Normal vector of hypersurface in Schwarzshild spacetime

Question

I've read that the following hypersurface is a null surface which means that the normal four-vector has a length of 0. I would however like to confirm this myself.

$$ 2 M G \ln \left|\frac{r}{2 M G}-1\right| + t+r= B $$

where $B$ is a constant and the surface is located in Schwarzshild spacetime.

How do I compute the normal four-vector and its length?

I suspect I should use the gradient to get a normal four-vector but what does the gradient look like here? Is it made up of partial derivatives or covariant derivatives?

Update: use the metric tensor?
It seems like the covariant derivative (which is used to get the four-gradient) reduces to the normal partial derivative in this case since it is a scalar function. Wikipedia says that

In GR, one must use the more general metric tensor $g^{\alpha \beta}$, and the tensor covariant derivative (en.wikipedia.org)

Answer 1

Your surface is defined implicitly by the equation $F(t,r,\theta,\phi) = B$. The natural object which defines the orientation of the surface is not a vector but rather a covector $n_\mu$, with components $$n_\mu = \partial_\mu F$$ The surface is null if $g^{\mu\nu} n_\mu n_\nu = 0$. If you insist on talking about vectors, then you can "raise the index" on $n_\mu$ to find its vector partner given by

$$\tilde n^\mu = g^{\mu \nu} n_\nu = g^{\mu\nu} \partial_\nu F$$

The surface is then null if $g_{\mu\nu} \tilde n^\mu \tilde n^\nu = 0$. Of course, going this route is pointless, because

$$g_{\mu\nu} \tilde n^\mu \tilde n^\nu = g_{\mu\nu} g^{\mu\alpha} n_\alpha g^{\nu\beta} n_\beta = \delta^\alpha_\nu g^{\nu \beta} n_\alpha n_\beta = g^{\alpha \beta}n_\alpha n_\beta$$

so converting $n_\mu$ to a vector first and then computing the inner product from there amounts to just circling the block a few times.