Deriving the Lorentz Boost for an Arbitrary Direction

[grindfreak]

Homework Statement

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

$$\[\begin{pmatrix}<br /> {ct}'\\ <br /> {x}'\\ <br /> {y}'\\ <br /> {z}'<br /> \end{pmatrix}=\begin{pmatrix}<br /> \gamma & -\gamma \beta _{x} & -\gamma \beta _{y} & -\gamma \beta _{z}\\ <br /> -\gamma \beta _{x}& 1+\alpha \beta ^{2}_{x} & \alpha \beta _{x}\beta _{y} & \alpha \beta _{x}\beta _{z} \\ <br /> -\gamma \beta _{y}& \alpha \beta _{y}\beta _{x} & 1+\alpha \beta ^{2}_{y} & \alpha \beta _{y}\beta _{z} \\ <br /> -\gamma \beta _{z}& \alpha \beta _{z}\beta _{x} & \alpha \beta _{z}\beta _{y} & 1+\alpha \beta ^{2}_{z}<br /> \end{pmatrix}\begin{pmatrix}<br /> ct\\ <br /> x\\ <br /> y\\ <br /> z<br /> \end{pmatrix}\]$$

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.

 

[vela]

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 (v_x, v_y, v_z) 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 v_y and v_z would be different in S', and so on.

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

 

[grindfreak]

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!