Understanding the Israel Junction Conditions

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/

• Why do you think this would always be zero? – Eletie Apr 14 at 18:42
• Because $h_{ab}$ do not change inside/outside the brane. So basically it commutes, and kill the $h^{ab}$ term. Leaving us with $\Delta K_{ab} - \Delta K_{ab} = 0$ – Johan Hansen Apr 14 at 19:42
• If it kills the $h^{ab}$ term aren't we left with $\Delta K_{ab} - \Delta K_{cd} h^{cd} h_{ab}$? – Eletie Apr 14 at 20:10

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 $$\Delta K_{ab} \neq h_{ab} h^{cd} \Delta K_{cd}$$.