A Radial, exterior, outgoing, null geodesics in Schwarzschild

[SonnetsAndMath]

I'm a little confused about the proper way to find these null geodesics. From the line element,
$$c^2 d{\tau}^2=\left(1-\frac{r_s}{r}\right) c^2 dt^2-\left(1-\frac{r_s}{r}\right)^{-1}dr^2-r^2(d{\theta}^2+\sin^2\theta d\phi^2),$$
I think we can set $d\tau$, $d\theta$ and $d\phi$ to $0$, and find $\frac{dr}{dt}$.

However, I wish I could find a reference for this, but I've also seen something like this done:
$$g_{\mu\nu}x^{\mu}x^{\nu}=0,$$
where $g_{\mu\nu}$ is the metric and I think $x^{\mu}$ is a position vector, maybe? Solving that equation should give the nulls. Setting the $\theta$ and $\phi$ components of $x^{\mu}$ to $0$ should give the radial nulls, then, I assume.

Are either of these correct? They seem to give two different answers, in this case. I'm ultimately trying to find a time coordinate delayed by the speed of the light, and the first method seems to give something in terms of a logarithm for a retarded time coordinate, while the second leaves me with me with something like a half integer power of $r$ over a square root, but I might just be blowing one or both. How do I find (specifically the radial, exterior, outgoing) null geodesics for Schwarzschild and for a general metric in general relativity?

 

[PeterDonis]

$$
g_{\mu\nu}x^{\mu}x^{\nu}=0,
$$

This is equivalent to setting $d\tau = 0$.

 

[Orodruin]

You can, but the resulting geodesic won't be null, it will be spacelike. In order to find a null geodesic you need to set $d\tau = 0$, and if you want a radial geodesic you also need to set $d\theta = d\phi = 0$.

How is this different from what he said in the OP?

I think we can set $d\tau$, $d\theta$ and $d\phi$ to $0$, and find $\frac{dr}{dt}$.

 

[SonnetsAndMath]

How is this different from what he said in the OP?

This is all extremely helpful, thank you! I'm confused about that too, though. So they should result in the same answer? I'm not sure if I'm blowing the algebra, or if I'm missing a fine point.

 

[PeterDonis]

How is this different from what he said in the OP?

Oops, you're right, I misread $d\tau = 0$ as $dt = 0$. I have gone ahead and deleted that post.

 

[PeterDonis]

So they should result in the same answer?

Yes.

 

[SonnetsAndMath]

Thank you. Last follow up: $x^{\mu}$ is a position vector? So my vector looks like $(t,r,\theta,\phi)$, except it's $(t,r,0,0)$ if I want radials?

EDIT: Or maybe it's a displacement vector, looking at what I just wrote?

 

[PeterDonis]

$x^{\mu}$ is a position vector? So my vector looks like $(t,r,\theta,\phi)$, except it's $(t,r,0,0)$ if I want radials?

Yes.

Or maybe it's a displacement vector

It's better thought of as a position vector--just the four coordinates put into a "vector" (4-tuple)--because its components--i.e., the coordinates--are not physical displacements from a particular point.

 

[Orodruin]

However, I wish I could find a reference for this, but I've also seen something like this done:
$$g_{\mu\nu}x^{\mu}x^{\nu}=0,$$
where $g_{\mu\nu}$ is the metric and I think $x^{\mu}$ is a position vector, maybe? Solving that equation should give the nulls. Setting the $\theta$ and $\phi$ components of $x^{\mu}$ to $0$ should give the radial nulls, then, I assume.

It is $g_{\mu\nu} \dot x^\mu \dot x^\nu = 0$. The $\dot x^\nu$ are the components of the tangent vector of the world-line. There is no such thing as a position vector in a general manifold.

 

[SonnetsAndMath]

It is gμν˙xμ˙xν=0gμνx˙μx˙ν=0g_{\mu\nu} \dot x^\mu \dot x^\nu = 0. The ˙xνx˙ν\dot x^\nu are the components of the tangent vector of the world-line.

Question fully answered, thanks again. (Having trouble typesetting the quote, though.)

There is no such thing as a position vector in a general manifold.

Had a feeling, by the way. I'm still relatively new to this.

 

[Orodruin]

Just a warning: Even if it works in this case, note that setting all but two coordinate differentials will generally give you a null world-line - not necessarily a null geodesic. Here you can argue that there should exist a purely radial null geodesic by symmetry and then it must be given by the described procedure. The same will not work for all but $d\theta$ and $dt$ set to zero (except for a particular value of $r$, try to figure out which!).

 

[SonnetsAndMath]

Just a warning: Even if it works in this case, note that setting all but two coordinate differentials will generally give you a null world-line - not necessarily a null geodesic.

That's also useful to me. I'm doing something similar for the Kerr metric, and I specifically chose to start with Boyer-Linquist coordinates because I think "frame-dragging" is "baked in" due to the lack of particular off-diagonals in the metric. Maybe you could comment on whether the lack of particular off-diagonals in Boyer-Linquist coordinates means I can think of the coordinates as "corotating with frame-dragging effects," if that makes any sense.

Here you can argue that there should exist a purely radial null geodesic by symmetry and then it must be given by the described procedure. The same will not work for all but dθdθd$\theta$ and dtdtdt set to zero (except for a particular value of rrr, try to figure out which!).

I think I maybe see why, given the curving of orbits. I'm guessing that critical point is one of infinity, the Schwarzschild radius, or the origin, but maybe I'm misunderstanding. In any case, it's not immediately important to what I'm trying to do, but let me know if and when you figure it out!

EDIT: Thinking about it, and, without actually cranking any actual math, I'm guessing that point is the Schwarzschild radius, because light would be trapped at the limit of a circular orbit, as the pop sci documentaries tell it.