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

S05. Spacetime - SOLUTIONS

  • Page ID
    3923
  • Exercise 5.1.1

    Answer

    We cheated a bit here and made a \(x\) vs. \(ct\) plot so that a particle moving at the speed of light has a slope of \(1\).

    Exercise 5.2.1

    Answer

    "at time \(t=t_1\)" so \(dt = 0\), so \(ds = a(t)dx\). Since the ruler is at rest in the given coordinate system its length is indeed given by \(ds\) at time \(t_1\). Therefore the length of the ruler is \(ds = a(t_1)dx_1\).

    Exercise 5.3.1

    Answer

    "constant \(x\)" so \(dx = 0\), and then \(ds = cdt\). Therefore the time elapsed on the clock is

    \[\begin{equation*}
      \begin{aligned}
            \int \frac{1}{c}\sqrt{-ds^2} =\int_{t_1}^{t_2} dt = t_2 - t_1.
      \end{aligned}
    \end{equation*}\]

    Exercise 5.4.1

    Answer

    See solution to Exercise 5.4.2

    Exercise 5.4.2

    Answer

    Since the amount of distance corresponding to a given \(dx\) is changing with time, the slope of the photon's world line is changing with time.

    Exercise 5.5.1

    Answer

    Substituting in \(dt = a(t)d\tau\) to Equation 5.2 and factoring out \(a^2(t)\) gives us

    \[\begin{equation*}
      \begin{aligned}
            ds^2 = -c^2a^2(t)[d\tau^2 + dx^2].
      \end{aligned}
    \end{equation*}\]

    Assuming a one-to-one correspondence between \(t\) and \(\tau\) (which one would have in an expanding universe given definition of \(d\tau\)) we can use \(a(\tau) \equiv a(t(\tau))\) in its place and write

    \[\begin{equation*}
      \begin{aligned}
            ds^2 = a^2(\tau)[-c^2d\tau^2 + dx^2]
      \end{aligned}
    \end{equation*}\]

    Exercise 5.6.1

    Answer