# What happens to the non-smooth (non differentiable) solutions to general relativity?

Since all the geodesic and differential geommetry is assumed to be smooth and differentiable. Does it mean that there are neither non-continous nor non-differentiable solutions to Einstein equations?

General relativity is fine up to $\mathcal C^2$ metrics, since the important quantities only require second derivatives at most, meaning the curvature will be $\mathcal C^0$ at worst.

It is still possible to have slightly less smooth metrics if you allow for weak derivatives, $\mathcal C^0$ metrics giving you discontinuous connections and distributions for the curvature. This is used in such domains as the thin shell approximation when the matter distribution is assumed to be infinitely thin (i.e. a delta function) and the study of gravitational shockwaves. This still works okay since the second derivative involved are usually linear, meaning the theory of distributions will work fine here. You might need to use some slightly more general distributions (such as the Colombeau algebra of distributions), but all quantities should still be roughly physical.

I don't think it's really a good idea to go further than $\mathcal C^0$. Once you start having products of delta functions, the non-linear distributions involved ceased to give meaningful answers (the product of two delta distributions is itself not a distribution, but a generalized function. It is "too large" to be a distribution).

There are important weak solutions in GR, where the metric is not continuous, e.g. thin shell solutions. The boundary conditions along the thin shell are described by Israel junction conditions, cf. e.g. Refs. 1-2.

References:

1. Eric Poisson, A Relativist's Toolkit, 2004.

2. Eric Poisson, An Advanced course in GR.

• The metric is continuous in the thin shell formalism, it is still $\mathcal C^0$. It is the matter distribution that isn't, hence the name. – Slereah Dec 22 '16 at 11:37
• IIRC the induced metric is indeed continuous across the thin shell/brane, cf. the first Israel junction condition, but the transversal/normal components of the bulk/ambient spacetime metric may jump. – Qmechanic Dec 22 '16 at 11:52

Since all the geodesic and differential geometry is assumed to be 'smooth' and differentiable .. does it mean that there are no non-continous nor differentiable solutions to Einstein equations ?

Only if spacetime is non continous, (which is what you are in effect asking), and which we have no evidence for, or (possibly) we want to explain Black Holes better than we do now, do we then run into trouble. If we avoid these issues we are, as far as I know, ok with GR in its present form.

Apologies if you know this already:

The starting point for the Einstein equation utilises Christoffel symbols, $\Gamma$:

$${\displaystyle R^{\rho }{}_{\sigma \mu \nu }=\partial _{\mu }\Gamma ^{\rho }{}_{\nu \sigma }-\partial _{\nu }\Gamma ^{\rho }{}_{\mu \sigma }+\Gamma ^{\rho }{}_{\mu \lambda }\Gamma ^{\lambda }{}_{\nu \sigma }-\Gamma ^{\rho }{}_{\nu \lambda }\Gamma ^{\lambda }{}_{\mu \sigma }}$$

This expression, the Riemann tensor, when suitably modified by the contraction of the indices from 4 to 2, produces the 10 element Ricci Tensor.

Each of the Christoffel symbols, $\Gamma$, is itself a derivative of the metric, given by:

$${\displaystyle \Gamma _{cab}={\tfrac {1}{2}}\left({\frac {\partial g_{ca}}{\partial x^{b}}}+{\frac {\partial g_{cb}}{\partial x^{a}}}-{\frac {\partial g_{ab}}{\partial x^{c}}}\right)}$$

So that is the basis for Danu's comment regarding the double derivative.

If you define smooth as differentiable to the nth value, we only need to proceed to the second derivatives to describe the curvature of spacetime.

A more important question is, at small enough scales, is spacetime discrete or continous, and so far we have no evidence that it is not continous.

But then, again apologies as you probably already know, at some stage we have to confront the fact that quantum mechanics does incorporate discreteness and we would like to explain black holes using Q.M. and just possibly the conjectured Planck scale effects, we then need to bring in a theory of Quantum Gravity