Skip to main content
[ "article:topic", "Box Solutions", "authorname:knoxl", "showtoc:no" ]
\(\require{cancel}\)
Physics LibreTexts

S04. Einstein Relativity - SOLUTIONS

  • Page ID
    3922
  • 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

    So we start with \(ds^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2\), where, due to the Lorentz transformation we have

    \[\begin{equation*}
    \begin{aligned}
    dt & = \gamma (dt'-vdx'/c^2) = \gamma/c (cdt'-vdx'/c), \\ dx & = \gamma (dx' - vdt'), \\ dy & = dy',\; {\rm and} \\ dz & = z'
    \end{aligned}
    \end{equation*}\]

    Therefore,

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

    Exercise 4.3.1

    Answer

    3_3_graphs.png

    For the first coordinate system we have

    \[\begin{equation*}
    \begin{aligned}
    (\Delta s)^2 = -c^2(t_2 - t_1)^2 + (x_2 - x_1)^2 = -c^2(\Delta t)^2 + (\Delta x)^2
    \end{aligned}
    \end{equation*}\]

    For the prime system we have

    \[\begin{equation*}
    \begin{aligned}
    (\Delta s')^2 = -c^2(t'_2 - t'_1)^2
    \end{aligned}
    \end{equation*}\]

    Now set them equal and solve for \(t'_2 - t'_1\)

    \[\begin{equation*}
    \begin{aligned}
    -c^2(t'_2 - t'_1)^2 & = -c^2(\Delta t)^2 + (\Delta x)^2 \\ \\ (t'_2 - t'_1)^2 & = (\Delta t)^2 - \frac{(\Delta x)^2}{c^2} \\ \\ {\rm note} \; & {\rm that,} \; \frac{(\Delta x)^2}{(\Delta t)^2} = v^2 \\ \\ (t'_2 - t'_1)^2 & = (\Delta t)^2\Big(1 - \frac{v^2}{c^2}\Big) \\ \\ t'_2 - t'_1 & = \gamma^{-1}\Delta t
    \end{aligned}
    \end{equation*}\]