# Lorentz Boosts in Group Representation (from Weinberg)

1. Jan 17, 2012

### Elwin.Martin

Alright, so excuse my ignorance, but I have no idea why the choice he uses for boosts is "convenient"

Just to make sure everyone is on the same metric etc etc.
Weinberg uses (-,+,+,+)
and God-given units

He requires that transformations..(oh my,,,how am I going to LaTeX this...)
$\Lambda^{\alpha}_{\gamma}\Lambda^{\beta}_{\delta} \eta_{\alpha \beta}\equiv \eta_{\gamma \delta}$

and he considers a particle in O frame at rest, that is at velocity v in fram O'
I understand how he arrives at
$\Lambda^{0}_{0}=\gamma$
and
$\Lambda^{i}_{0}=\gamma v_{i}$

(nevermind, that wasn't so bad for LaTeX-ing)

but he then says that it is convenient to use
$\Lambda^{i}_{j} = \delta_{ij}+ v_{i}v_{j}\frac{\gamma - 1}{\textbf{v}^{2}}$
and
$\Lambda^{0}_{j}=\gamma v_{j}$

Why is this convenient? I get that we have multiple representations for said boost because of the arbitrary rotations we may perform, but why is this helpful?

Thanks for any help

2. Jan 18, 2012

### Fredrik

Staff Emeritus
3. Jan 18, 2012

### torquil

Eq. 2.1.20 and 2.1.21 in his 1972-book "Gravitation", but it also appears expressed in terms of a massive particle momentum in Eq. 2.5.24 in his first QFT-book (1996). There are more details about the choice in the latter book, where I think he writes it as a decomposition $RBR^{-1}$ and uses this in the subsequent development.

4. Jan 18, 2012

### Fredrik

Staff Emeritus
Let B be the 3×3 matrix with $\Lambda^i{}_j$ on row i, column j. When $\Lambda$ is a pure boost in the 1 direction, we have
$$B=\begin{pmatrix}\gamma & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{pmatrix}.$$ Because of this, it seems likely that when $\Lambda$ is a pure boost with arbitrary velocity v, B will be such that for all x,
$$Bx=\gamma x_\parallel+x_\perp,$$ where $x_\parallel$ and $x_\perp$ are the unique vectors such that $x_\parallel$ is parallel to v, $x_\perp$ is orthogonal to v, and $x=x_\parallel+x_\perp$. We have
\begin{align}
x_\parallel &=\Big\langle\frac{v}{\|v\|},x\Big\rangle \frac{v}{\|v\|} =\frac{v^Tx}{\sqrt{v^Tv}}\frac{v}{\sqrt{v^Tv}} =\frac{vv^T}{v^Tv}x\\
x_\perp &= x-x_\parallel=\Big(I-\frac{vv^T}{vv^T}\Big)x.
\end{align} So
$$Bx =\gamma x_\parallel+x_\perp =\gamma\frac{vv^T}{v^Tv}x +\Big(I-\frac{vv^T}{vv^T}\Big)x=(\gamma-1)\frac{vv^T}{v^Tv}x+x.$$
Since this holds for all x, we have
$$B=I+(\gamma-1)\frac{vv^T}{v^Tv}.$$ If anyone can think of a way to make the "seems likely" part rigorous, let me know.

5. Jan 19, 2012

### Elwin.Martin

Thank you!

That was very clear, I feel kind of silly not seeing that at first, haha.

I will be sure to give people page numbers for references to standard texts in the future to help them in helping me (and making this clearer for anyone who might search this later).

Thanks again,
Elwin