# 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.