- #1
grindfreak
- 39
- 2
Homework Statement
So, I'm working through a relativity book and I'm having trouble deriving the Lorentz transformation for an arbitrary direction [tex]v=(v_{x},v_{y},v_{z})[/tex]:
[tex]\[\begin{pmatrix}
{ct}'\\
{x}'\\
{y}'\\
{z}'
\end{pmatrix}=\begin{pmatrix}
\gamma & -\gamma \beta _{x} & -\gamma \beta _{y} & -\gamma \beta _{z}\\
-\gamma \beta _{x}& 1+\alpha \beta ^{2}_{x} & \alpha \beta _{x}\beta _{y} & \alpha \beta _{x}\beta _{z} \\
-\gamma \beta _{y}& \alpha \beta _{y}\beta _{x} & 1+\alpha \beta ^{2}_{y} & \alpha \beta _{y}\beta _{z} \\
-\gamma \beta _{z}& \alpha \beta _{z}\beta _{x} & \alpha \beta _{z}\beta _{y} & 1+\alpha \beta ^{2}_{z}
\end{pmatrix}\begin{pmatrix}
ct\\
x\\
y\\
z
\end{pmatrix}\][/tex]
where [tex]\[\beta =\frac{v}{c}\][/tex], [tex]\[\gamma =(1-\beta ^{2})^{-1/2}\][/tex] and [tex]\[\alpha =\frac{\gamma -1}{\beta ^{2}}\][/tex]
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
[tex]\[{ct}'=\gamma _{x}(ct-\beta _{x}x) \][/tex]
[tex]\[{x}'=\gamma _{x}(x-\beta _{x}ct) \][/tex]
[tex]\[{y}'=y \][/tex]
[tex]\[{z}'=z \][/tex]
For S''->S'
[tex]\[{ct}''=\gamma _{y}({ct}'-\beta _{y}{y}') \][/tex]
[tex]\[{x}''={x}' \][/tex]
[tex]\[{y}''=\gamma _{y}({y}'-\beta _{y}{ct}') \][/tex]
[tex]\[{z}''={z}' \][/tex]
For S'''->S''
[tex]\[{ct}'''=\gamma _{z}({ct}''-\beta _{z}{z}'') \][/tex]
[tex]\[{x}'''={x}'' \][/tex]
[tex]\[{y}'''={y}'' \][/tex]
[tex]\[{z}'''=\gamma _{z}({z}''-\beta _{z}{ct}'') \][/tex]
But when I substitute in I get:
[tex]\[{ct}'''=\gamma _{x} \gamma _{y}\gamma _{z}ct-\gamma _{x} \gamma _{y}\gamma _{z}\beta _{x}x-\gamma _{y}\gamma _{z}\beta _{y}y-\gamma _{z}\beta _{z}z\][/tex]
[tex]\[{x}'''=-\gamma _{x}\beta _{x}ct+\gamma _{x}x\][/tex]
[tex]\[{y}'''=-\gamma _{x}\gamma _{y}\beta _{y}ct+\gamma _{x}\gamma _{y}\beta _{x}\beta_{y}x+\gamma _{y}y\][/tex]
[tex]\[{z}'''=\gamma _{x}\gamma _{y}\beta _{z}ct+\gamma _{x}\gamma _{y}\beta _{x}\beta _{z}x+\beta _{y}\beta _{z}y+\gamma _{z}z\][/tex]
I suppose I would need to eliminate [tex]\[\gamma _{x}\][/tex], [tex]\[\gamma _{y}\][/tex] and [tex]\[\gamma _{z}\][/tex] in favor of [tex]\[\gamma\][/tex] but I guess I would just like to make sure I'm headed in the right direction before I undertake the calculation since:
[tex]\[\gamma _{x}\gamma _{y}\gamma _{z}=\frac{1}{\sqrt{(1-\beta _{x}^{2})(1-\beta _{y}^{2})(1-\beta _{z}^{2})}}=(\frac{1}{\gamma ^{2}}+\beta _{x}^{2}\beta _{y}^{2}+\beta _{x}^{2}\beta _{z}^{2}+\beta _{y}^{2}\beta _{z}^{2}-\beta _{x}^{2}\beta _{y}^{2}\beta _{z}^{2})^{-1/2}\][/tex]
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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