Killing vector in kruskal coordinates

In summary, the expression for the vector \xi in exterior Schwarzschild coordinates is given by: -U\partial_U+V\partial_V=-U\left[\frac{-4m^2V}{r}e^{-r/2m}\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]
  • #1
cristo
Staff Emeritus
Science Advisor
8,146
75
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:
Physics news on Phys.org
  • #2
cristo said:
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!

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?
 
  • #3
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:
  • #4
Your derivatives of [itex]t[/itex] with respect to [itex]U[/itex] and [itex]V[/itex] don't look right.
 
  • #5
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!
 

FAQ: Killing vector in kruskal coordinates

What is a Killing vector in Kruskal coordinates?

A Killing vector in Kruskal coordinates is a vector field that preserves the metric of a spacetime in Kruskal coordinates. It is a vector field that generates an isometry, meaning it preserves the distance and angles between points in the spacetime.

How is a Killing vector related to symmetries in a spacetime?

A Killing vector is related to symmetries in a spacetime because it generates an isometry, which is a transformation that preserves the symmetries of a spacetime. In other words, a spacetime has a Killing vector if there exists a symmetry in the spacetime that is preserved by the vector field.

What are the properties of a Killing vector in Kruskal coordinates?

Some properties of a Killing vector in Kruskal coordinates include: it is a solution to the Killing equation, it generates an isometry, it preserves the metric of the spacetime, and it is conserved along geodesics.

How are Killing vectors used in the study of black holes?

Killing vectors are used in the study of black holes to understand the symmetries and properties of the spacetime around black holes. They can also be used to calculate important quantities such as the event horizon and the angular momentum of a black hole.

Can Killing vectors be expressed in other coordinates besides Kruskal coordinates?

Yes, Killing vectors can be expressed in any coordinate system. However, in Kruskal coordinates, the Killing vectors have particularly simple expressions, making them useful for studying black holes and other spacetimes with high degrees of symmetry.

Similar threads

Replies
44
Views
2K
Replies
4
Views
494
Replies
8
Views
2K
Replies
8
Views
1K
Replies
5
Views
1K
Back
Top