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

  1. Mar 19, 2015 #1
    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. jcsd
  3. Mar 19, 2015 #2

    Orodruin

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    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.
     
  4. Mar 19, 2015 #3

    PeterDonis

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    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.
     
  5. Mar 19, 2015 #4
    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)##
     
  6. Mar 19, 2015 #5

    Orodruin

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    To figure this out, simply apply the chain rule:
    $$
    \partial_v = \frac{\partial r}{\partial v} \partial_r + \frac{\partial t}{\partial v} \partial_t.
    $$
     
  7. Mar 19, 2015 #6
    so it is?
    And if I use coordinates ##(v,r,\theta,\phi)## to specify the partial derivatives I would get something trivial?
     
  8. Mar 19, 2015 #7

    PeterDonis

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    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.
     
  9. Mar 19, 2015 #8
    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?
     
  10. Mar 19, 2015 #9

    PeterDonis

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    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.
     
  11. Mar 19, 2015 #10
    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.
     
  12. Mar 19, 2015 #11

    PeterDonis

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    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)##.
     
  13. Mar 20, 2015 #12
    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.
     
  14. Mar 20, 2015 #13
    "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.
     
  15. Mar 21, 2015 #14
    thanks !
     
  16. Apr 13, 2015 #15
    Sorry why is ##\partial/\partial x^{\mu}=V^{\mu}## and not ##V_{\mu}##? thanks.
     
  17. Apr 13, 2015 #16

    PeterDonis

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    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.
     
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