# Show that this vector is timelike?

Hi guys,

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

$v^{\mu}=t^{\mu}-\frac{t^{\mu}\cdot X^{\mu\prime}}{X^{\mu\prime}\cdot X^{\mu\prime}}X^{\mu\prime}$,

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

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

$v_{\mu}v^{\mu}=t^{2}-2\frac{(t\cdot X')^{2}}{X^{\prime}\cdot X^{\prime}}+(t\cdot X^{\prime})^{2}$

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

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.

Yes doctor that is wrong - fixed it and now I get this:

$v_{\mu}v^{\mu}=t^{2}-\frac{(t\cdot X^{\prime})^{2}}{X^{\prime}\cdot X^{\prime}}$ and... is this equal to 0? :O

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.

I dont know how the signs could change. But is the maths right? I mean is that the true expression?

Nevermind I think i got it -- because X is spacelike, its square is negative right :O so i get something positive overall?

Orodruin
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Correct. Also ##(t\cdot X)^2## is just the square of a number and thus also positive.

Thank you doctor once again :) while we're at it could you please also confirm / deny the following statement:

$(t\cdot X^{\prime})^{2} = t^{2}X^{\prime 2}$?

Orodruin
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Not true. The left hand side is positive and the right hand side is negative.

I thought so. Thank you doctor :D you're a life saver lol ! (no pun intended)