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For a perfect fluid $$T^{\mu\nu} = (\rho+p)U^\mu U^\nu + pg^{\mu\nu}.$$ I am trying to show, as is claimed possible in A First Course in General Relativity (Schutz), that the conservation of four-momentum law $$T^{\mu\nu}_{\ \ \ \ \ \ \ ;\nu}=0$$ implies the "relativistic equation of hydrostatic equilibrium": $$p_{,i} + (\rho+p)\left[\frac{1}{2}\ln\left(-g_{00}\right)\right]_{,i} = 0.$$ According to Schutz, this can be easily derived by taking the fluid to be static: $g_{i0}=0,\ \ g_{\alpha\beta,0}=0,\ \ U^i = 0, \ \ p_{,0} = 0$, etc.

The tensor algebra seems simple enough, but by evaluating the conservation law for the spatial component $i$, I keep arriving at $$(\rho+p)\Gamma^i_{\ \ 00}+p_{,\nu}g^{i\nu} = 0$$ where (I think) $\Gamma^i_{\ \ 00} = \frac{1}{2}g^{ij}g_{00,j}$ (sum over $j = 1,2,3$).

My result is very close to Schutz's, it seems, but I cannot figure out where we diverged.

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  • $\begingroup$ I figured Schutz's form of the "relativistic equation of hydrostatic equilibrium" would be an equation I could easily find online, but I still have not seen it anywhere. Does anyone have an online reference? $\endgroup$ – Doubt Mar 27 '17 at 23:44
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I would call that equation "general relativistic Bernoulli theorem" in the case of a static, ideal, relativistic fluid, since its Newtonian limit is the classical Bernoulli theorem for the case of a non-relativistic, staticc Newtonian fluid; see e.g. [Eric Gourgoulhon, 2010, An introduction to the theory of rotating relativistic stars] sec. 3.4.2. The equation of motion you gave can be rewritten and integrated using the fluid log-enthalpy $h$ defined as $dh/dP=1/(P+\rho)$:

\begin{align}(p+\rho)\frac{dP}{dr}+\frac{1}{2}\frac{d \ln(-g_{tt}(r))}{dr}&=0\\ \frac{dh}{dr}+\frac{1}{2}\frac{d \ln(-g_{tt}(r))}{dr}&=0\\ h(r)+\frac{1}{2}\ln(-g_{tt}(r))=\mathrm{const.}\tag{1} \end{align}

The Newtonian limit of eq. (1) is the classical Bernoulli theorem for a static ideal fluid: $$h_N(r)+\phi(r)=\mathrm{const.},$$ with $h_N(r)$ as classical enthalpy per unit mass and Newtonian gravitational potential $\phi(r)$.

For a derivation I would recommend $\nabla_\nu T_\mu^{~\nu}=0$ because $T_\mu^{~\nu}$ is very simple in the present case since $u_\mu u^\nu=-\delta_\mu^t \delta^t_\nu$ but $\nabla_\nu T^{\mu\nu}=0$ will work fine too, using $u^t=\sqrt{-1/g_{tt}(r)}$. For the metric the: present problem is a spherical symmetric problem so $(g_{\mu \nu})=\mathrm{diag}(g_{tt}(r),g_{tt}(r),r^2,r^2\sin^2 \theta)$ is best suited. If you should get the desired result.

Especially in its GR integral form (1) it is a very underused but extremely powerful identity.

EDIT: Upon request here the computation to arrive at (1) starting from $\nabla_\nu T^{\mu\nu}=0$:

$$(T^{\mu \nu})=\mathrm{diag}(-1/g_{tt}(r)\rho(r),1/g_{rr}(r)P(r),P(r)/r^2,P(r)/(r^2\sin^2 \theta)). \tag{2}$$

