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

[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.

 

[Orodruin]

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.

 

[PeterDonis]

I have never heard of a partial derivative of a coordinate..

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 $(0, 1, 0, 0)$.

do I do the same check as you do for normal vectors classification - checking whether the pseudo scalar product is $>0, <0$

Yes.

 

[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 $\frac{\partial}{\partial v}=(1, \frac{r}{r-2m},0,0)$

 

[Orodruin]

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

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

 

[binbagsss]

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

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

 

[PeterDonis]

In $(t,r,\theta,\phi)$ coordinates

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.

 

[binbagsss]

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.

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?

 

[PeterDonis]

So my post #4 is wrong as I haven't used the metric?

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.

 

[binbagsss]

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.

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.

 

[PeterDonis]

Ok in $(v',u',\theta,\phi)$ coordinates $\partial / \partial v'=(1,0,0,0)$ ?

Yes.

So its pseduo scalar $=(1)^{2}-(0)^{2}-(0)^{2}-(0)^{2}=1$

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 $V^{\mu} = (1, 0, 0, 0)$.

 

[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.

 

[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, doesn't 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.

 

[binbagsss]

"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, doesn't 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.

thanks !

 

[binbagsss]

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 $V^{\mu} = (1, 0, 0, 0)$.

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

 

[PeterDonis]

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

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.