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With respect to the coordinates $(x^{0},x^{1},x^{2},x^{3})=(v,r,\theta,\phi)$, we have the following components of the metric tensor:

$\begin{bmatrix} g_{00} & g_{01} & g_{02} & g_{03} \\[0.3em] g_{10} & g_{11} & g_{12} & g_{13} \\[0.3em] g_{20} & g_{21} & g_{22} & g_{23} \\[0.3em] g_{30} & g_{31} & g_{32} & g_{33} \end{bmatrix}$ $=\begin{bmatrix} 1-\frac{2M}{r} & 1 & 0 & 0 \\[0.3em] 1 & 0 & 0 & 0 \\[0.3em] 0 & 0 & r^{2} & 0 \\[0.3em] 0 & 0 & 0 & r^{2}sin^{2}\theta \end{bmatrix}$

Suppose we've got a family of hypersurfaces defined by $v=\mathrm{constant}$. I've been asked to characterize the normal vectors to these hypersurfaces (whether they are timelike, spacelike or null).

If I would only know the form of the normal vectors I would have no trouble determining whether they are TL, SL or null. However, I have no idea how to compute the normal vectors.

My logic so far has been to associate the family of hypersurfaces with a vector:

$h=\begin{bmatrix} d\\[0.3em] r\\[0.3em] \theta \\[0.3em] \phi \end{bmatrix}$

Where $r$, $\theta$, and $\phi$ are free, and $d\in \mathbb{R}$ is a constant (in the place of $v$).

I know that in regular flat Euclidean space $\mathbb{R}^{n}$, if you've got a function $\ f$ characterizing a hypersurface (a $\mathbb{R}^{n-1}$ object), you can find the normal vectors to the hypersurface by playing with the gradient of $\ f$.

However, this Lorentzian manifold is not flat, and I don't even know how to find a function like $\ f$ in this case - all I know is that $v$ is constant.

Can someone prod me in the right direction?

EDIT: I should mention that the above uses Eddington-Finkelstein coordinates in the Schwarzschild spacetime where $v=t+r+2m\mathrm{log}(\frac{r}{2m}-1)$.

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    $\begingroup$ The correct generalization of the gradient in this context is the differential (exterior derivative) $d$, so you want to look at $dv$. $\endgroup$ – Robin Ekman Mar 14 '15 at 15:32
  • $\begingroup$ So since $v$ is a constant, then $dv=0$? Is this the kind of approach I should take? $\endgroup$ – Greg.Paul Mar 14 '15 at 15:39
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    $\begingroup$ The projection of $dv$ onto the hypersurface must be zero. $\endgroup$ – Robin Ekman Mar 14 '15 at 17:45
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I don't think your confusion is related to curvature, and I think your question would persist even in the following flat space example.

Consider a surface in Minkowski space $$\{(t,x,y,z): t,x,y,z, \in \mathbb{R}\},$$ with metric determined by $$\begin{bmatrix} -c^2 & 0 & 0 & 0 \\[0.3em] 0 & 1 & 0 & 0 \\[0.3em] 0 & 0 & 1 & 0 \\[0.3em] 0 & 0 & 0 & 1 \end{bmatrix}.$$

Consider the vector $n=(1,1,0,0)$, and the hyperplane orthogonal to it:

$$\{(t,x,y,z): t,x,y,z, \in \mathbb{R}, \text{ with } x=ct\}.$$

Note that $n$ itself is actually in the hyperplane, a vector orthogonal to a hyperplane can actually be in the hyperplane. This happens in your situation, you need to explicitly look for a vector orthogonal to every vector tangent to your surface rather than (wrongly in this case) assuming that this vector will point out of your surface.

In fact, you can see with little effort that when $\Delta v=0$ one of your basis vectors for the tangent space is actually orthogonal to all of the tangent vectors.

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Suppose that the vector field $X$ is tangent to the hypersurface. Then $X\cdot v$ must vanish, since $v$ is a constant on the surface. But $X \cdot v$ is precisely what is meant by $dv(X)$. So the tangent space is precisely the null space of $dv$. Now since we have a metric, we can "raise the index", $$dv \mapsto g^\sharp dv = g^{\mu\nu} (dv)_\nu = V^\mu$$ and the condition that $X$ be in the nullspace of $dv$ is $$g_{\mu\nu} V^\mu X^\nu = 0$$ that is, that $V^\mu$ is orthogonal to $X^\mu$.

