Show that this vector is timelike?

  • Thread starter Dixanadu
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  • #1
Dixanadu
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Hi guys,

So I have a vector which I need to show is timelike. The vector is

[itex] v^{\mu}=t^{\mu}-\frac{t^{\mu}\cdot X^{\mu\prime}}{X^{\mu\prime}\cdot X^{\mu\prime}}X^{\mu\prime} [/itex],

where [itex]t^{\mu}[/itex] is a timelike vector and [itex]X^{\mu\prime}[/itex] is spacelike, however these two vectors are not perpendicular so their dot product does not vanish.

I understand that in order to show that [itex]v^{\mu}[/itex] is timelike, I need to find [itex]v_{\mu}v^{\mu}[/itex] and show that this is greater than 0. So:

[itex]v_{\mu}v^{\mu}=t^{2}-2\frac{(t\cdot X')^{2}}{X^{\prime}\cdot X^{\prime}}+(t\cdot X^{\prime})^{2}[/itex]

So first of all I dont know if this is correct (yes I dont know how to take a dot product obviously lol) and secondly, even if it is, how do I show that this is greater than 0?

Thanks guys!
 

Answers and Replies

  • #2
Orodruin
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Your last term is wrong. Try to redo the inner product.

You should also look over your use of indices in the original problem statement.
 
  • #3
Dixanadu
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Yes doctor that is wrong - fixed it and now I get this:

[itex]v_{\mu}v^{\mu}=t^{2}-\frac{(t\cdot X^{\prime})^{2}}{X^{\prime}\cdot X^{\prime}}[/itex] and... is this equal to 0? :O
 
  • #4
Orodruin
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No, if it was zero then v would be light-like. I suggest you try to figure out the signs of both of these contributions.
 
  • #5
Dixanadu
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I dont know how the signs could change. But is the maths right? I mean is that the true expression?
 
  • #6
Dixanadu
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Nevermind I think i got it -- because X is spacelike, its square is negative right :O so i get something positive overall?
 
  • #7
Orodruin
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Correct. Also ##(t\cdot X)^2## is just the square of a number and thus also positive.
 
  • #8
Dixanadu
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Thank you doctor once again :) while we're at it could you please also confirm / deny the following statement:

[itex](t\cdot X^{\prime})^{2} = t^{2}X^{\prime 2}[/itex]?
 
  • #9
Orodruin
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Not true. The left hand side is positive and the right hand side is negative.
 
  • #10
Dixanadu
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I thought so. Thank you doctor :D you're a life saver lol ! (no pun intended)
 

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