General Relativity - Killing Vectors and Geodesics

In summary, the author is having trouble understanding how to find the geodesic equations from a question given. They are given the metric ds^2=-du^2+u^2dv^2 and are asked to use equation (*) to find the geodesic equations. They are told to use V^a=\dot{x}^a the tangent vector to the geodesic and presumably use the three killing vectors given, but when they try to partial differentiate w.r.t to u, they get an equation that does not seem right to them. They are asked to check where they got the question from, and it is from a 2010 Part II paper. They are told to
  • #1
Tangent87
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0
Hi, I'm stuck on the last bit the attached question where we're given the metric [tex]ds^2=-du^2+u^2dv^2[/tex] and have to use equation (*) to find the geodesic equations.

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

[tex]\left(V^ak_a\right)_{,b}V^b=\left(\dot{x}^ak^cg_{ac}\right)_{,b}\dot{x}^b=0[/tex]
But then using the killing vector (0,1) and the metric I get the equation [tex]2u\dot{v}\dot{u}=0[/tex] 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 [tex]\dot{u}[/tex] term alone right?

Thanks.
 

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  • #2
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 [tex] \dot{u} = \frac{du}{dv} \dot{v} [/tex]
 
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  • #3
sgd37 said:
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 [tex] \dot{u} = \frac{du}{dv} \dot{v} [/tex]

Thank you for replying.

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

[tex]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}[/tex]

So now are you saying that we when we partial differentiate w.r.t u say we have to differentiate v terms as well?
 
  • #4
and vice versa yes. Can i ask where did you get this question from because it doesn't look like a past paper
 
  • #5
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 [Broken]
 
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  • #6
sgd37 said:
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:

[tex]\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=0[/tex]Is that correct? Have you worked through it to the end sgd37?
 
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  • #7
No sorry I have my own exams to worry about but of the little I did I will share

so we start from

[tex] \partial_u ( \dot{v} u^2 ) \dot{u} + \partial_v ( \dot{v} u^2 ) \dot{v} [/tex]

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

[tex] \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 [/tex]

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

using the same procedure for the other vectors I suspect you may arrive at the right equation but I'm not certain
 
  • #8
sgd37 said:
No sorry I have my own exams to worry about but of the little I did I will share

so we start from

[tex] \partial_u ( \dot{v} u^2 ) \dot{u} + \partial_v ( \dot{v} u^2 ) \dot{v} [/tex]

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

[tex] \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 [/tex]

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

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.
 
  • #9
glad i could help and good luck
 

1. What is the concept of Killing vectors in general relativity?

Killing vectors are vector fields that generate infinitesimal isometries in a spacetime. In other words, they represent the symmetries of a spacetime that leave the metric unchanged. These vectors play a crucial role in understanding the properties of a spacetime, such as its conserved quantities and the existence of symmetries.

2. How are Killing vectors related to geodesics in general relativity?

Killing vectors are closely related to geodesics, which are the paths that particles follow in a spacetime under the influence of gravity. In fact, the existence of Killing vectors is directly related to the existence of conserved quantities along geodesics. This allows us to study the motion of particles in a spacetime using the symmetries provided by Killing vectors.

3. Can Killing vectors be used to solve Einstein's field equations?

Yes, Killing vectors can be used to find solutions to Einstein's field equations. In particular, they can help us find symmetries in a given spacetime, which can then be used to simplify the equations and find solutions. This is a powerful tool in understanding the properties of different spacetimes and their physical implications.

4. Are there any limitations to using Killing vectors in general relativity?

While Killing vectors are a useful tool in studying the properties of spacetime, they do have limitations. One limitation is that they can only be used in stationary spacetimes, where the metric does not change with time. Additionally, Killing vectors cannot be used to study the effects of matter and energy in a spacetime, as they only provide information about the geometry itself.

5. How do Killing vectors and geodesics relate to the curvature of spacetime?

Killing vectors and geodesics are both related to the curvature of spacetime, which is a fundamental concept in general relativity. Killing vectors represent the symmetries of a spacetime, while geodesics represent the paths that particles follow in that spacetime. The presence of curvature in a spacetime affects both Killing vectors and geodesics, and studying these concepts can help us understand the effects of gravity in our universe.

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