Killing Vector Solutions for General Relativity Metric | Self-Study Tips

In summary: That makes a lot of sense. So just to clarify, the scalar curvature being dependent on x and \xi_x=0 means that the vector \vec{K}=K_y\partial_y is the only possible Killing vector. And since it satisfies the Killing equation, it is indeed a Killing vector. In summary, the conversation discusses finding Killing Vector solutions for a given metric and the steps involved in solving for them. It is shown that in the case of the given metric, the only possible Killing vector is a scalar multiple of the vector (del/del y). This solution is confirmed to satisfy the Killing equation and is therefore a valid Killing vector.
  • #1
purakanui
10
0
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
 
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  • #2
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
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
Whoops, I accidentally posted a half-finished reply.

(The latex does not work well in itex tags

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

is unreadable )
 
  • #5
purakanui said:
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.
 
  • #6
Instead of using of using 7.52, set 7.51 equal to zero, which will
bcrowell said:
give some differential equations that you'd have to solve.
and which I have solved. For another example of Killing vectors, see

https://www.physicsforums.com/showthread.php?t=454237.
 
  • #7
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
Thanks!
 

1. What is a Killing vector in general relativity?

A Killing vector is a vector field that satisfies the Killing equation, which states that the Lie derivative of the metric tensor along the vector field is equal to zero. In simpler terms, it is a vector that preserves the geometry of a spacetime metric, meaning that it does not change the distance between two points in the spacetime.

2. What is the significance of Killing vectors in general relativity?

Killing vectors play a crucial role in understanding the symmetries and conservation laws in general relativity. They represent the infinitesimal generators of isometries, which are transformations that preserve the metric of a spacetime. Additionally, the existence of Killing vectors can simplify the calculations of certain physical quantities, such as energy and momentum.

3. How are Killing vectors used to solve the general relativity metric?

Killing vectors can be used to determine the form of the metric tensor for a given spacetime. By finding the Killing vectors of a spacetime, one can determine the symmetries and isometries of the spacetime and use them to construct the metric tensor. This can greatly simplify the process of solving the Einstein field equations.

4. What are some common applications of Killing vectors in general relativity?

Killing vectors have various applications in general relativity, including finding and classifying exact solutions to the Einstein field equations, calculating conserved quantities in spacetimes with symmetries, and studying the global structure of spacetimes.

5. How can I improve my understanding of Killing vector solutions in general relativity?

To improve your understanding of Killing vector solutions in general relativity, it is important to practice solving problems and working through examples. Additionally, exploring the geometrical interpretations of Killing vectors and their applications in different spacetimes can also deepen your understanding. Self-study resources such as textbooks, online lectures, and practice problems can also be helpful in improving your understanding of this topic.

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