Timelike, Spacelike and Null

1. Feb 15, 2012

latentcorpse

I've forgotten a lot of material and have been trying to go over it to get back up to speed. However, I can't figure out this one:

1, If $P^\mu$ is timelike and $P^\mu S_\mu=0$ then $S^\mu$ is spacelike.

I said that taking $\eta_{\mu \nu} = (-1,+1,+1,+1),$ we could deduce that $P^2=P^0P^0-P^iP^i.$ For this to be timelike, P^2 must be negative and so we can deduce that $P^i>P^0.$

However, $P^\mu S_\mu=P^0S^0-P^iS^i=0.$

Given that $P^i>P^0$, the only way for this to be true is if $S^0>S^i \Rightarrow S^0S^0-S^iS^i>0$ and so $S^\mu$ would be spacelike.

The one problem is that I took $P^0P^0-P^iP^i<0$ to imply $P^0-P^i>0$ which I don't think it does since $P^\mu=(-4,1,1,1)$ contradicts this and there is no requirement in the question that either of the vectors be future directed.

What am I doing wrong?

2, If $P^\mu$ and $Q^\mu$ are timelike and $P^\mu Q_\mu<0$ then either both are future directed or both are past directed. Again I proceeded in a similar fashion to above but seeing as I am unconvinced about that, I would like to seek some help for this also.

Thanks a lot.

P.S. How come no combination of LaTeX tags that I try ever works in this forum anymore?

Last edited by a moderator: Feb 15, 2012
2. Feb 15, 2012

vela

Staff Emeritus
Try calculating $(P^\mu+S^\mu)^2$. Your attempt fails for the reason you already noted. The implication you relied on isn't true.

Fixed your LaTeX for you. I changed the TEX tags to ITEX so it wasn't so spread out too.

3. Feb 15, 2012

well that gives $P^\mu P_\mu + 2P^\mu S_\mu + S^\mu S_\mu[\itex] but P^2<0 and P.S=0 I don't know what (P+S)^2 is though so I can't say anything constructive about the nature of S^2 Does (P+S)^2 have to equal zero? If so, why? 4. Feb 15, 2012 vela Staff Emeritus Sorry, brain fart on my part. Let me think about it. 5. Feb 16, 2012 vela Staff Emeritus Since $P^\mu$ is timelike, you should be able to find a frame where $P'^\mu = \Lambda^\mu{}_\nu P^\nu$ has the form (P'0, 0, 0, 0). Try using that. 6. Feb 16, 2012 latentcorpse Hmmm. Afraid I'm not seeing it still. I get (P+S)^2=P^2+2P.S+S^2 = P^2+S^2 since the cross term vanishes (from the info in the question) This means (P+S)^2=P^2+S^2 What is [itex](P+S)^\mu$?

Well, we evaluate in the primed frame you suggest and find

$(P+S)^\mu=(P^0+S^0,S^1,S^2,S^3)$ where I've suppressed all the primes

Now we square it wrt $\eta_{\mu \nu} = (-1,1,1,1)$

$(P+S)^2=-(P^0+S^0)^2 +\displaystyle\sum_{i=1}^3 (S^i)^2$

Even expanding the bracket, I cannot arrive at anything helpful ???

7. Feb 16, 2012

vela

Staff Emeritus
Oh, sorry, I meant forget my former suggestion and just analyze $P'^\mu S'_\mu=0$.

8. Feb 16, 2012

latentcorpse

ok so we know we can find the frame that makes $P'^\mu=(p'^0,0,0,0)$ because it is timelike and therefore there exists a coordinate system such that any two points along the integral curve of P will be causally connected and therefore have no spatial separation.

This means that $P'^\mu S'_\mu = -P'^0 S'^0 =0$

In order for this to vanish, we must have $S'^0=0$

However, in order for S' to be a non-zero vector, we must have $S'^i \neq 0$ for some $i \in \{ 1,2,3 \}$

This means S' will be spacelike in the primed frame and since the causal nature of a vector is unaffected by Lorentz transformations, S will still be spacelike in the unprimed frame, yes?

Last edited: Feb 16, 2012
9. Feb 16, 2012

vela

Staff Emeritus
Yup.