A question on an assumption of space-time

"A four-dimensional differentiable (Hausdorff and paracompact) manifold $M$ will be called a space time if it possesses a pseudo-Riemannian metric of hyperbolic normal signature $(+,-,-,-)$ and a time orientation. There will be no real loss of generality in physical applications if we assume that $M$ and its metric are both $\mathcal{C}^{\infty}$ ."

The above is an excerpt is taken from this paper. I'd like to know how the assumption that $M$ and its metric are both $\mathcal{C}^{\infty}$ be made with out any real loss of generality in physical applications. Any intuitive answer is also appreciated.

• May 24, 2011 at 18:40

Dear Rajesh, in reality, physics sometimes works with continuous functions that are not infinitely differentiable - for example look at the energy of the beam at atlas.ch (click at the "Status" button in the middle) when they ramp it up - there are all kinds of discontinuities.

But an arbitrary function that is smooth almost everywhere - and this is a description of functions that really covers everything that a physicist would use - may be approximated by infinitely differentiable functions with an arbitrary accuracy. So it doesn't really hurt if your theorems assume that all the spacetime fields including the metric tensor are infinitely differentiable; you may solve the situations involving functions that are not infinitely differentiable by taking a limit of the infinitely differentiable ones.

May we ask a physics question whether the fields in electromagnetism are infinitely differentiable? Well, we may but it is a meaningless question because the real world is not described by classical physics. So classical physics itself is just an approximation, so both "classical physics with all differentiable functions" and "classical physics with infinitely differentiable functions only" are just approximations of the reality, with none of them being more "physically real".

In quantum physics, we don't use classical fields but we may use wave functions. The time evolution is dictated by Schrödinger's equations - but may we ask whether $\psi(x,y,z)$ should be an infinitely differentiable function of $x,y,z$? Well, in this case, we usually don't make this assumption. Instead, we use all $L^2$ functions which are much more natural at the level of the Hilbert space, Fourier transforms, and so on. The $L^2$ condition surely doesn't require infinite differentiability.

On the other hand, a finite expectation value of the kinetic energy does force $\psi(x,y,z)$ to be a continuous function. Nevertheless, the infinite differentiability condition isn't ever natural in quantum mechanics.

I want to emphasize that all these extra conditions - whether something should be smooth or infinitely smooth etc. - are only a matter of mathematical taste. There can't exist an operational "physics" way to determine whether the world allows functions that are not infinitely differentiable. It's because the infinitely smooth and not-infinitely smooth functions can mimic each other with an arbitrary accuracy but the accuracy of any measurement in physics is always limited. So the restrictions on the "nice behavior" of the mathematical functions is always a matter of mathematical taste.

Exotic differentiable structures could be counterexamples to what I just wrote - because they're "qualitatively" different and their pathological behavior is the very point of their existence (so in some important sense, the ability of them to mimic the normal functions disappears) - but their role in physics remains very confusing and limited as of today.