Problem 2.2a

A First Course in String Theory

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2.2 Lorentz transformations for light-cone coordinates.

Consider coordinates \displaystyle{x^\mu = ( x^0, x^1, x^2, x^3 )} and the associated light-cone coordinates \displaystyle{x^\mu = ( x^+, x^-, x^2, x^3 )}. Write the following Lorentz transformations in terms of the light-cone coordinates.

(a) A boost with velocity parameter \displaystyle{\beta} in the \displaystyle{x^1} direction.

\displaystyle{ \begin{aligned} \begin{bmatrix} c t' \\ x' \\ y' \\ z' \end{bmatrix}  &= \begin{bmatrix} \gamma&-\beta \gamma&0&0\\ -\beta \gamma&\gamma&0&0\\ 0&0&1&0\\ 0&0&0&1\\ \end{bmatrix} \begin{bmatrix} c\,t \\ x \\ y \\ z \end{bmatrix} \\   \end{aligned}}

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\displaystyle{ \begin{aligned} \begin{bmatrix} x^+ \\ x^- \end{bmatrix}  &= \frac{1}{\sqrt{2}}     \begin{bmatrix}        1 & 1 \\        1 & -1 \\     \end{bmatrix}     \begin{bmatrix} x^0 \\ x^1 \end{bmatrix}    \end{aligned}}

The matrix is its own inverse.

\displaystyle{ \begin{aligned} \begin{bmatrix} x^0 \\ x^1 \end{bmatrix}  &=  \frac{1}{\sqrt{2}}     \begin{bmatrix}        1 & 1 \\        1 & -1 \\     \end{bmatrix}  \begin{bmatrix} x^+ \\ x^- \end{bmatrix} \\   \end{aligned}}

\displaystyle{ \begin{aligned} \begin{bmatrix} (x^0)' \\ (x^1)' \end{bmatrix}  &=     \begin{bmatrix}        \frac{1}{\sqrt{2}}  & \frac{1}{\sqrt{2}}  \\        \frac{1}{\sqrt{2}}  & -\frac{1}{\sqrt{2}}  \\     \end{bmatrix}  \begin{bmatrix} (x^+)' \\ (x^-)' \end{bmatrix} \\   \end{aligned}}

Apply the result to the original transformation:

\displaystyle{ \begin{aligned} \begin{bmatrix} (x^0)' \\ (x^1)' \\ (x^2)' \\ (x^3)' \end{bmatrix}  &= \begin{bmatrix} \gamma&-\beta \gamma&0&0\\ -\beta \gamma&\gamma&0&0\\ 0&0&1&0\\ 0&0&0&1\\ \end{bmatrix} \begin{bmatrix} x^0 \\ x^1 \\ x^2 \\ x^3 \end{bmatrix} \\   \end{aligned}}

\displaystyle{ \begin{aligned}    \begin{bmatrix}        \frac{1}{\sqrt{2}}  & \frac{1}{\sqrt{2}} & 0 & 0 \\        \frac{1}{\sqrt{2}}  & -\frac{1}{\sqrt{2}} & 0 & 0 \\                         0  & 0 & 1 & 0 \\                         0  & 0 & 0 & 1 \\     \end{bmatrix}  \begin{bmatrix} (x^+)' \\ (x^-)' \\ y' \\ z' \end{bmatrix}  &= \begin{bmatrix} \gamma&-\beta \gamma&0&0\\ -\beta \gamma&\gamma&0&0\\ 0&0&1&0\\ 0&0&0&1\\ \end{bmatrix}     \begin{bmatrix}        \frac{1}{\sqrt{2}}  & \frac{1}{\sqrt{2}} & 0 & 0 \\        \frac{1}{\sqrt{2}}  & -\frac{1}{\sqrt{2}} & 0 & 0 \\                         0  & 0 & 1 & 0 \\                         0  & 0 & 0 & 1 \\     \end{bmatrix}  \begin{bmatrix} x^+ \\ x^- \\ y \\ z \end{bmatrix}  \end{aligned}}

\displaystyle{ \begin{aligned}    \begin{bmatrix}        1  & 1 \\        1  & -1 \\     \end{bmatrix}  \begin{bmatrix} (x^+)' \\ (x^-)' \end{bmatrix}  &= \begin{bmatrix}       \gamma & -\beta \gamma \\       -\beta \gamma &\gamma       \end{bmatrix}     \begin{bmatrix}        1  & 1 \\        1  & -1 \\     \end{bmatrix}  \begin{bmatrix} x^+ \\ x^- \end{bmatrix}  \end{aligned}}

\displaystyle{ \begin{aligned} \begin{bmatrix} (x^+)' \\ (x^-)' \end{bmatrix}  &= \begin{bmatrix}        \gamma (1-\beta) & 0 \\        0 & \gamma (1+\beta) \\     \end{bmatrix}  \begin{bmatrix} x^+ \\ x^- \end{bmatrix}  \end{aligned}}

— Me@2020-02-27 07:14:19 PM

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2020.02.27 Thursday (c) All rights reserved by ACHK