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Killing vector in kruskal coordinates

  1. Feb 26, 2007 #1

    cristo

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    Let [itex](U,V,\theta, \phi)[/itex] be Kruskal coordinates on the Kruskal manifold, where [tex]-UV=\left(\frac{r}{2m}-1\right)e^{r/2m},\hspace{1cm} t=2m\ln\left(\frac{-V}{U}\right)[/tex] and [itex]\theta[/itex] and [itex]\phi[/itex] are the usual polar angles. The metric is [tex]ds^2=\frac{-32m^3}{r}e^{\frac{-r}{2m}}dUdV+r^2d\Omega^2[/tex]. The vector [tex]\xi=-U\partial_U+V\partial_V[/tex] is a Killing vector.

    I need to express [itex]\xi[/itex] in exterior Schwarzschild coordinates, however I'm not sure how to go about doing this. I guess I need to transform the basis vectors [itex]\partial_U[/itex] and [itex]\partial_V[/itex] into basis vectors in the Schwarzschild coordinates, but can't see how to, as U and V are defined implicitly.

    Any help would be much appreciated!
     
    Last edited: Feb 26, 2007
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  3. Feb 26, 2007 #2

    George Jones

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    Does it work out if you use the chain rule, for example,

    [tex]\frac{\partial}{\partial U} = \frac{\partial r}{\partial U} \frac{\partial}{\partial r} + \frac{\partial t}{\partial U} \frac{\partial}{\partial t},[/tex]

    and implicit differentiation?
     
  4. Feb 26, 2007 #3

    cristo

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    Well, I've tried that; here's my attempt: [tex]-UV=\left(\frac{r}{2m}-1\right)e^{r/2m},\hspace{1cm} t=2m\ln\left(\frac{-V}{U}\right)[/tex]. Differentiating the first wrt U gives [tex]-V=\frac{\partial}{\partial U}\left[\frac{r}{2m}-1\right]e^{r/2m}=\frac{r}{4m^2}e^{r/2m}\frac{\partial r}{\partial U} \boxed{\Rightarrow \frac{\partial r}{\partial U}=\frac{-4m^2V}{r}e^{-r/2m}}[/tex]. Differentiating the second wrt U gives [tex] \boxed{\frac{\partial t}{\partial U}=\frac{2m}{UV}}[/tex]. Doing a similar thing for V gives [tex]\boxed{\frac{\partial r}{\partial V}=\frac{-4m^2U}{r}e^{-r/2m}}[/tex] and [tex]\boxed{\frac{\partial t}{\partial V} =\frac{-2mU}{V}}[/tex]. And so, [tex]-U\partial_U+V\partial_V=-U\left[\frac{-4m^2V}{r}e^{-2m/r}\frac{\partial}{\partial r}+\frac{2m}{UV}\frac{\partial}{\partial t}\right]+V\left[\frac{-4m^2U}{r}e^{-r/2m}\frac{\partial}{\partial r}-\frac{2mU}{V}\frac{\partial}{\partial t}\right] [/tex] and this simplifies to give [tex] \xi =\frac{4m^2}{r}e^{-r/2m}(V-U)\frac{\partial}{\partial r}-2m\left(\frac{1-UV}{V}\right)\frac{\partial}{\partial t}[/tex]

    However, now I can't get rid of the U's and V's in the answer, since I don't have explicit definitions for them (well, obviously, the UV in the last term could be eliminated, but this won't help the fact that there are other U's and V's in the first term, and in the denominator of the second). Have I made a mistake somewhere? Thanks for helping, by the way!
     
    Last edited: Feb 26, 2007
  5. Feb 26, 2007 #4

    George Jones

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    Your derivatives of [itex]t[/itex] with respect to [itex]U[/itex] and [itex]V[/itex] don't look right.
     
  6. Feb 26, 2007 #5

    cristo

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    Ok, I've recalculated the derivatives of time, and get [tex]\frac{\partial t}{\partial U}=-\frac{2m}{U}[/tex] and [tex]\frac{\partial t}{\partial V}=-\frac{2m}{V}[/tex]. I also noted a mistake when calculating the last line; namely that I didn't multiply U and V (respectively) into the brackets. Doing so, causes the term in r to disappear, so we end up with [tex]\xi=4m\frac{\partial}{\partial t}[/tex]. I think this makes sense, since [itex]\partial_t[/itex] is a Killing vector in the Schwarzschild metric.

    Thanks a lot for your help, George!
     
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