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The well known and frequently used Israel Junction conditions are the equivalent of Einstein's field equation on a membrane in the brane-world picture.

All the sources have this notation: $K_{ab}^{(i)}$ is the extrinsic curvature and $i$ can be either 1 (2) telling us if we are inside (outside) the bubble. $K$ is the trace of the extrinsic curvature. $S_{ab}$ is the energy-momentum tensor. $h_{ab}$ is the induced metric on the membrane.

Now, the Israel Junction Condition are on the form: $$S_{ab} = \Delta K_{ab} - \Delta K h_{ab}.$$

To me, this equation makes absolutely no sense what so ever. Because $K$ is defined as $K = h^{ab} K_{ab}$.

So the right-hand side will always be zero? Just pick eg. (a,b) = (1,1). How is this not always just zero? What am I missing? I just can't see it.

See e.g. eq.(2.13) here: https://cds.cern.ch/record/450123/

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  • $\begingroup$ Why do you think this would always be zero? $\endgroup$ – Eletie Apr 14 at 18:42
  • $\begingroup$ Because $h_{ab}$ do not change inside/outside the brane. So basically it commutes, and kill the $h^{ab}$ term. Leaving us with $\Delta K_{ab} - \Delta K_{ab} = 0$ $\endgroup$ – Johan Hansen Apr 14 at 19:42
  • $\begingroup$ If it kills the $h^{ab}$ term aren't we left with $\Delta K_{ab} - \Delta K_{cd} h^{cd} h_{ab}$? $\endgroup$ – Eletie Apr 14 at 20:10
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From what I've understood, you seem to be making a mistake with the $h_{ab} \Delta K$ term, with $K$ defined as $K= h^{ab} K_{ab}$. The indices in $K$ are summed, not the same as free indices of $S_{ab}$, so you need to make them different. For example, witing the Israel Junction Condition in full and using $\Delta h_{ab} = 0$ gives $$ S_{ab} = \Delta K_{ab} -h_{ab} \Delta K= \Delta K_{ab} - h_{ab} h^{cd} \Delta K_{cd} \ , $$ which doesn't appear to be zero in general because $\Delta K_{ab} \neq h_{ab} h^{cd} \Delta K_{cd} $.

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    $\begingroup$ Thank you for pointing this out. You are right and there is probably where I got it wrong. I will try to compute a known result using this. $\endgroup$ – Johan Hansen Apr 15 at 11:27
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The two bulk 4D spacetime metrics on each side of the 3D hypersurface/brane only have to induce the same induced 3D metric on the brane, cf. the 1st Israel junction condition. This still leaves plenty of room (in the dependence of the normal direction) that the extrinsic curvature of the 2 sides could be different. In fact the discontinuity in the extrinsic curvature is related to the stress-energy-momentum tensor of the brane, cf. the 2nd Israel junction condition.

References:

  1. Eric Poisson, A Relativist's Toolkit, 2004; Section 3.7.
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  • $\begingroup$ What you are saying is true. And I am not arguing against that. The extrinsic curvature contributions must be different inside/outside to get any interesting results at all. What I am saying is that the (in your example) 3D induced metric, is what does not change inside/outside. So in my equation, the two terms on the right-hand side become the same. $\endgroup$ – Johan Hansen Apr 14 at 19:46

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