So the normal vector you are looking for is $V^\mu$ (or any non-zero scalar multiple of it).

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If you have coordinates $(v,r,\theta,\phi)$ on $M$, and your hypersurfaces $\Sigma_v$ are characterized by $v=const$, then $$ \frac{\partial}{\partial r},\ \frac{\partial}{\partial \theta}\ \mathrm{and}\ \frac{\partial}{\partial \phi} $$ are coordinate basis fields on $\Sigma_v,$ and $\left.\partial/\partial v\right|_v$ is a vector field that pointwise intersects $\Sigma_v$, but is not necessarily orthogonal to it.

You can then project $\left.\partial/\partial v\right|_v(p)$ onto the $T_p\Sigma_v$ tangent space for all points $p\in\Sigma_v$ (execuse my rather unrigorous notation here, I should make a difference between the hypersurface as a manifold and its image via the embedding, but I believe I am clear though), for example by Gram-Schmidt orthogonalizing your coordinate vector fields on $\Sigma_v$ to obtain an orthonormal frame.

If the result of G-S on $$ \frac{\partial}{\partial r},\ \frac{\partial}{\partial \theta},\ \frac{\partial}{\partial \phi} $$ are$$ \hat{r},\ \hat{\theta},\ \mathrm{and}\ \hat{\phi}, $$ then you can project $\partial/\partial v$ onto the hypersurface by $$ \mathrm{Proj}(\partial/\partial v)=g(\hat{r},\partial/\partial v)\hat{r}+g(\hat{\theta},\partial/\partial v)\hat{\theta}+g(\hat{\phi},\partial/\partial v)\hat{\phi}, $$ then substract the projection onto the hypersurface from $\partial/\partial v$ to get a vector $$ n=\frac{\partial}{\partial v}-\mathrm{Proj}\left(\frac{\partial}{\partial v}\right) $$ that is normal to $\Sigma_v$.

There might be simpler methods but right now this is what I can think of.

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    $\begingroup$ You can't Gram Schmidt a linearly independent set of vectors in the hypersurface to get an orthonormal basis for the tangent space because there is no orthonormal basis for the tangent space, the metric applied to the hypersurface is degenerate. $\endgroup$ – Timaeus Mar 14 '15 at 18:03
  • $\begingroup$ Only if the hypersurface is null, which might be the case here, I didn't check, I merely used a generic example. If the hypersurface is not null, then I'm pretty sure you can orthogonalize all you wish. $\endgroup$ – Bence Racskó Mar 14 '15 at 18:24
  • $\begingroup$ Since the OP was asked to show whether the normal was TL, SL or null it was a possibility that cannot be ignored, and for a $v=const$ surface you just look at the three by three matrix in the lower right which is diagonal so its type is pretty obvious. You can orthogonalize, but not uniquely, and you can't normalize. $\endgroup$ – Timaeus Mar 14 '15 at 18:29
  • $\begingroup$ I guess yer right. I'd edit my post, but basically all other solutions I know have been posted by others, so... $\endgroup$ – Bence Racskó Mar 14 '15 at 19:08
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Curvature isn't really a problem -- find the normal vector at an arbitrary point, since then this is just a linear algebra problem.

Lorentzian signature is also not a problem -- as long as we have an inner product, we're good.

The procedure is analogous to the "3+1 split" of spacetime used, for instance, in the ADM formalism. Often we foliate spacetime with a 1-parameter family ("1") of everywhere spacelike 3-hypersurfaces ("3"). The normal vector field will naturally be timelike in that case, often denoted $\hat{n}$ when normalized.

You should seek a vector with components $n^\mu$ that is orthogonal to the surface. What does orthogonality to a surface mean, though? Well, if you have a set of 3 (because it is 3-dimensional) linearly independent vectors tangent to the surface, orthogonality to the surface is equivalent to orthogonality to all 3 such vectors. And you have such vectors. Each orthogonality condition provides 1 constraint on your 4 unknown components; the remaining degree of freedom is of course your normalization.

Also, your life will often be easier if you become adept at utilizing upper/lower indices to your advantage. For instance, upper-indexed forms of coordinate vectors only have a single nonzero component, and taking the inner product of such a vector with another is particularly easy if the other one is written in lower-indexed form.

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