S04. Einstein Relativity - SOLUTIONS
- Page ID
- 3922
Exercise 4.1.1
- Answer
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\[\begin{equation*}
\begin{aligned}
\frac{dx'}{d\lambda} &= \gamma(c - v), \; {\rm and} \\ \\ \frac{dt'}{d\lambda} &= \frac{\gamma}{c}(c -v)
\end{aligned}
\end{equation*}\]Therefore,
\[\begin{equation*}
\begin{aligned}
\frac{dx'}{d\lambda}\left(\frac{dt'}{d\lambda}\right)^{-1} = \gamma(c - v)\Big(\frac{\gamma}{c}(c -v)\Big)^{-1} = c
\end{aligned}
\end{equation*}\]
Exercise 4.2.1
- Answer
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\[\begin{equation*}
\begin{aligned}
ds'^2 & = -c^2 dt'^2+dx'^2+dy'^2+dz'^2 = -\gamma^2\Big(cdt - \frac{vdx}{c}\Big)^2 + \gamma^2(dx - vdt)^2 + dy^2 + dz^2 \\ \\ & = -\gamma^2(c^2 - v^2)dt^2 + \gamma^2\Big(1 - \frac{v^2}{c^2}\Big)dx^2 + dy^2 + dz^2 \\ \\ & = -\gamma^2 c^2\Big(1-\frac{v^2}{c^2}\Big)dt^2 + \gamma^2\Big(1-\frac{v^2}{c^2}\Big)dx^2 + dy^2 + dz^2 \\ \\ & = -c^2dt^2 + dx^2 + dy^2 + dz^2 =ds^2
\end{aligned}
\end{equation*}\]That completes the exercise, but note that we could go the opposite direction and start with \(ds^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2\). In this case we need the inverse Lorentz transformation, which is the same as the Lorentz transformation but with \(v \rightarrow -v\). Therefore we have
\[\begin{equation*}
\begin{aligned}
dt & = \gamma (dt'+vdx'/c^2) = \gamma/c (cdt'+vdx'/c), \\ dx & = \gamma (dx' + vdt'), \\ dy & = dy',\; {\rm and} \\ dz & = dz'
\end{aligned}
\end{equation*}\]Therefore,
\[\begin{equation*}
\begin{aligned}
ds^2 & = -\gamma^2\Big(cdt' + \frac{vdx'}{c}\Big)^2 + \gamma^2(dx' + vdt')^2 + dy'^2 + dz'^2 \\ \\ & = -\gamma^2(c^2 - v^2)dt'^2 + \gamma^2\Big(1 - \frac{v^2}{c^2}\Big)dx'^2 + dy'^2 + dz'^2 \\ \\ & = -\gamma^2 c^2\Big(1-\frac{v^2}{c^2}\Big)dt'^2 + \gamma^2\Big(1-\frac{v^2}{c^2}\Big)dx'^2 + dy'^2 + dz'^2 \\ \\ & = -c^2dt'^2 + dx'^2 + dy'^2 + dz'^2 =ds'^2
\end{aligned}
\end{equation*}\]
Exercise 4.3.1
- Answer
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The speed \(v\) is introduced as the speed of the clock in the unprimed frame. By definition, speed is change in location divided by change in time.