7

Update: Trimok and MBN helped me solve most of my confusion. However, there is still an extra term $-(2/r)T$ in the final result. Brown doesn't write this term, and it seems physically wrong.

Update #2: Possible resolution of the remaining issue. See comment on MBN's answer.

Suppose we have a rope hanging statically in a Schwarzschild spacetime. It has constant mass per unit length $\mu$, and we want to find the varying tension $T$. Brown 2012 gives a slightly more general treatment of this, which I'm having trouble understanding. Recapitulating Brown's equations (3)-(5) and specializing them to this situation, I have in Schwarzschild coordinates $(t,r,\theta,\phi)$, with signature $-+++$, the metric

$$ ds^2=-f^2 dt^2+f^{-2}dr^2+... \qquad , \text{ where} f=(1-2M/r)^{1/2} $$

and the stress-energy tensor

$$ T^\kappa_\nu=(4\pi r^2)^{-1}\operatorname{diag}(-\mu,-T,0,0) \qquad .$$

He says the equation of equilibrium is:

$$ \nabla_\kappa T^\kappa_r=0 $$

He then says that if you crank the math, the equation of equilibrium becomes something that in my special case is equivalent to

$$ T'+(f'/f)(T-\mu)=0 \qquad ,$$

where the primes are derivatives with respect to $r$. This makes sense because in flat spacetime, $f'=0$, and $T$ is a constant. The Newtonian limit also makes sense, because $f'$ is the gravitational field, and $T-\mu\rightarrow -\mu$.

There are at least two things I don't understand here.

First, isn't his equation of equilibrium simply a statement of conservation of energy-momentum, which would be valid regardless of whether the rope was in equilibrium?

Second, I don't understand how he gets the final differential equation for $T$. Since the upper-lower-index stress-energy tensor is diagonal, the only term in the equation of equilibrium is $\nabla_r T^r_r=0$, which means $\mu$ can't come in. Also, if I write out the covariant derivative in terms of the partial derivative and Christoffel symbols (the relevant one being $\Gamma^r_{rr}=-m/r(r-2m)$), the two Christoffel-symbol terms cancel, so I get

$$ \nabla_r T^r_r = \partial _r T^r_r + \Gamma^r_{rr} T^r_r - \Gamma^r_{rr} T^r_r \qquad , $$

which doesn't involve $f$ and is obviously wrong if I set it equal to 0.

What am I misunderstanding here?

References

Brown, "Tensile Strength and the Mining of Black Holes," http://arxiv.org/abs/1207.3342

Qmechanic
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1 Answers1

2

From that you have $\nabla_r T^r_r=T'$, but there is also

$\nabla_t T^t_r=\frac{\partial}{\partial t}T^t_r+\Gamma^t_{\alpha t}T^\alpha_r-\Gamma^\alpha_{r t}T^t_\alpha=\Gamma^t_{r t}T^r_r-\Gamma^t_{r t}T^t_t=-\Gamma^t_{rt}(T-\mu)$.

So

$\nabla_k T^k_r=\nabla_rT^r_r+\nabla_tT^t_r=-T'-(f'/f)(T-\mu).$

This should be a comment, but the symbols didn't work.

My guess is that it is called equation of equilibrium because the T is the stress energy of the rope, not a stress energy that affects the space time geometry. The background is fixed and the rope lives on it.

MBN
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