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