# In GR, why should the spacetime manifold be differentiable?

In general relativity (GR), spacetime is viewed as a differentiable manifold of dimension $D$ with a metric of Lorentzian signature $(-,+,+,...,+)$.

My question is why differentiable?

• Naively because one needs a well defined local tangent space on which one can do the usual physics in a uniquely way. Even trivial cases of non-differential manifolds will lead to multi-valued or ill-defined solutions. Commented May 21, 2015 at 19:22
• And because the Riemann tensor would be ill defined (or at least a complicated object), if the manifold were not differentiable. Commented May 21, 2015 at 19:25
• Related: physics.stackexchange.com/q/1324/2451 and links therein. Commented May 21, 2015 at 19:36
• Uh...because almost nothing in GR works if you can't differentiate? What kind of answer are you looking for? Commented May 21, 2015 at 19:48

Because if your manifold is not differentiable (and even then, at least $C^3$), you end up doing non-linear distribution theory and having to use Colombeau algebras, and trust me, you do not want that.

The basic problem with non-differentiable manifolds is that, unlike say, electromagnetism, general relativity isn't linear, making it difficult to make sense of distributions. You end up with questions like "What is $\delta(x)^2$?", which do not have answers in the basic theory of distributions. Colombeau algebras are a modern framework for dealing with such problems, and you can find applications for it here, for instance.

In order to be able to even define a metric, you need tangent vectors, since these are the arguments to the metric, and to have tangent vectors you need differentiability.

A very short answer might be that in general relativity, spacetime can be curved. To estimate how much it's curved, you need to be able to calculate the rate of change, that is done by differentiating the co-ordinate system you are using to map each region of spacetime you are dealing with.

There is a simple derivation:

Spacetime curvature is a product of gravitation. Gravitation is a force. The force the gravitational field is declining continually with increasing radius, proportionally to the squared radius. The corresponding function is continuous and differentiable.

Once we saw that the function of gravitational force (and thus of spacetime curvature) is differentiable with regard to one mass object and with regard to one direction (one dimension), we can extend this insight without problem to a universe with several mass objects and several space dimensions. In short: Fields in general are differentiable, and spacetime curvature may be assimilated.

If you had just one coordinate chart, then you wouldn't need to ever transition from one chart to another. So you wouldn't have to worry about whether your transition was continuous, differentiable, $C^2$ or smooth.

But if you have two charts and need two charts, then at some events they will be in the intersection of the two charts. And in the one chart you might have a curve parameterized by proper time. And have a vector field. And the vector field might agree with the tangent of the curve and have a metric length equal to some mass $m.$ And everything would be fine, its all in one chart and all the derivatives and the lengths are computing using that one coordinate system and the metric in the one coordinate system.

But you could imagine doing a similar procedure in the other coordinate system. Everything would be just as nice.

But what about in the overlap? If one cooridinate system for the overlap says that the vector field is tangent to the world line would the other one agree? No.

In general it would not. It might not even agree that the world line has a tangent at all the events the other one said it did. But if the transition maps are differentiable then they agree.

If the transition maps are second differentiable then the two charts could agree on whether second derivatives match up. That's essential for GR because there are physical tensors such as the Einstein Tensor that correspond to second derivatives. You want two charts to agree on whether a rank two tensor field exists and is the Einstein Tensor.

• "What it really does is eliminate possible models." Are you sure about that? For $k\ge 1$ any maximal $C^k$-atlas contains smooth atlases by Whitney's theorem, and all of these are diffeomorphic (en.wikipedia.org/wiki/…), but maybe I am misunderstanding what you are saying. Commented Jan 14, 2016 at 9:14