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Space-like and time-like vectors

  1. Dec 27, 2015 #1
    1. The problem statement, all variables and given/known data

    Prove the following:

    a) If ##P^a## is time-like and ##P^{a}S_{a}=0##, then ##S^{a}## is space-like.

    b) If ##P^a## is null and ##P^{a}S^{a}=0##, then ##S_{a}## is space-like or ##S^{a} \propto P^{a}##.

    2. Relevant equations

    Using the 'mostly minus' convention, ##A^a## is time-like, null and space-like if ##A^{a}A_{a}## is ##>0,=0, <0## respectively.

    3. The attempt at a solution

    a) ##S^{a}S_{a} = S^{a}S_{a} + P^{a}S_{a} = (S^{a} + P^{a})S_{a} = (S^{a} + P^{a})(S_{a} + P_{a}) = S^{a}S_{a} + P^{a}S_{a} + P_{a}S^{a} + P^{a}P_{a} = S^{a}S_{a} + P^{a}P_{a} = ???##
     
  2. jcsd
  3. Dec 27, 2015 #2

    TSny

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    How do you justify the last equality shown above?

    Note that ##P^{a}S_{a}## is an invariant. It has the same value in all reference frames. Try to pick a nice reference frame using what you know about ##P^{a}##.
     
  4. Dec 28, 2015 #3
    Well, in the 'mostly minus' convention, ##P^{a} = \{1,0,0,0 \}## is one possibility.

    Therefore, ##P^{a}S_{a} = 0## implies ##S_{0} = 0##.

    Therefore, ##S^{a}S_{a} = S^{0}S_{0} + S^{1}S_{1} + S^{2}S_{2} + S^{3}S_{3} = S^{1}S_{1} + S^{2}S_{2} + S^{3}S_{3} = - S_{1}S_{1} - S_{2}S_{2} - S_{3}S_{3}##.

    ##S_{i}S_{i} >0##, so that ##S^{a}S_{a} < 0##, therefore ##S^{a}## is space-like.
     
  5. Dec 28, 2015 #4
    b) In the 'mostly minus' convention, ##P^{a} = \{-1,-1,0,0 \}## is one possibility.

    Therefore, ##P^{a}S_{a} = 0## implies ##S_{0}+S_{1} = 0##, so that ##S^{0}=S_{0}=-S_{1}##.

    Therefore, ##S^{a}S_{a} = S^{0}S_{0} + S^{1}S_{1} + S^{2}S_{2} + S^{3}S_{3} = S_{1}S_{1} - S_{1}S_{1} - S_{2}S_{2} - S_{3}S_{3} = - S_{2}S_{2} - S_{3}S_{3}##.

    ##S_{i}S_{i} >0##, so that ##S^{a}S_{a} < 0##, therefore ##S^{a}## is space-like.

    The other possibility is that ##P^{a} \propto S^{a}## for obvious reasons.
     
  6. Dec 28, 2015 #5
    Are my solutions correct?
     
  7. Dec 28, 2015 #6

    TSny

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    Yes, it's a possibility. But you want a general proof. You have the right idea, but you can't assume that P0 = 1 in the frame where all the Pi = 0. Similarly for part (b) where you assumed that P0 and P1 are both equal to -1.
     
  8. Dec 28, 2015 #7
    In that case, does ##P^{a} = \{p,0,0,0\}## for part (a) and ##P^{a}=\{p,-p,0,0\}## for part (b) qualify as four-vectors sufficiently general to give a general proof?
     
  9. Dec 28, 2015 #8

    TSny

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    OK. For part (b), you could rotate the spatial coordinate axes to get ##P^{a}=\{p,p,0,0\}## (without any negative signs). But I think the way you wrote it is OK also.

    In post 4 you had
    Logically, can you really claim that ##S^{a}S_{a} < 0## follows from what you have written? Or should you conclude only that ##S^{a}S_{a} \leq 0##?
     
  10. Dec 28, 2015 #9
    Yes, that proves that ##S^{a}## is space-like or null for part (b)?

    But, by the same token, can we also not conclude that ##S^{a}## is space-like or null (and not just space-like) for part (a)?
     
  11. Dec 28, 2015 #10

    TSny

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    Yes. Also, you should go on to show that if S is null, then it must be proportional to P.

    No, you should be able to prove that in (a), S must be space-like. It can't be null. [Of course, S is assumed to be a nonzero 4-vector in both (a) and (b).]
     
  12. Dec 28, 2015 #11
    For the case when both ##S^{a}S_{0}=0## and ##P^{a}P_{a}=0##, there are probably many solutions that relate ##S^{a}## and ##P^{a}##, but one of them is surely ##S^{a}=cP^{a}##. What are the other solutions that could arise?

    Ah! If ##S^{a}## is a non-zero four vector in (a), then ##S_{i}S_{i}>0## in (a). Got it!
     
  13. Dec 28, 2015 #12

    TSny

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    You need to show that there are no other solutions when S is null. That is, if S and P are null and ##P^{a}S_{a}=0##, then show that S must be proportional to P.
     
