# Show that this vector is timelike?

1. Feb 16, 2015

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!

2. Feb 16, 2015

### Orodruin

Staff Emeritus
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. Feb 16, 2015

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

4. Feb 16, 2015

### Orodruin

Staff Emeritus
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. Feb 16, 2015

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

6. Feb 16, 2015

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

7. Feb 16, 2015

### Orodruin

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

8. Feb 16, 2015

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

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

9. Feb 16, 2015

### Orodruin

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

10. Feb 16, 2015