# Deriving the Lorentz Boost for an Arbitrary Direction

## Homework Statement

So, I'm working through a relativity book and I'm having trouble deriving the Lorentz transformation for an arbitrary direction $$v=(v_{x},v_{y},v_{z})$$:

$$$\begin{pmatrix} {ct}'\\ {x}'\\ {y}'\\ {z}' \end{pmatrix}=\begin{pmatrix} \gamma & -\gamma \beta _{x} & -\gamma \beta _{y} & -\gamma \beta _{z}\\ -\gamma \beta _{x}& 1+\alpha \beta ^{2}_{x} & \alpha \beta _{x}\beta _{y} & \alpha \beta _{x}\beta _{z} \\ -\gamma \beta _{y}& \alpha \beta _{y}\beta _{x} & 1+\alpha \beta ^{2}_{y} & \alpha \beta _{y}\beta _{z} \\ -\gamma \beta _{z}& \alpha \beta _{z}\beta _{x} & \alpha \beta _{z}\beta _{y} & 1+\alpha \beta ^{2}_{z} \end{pmatrix}\begin{pmatrix} ct\\ x\\ y\\ z \end{pmatrix}$$$

where $$$\beta =\frac{v}{c}$$$, $$$\gamma =(1-\beta ^{2})^{-1/2}$$$ and $$$\alpha =\frac{\gamma -1}{\beta ^{2}}$$$

## The Attempt at a Solution

I thought the best way to approach it would be to define four reference frames: S, S', S'' and S'''. Where S' is related to S by a boost in the x direction, S'' is related to S' by a boost in the y' direction and S''' is related to S'' by a boost in the z'' direction. This produces the transformations:

For S'->S
$$${ct}'=\gamma _{x}(ct-\beta _{x}x)$$$
$$${x}'=\gamma _{x}(x-\beta _{x}ct)$$$
$$${y}'=y$$$
$$${z}'=z$$$

For S''->S'
$$${ct}''=\gamma _{y}({ct}'-\beta _{y}{y}')$$$
$$${x}''={x}'$$$
$$${y}''=\gamma _{y}({y}'-\beta _{y}{ct}')$$$
$$${z}''={z}'$$$

For S'''->S''
$$${ct}'''=\gamma _{z}({ct}''-\beta _{z}{z}'')$$$
$$${x}'''={x}''$$$
$$${y}'''={y}''$$$
$$${z}'''=\gamma _{z}({z}''-\beta _{z}{ct}'')$$$

But when I substitute in I get:

$$${ct}'''=\gamma _{x} \gamma _{y}\gamma _{z}ct-\gamma _{x} \gamma _{y}\gamma _{z}\beta _{x}x-\gamma _{y}\gamma _{z}\beta _{y}y-\gamma _{z}\beta _{z}z$$$
$$${x}'''=-\gamma _{x}\beta _{x}ct+\gamma _{x}x$$$
$$${y}'''=-\gamma _{x}\gamma _{y}\beta _{y}ct+\gamma _{x}\gamma _{y}\beta _{x}\beta_{y}x+\gamma _{y}y$$$
$$${z}'''=\gamma _{x}\gamma _{y}\beta _{z}ct+\gamma _{x}\gamma _{y}\beta _{x}\beta _{z}x+\beta _{y}\beta _{z}y+\gamma _{z}z$$$

I suppose I would need to eliminate $$$\gamma _{x}$$$, $$$\gamma _{y}$$$ and $$$\gamma _{z}$$$ in favor of $$$\gamma$$$ but I guess I would just like to make sure I'm headed in the right direction before I undertake the calculation since:
$$$\gamma _{x}\gamma _{y}\gamma _{z}=\frac{1}{\sqrt{(1-\beta _{x}^{2})(1-\beta _{y}^{2})(1-\beta _{z}^{2})}}=(\frac{1}{\gamma ^{2}}+\beta _{x}^{2}\beta _{y}^{2}+\beta _{x}^{2}\beta _{z}^{2}+\beta _{y}^{2}\beta _{z}^{2}-\beta _{x}^{2}\beta _{y}^{2}\beta _{z}^{2})^{-1/2}$$$

Btw, I don't need a complete solution, I would just like to see if I have the right concept down to solve this problem.

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vela
Staff Emeritus
Homework Helper
I suppose I would need to eliminate $$$\gamma _{x}$$$, $$$\gamma _{y}$$$ and $$$\gamma _{z}$$$ in favor of $$$\gamma$$$ but I guess I would just like to make sure I'm headed in the right direction
I'm pretty sure this isn't the correct approach. A single boost to (vx, vy, vz) isn't the same as the product of the separate three boosts. After the first boost, for instance, you no longer have t'=t, so vy and vz would be different in S', and so on.

I'd try a single boost followed by a spatial rotation.

Yeah I think I figured it out, I don't know what I was thinking since boosts aren't commutative or even closed. Thanks a lot for replying!