# Weak Solutions to the Einstein Equation across a Junction

Consider the principle part, i.e., the part which contains the highest derivatives of the metric (which is the $$2^{nd}$$ derivative) is $$\mathcal{P}\{R_{ab}\}=\frac{1}{2}g^{cd}\left(\partial_{a}\partial_{c}g_{db}+\partial_{b}\partial_{c}g_{da}-\partial_{a}\partial_{b}g_{cd}-\partial_{c}\partial_{d}g_{ab} \right)\tag{1}.$$

The metric tensor can be expressed in terms of the Heaviside function on either side of spacetime as follows $$g_{ab} = \Theta(l)g_{ab}^{+}+\Theta(-l)g_{ab}^{-}\tag{2}.$$ There exists a unique $$C^{1}$$ atlas on the spacetime manifold $$\mathcal{M}$$, which is expressed as the disjoint union of two $$C^{3}$$ spacetimes $$\mathcal{M}^{+}$$ and $$\mathcal{M}^{-}$$ (each with boundary $$\Sigma^{+}$$ and $$\Sigma^{-}$$),which induces the given $$C^{3}$$ structures on $$\mathcal{M}^{+}$$ and $$\mathcal{M}^{-}$$ sucgh that the metric $$g$$ admist a continuous extension on the whole $$\mathcal{M}$$ iff their first fundamental forms $$g^{+}$$ and $$g^{-}$$ agree, i.e., $$g^{+}=g^{-}$$, or in component form, $$\left[g_{ab}\right]=g_{ab}^{+}-g_{ab}^{-}=0\tag{3}.$$ This is basically the first junction condition. The problem with using a metric of the form mentioned in equation $$(2)$$ is that when differentiated it yields

$$g_{ab,c}=\Theta(l)g_{ab,c}^{+} +\Theta(-l)g_{ab,c}^{-}+\epsilon\delta(l)\left[g_{ab}\right]n_{c} \tag{4},$$

where $$\epsilon=n^{c}n_{c}$$ and I have used the fact that $$\Theta'(l) = \delta(l)$$. Now, the problem arises in the last term of the equation $$(4)$$ which is singular and generates terms which are proportional to $$\Theta(l)\delta(l)$$ which are not distributions and hence, causes problems in computing the to used in the Christoffel symbols. To fix this the first junction condition mentioned in equation $$(3)$$ is employed. A similar problem arises when we compute the Riemann tensor and the Ricci tensor. To fix this we then employ the second junction condition.

Now, using the metric of type $$(2)$$ in $$(1)$$ and employing junction conditions $$(3)$$, $$\mathcal{P}\{R_{ab}\}=\mathcal{P}\{R_{ab}^{+}\}+\mathcal{P}\{R_{ab}^{-}\}\tag{5},$$ and hence, $$\mathcal{P}\{g^{ab}R_{ab}\}=\mathcal{P}\{R\}=\mathcal{P}\{R^{+}\}+\mathcal{P}\{R^{-}\}\tag{6}.$$ Using equations $$(5)$$ and $$(6)$$, we can write the Einstein equation in terms of the principle parts, for a source placed on a shell, as follows $$\mathcal{P}\{G_{ab}\}=T_{ab}\tag{7}.$$

It is now possible to express the equations in the form of a linear operator (since only the principle parts were considered) acting on the metric tensor on the LHS equal to the stress-energy tensor of the source on the RHS. Firstly, is this correct? And can this problem now be intepreted as an electrostatics Poisson-like equation whose solution could be intepreted as a weak solution to Einstein's equations?