# How to classify coordinate as S-L,Null,T-L?

1. Mar 19, 2015

### binbagsss

I'm looking at the extension of the Schwarzschild metric using Kruskal coordinates defined as $u'=(\frac{r}{2M}-1)^{\frac{1}{2}e^{\frac{(r+t)}{4M}}}$
$v'=(\frac{r}{2M}-1)^{\frac{1}{2}e^{\frac{(r-t)}{4M}}}$

In these coordinates the metric is given by:
$ds^{2}=-\frac{16M^{3}}{r}e^{-\frac{r}{2M}}(du'dv'+dv'du')+r^{2}d\Omega^{2}$

Question

The text says (lecture notes on GR, Sean M.Carroll) $u'$ and $v'$ are null coordinates in the sense that their partial derivatives $\frac{\partial}{du'} ,\frac{\partial}{dv'}$ are null vectors.

I've had a google on can't seem to find anything on this.
I have no idea what he means here and what is meant by $\frac{\partial}{du}$.
I have never heard of a partial derivative of a coordinate..

Once I know what this is, do I do the same check as you do for normal vectors classification - checking whether the pseudo scalar product is $>0$, $<0$ etc?

Thanks.

2. Mar 19, 2015

### Orodruin

Staff Emeritus
It is not the partial derivatives of the coordinates, it is the partial derivatives with respect to the coordinates. These are the usual partial derivatives. In differential geometry, tangent vectors at a point in a manifold may be seen (or defined) as linear combinations of partial derivatives.

3. Mar 19, 2015

### Staff: Mentor

It's the partial derivative with respect to a coordinate, which is just the directional derivative along that coordinate direction. For example, $\partial / \partial x$ in ordinary Cartesian coordinates is just the derivative along the $x$ direction.

The key fact Carroll is using here is that, at any point in spacetime, there is a one-to-one mapping between vectors and directional derivatives. The directional derivatives $\partial / \partial x^{\mu}$ correspond to the coordinate basis vectors $\hat{e}_{\mu}$, i.e., the basis vectors in the directions $x^{\mu}$. If we write out vectors as 4-tuples of components, then the basis vectors are the ones with components $(1, 0, 0, 0)$, $(0, 1, 0, 0)$, etc. So in coordinates $(u, v, \theta, \phi)$, the directional derivative $\partial / \partial u$ is just the basis vector $(1, 0, 0, 0)$, and $\partial / \partial v$ is just the basis vector $(0, 1, 0, 0)$.

Yes.

4. Mar 19, 2015

### binbagsss

mmm thanks. I'm not sure I've understood properly.
$v=t+r+2M In (r-2M)$,
In $(t,r,\theta,\phi)$ coordinates is $\frac{\partial}{\partial v}=(1, \frac{r}{r-2m},0,0)$

5. Mar 19, 2015

### Orodruin

Staff Emeritus
To figure this out, simply apply the chain rule:
$$\partial_v = \frac{\partial r}{\partial v} \partial_r + \frac{\partial t}{\partial v} \partial_t.$$

6. Mar 19, 2015

### binbagsss

so it is?
And if I use coordinates $(v,r,\theta,\phi)$ to specify the partial derivatives I would get something trivial?

7. Mar 19, 2015

### Staff: Mentor

You don't need to use those coordinates. You can compute the scalar product of $\partial / \partial u$ with itself, and $\partial / \partial v$ with itself, in any coordinate chart in which you have an expression for the metric. You have one for $(u, v, \theta, \phi)$ coordinates, so you can just use that. That makes the computation easy because $\partial / \partial u$ and $\partial / \partial v$ are basis vectors in this chart, so their components are very simple, as I said in post #3.

8. Mar 19, 2015

### binbagsss

So my post #4 is wrong as I haven't used the metric?
In what way do you use the metric in computing the partial derivative?

