Deriving the Lorentz Boost for an Arbitrary Direction

In summary, the conversation discusses the difficulty in deriving the Lorentz transformation for an arbitrary direction of velocity. The attempt at a solution involves defining four reference frames and using boosts and rotations, but it is realized that this approach is incorrect due to non-commutativity of boosts. The correct approach is to use a single boost followed by a spatial rotation.
  • #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.
 
Last edited:
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  • #2
grindfreak said:
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
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!
 

Related to Deriving the Lorentz Boost for an Arbitrary Direction

1. What is the Lorentz Boost for an Arbitrary Direction?

The Lorentz Boost for an Arbitrary Direction is a mathematical formula used to transform coordinates and velocities between two frames of reference in special relativity.

2. Why is it important to derive the Lorentz Boost for an Arbitrary Direction?

Deriving the Lorentz Boost for an Arbitrary Direction is important because it allows us to understand how objects appear to move and behave differently in different frames of reference, and how this relates to the principles of special relativity.

3. What are the key components of the Lorentz Boost for an Arbitrary Direction?

The key components of the Lorentz Boost for an Arbitrary Direction include the relative velocity between the two frames of reference, the Lorentz factor, and the direction of motion in each frame.

4. How is the Lorentz Boost for an Arbitrary Direction derived?

The Lorentz Boost for an Arbitrary Direction is derived using mathematical equations and principles from special relativity, including the Lorentz transformation equations and the concept of time dilation.

5. Can the Lorentz Boost for an Arbitrary Direction be applied to any situation?

The Lorentz Boost for an Arbitrary Direction can be applied to any situation where objects are moving at relativistic speeds and there are multiple frames of reference involved. However, it is most commonly used in situations involving linear motion.

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