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I have to apply the Israel junction conditions in a region in which a hypersurface with O(3) symmetry separates two spacetime with Schwarzschild metric (with masses $M_+$, the exterior one, and $M_-$, the interior one). The hypersurface is: $$\Sigma_{\pm}=\{(t_{\pm}, r_{\pm}, \theta_{\pm}, \phi_{\pm})| F_{\pm}(t_{\pm},r_{\pm})=r_{\pm}-R(\tau(t_{\pm}))=0\}$$ where $\tau$ is the proper time on $\Sigma$. Now the question: the result given by the article for the normal unit vector is $$n_{\pm\mu}=\frac{\partial_{\mu}F_{\pm}}{\sqrt{|g^{(\pm)\mu\nu}\partial_{\mu}F_{\pm}\partial_{\nu}F_{\pm}|}}=(-\dot{R},\dot{t_{\pm}},0,0) $$ How can I find that $n_{\pm\mu}=(-\dot{R},\dot{t_{\pm}},0,0)$?

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  • $\begingroup$ The normal is the gradient with the denominator to normalize it... $\endgroup$ – Jan Bos Oct 3 '19 at 15:50
  • $\begingroup$ Ok, but I need to understand the calculation to get the result $n_{\pm\mu}=(-\dot{R},\dot{t_{\pm}},0,0) $ in detail $\endgroup$ – Priuk Oct 3 '19 at 15:55
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If your embedding of the hypersurface is given by a level-set function $F$, the normal vector to that hypersurface is given by the exterior derivative of that level-set function, ie

\begin{equation} n = dF \end{equation}

Any vector tangent to your surface will be obtained by a path entirely within $\Sigma$. For a path $\gamma \in \Sigma$, the tangent vector to that path is $\dot{\gamma}$, but we also have that $F(\gamma(\tau)) = 0$ for every $\tau$, therefore

\begin{equation} \dot{F}(\gamma(\tau)) = \dot{\gamma}(\tau) dF(\gamma(\tau)) = 0 \end{equation}

This is the $1$-form $dF$ and the vector $\dot{\gamma}$. If we put this in coordinate form, this is simply (with $u = \dot{\gamma}$ our tangent vector)

\begin{equation} g^{\mu\nu} u_\mu \partial_\mu F(\gamma(\tau)) = 0 \end{equation}

In other words, this vector is indeed always orthogonal to any tangent vector.

To make it a unit vector, we simply divide it by its norm, as usual :

\begin{equation} n = \frac{dF}{\|dF\|} \end{equation}

Or again, in coordinates,

\begin{equation} n_\mu = \frac{\partial_\mu F}{\sqrt{g^{\mu\nu} \partial_\mu F \partial_\mu F}} \end{equation}

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  • $\begingroup$ Yes, but how can I find the result $n_{\pm \mu}=(-\dot{R}, \dot{t_{\pm}},0,0)$? $\endgroup$ – Priuk Oct 3 '19 at 13:23
  • $\begingroup$ Is $R$ a specific quantity here? $\endgroup$ – Slereah Oct 3 '19 at 13:46
  • $\begingroup$ It's the trajectory of the hypersurface (look at the definition I gave) $\endgroup$ – Priuk Oct 3 '19 at 13:55
  • $\begingroup$ So can you help me? $\endgroup$ – Priuk Oct 4 '19 at 7:23

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