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S05. Spacetime - SOLUTIONS

[ "article:topic", "Box Solutions", "authorname:knoxl" ]
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    Box \(\PageIndex{1}\)

    Exercise 5.1.1: For the spacetime specified by Equation 5.1: On a plot of \(x\) vs. \(t\) (what we call a spacetime diagram) draw the trajectory of a particle that is not moving, one that is moving slowly, and then of one that is moving at the speed of light. Place the \(x\)-coordinate on the horizontal axis, as is the usual convention.

    Solution \(\PageIndex{1.1}\)

    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 \(1\).


    Box \(\PageIndex{2}\)

    Exercise 5.2.1: Imagine a very small ruler instantaneously at rest in the \(x,t\) coordinate system of Equation 5.2 at time \(t=t_1\), with one end at location \(x=x_1\) and its other end at \(x=x_1+dx_1\). How long is the ruler?

    Solution \(\PageIndex{2.1}\)

    "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 \(ds = a(t_1)dx_1\).


    Box \(\PageIndex{3}\)

    Exercise 5.3.1: How much time elapses on a clock on a trajectory of constant \(x\), from \(t=t_1\) to \(t=t_2\)?

    Solution \(\PageIndex{3.1}\)

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

            \int \frac{1}{c}\sqrt{-ds^2} =\int_{t_1}^{t_2} dt = t_2 - t_1.


    Box \(\PageIndex{4}\)

    Exercise 5.4.1:  Still assuming Equation 5.2, draw the paths through spacetime of a pair of particles that are separated from each other and that are not "moving" -- that is, their \(x\) coordinate value is not changing over time. Assume \(a(t)\) is an increasing function of time. What do you notice about the distance between them and how it evolves over time? Be careful not to confuse "distance between them" with the difference in the values of their spatial coordinates. 

    Exercise 5.4.2: Now, add in the trajectory of a light ray passing from one of these particles to the other. While sketching it out, remember that \(a(t)dx\) is the distance traversed (as measured by an observer at rest in the \(x,t\) coordinate system) as the time coordinate changes by \(dt\), which is the time elapsed as measured by an observer at rest in the \(x,t\) coordinate system.  In this \(x\) vs. \(t\) diagram, does light travel in a straight line?

    Solution \(\PageIndex{4.1}\) and  \(\PageIndex{4.2}\)

    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.


    Box \(\PageIndex{5}\)

    Exercise 5.5.1: Given a spacetime described by Equation 5.2, work out the invariant distance specified for \(\tau,x\) labeling instead of \(t,x\) labeling.

    Solution \(\PageIndex{5.1}\)

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

            ds^2 = -c^2a^2(t)[d\tau^2 + dx^2].

    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

            ds^2 = a^2(\tau)[-c^2d\tau^2 + dx^2]


    Box \(\PageIndex{6}\)

    Exercise 5.6.1: Draw how light rays move on a plot of \(x\) vs. \(\tau\). Start from \(ds^2=0\) to find the relationship between \(d\tau\) and \(dx\), then draw a trajectory consistent with that relationship.

    Solution \(\PageIndex{6.1}\)