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S04. Einstein Relativity - SOLUTIONS

  • Page ID
    3922
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    Exercise 4.1.1

    Answer

    \[\begin{equation*}
    \begin{aligned}
    \frac{dx'}{d\lambda} &= \gamma(c - v), \; {\rm and} \\ \\ \frac{dt'}{d\lambda} &= \frac{\gamma}{c}(c -v)
    \end{aligned}
    \end{equation*}\]

    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

    \[\begin{equation*}
    \begin{aligned}
    ds'^2 & = -c^2 dt'^2+dx'^2+dy'^2+dz'^2 = -\gamma^2\Big(cdt - \frac{vdx}{c}\Big)^2 + \gamma^2(dx - vdt)^2 + dy^2 + dz^2 \\ \\ & = -\gamma^2(c^2 - v^2)dt^2 + \gamma^2\Big(1 - \frac{v^2}{c^2}\Big)dx^2 + dy^2 + dz^2 \\ \\ & = -\gamma^2 c^2\Big(1-\frac{v^2}{c^2}\Big)dt^2 + \gamma^2\Big(1-\frac{v^2}{c^2}\Big)dx^2 + dy^2 + dz^2 \\ \\ & = -c^2dt^2 + dx^2 + dy^2 + dz^2 =ds^2
    \end{aligned}
    \end{equation*}\]

    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

    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.


    This page titled S04. Einstein Relativity - SOLUTIONS is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Lloyd Knox.