General Relativity - Killing Vectors and Geodesics

[Tangent87]

Hi, I'm stuck on the last bit the attached question where we're given the metric ds^2=-du^2+u^2dv^2 and have to use equation (*) to find the geodesic equations.

They tell us to use V^a=\dot{x}^a the tangent vector to the geodesic and presumably we use the three killing vectors they gave us, so then from (*) we have:

\left(V^ak_a\right)_{,b}V^b=\left(\dot{x}^ak^cg_{ac}\right)_{,b}\dot{x}^b=0
But then using the killing vector (0,1) and the metric I get the equation 2u\dot{v}\dot{u}=0 which doesn't seem right to me. Am I correct in thinking that when we partial differentiate w.r.t to u say we leave the \dot{u} term alone right?

Thanks.

 

[sgd37]

From the question itself we cannot assume that u is independent from v so that will change the equations a bit and you can also have \dot{u} = \frac{du}{dv} \dot{v}

 

[Tangent87]

From the question itself we cannot assume that u is independent from v so that will change the equations a bit and you can also have \dot{u} = \frac{du}{dv} \dot{v}

Thank you for replying.

Ok so when we put k^c=(0,1) in we get:

0=\left(\dot{x}^ag_{av}\right)_{,b}\dot{x}^b=\left(\dot{v}g_{vv}\right)_{,b}\dot{x}^b=\left(\dot{v}u^2\right)_{,b}\dot{x}^b=\left(\dot{v}u^2\right)_{,u}\dot{u}+\left(\dot{v}u^2\right)_{,v}\dot{v}

So now are you saying that we when we partial differentiate w.r.t u say we have to differentiate v terms as well?

 

[sgd37]

and vice versa yes. Can i ask where did you get this question from because it doesn't look like a past paper

 

[Tangent87]

It is, I got it from 2010 Part II paper 2 page 22:

http://www.maths.cam.ac.uk/undergrad/pastpapers/2010/Part_II/PaperII_2.pdf

 

[Tangent87]

and vice versa yes. Can i ask where did you get this question from because it doesn't look like a past paper

Hmm okay, so now on the next line I get:

\left(\frac{dv}{du}u^2\dot{u}\right)_{,u}\dot{u} + \left(\dot{v}u^2\right)_{,v}\dot{v}=\frac{d^2v}{du^2}u^2\dot{u}^2 + 2\frac{dv}{du}u\dot{u}^2 + \frac{dv}{du}u^2\ddot{u} + \ddot{v}u^2=0Is that correct? Have you worked through it to the end sgd37?

 

[sgd37]

No sorry I have my own exams to worry about but of the little I did I will share

so we start from

\partial_u ( \dot{v} u^2 ) \dot{u} + \partial_v ( \dot{v} u^2 ) \dot{v}

assuming u is a function of v we can reduce this to

\frac{dv}{du} \partial_v ( \dot{v} u^2 ) \frac{du}{dv} \dot{v} + \partial_v ( \dot{v} u^2 ) \dot{v} = 2\partial_v ( \dot{v} u^2 ) \dot{v} = 0

so that we have sensible tangent vector \dot{v} \neq 0 thus \dot{v} u^2 = constant

using the same procedure for the other vectors I suspect you may arrive at the right equation but I'm not certain

 

[Tangent87]

No sorry I have my own exams to worry about but of the little I did I will share

so we start from

\partial_u ( \dot{v} u^2 ) \dot{u} + \partial_v ( \dot{v} u^2 ) \dot{v}

assuming u is a function of v we can reduce this to

\frac{dv}{du} \partial_v ( \dot{v} u^2 ) \frac{du}{dv} \dot{v} + \partial_v ( \dot{v} u^2 ) \dot{v} = 2\partial_v ( \dot{v} u^2 ) \dot{v} = 0

so that we have sensible tangent vector \dot{v} \neq 0 thus \dot{v} u^2 = constant

using the same procedure for the other vectors I suspect you may arrive at the right equation but I'm not certain

Ah I see thanks. I've worked it through now and it does come out, the key is to not try and differentiate out the brackets but just deduce that the stuff inside the brackets must be a constant.

 

[sgd37]

glad i could help and good luck