# Killing Vector Question

purakanui
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

Mentz114
I'm learning this sort of thing myself, so I thought I'd give it try.

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

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

The Killing equations are

$$\nabla_1 X_1+\nabla_1 X_1=0 \ \ \rightarrow \ \ \nabla_1 X_1=\frac{\left( \frac{\partial K_x}{\partial \,x}\,\right) \,x-K_x}{x}=0$$

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

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

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

But I think I must have made a mistake because I don't think $\vec{K}=\partial_y$ is a solution.

Last edited:
purakanui
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

Mentz114
Whoops, I accidentally posted a half-finished reply.

(The latex does not work well in itex tags

$$\vec{K}$$

Staff Emeritus
Gold Member
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.

Staff Emeritus
Gold Member
Staff Emeritus
Gold Member
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 $3/2x^4$. 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 $\xi$ must have $\xi_x=0$. The Killing equation then becomes $\nabla_x\xi_y=\nabla_y\xi_y=0$. These equations constrain both $\partial_x\xi_y$ and $\partial_y\xi_y$, which means that given a value of $\xi_y$ 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.

purakanui
Thanks!