2.5: S05. Spacetime - SOLUTIONS
Exercise 5.2.1
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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 5.3.1
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"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 5.4.1
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"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 5.5.1
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No solution available yet
Exercise 5.5.2
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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 5.6.1
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Substituting in \(dt = a(t)d\eta\) to Equation 5.2 and factoring out \(a^2(t)\) gives us
\[\begin{equation*}
\begin{aligned}
ds^2 = -c^2a^2(t)[d\eta^2 + dx^2].
\end{aligned}
\end{equation*}\]Assuming a one-to-one correspondence between \(t\) and \(\eta\) (which one would have in an expanding universe given definition of \(d\eta\)) we can use \(a(\eta) \equiv a(t(\eta))\) in its place and write
\[\begin{equation*}
\begin{aligned}
ds^2 = a^2(\eta)[-c^2d\eta^2 + dx^2]
\end{aligned}
\end{equation*}\]
Exercise 5.6.2
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