1
$\begingroup$

I have been trying to calculate the surface gravity of analogue black hole formed in a classical fluid, following Visser (https://arxiv.org/abs/gr-qc/9712010, p19, section 9.1). The line element in acoustic spacetime is given by, \begin{align} ds^2 = \frac{\rho}{c}\Big[ - (c^2-v^2)dt^2 + 2vdtdx + dx^2 + dy^2 + dz^2\Big] \end{align} Here, $c$ is the speed of sound in the fluid and $v, \rho$ are the velocity and density fields of the fluid respectively. The variables are functions of space and not time. This is a static geometry. The time translation Killing vector field $K^a = (1,0,0,0)^a$ and $g(K,K)= g_{00}= - \frac{\rho}{c}(c^2-v^2) = - ||K||^2$. He sets up `fiducial observers' with velocity \begin{align} V_{FIDO}= \frac{K}{||K||} \end{align}

I'll omit the subscript `FIDO' from here on. The acceleration of the fiducial observer is given by \begin{align} A^a & \ =(\nabla_V V)^a , \\ A_a & \ = (V^m \nabla_m V)_a \\ & \ = - \frac{1}{2}\nabla_a (V^m V_m)~. \end{align} The last expression is obtained using Killing equation. Now, $g(V,V)=\frac{g_{00}}{||K||^2} = -\frac{||K||^2}{||K||^2} = -1$. So, \begin{align} A_a = \frac{1}{2}\nabla_a \left(\frac{||K||^2}{||K||^2}\right)=0. \end{align} However, Visser writes in eq. (67) that \begin{align} A = \frac{1}{2}\frac{\nabla ||K||^2}{||K||^2}~. \end{align}

I am not able to determine where I am going wrong. How can I get the above expression given in Visser's paper. Another confusion I have is that the normalisation used in the paper for $V$ is ill defined on the horizon located at the 2-surface where $c=v$. Is this a proper normalisation to choose then?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.