# Killing tensor in Minkowski space

I'm trying to solve the Killing tensor equation $\nabla_{(a}K_{bc)} = 0$ in Minkowski space.

I'd like to generalise the method we use to find Killing tensors in Minkowski space. We can take $\nabla_c$ derivatives of $\nabla_{(a} \xi_{b)}$ and then write out permutations of $a,b,c$ to find $\nabla_a \nabla_b \xi_c = 0$. I wonder if this is a good approach to solve the Killing tensor equation as well - unfortunately I wasn't able to make much progress.

As mentioned in the comments I'd also be interested to see how such Killing tensors decompose as tensor products of the known Killing vectors on Minkowski space.

• What do you mean by "right approach"? The right approach to solve an equation is the one that yields a solution. What other measure of "right" could you use? What is the conceptual physics question here? – ACuriousMind Jun 4 '16 at 9:28
• Okay, "right" is probably the wrong word, of course I mean is this a good/possible approach. The physics content is to understand what the Killing tensors are in Minkowski space and whether we get anything new, or if perhaps these Killing tensors just decompose as Killing vectors. I think that is an interesting question. – Wooster Jun 4 '16 at 10:06
• They do decompose as symmetrized tensor products of Killing vectors. So nothing new here. – Blazej Jun 4 '16 at 10:45
• Okay great! Well the question now is how do we show that then? – Wooster Jun 4 '16 at 10:46

## 1 Answer

They do decompose as symmetrized tensor products of Killing vectors. So nothing new here. I don't really remember the details but there are inequalities which tell how many "Killing objects" can a spacetime have, and this number is saturated for Minkowski and other maximally symmetric spaces. Class of spacetimes where you do expect to see nontrivial Killing tensors are Petrov type D spacetimes with less than four Killing vectors (so not stationary and spherically symmetric). This was shown in papers by Penrose and Wheeler. I think also Carter made some contributions to this so you can look in his papers also. You might also want to look at the books "Spinors and spacetime" by Penrose.

Standard example of spacetime with nontrivial Killing tensor is of course Kerr spacetime, but it is not the only one.

• Thank you for your answer - I appreciate the references, I will look through those, that is very helpful. I would still be interested in seeing details of how we show this though. – Wooster Jun 4 '16 at 11:12
• In general, Killing tensors are not given by symmetrizing Killing vectors. There are more general kind of Killing tensors. For example, Kerr geometry contains a Killing tensor which is not related to Killing vectors. – Frame Jun 4 '16 at 12:54