Killing Vector Question

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Hi. Currently I am self-studying a book on general relativity (Introducing Einstein's Relativity by Ray D'Inverno), I am stuck trying to find a Killing Vector solution to the following problem.

ds^2 = (x^2)dx^2 + x(dy)^2

You can easily obtain the metric from the above.

Now the question is find all Killing Vector solutions of the metric.

I know to solve this the lie derivative must equal 0. ie

Covariant derivative of Xa with respect to b + the covariant derivative of Xb with respect to a = 0. (1)

The answer in the back is the partial derivative with respect to y. (del/del y).

Basically I get to expanding (1) so now there are christoffel symbols but don't know where to go from there.

Thanks,

Chris
 

Answers and Replies

  • #2
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I'm learning this sort of thing myself, so I thought I'd give it try.

Suppose we have a vector [itex]\vec{K}=K_x\partial_x+K_y \partial_y[/itex] where Kx and Ky are functions of x and y. Using the metric above I find that

[tex]
\nabla_\nu X_\mu=
\left[ \begin{array}{cc}
\frac{\left( \frac{\partial K_x}{\partial \,x}\,\right) \,x-K_x}{x} & \frac{2\,\left( \frac{\partial K_x}{\partial \,y}\,\right) \,x-K_y}{2\,x} \\\
\frac{2\,\left( \frac{\partial K_y}{\partial \,x}\,\right) \,x-K_y}{2\,x} & \frac{2\,\left( \frac{\partial K_y}{\partial \,y}\,\right) \,{x}^{2}+K_x}{2\,{x}^{2}} \end{array} \right]
[/tex]

The Killing equations are

[tex]
\nabla_1 X_1+\nabla_1 X_1=0 \ \ \rightarrow \ \ \nabla_1 X_1=\frac{\left( \frac{\partial K_x}{\partial \,x}\,\right) \,x-K_x}{x}=0
[/tex]

[tex]
\nabla_2 X_2+\nabla_2 X_2=0 \ \ \rightarrow \ \ \nabla_2 X_2=\frac{2\,\left( \frac{\partial K_y}{\partial \,y}\,\right) \,{x}^{2}+K_x}{2\,{x}^{2}}=0
[/tex]

[tex]
\nabla_1 X_2+\nabla_2 X_1=\frac{2\,\left( \frac{\partial K_y}{\partial \,y}\,\right) \,x-K_y}{2\,x}+\frac{2\,\left( \frac{\partial K_x}{\partial \,y}\,\right) \,x-K_y}{2\,x}=0
[/tex]

which come from the diagonal and off-diagonal elements respectively.

But I think I must have made a mistake because I don't think [itex]\vec{K}=\partial_y[/itex] is a solution.
 
Last edited:
  • #3
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I have also thought about it some more. It is obvious that (del/del y) is a solution because the metric tensor components are either 0 or functions of x. Thus if you use that solution you will get 0 for the lie derivative, making it a Killing vector. But I would like to know how you get there mathematically?

Like if you were finding the killing vectors of euclidian 3 space. Ie with a line element (ds)^2 = (dx)^2 + (dy)^2 + (dz)^2.

Here you know straight away three solutions are (del/del x), (del/del y) and (del/del z). But there are 3 more. How do you get the rest????

Thanks
 
  • #4
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292
Whoops, I accidentally posted a half-finished reply.

(The latex does not work well in itex tags

[tex]
\vec{K}
[/tex]

is unreadable )
 
  • #5
bcrowell
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I have also thought about it some more. It is obvious that (del/del y) is a solution because the metric tensor components are either 0 or functions of x. Thus if you use that solution you will get 0 for the lie derivative, making it a Killing vector. But I would like to know how you get there mathematically?

I don't see any problem with this argument that d/dy is a Killing vector. What's a little harder is to prove that there are not any more Killing vectors.

The regions x<0, x=0, and x>0 have intrinsic properties that differ from one another, since the signature of the metric is an intrinsic property. This means that it has a lower symmetry than the Euclidean plane, so it definitely can't have three Killing vectors.

But I don't know how to prove that there aren't two, without doing the grotty Lie derivatives, which would presumably give some differential equations that you'd have to solve.
 
  • #7
bcrowell
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As an alternative to solving the diffeqs, here's a fairly simple way to prove the result.

Here is some maxima code that calculates the scalar curvature for this metric:
Code:
load(ctensor);
dim:2;
ct_coords:[x,y];
lg:matrix([x^2,0],[0,x]);
cmetric();
R:scurvature(); /* scalar curvature */
The result is that the scalar curvature is [itex]3/2x^4[/itex]. A Killing vector flow can never flow from a point that has certain intrinsic properties to a point that has different intrinsic properties. Since the scalar curvature depends on x and is intrinsic, this implies that any Killing vector [itex]\xi[/itex] must have [itex]\xi_x=0[/itex]. The Killing equation then becomes [itex]\nabla_x\xi_y=\nabla_y\xi_y=0[/itex]. These equations constrain both [itex]\partial_x\xi_y[/itex] and [itex]\partial_y\xi_y[/itex], which means that given a value of [itex]\xi_y[/itex] at some point in the plane, its value everywhere else is determined. Therefore the only possible Killing vectors are scalar multiples of the Killing vector already found.
 
  • #8
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Thanks!
 

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