Deriving the Lorentz Boost for an Arbitrary Direction

  • Thread starter grindfreak
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  • #1
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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.
 
Last edited:

Answers and Replies

  • #2
vela
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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
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.
 
  • #3
39
2
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!
 

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