Killing vector in kruskal coordinates

[cristo]

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

I need to express \xi in exterior Schwarzschild coordinates, however I'm not sure how to go about doing this. I guess I need to transform the basis vectors \partial_U and \partial_V 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!

 

[George Jones]

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

I need to express \xi in exterior Schwarzschild coordinates, however I'm not sure how to go about doing this. I guess I need to transform the basis vectors \partial_U and \partial_V 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!

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

\frac{\partial}{\partial U} = \frac{\partial r}{\partial U} \frac{\partial}{\partial r} + \frac{\partial t}{\partial U} \frac{\partial}{\partial t},

and implicit differentiation?

 

[cristo]

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

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!

 

[George Jones]

Your derivatives of t with respect to U and V don't look right.

 

[cristo]

Ok, I've recalculated the derivatives of time, and get \frac{\partial t}{\partial U}=-\frac{2m}{U} and \frac{\partial t}{\partial V}=-\frac{2m}{V}. 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 \xi=4m\frac{\partial}{\partial t}. I think this makes sense, since \partial_t is a Killing vector in the Schwarzschild metric.

Thanks a lot for your help, George!