Find the function phi(r,t) given its partial derivatives.

[stedwards]

I would like to define $t^*= \phi(r, t)$ given $dt^* = \left( 1-\frac{k}{r} \right) dt + 0dr$ where k is a constant.

Perhaps it doesn't exist. It appears so simple, yet I've been running around in circles. Any hints?

 

[fzero]

You want $dt^*$ to be an exact form, so that you can integrate it. An exact form can be expressed as

$$ df = M dx + N dy,~~~M = \frac{\partial f}{\partial x},~~~N = \frac{\partial f}{\partial y},$$

therefore a necessary condition that $df$ be exact is that

$$ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}.$$

The corresponding relation for $dt^*$ fails, so we conclude that it is not exact and the corresponding $t^*(r,t)$ does not exist.

 

[stedwards]

Nicely done. Thank you!