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Timelike, Spacelike and Null

  1. Feb 15, 2012 #1
    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 [itex]P^\mu[/itex] is timelike and [itex]P^\mu S_\mu=0[/itex] then [itex]S^\mu[/itex] is spacelike.

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

    However, [itex]P^\mu S_\mu=P^0S^0-P^iS^i=0.[/itex]

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

    The one problem is that I took [itex]P^0P^0-P^iP^i<0[/itex] to imply [itex]P^0-P^i>0[/itex] which I don't think it does since [itex]P^\mu=(-4,1,1,1)[/itex] 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 [itex]P^\mu[/itex] and [itex]Q^\mu[/itex] are timelike and [itex]P^\mu Q_\mu<0[/itex] 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. jcsd
  3. Feb 15, 2012 #2

    vela

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    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.
     
  4. Feb 15, 2012 #3
    well that gives [itex]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?
     
  5. Feb 15, 2012 #4

    vela

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    Sorry, brain fart on my part. Let me think about it.
     
  6. Feb 16, 2012 #5

    vela

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    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.
     
  7. Feb 16, 2012 #6
    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[/itex]?

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

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

    Now we square it wrt [itex]\eta_{\mu \nu} = (-1,1,1,1)[/itex]

    [itex](P+S)^2=-(P^0+S^0)^2 +\displaystyle\sum_{i=1}^3 (S^i)^2 [/itex]

    Even expanding the bracket, I cannot arrive at anything helpful ???
     
  8. Feb 16, 2012 #7

    vela

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    Oh, sorry, I meant forget my former suggestion and just analyze ##P'^\mu S'_\mu=0##.
     
  9. Feb 16, 2012 #8
    ok so we know we can find the frame that makes [itex]P'^\mu=(p'^0,0,0,0)[/itex] 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 [itex]P'^\mu S'_\mu = -P'^0 S'^0 =0[/itex]

    In order for this to vanish, we must have [itex]S'^0=0[/itex]

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

    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
  10. Feb 16, 2012 #9

    vela

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    Yup.
     
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