I am actually having a little trouble with the anti-symmetric part of your answer. I understand that T_{[ab]} = \frac{1}{2}(T_{ab}-T_{ba}). But how do you expand when the square brackets go over more than one tensor? I.e. in K^aF_{a[v}K_{\sigma]} = 0?
Thanks again
I am starting my honours project on colliding plane gravitational waves and I am learning about the Petrov-Penrose classification of the Weyl tensor. I can't find any good explanation on what a principal null direction is.
Thanks
Chris
I have another problem that I am stuck on.
Show that a timelike vector cannot be orthogonal to a null vector.
Timelike:
X^{2} = g(sub a b)X^{a}X^{b} > 0
Null:
X^{2} = g(sub a b)X^{a}X^{b} = 0
In order for them to be orthogonal...
g(sub a b)X^{a}Y^{a} = 0
I know from the line...
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...
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...