To get (2) one can use the fundamental definition of the energy-momentum tensor of an ideal fluid with the metric and four velocity I described previously. The only non trivial component of $\nabla_\nu T^{\mu\nu}$ is $\nabla_\nu T^{r\nu}$ so lets compute it:

\begin{align}\nabla_t T^{rt}&=\Gamma^r_{~tt}T^{tt}+\Gamma^t_{~tr}T^{rr} \\ &=\frac{\rho g_{tt}'}{2g_{rr}g_{tt}}+\frac{P g_{tt}'}{2g_{rr}g_{tt}}\\ \nabla_r T^{rr}&=\partial_r T^{rr}+\Gamma^r_{~rr}T^{rr}+\Gamma^r_{~rr}T^{rr}\\ &=\frac{P'}{g_{rr}}\\ \nabla_\theta T^{r\theta}&=0\\ \nabla_\phi T^{r\phi}&=0\\ \Rightarrow \nabla_\nu T^{r\nu} &= \frac{\rho g_{tt}'}{2g_{rr}g_{tt}}+\frac{P g_{tt}'}{2g_{rr}g_{tt}} + \frac{P'}{g_{rr}}\tag{3} \end{align}

I suppressed the $r$-dependency to improve readability. Since $\nabla_\nu T^{r\nu}=0$ we can multiply (3) with $g_{rr}$ to arrive at:

$$0=P'(r)+ (\rho(r)+P(r))\frac{g_{tt}'(r)}{2g_{tt}(r)} \tag{4}.$$

Using the chain rule leads to:

$$\frac{g_{tt}'(r)}{2g_{tt}(r)}=\frac{-g_{tt}'(r)}{-2g_{tt}(r)}=\partial_r\log(-g_{tt}(r))/2 \tag{5}. $$ The step where I multiplied with $1=-1/(-1)$ was done to get to $\log(-g_{tt}(r))$ more naturally. $\log(-g_{tt}(r))$ is a more pleasing notation since $g_{tt}<0$ and the $\log$ by itself would become complex: BUT we are not interested in the $\log$: we are interested in the logarithmic derivative and

$$ \partial_x \log(f(x))=\partial_x \log(-f(x))=\partial_x \log(42 f(x))=\frac{f'(x)}{f(x)}.$$

Using eq. (5) in (4) leads to the result in the desired form:

$$0=P'(r)+ (\rho(r)+P(r))\partial_r\log(-g_{tt}(r))/2. \tag{6}$$

but my favorite notation of the differential identity is clearly

$$0=P'(r)+ (\rho(r)+P(r))\partial_r\log(42 g_{tt}(r))/2$$ for obvious reasons.

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  • $\begingroup$ Can you show the steps by which $\nabla_\nu T^{\mu\nu}$ leads to (1)? It seems extremely straightforward, but I keep arriving at $p_{,i}-\frac{1}{2}(\rho+p)g_{00,i}=0$ instead of your/Schutz's result. $\endgroup$ – Doubt Apr 3 '17 at 18:55
  • $\begingroup$ I added the computation starting from $\nabla_\nu T^{\mu\nu}=0$. $\endgroup$ – N0va Apr 4 '17 at 11:27
  • $\begingroup$ Thank you! From your work, I was finally able to see my own error. It is not necessary to work in spherically-symmetric coordinates: it is enough to keep $g_{\alpha\beta}$ general (though demanding $g_{i\alpha}=g_{\alpha\beta,0}=0$). My problem was assuming $U^\alpha U^\alpha = -1$, which is only true in Minkowski coordinates. In reality, $U^\alpha = U^0 = \sqrt{-g_{00}}$. This explains my missing factor of $g_{00}$, which in my original formula was replaced by $-1$. $\endgroup$ – Doubt Apr 11 '17 at 0:51
  • $\begingroup$ Yes you can work it through without specifying specific coordinates. $u_\mu u^\mu=-1$ holds coordinate independent and for vanishing spatial velocity components it leads to $u^0=\sqrt{-1/g_{00}}$ since $u_\mu u^\mu=u_0 u^0= g_{00} u^0 u^0 = g_{00}(\frac{-1}{g_{00}})=-1 $. $\endgroup$ – N0va Apr 11 '17 at 11:02

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