- #1

unified

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**Summary:**The problem is to generalize the Lorentz transformation to two dimensions.

Relevant Equations

Lorentz Transformation along the positive x-axis:

$$ \begin{pmatrix}

\bar{x^0} \\

\bar{x^1} \\

\bar{x^2} \\

\bar{x^3} \\

\end{pmatrix} =

\begin{pmatrix}

\gamma & -\gamma \beta & 0 & 0 \\

-\gamma \beta & \gamma & 0 & 0 \\

0 & 0 & 1 & 0 \\

0 & 0 & 0 & 1

\end{pmatrix}

\begin{pmatrix}

{x^0} \\

{x^1} \\

{x^2} \\

{x^3} \\

\end{pmatrix}

$$

Lorentz Transformation along the positive y-axis:

$$ \begin{pmatrix}

\bar{x^0} \\

\bar{x^1} \\

\bar{x^2} \\

\bar{x^3} \\

\end{pmatrix} =

\begin{pmatrix}

\gamma & 0 & -\gamma \beta & 0 \\

0 & 1 & 0 & 0 \\

-\gamma \beta & 0 & \gamma & 0 \\

0 & 0 & 0 & 1

\end{pmatrix}

\begin{pmatrix}

{x^0} \\

{x^1} \\

{x^2} \\

{x^3} \\

\end{pmatrix}

$$

Velocity transformations from S to S' where S' moves along the positive x-axis at speed v relative to S:

$$

u'_x = \frac{u_x - v}{1 - \frac{u_xv}{c^2}} \\

u'_y = \frac{u_y}{\gamma(1 - \frac{u_xv}{c^2})}

$$

Problem Statement

$$\bar{S} \text{ moves at velocity } \\ \vec{v} = \beta c(Cos\phi \hat{x} + Sin\phi \hat{y}) \text{ relative to S, with axes parallel and origins coinciding at time } \\ t = \bar{t} = 0. \text{ Find the Lorentz transformation from S to } \bar{S}$$

Attempt at a solution

$$ \text{Let S' move at velocity } \beta cCos\phi\hat{x} \text{ relative to S. We can find the velocity of } \bar{S} \text { relative to S' using the velocity transformations. } \\ u'_x = 0 \\ u'_y = \gamma_x\beta cSin\phi \\ where \, \gamma_x = \frac{1}{(1-\beta^2Cos^2\phi)^\frac{1}{2}}$$

$$ \text{We can express the coordinates in } \bar{S} \text{ given the coordinates in S' by the y-axis Lorentz transformation, with } \\ \bar{\beta} = \beta Sin\phi \gamma_x \, \\ \bar{\gamma} = \frac{\gamma}{\gamma_x} \\ where \, \gamma = \frac{1}{(1 - \beta^2)^\frac{1}{2}}$$$$ \text{And we can express the coordinates in S' given the coordinates in S by the x-axis transformation with } \\ \beta' = \beta Cos\phi \\ \gamma' = \gamma_x$$

$$ \text{ Therefore we can express the coordinates in } \bar{S} \text { given the coordinates in S as follows } $$

$$ \begin{pmatrix}

\bar{x^0} \\

\bar{x^1} \\

\bar{x^2} \\

\bar{x^3} \\

\end{pmatrix} =

\begin{pmatrix}

\frac{\gamma}{\gamma_x} & 0 & -\frac{\gamma}{\gamma_x}\beta Sin\phi \gamma_x & 0 \\

0 & 1 & 0 & 0 \\

-\frac{\gamma}{\gamma_x}\beta Sin\phi \gamma_x & 0 & \frac{\gamma}{\gamma_x} & 0 \\

0 & 0 & 0 & 1

\end{pmatrix}

\begin{pmatrix}

\gamma_x & -\gamma_x\beta Cos\phi & 0 & 0 \\

-\gamma_x \beta Cos\phi & \gamma_x & 0 & 0 \\

0 & 0 & 1 & 0 \\

0 & 0 & 0 & 1

\end{pmatrix}

\begin{pmatrix}

x^0 \\

x^1 \\

x^2 \\

x^3

\end{pmatrix}

=

$$

\begin{pmatrix}

\gamma & -\gamma \beta Cos\phi & -\gamma \beta Sin\phi & 0 \\

-\gamma_x\beta Cos\phi & \gamma_x & 0 & 0 \\

-\gamma \gamma_x\beta Sin\phi & \gamma \gamma_x\beta^2Sin\phi Cos\phi & \frac{\gamma}{\gamma_x} & 0 \\

0 & 0 & 0 & 1 \end{pmatrix}

\begin{pmatrix}

x^0 \\

x^1 \\

x^2 \\

x^3

\end{pmatrix}

This final matrix represents the solution to the problem. But, this solution is wrong, as the correct matrix is given an follows.

\begin{pmatrix}

\gamma & -\gamma \beta Cos\phi & -\gamma \beta Sin\phi & 0 \\

-\gamma \beta Cos\phi & (\gamma Cos^2\phi + Sin^2\phi) & (\gamma - 1)Sin\phi Cos\phi & 0 \\

-\gamma \beta Sin\phi & (\gamma - 1)Sin\phi Cos\phi & (\gamma Sin^2\phi + Cos^2\phi) & 0 \\

0 & 0 & 0 & 1

\end{pmatrix}

I am unable to determine what is wrong with this solution, any suggestions would be appreciated.