# 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**