### WannabeNewton

You can't do that because $H^{\mu\nu\alpha\beta}$ is itself defined in terms of the asymptotic gravitational field with space-time geometry $g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu}$ far away from the isolated source (c.f. expression 20.3) so it assumes an asymptotically Minkowskian coordinate system.

You can't use the ADM energy-momentum expressions (20.6) if you aren't in an asymptotically Minkowskian coordinate system. See, for example, problem 3 of the following problem set: http://phys.columbia.edu/~nicolis/HW1_solutions.pdf

If you want to explicitly calculate the energy of Schwarzschild space-time in an arbitrary coordinate system then use the Komar integral for energy. If you want I can show you the explicit calculation of the Komar energy in the regular Schwarzschild coordinates.

21. Jan 18, 2014

### sergiokapone

Yes, of course. It would be great.

22. Jan 18, 2014

### WannabeNewton

Sorry for the delay! I got distracted by a video game :p

So recall that the definition of the Komar energy is $E = -\frac{1}{8\pi}\int _{S}\epsilon_{\mu\nu\alpha\beta}\nabla^{\alpha}\xi^{\beta}$ where $S$ is a 2-sphere taken at spatial infinity and $\epsilon_{\mu\nu\alpha\beta}$ is the volume element on Schwarzschild space-time.

Now in Schwarzschild coordinates we have $\xi^{\mu} = \delta^{\mu}_t$ so the following two results obtain:

$\nabla^{\alpha}\xi^{\beta} = g^{\alpha\gamma}\nabla_{\gamma}\delta^{\beta}_{t} \\= g^{\alpha \gamma}\Gamma ^{\beta}_{\gamma t} \\= g^{r r}\Gamma ^{t}_{r t}\delta^{\alpha}_{r}\delta^{\beta}_{t} + g^{t t}\Gamma ^{r}_{t t}\delta^{\alpha}_{t}\delta^{\beta}_{r}$

$\epsilon_{\mu\nu\alpha\beta}\nabla^{\alpha}\xi^{\beta} \\= \epsilon_{\mu\nu tr}\nabla^{t}\xi^{r} + \epsilon_{\mu\nu rt}\nabla^{r}\xi^{t} \\= 2\epsilon_{\theta\phi tr}\nabla^{t}\xi^{r} = 2r^2 \sin\theta \nabla^{t}\xi^{r}d\theta d\phi$

where I have used the antisymmetry of $\epsilon_{\mu\nu\alpha\beta}$ and $\nabla^{\alpha}\xi^{\beta}$ (recall that a Killing field satisfies $\nabla^{\alpha}\xi^{\beta} = -\nabla^{\beta}\xi^{\alpha}$).

Next, $\Gamma ^{r}_{t t} = -\frac{1}{2}g^{r \mu}\partial_{\mu}g_{tt} = (1 - \frac{2M}{r}) \frac{M}{r^2}$ so using our expression for $\nabla^{\alpha}\xi^{\beta}$ above, we have $\nabla^t \xi^r = -\frac{M}{r^2}$.

So finally $E = \frac{M}{4\pi}\int _{S}\sin\theta d\theta d\phi = \frac{M}{4\pi}\int _{0}^{2\pi}\int _{0}^{\pi}\sin\theta d\theta d\phi = M$.

EDIT: And make sure you take a look at the explicit calculation of the ADM energy of Schwarzschild space-time given in the link in post #20. It shows how to go from the regular Schwarzschild coordinate system to a coordinate system that is manifestly asymptotically Minkowskian, and subsequently calculates the ADM energy in this asymptotically Minkowskian coordinate system.

Last edited: Jan 18, 2014