What is the relationship between timelike and spacelike vectors?

[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?

 

[vela]

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.

 

[latentcorpse]

well that gives P^\mu P_\mu + 2P^\mu S_\mu + S^\mu S_\mu[\itex]<br /> <br /> but P^2&lt;0 and P.S=0 <br /> <br /> I don&#039;t know what (P+S)^2 is though so I can&#039;t say anything constructive about the nature of S^2<br /> <br /> Does (P+S)^2 have to equal zero? If so, why?

 

[vela]

Sorry, brain fart on my part. Let me think about it.

 

[vela]

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.

 

[latentcorpse]

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.

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 (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 ?

 

[vela]

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

 

[latentcorpse]

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

ok so we know we can find the frame that makes P&#039;^\mu=(p&#039;^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&#039;^\mu S&#039;_\mu = -P&#039;^0 S&#039;^0 =0

In order for this to vanish, we must have S&#039;^0=0

However, in order for S' to be a non-zero vector, we must have S&#039;^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?

 

[vela]

Yup.