Killing vectors in KS coordinates

[pervect]

I want to prove (or disprove) that the vector with components
$$\xi^u = \frac{v}{2 r_s}\hspace{5 mm} \xi^v = \frac{u}{2 r_s}$$

is a Killing vector of the KS space-time with line element

$$\frac{4 r_s^3}{r} e^{-\frac{r}{r_s}} \left( du^2 -dv^2\right) + r^2 \left( d\theta^2 + sin^2 \theta d\phi^2\right)$$

Here r is an implicit function of u,v, which is handled by a constraint equation.

However, I seem to be having a hard time getting GRT to handle the contraints. I try

$$\left( \frac{r}{r_s} - 1 \right) e^{\frac{r}{r_s}} = u^2 - v^2$$ directly, but it doesn't simplify.

I try to feed it the following
$$<br /> {\frac {\partial }{\partial u}}r \left( u,v \right) =2\,{{\it r\_s}}^{<br /> 2}u \left( r \left( u,v \right) \right) ^{-1} \left( {e^{{\frac {r<br /> \left( u,v \right) }{{\it r\_s}}}}} \right) ^{-1}<br />$$

but it complains about "illegal use of object as a name", I can't see what it's objecting to.

So a) -does this look like the right expression for the Killing Vector? And b) - how does one successfully get the constraints into GrTensor?

 

[Bill_K]

Don't need any complicated algebra, it's true almost by inspection. Just show directly that the Lie derivative of g_μν with respect to ξ^μ is zero. The coordinate transformation is u → u + εv, v → v + εu. Then u² - v² → u² - v² so it obeys the constraint, and du² - dv² → du² - dv² so it's an isometry, and that's really all there is to it!

 

[pervect]

I set r_s to 1, and renamed r to rr,and finally got GRT to crunch through it,but your way is both much easier and more insightful. Thanks!