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}):

\[\begin{pmatrix}<br /> {ct}&#039;\\ <br /> {x}&#039;\\ <br /> {y}&#039;\\ <br /> {z}&#039;<br /> \end{pmatrix}=\begin{pmatrix}<br /> \gamma &amp; -\gamma \beta _{x} &amp; -\gamma \beta _{y} &amp; -\gamma \beta _{z}\\ <br /> -\gamma \beta _{x}&amp; 1+\alpha \beta ^{2}_{x} &amp; \alpha \beta _{x}\beta _{y} &amp; \alpha \beta _{x}\beta _{z} \\ <br /> -\gamma \beta _{y}&amp; \alpha \beta _{y}\beta _{x} &amp; 1+\alpha \beta ^{2}_{y} &amp; \alpha \beta _{y}\beta _{z} \\ <br /> -\gamma \beta _{z}&amp; \alpha \beta _{z}\beta _{x} &amp; \alpha \beta _{z}\beta _{y} &amp; 1+\alpha \beta ^{2}_{z}<br /> \end{pmatrix}\begin{pmatrix}<br /> ct\\ <br /> x\\ <br /> y\\ <br /> z<br /> \end{pmatrix}\]

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
\[{ct}&#039;=\gamma _{x}(ct-\beta _{x}x) \]
\[{x}&#039;=\gamma _{x}(x-\beta _{x}ct) \]
\[{y}&#039;=y \]
\[{z}&#039;=z \]

For S''->S'
\[{ct}&#039;&#039;=\gamma _{y}({ct}&#039;-\beta _{y}{y}&#039;) \]
\[{x}&#039;&#039;={x}&#039; \]
\[{y}&#039;&#039;=\gamma _{y}({y}&#039;-\beta _{y}{ct}&#039;) \]
\[{z}&#039;&#039;={z}&#039; \]

For S'''->S''
\[{ct}&#039;&#039;&#039;=\gamma _{z}({ct}&#039;&#039;-\beta _{z}{z}&#039;&#039;) \]
\[{x}&#039;&#039;&#039;={x}&#039;&#039; \]
\[{y}&#039;&#039;&#039;={y}&#039;&#039; \]
\[{z}&#039;&#039;&#039;=\gamma _{z}({z}&#039;&#039;-\beta _{z}{ct}&#039;&#039;) \]

But when I substitute in I get:

\[{ct}&#039;&#039;&#039;=\gamma _{x} \gamma _{y}\gamma _{z}ct-\gamma _{x} \gamma _{y}\gamma _{z}\beta _{x}x-\gamma _{y}\gamma _{z}\beta _{y}y-\gamma _{z}\beta _{z}z\]
\[{x}&#039;&#039;&#039;=-\gamma _{x}\beta _{x}ct+\gamma _{x}x\]
\[{y}&#039;&#039;&#039;=-\gamma _{x}\gamma _{y}\beta _{y}ct+\gamma _{x}\gamma _{y}\beta _{x}\beta_{y}x+\gamma _{y}y\]
\[{z}&#039;&#039;&#039;=\gamma _{x}\gamma _{y}\beta _{z}ct+\gamma _{x}\gamma _{y}\beta _{x}\beta _{z}x+\beta _{y}\beta _{z}y+\gamma _{z}z\]

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!