Killing vectors in KS coordinates

pervect
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Messages
10,400
Reaction score
1,579
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 /> <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 /> <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?
 
Last edited:
Physics news on Phys.org
Don't need any complicated algebra, it's true almost by inspection. Just show directly that the Lie derivative of gμν with respect to ξμ 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!
 
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!
 
I started reading a National Geographic article related to the Big Bang. It starts these statements: Gazing up at the stars at night, it’s easy to imagine that space goes on forever. But cosmologists know that the universe actually has limits. First, their best models indicate that space and time had a beginning, a subatomic point called a singularity. This point of intense heat and density rapidly ballooned outward. My first reaction was that this is a layman's approximation to...
Thread 'Dirac's integral for the energy-momentum of the gravitational field'
See Dirac's brief treatment of the energy-momentum pseudo-tensor in the attached picture. Dirac is presumably integrating eq. (31.2) over the 4D "hypercylinder" defined by ##T_1 \le x^0 \le T_2## and ##\mathbf{|x|} \le R##, where ##R## is sufficiently large to include all the matter-energy fields in the system. Then \begin{align} 0 &= \int_V \left[ ({t_\mu}^\nu + T_\mu^\nu)\sqrt{-g}\, \right]_{,\nu} d^4 x = \int_{\partial V} ({t_\mu}^\nu + T_\mu^\nu)\sqrt{-g} \, dS_\nu \nonumber\\ &= \left(...
In Philippe G. Ciarlet's book 'An introduction to differential geometry', He gives the integrability conditions of the differential equations like this: $$ \partial_{i} F_{lj}=L^p_{ij} F_{lp},\,\,\,F_{ij}(x_0)=F^0_{ij}. $$ The integrability conditions for the existence of a global solution ##F_{lj}## is: $$ R^i_{jkl}\equiv\partial_k L^i_{jl}-\partial_l L^i_{jk}+L^h_{jl} L^i_{hk}-L^h_{jk} L^i_{hl}=0 $$ Then from the equation: $$\nabla_b e_a= \Gamma^c_{ab} e_c$$ Using cartesian basis ## e_I...

Similar threads

Back
Top