Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I want to derive the Weyl tensor along the lines of this derivation, but I am unable to complete it. (I am only interested in $4$ dimension for now.)

Every contraction I perform gives either $0=R + 3 \lambda L$ or $0=0$ but not an additional equation to calculate $L$ and $\lambda$.

Does anyone know a source where I can find this derivation in detail or point me to the crucial point I am missing?

share|cite|improve this question
up vote 1 down vote accepted

First, drop $\lambda$, because it is useless (can be absorbed into $L$). Second, try contracting in $a$ and $c$ only. This contraction should also be zero.

share|cite|improve this answer
Ok, then I obviously get $0=R + 3 L$, which i can write as $0=R_{bd}g^{bd} + 3 L_{bd}g^{bd}$, but how to continue now? – Michal May 17 '13 at 21:55
Don't do double contractions. You can get more information by just doing the single contraction $C_{abcd}g^{ac}=0$. – Edward Hughes May 17 '13 at 22:10
You mean, that I should stop at $0=R_{bd} + \frac{1}{2}\bigl(2 L_{bd} + g_{bd}L\bigr)$ and not contract then with $g^{bd}$? – Michal May 18 '13 at 10:04
@Michal, yes. If you now recall your $R+3L=0$, you should be able to find $L_{bd}$. – Peter Kravchuk May 18 '13 at 19:42
Thanks, I should have seen that! – Michal May 19 '13 at 9:04

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.