- #1

radonballoon

- 21

- 0

## Homework Statement

Show that [tex]a^{\mu}b_{\mu} \equiv -a^0b^0 + \vec{a} \bullet \vec{b} [/tex] is invariant under Lorentz transformations.

## Homework Equations

[tex]\Lambda_{\nu}^{\mu} \equiv \left(

\begin{array}{cccc}

\gamma & -\beta \gamma & 0 & 0 \\

-\beta \gamma & \gamma & 0 & 0 \\

0 & 0 & 1 & 0 \\

0 & 0 & 0 & 1

\end{array} \right)

[/tex]

[tex]

b^0 = -b_0

[/tex]

## The Attempt at a Solution

So the only way I could get it to be invariant was performing a lorentz transformation on each [tex]a^{\mu}b_{\mu}[/tex], squaring each, and then setting the first term negative and adding them. I don't know why this works, however.:

[tex]

\Lambda_{\nu}^{\mu}a^{\mu}b_{\mu} = \left(

\begin{array}{cccc}

\gamma & -\beta \gamma & 0 & 0 \\

-\beta \gamma & \gamma & 0 & 0 \\

0 & 0 & 1 & 0 \\

0 & 0 & 0 & 1

\end{array} \right) \left(

\begin{array}{c}

a^0b_0 \\

a^1b_1 \\

a^2b_2 \\

a^3b_3

\end{array} \right) = \left(

\begin{array}{c}

\gamma (a^0b_0 - \beta a^1b_1) \\

\gamma (a^1_b1 - \beta a^0b_0) \\

a^2b_2 \\

a^3b_3

\end{array} \right)

[/tex]

Squaring each term, negating first term squared, and summing:

[tex]

\gamma^2(-(a^0b_0)^2-\beta^2 (a^1b_1)^2 + 2\beta a^1b_1a^0b_0 + (a^1b_1)^2+\beta^2 (a^0b_0)^2 - 2\beta a^1b_1a^0b_0) + (a^2b_2)^2 + (a^3b_3)^2

[/tex]

Doubled terms cancel, and the others can be grouped so that they are multiplied by [tex]1-\beta^2[/tex] which cancels the [tex]\gamma^2[/tex], and you are left with the original equation again, albeit with terms squared.

Any help would be great!