9. Mar 19, 2015

### Staff: Mentor

I'm not saying it's wrong; I'm saying that there's a quicker way to the answer you are ultimately trying to get, which is whether or not the vectors $\partial / \partial u'$ and $\partial / \partial v'$ are null. You can get that answer in the coordinates you're already using, the $u', v'$ coordinates in which you wrote the metric in your OP. (I see that you put primes on the coordinates in the OP, so I'm doing it here; I left them out in previous posts.) And doing it that way will be simpler than trying to do it in $(v', r, \theta, \phi)$ or $(u', r, \theta, \phi)$ coordinates.

10. Mar 19, 2015

### binbagsss

Ok in $(v',u',\theta,\phi)$ coordinates $\partial / \partial v'=(1,0,0,0)$ ?
So its pseduo scalar $=(1)^{2}-(0)^{2}-(0)^{2}-(0)^{2}=1$
I conclude it's not null, but this is the wrong answer.

11. Mar 19, 2015

### Staff: Mentor

Yes.

No. To get the squared length of a vector, which is what you need to see if it is timelike, spacelike, or null, you have to use the metric: a vector $V^{\mu}$ has squared length $g_{\mu \nu} V^{\mu} V^{\nu}$. The line element you give in your OP gives you $g_{\mu \nu}$, and you know $V^{\mu} = (1, 0, 0, 0)$.

12. Mar 20, 2015

### binbagsss

Ofc, apologies !
So $g_{ab}V^{a}V^{b}=V^{a}V_{a}$
Where $g_{ab}=$
$\left(\begin{array}{cccc} 0 & \frac{-16M^{3}}{r}e^{-\frac{r}{2M}} & 0 & 0\\ \frac{-16M^{3}}{r}e^{-\frac{r}{2M}}& 0 & & 0 & 0 \\ 0 & 0 & r^{2} & 0 \\ 0 & 0 & 0 & r^{2}\\ \end{array}\right)$

And so $g^{ab}= \left(\begin{array}{cccc} 0 & \frac{r}{-16M^{3}}e^{\frac{r}{2M}} & 0 & 0\\ \frac{r}{-16M^{3}}e^{\frac{r}{2M}}& 0 & & 0 & 0 \\ 0 & 0 & 1/r^{2} & 0 \\ 0 & 0 & 0 & 1/r^{2}\\ \end{array}\right)$

$V^{a}=g^{ab}V_b$

Then looking at the non-zero terms of the metric
Sp $V^{0}=g^{01}V_{1}=0$
$V^{1}=g^{10}V_{0}$
$V^{2}=g^{22}V_{2}=0$
$V^{3}=g^{33}V_{3}=0$

And so $V^{a}=(0,-\frac{r}{16M^{3}}e^{\frac{r}{2M}},0,0)$
gives the result.

Question: When we get $\partial / \partial_{v'} = (1,0,0,0)$, we assume no dependence between $u'$ and $v'$? But since they both depend on $r$, how is this ok?

Thanks.

13. Mar 20, 2015

### Roy_1981

"Question: When we get ∂/∂v′=(1,0,0,0), we assume no dependence between u′ and v′? But since they both depend on r, how is this ok?"

They both might depend on r, but the dependence on r does not turn u' into a function v'. It is very easy to see this in a toy example. Say you define, u = r-t, v = r+t. Given a value of v, say v=0, doesnt determine or better doesn't constrain the value of u in any manner. All you can say is for v=0, u = 2r. But this r can assume any (real) value, from -∞ to ∞. In this sense v and u are totally independent.

14. Mar 21, 2015

### binbagsss

thanks !

15. Apr 13, 2015

### binbagsss

Sorry why is $\partial/\partial x^{\mu}=V^{\mu}$ and not $V_{\mu}$? thanks.

16. Apr 13, 2015

### Staff: Mentor

The short answer is that there is a one-to-one correspondence between tangent vectors and directional derivatives, as has already been noted.

The other fairly short answer is that putting the index on $x^{\mu}$ "upstairs" instead of "downstairs" is somewhat of an abuse of notation, because $x^{\mu}$ is not really a vector--more precisely, it's not a tangent vector, and when we are dealing with curved spacetime the only meaningful notion of "vector" is "tangent vector". So the partial derivative operator $\partial / \partial x^{\mu}$ is not a 1-form, even though the (abused) notation suggests that it should be.