It is often said that given the metrics $g^+$, $g^-$ on two sides of a hypersurface $\Sigma$, then, with a level-set function $\phi$ such that $\Sigma = \phi^{-1}(0)$, we can describe the metric on the whole manifold by

\begin{equation} g = \theta(\phi) g^+ + (1 - \theta(\phi)) g^- \tag{1} \end{equation}

And then, the derivatives of the components are simply

\begin{equation} g_{ab,c} = \partial_c \theta(\phi) (g^+ - g^-) + \theta(\phi) g^+_{ab,c} + (1 - \theta(\phi)) g^-_{ab,c}\tag{2} \end{equation}

and since it is assumed that $g$ is continuous,

\begin{equation} g_{ab,c} = \theta(\phi) g^+_{ab,c} + (1 - \theta(\phi)) g^-_{ab,c}\tag{3} \end{equation}

The discontinuity in the derivatives is then said to be

\begin{equation} [g_{ab,c}] = \gamma_{ab} n_c\tag{4} \end{equation}

for $n$ a normal form to $\Sigma$ and $\gamma_{ab}$ some tensor, and the notation corresponds to

\begin{equation} [F] = \lim_{p \in M^+ \to \Sigma} F(p) - \lim_{p \in M^- \to \Sigma} F(p)\tag{5} \end{equation}

The proof for this seems rather elusive, but according to Clarke and Dray, this stems from the fact that for $v$ some vector field such that $g(v, n) = 0$, with $n$ some extension of the normal form (I'm guessing via the normal bundle of the surface), we have

\begin{equation} v^c[g_{ab,c}] = v^c [g_{ab}]_{,c} = 0\tag{6} \end{equation}

which then implies that $[g_{ab,c}] = \gamma_{ab} n_c$. I'm not quite sure how to show this. Expanding everything, I get

\begin{equation} (\lim_{p \in M^+ \to \Sigma} \theta v^c g^+_{ab,c} - \lim_{p \in M^- \to \Sigma} (1 - \theta) v^c g^-_{ab,c})\tag{7} \end{equation}

given coordinates with tangent vectors $(n, \partial_\alpha)$, we can decompose this as

\begin{equation} v^c g^\pm_{ab,c} = v^\alpha g^\pm_{ab,\alpha}\tag{8} \end{equation}

since $v$ has no $n$ component. How to show that this quantity is then continuous upon crossing the boundary? Do I need to define the first fundamental form for every hypersurface $\Sigma_\varepsilon$ along the normal bundle of coordinate $\varepsilon$ and show that this is continuous?

  • $\begingroup$ Is $n$ assumed to be null? non-null? $\endgroup$ – Qmechanic Feb 12 '18 at 14:48
  • $\begingroup$ This is for strictly timelike hypersurfaces, so $n$ is always spacelike. $\endgroup$ – Slereah Feb 12 '18 at 14:49

FWIW, interestingly, the Israel junction conditions are born out of mathematical necessity to avoid ill-defined products$^1$ of distributions rather than actual physical considerations. See e.g. Refs. 1 & 2 for details.


  1. Eric Poisson, A Relativist's Toolkit, 2004; Section 3.7.

  2. Eric Poisson, An Advanced course in GR; Section 3.7.


$^1$ We ignore Colombeau theory. See also this Phys.SE post.


I am you from the future. Here's a probably fine argument for this.

Take the normal bundle of the hypersurface $N = (-1, 1) \times \Sigma$, with the adapted coordinates $(r, y)$ such that $\partial_r$ is a vector normal to the hypersurface. We can decompose the metric tensor as

$$g = g(\partial_r, \partial_r) dr \otimes dr + g^r_{\Sigma}$$

where $g_{\Sigma, r}$ is the first fundamental form on the surface defined by the projection of the normal bundle at $r$. The derivative with respect to the $\Sigma$ coordinates at $r = \varepsilon$

$$\partial_c \bar g^\varepsilon_{ab} = \lim_{h \to 0}\frac{ \bar g^\varepsilon_{ab}(x + h) - \bar g^\varepsilon_{ab}(x)}{h}$$

The discontinuity is then gonna be

$$\lim_{\varepsilon \to 0} \lim_{h \to 0} [\frac{ \bar g^\varepsilon_{ab}(x + h) - \bar g^\varepsilon_{ab}(x) - \bar g^{-\varepsilon}_{ab}(x + h) + \bar g^{-\varepsilon}_{ab}(x)}{h} ]$$

I think we can assume uniform convergence ($g$ should be bounded on the neighbourhood), in which case we can switch the limits and, since $g$ is continuous,

\begin{eqnarray} [\partial_c g_{ab}] &=& \lim_{h \to 0} \lim_{\varepsilon \to 0} [\frac{ \bar g^\varepsilon_{ab}(x + h) - \bar g^\varepsilon_{ab}(x) - \bar g^{-\varepsilon}_{ab}(x + h) + \bar g^{-\varepsilon}_{ab}(x)}{h} ] \\ &=& \lim_{h \to 0} [\frac{ \bar g^0_{ab}(x + h) - \bar g^0_{ab}(x) - \bar g^0_{ab}(x + h) + \bar g^{0}_{ab}(x)}{h} ]\\ &=& 0 \end{eqnarray}

Since $v$ only depends on this part of the metric, we have indeed $v^\sigma [g_{\mu\nu,\sigma}] = 0$.


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