# I Differential forms in GR

1. Nov 2, 2016

### kent davidge

The differential form of a function is
$\partial{f(x^1,...,x^n)}=\frac{\partial{f(x^1,...,x^n)}}{\partial{x^1}}dx^1+...+\frac{\partial{f(x^1,...,x^n)}}{\partial{x^n}}dx^n$

Is there (especially in General Relativity) differential of higher orders, like $\partial^2{f(x^1,...,x^n)}$? If so, how is it computed?

2. Nov 2, 2016

### Staff: Mentor

This paper discusses Differential forms, Tensors and uses in General Relativity so I would say yes higher order forms are used and are useful in General Relativity.

http://www.aei.mpg.de/~gielen/diffgeo.pdf

3. Nov 3, 2016

### Orodruin

Staff Emeritus
But at the same time no, there are no forms that are the second derivative of a function (as the OP suggests) as the exterior derivative applied twice gives zero (the any-dimensional equivalent of curl(grad(f))=0).

Of course there are other forms of higher order.

4. Nov 3, 2016

### stevendaryl

Staff Emeritus
Yes, $d^2$ always produces zero. However, you can get something sort of conceptually similar to $d^2$:
1. Operate on $F$ with $d$ to produce $dF$.
2. Take the Hodge dual, $*dF$.
3. Operate on THAT with $d$, to produce $d * dF$
This isn't necessarily zero, and is sort of like a higher-order derivative. In 3-D, if $F$ is a scalar function, then $* d * d F = \nabla^2 F$.

5. Nov 3, 2016

### kent davidge

In General Relativity one frequently deal with differential forms, say W. What actually is it? I know it has to be a completely antisymmetric (0,p) tensor. But what are its components Wμ1...μp? Would it be some array of derivatives of a function?

6. Nov 4, 2016

### haushofer

The components don't have to be derivatives. They are just antisymmetric, which makes threm taylor-fit for integration over manifolds.