  14. Dec 28, 2015 #13
    Well, ##S^{a}S_{a}=0## means that ##(S_{0})^{2}=(\vec{S})^{2}## and ##P^{a}P_{a}=0## means that ##(P_{0})^{2}=(\vec{P})^{2}##.

    Now, let's divide and obtain ##\frac{(S_{0})^{2}}{(P_{0})^{2}} = \frac{(\vec{S})^{2}}{(\vec{P})^{2}}##.

    Call the ratio ##\frac{(S_{0})^{2}}{(P_{0})^{2}} = c^{2}##.

    Therefore, ##c^{2} = \frac{(\vec{S})^{2}}{(\vec{P})^{2}} \implies (\vec{S})^{2} = c^{2} (\vec{P})^{2}##.

    Therefore, component-wise, the four-vectors are proportional.

    Is this correct?
     
  15. Dec 28, 2015 #14

    TSny

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    OK.

    You have not shown that the individual components of S are proportional to the corresponding individual components of P (with the same proportionality constant). For example, everything you stated above could be satisfied with ##(P^0, P^1, P^2, P^3) = (5, 0, 3, 4)## and ##(S^0, S^1, S^2, S^3) = (5, 0, 5, 0)##.

    Note that you did not use ##P^aS_a = 0## in your argument above.
     
  16. Dec 30, 2015 #15
    Alright, let me start from scratch.

    I need to show that if ##S^{a}## and ##P^{a}## are null, and if ##P^{a}S_{a} = 0##, then ##S^{a} \propto P^{a}##.

    Here's my proof.

    ##S^{a}## is null ##\implies (S_{0})^{2}=(\vec{S})^{2}##.
    ##P^{a}## is null ##\implies (P_{0})^{2}=(\vec{P})^{2}##.

    ##P^{a}S_{a}=0 \implies P_{0}S_{0}=(\vec{P}\cdot{\vec{S}}) \implies \sqrt{(\vec{P})^{2}(\vec{S})^{2}}=(\vec{P}\cdot{\vec{S}})##.

    Alluding to the rule for dot product in Euclidean geometry, ##P^{i} = k S^{i}## for ##i=1,2,3##, where ##k## is a positive real number.

    Therefore, ##(\vec{P})^{2} = k^{2} (\vec{S})^{2} \implies (P_{0})^{2} = k^{2} (S_{0})^{2} \implies P_{0}=kS_{0}##.

    What do you think?
     
  17. Dec 30, 2015 #16

    TSny

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    In getting to the last equation, can you assume that ##P_{0}S_{0} = \sqrt{(\vec{P})^{2}(\vec{S})^{2}}## or only that ##P_{0}S_{0} = \pm \sqrt{(\vec{P})^{2}(\vec{S})^{2}}##?

    Does ##k## necessarily have to be positive?

    From this argument, you can only conclude that ##P_{0}= \pm kS_{0}##.

    I like this approach. It does not require going to a specific reference frame. You just need to take care of the sign issue.

    You can use this approach to also prove part (a) without using a specific frame.
     
  18. Dec 30, 2015 #17
    Well, ##(S_{0})^{2}=(\vec{S})^{2} \implies S_{0} = \pm \sqrt{\vec{S}^{2}}## and ##(P_{0})^{2}=(\vec{P})^{2} \implies P_{0} = \pm \sqrt{\vec{P}^{2}}##.

    Therefore, ##P_{0}S_{0} = \sqrt{(\vec{P})^{2}(\vec{S})^{2}}##, don't you think?
     
  19. Dec 30, 2015 #18

    TSny

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    The ##\pm## sign in ##S_{0} = \pm \sqrt{\vec{S}^{2}}## is not necessarily "correlated" with the ##\pm## sign in ##P_{0} = \pm \sqrt{\vec{P}^{2}}## .

    There is no reason why ##P_{0}S_{0}## must be positive.
     
  20. Dec 30, 2015 #19
    Let ##P^{i}=cS^{i}##, where ##c## is a non-zero real number.

    Therefore, ##P^{a}S_{a}=0 \implies P^{0}S_{0}+P^{i}S_{i} =0 \implies P^{0}S_{0}+cS^{i}S_{i} = 0 \implies P^{0}S_{0}-cS_{i}S_{i} = 0 \implies P^{0}S_{0}-c(S_{0})^{2}=0 \implies S_{0}(P_{0}-cS_{0})=0 \implies S_{0}=0\ \text{or}\ P_{0}=cS_{0} \implies S_{0}=0\ \text{or}\ P^{0}=cS^{0}##.

    Now, ##S_{0}=0 \implies (\vec{S})^{2} = (S_{0})^{2}=0##, but ##S^{a}## cannot be a zero vector.

    Therefore, ##P^{a} \propto S^{a}##.

    Is this correct?
     
  21. Dec 30, 2015 #20

    TSny

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    That looks good to me.
     
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