# Problem 2.8

A First Course in String Theory

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2.8 Spacetime diagrams and Lorentz transformations

Show that the $\displaystyle{{x'}^0}$ and $\displaystyle{{x'}^1}$ axes … appear in the original spacetime diagram as oblique axes.

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The Lorentz transform:

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

The inverse Lorentz transform:

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

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The $\displaystyle{(x')^0}$-direction unit vector is $\displaystyle{((x')^0, (x')^1) = (1, 0)}$.

When $\displaystyle{((x')^0, (x')^1) = (1,0)}$,

$\displaystyle{(x^0, x^1) = (\gamma, \gamma \beta)}$

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When $\displaystyle{\beta > 0}$,

\displaystyle{ \begin{aligned} \tan \phi &= \frac{\gamma \beta}{\gamma} \\ &= \beta \\ \end{aligned}}

where $\displaystyle{\phi}$ is the angle between $\displaystyle{x^0}$-axis and $\displaystyle{(x')^0}$-axis.

— Me@2021-03-17 03:42:23 PM

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2021.03.17 Wednesday (c) All rights reserved by ACHK