Proving the dor product of 4-vectors is Lorentz invariant

[chipotleaway]

Homework Statement

Let A and B be 4-vectors. Show that the dot product of A and B is Lorentz invariant.

The Attempt at a Solution

Should I be trying to show that A.B=\gamma(A.B)?

Thanks

 

[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

 

[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

 

[tiny-tim]

hi chipotleaway!

sorry, I'm completely confused …

what is t_a ? ​

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

 

[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 .

 

[tiny-tim]

hi chipotleaway!

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 .

yes that's ok …

though you really ought to say that you're choosing the "1" coordinate to be in the direction of u !

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)

yes, using ct_a instead a_o is probably a good idea, since it makes the dot product easier

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

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)

 

[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 A'?

 

[vela]

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)

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.