# Proving the dor product of 4-vectors is Lorentz invariant

1. Feb 13, 2014

### chipotleaway

1. The problem statement, all variables and given/known data
Let A and B be 4-vectors. Show that the dot product of A and B is Lorentz invariant.

3. The attempt at a solution
Should I be trying to show that $A.B=\gamma(A.B)$?

Thanks

2. Feb 13, 2014

### tiny-tim

hi chipotleaway!

no, you should be applying the lorentz transformation to A and B, and showing that it doesn't affect A.B

3. Feb 18, 2014

### chipotleaway

Thanks tiny-tim, I've got it now but I had to make one assumption to get the result I wanted, namely if $A=(a_0, a_1, a_2, a_3)$ and $B=(b_0, b_1, b_2, b_3)$, I had to assume $a_0=ct_a$ and $b_0=ct_b$ which is only for 4-position vectors, so I think my result only applies for 4-vectors!

EDIT: Oh wait, I guess I must've assumed I was dealing with 4-position vectors because I started with
$A'=\gamma (a_0-\beta a_1, a_1 - ut_a, a_2, a_3)$ and $B'=\gamma (b_0-\beta b_1, b_1 - ut_b, b_2, b_3)$

EDIT 2: Added 3rd and 4th components to above expressions

Last edited: Feb 18, 2014
4. Feb 18, 2014

### tiny-tim

hi chipotleaway!

sorry, i'm completely confused …

what is ta ?

(and why do your final expressions only have two components instead of four?)

5. Feb 18, 2014

### chipotleaway

My bad, forgot to say what they were! $t_a$ is the time of the event of vector A (so, I think I've only done it for 4-position vectors, $A=(ct_a, a_1, a_2, a_3)$

I forgot to put the other two components in as I left it out in my working to save space (fixed now)...but we don't have to apply the transformation to those as well directions do we? Because I think we could just arrange out coordinate system so that the motion takes place in one direction .

6. Feb 18, 2014

### tiny-tim

hi chipotleaway!
yes that's ok …

though you really ought to say that you're choosing the "1" coordinate to be in the direction of u !
yes, using cta instead ao is probably a good idea, since it makes the dot product easier

but you can't then mix them both in the same expression …

7. Feb 18, 2014

### chipotleaway

Yeah, I should state everything I assume!

When you say 'mix them both in the same expression'...do you mean the timelike and spacelike components, like
$a_1-ut_a[[/tex] in [itex]A'$?

8. Feb 18, 2014

### vela

Staff Emeritus
Why do you have $a'_1 = \gamma(a_1-u t_a)$ instead of $a'_1 = \gamma(a_1-\beta a_0)$? Also, you shouldn't have $a'_i =\gamma a_i$ and $b'_i = \gamma b_i$ for i=2 and i=3.