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# S03. Einstein Relativity - SOLUTIONS

Box $$\PageIndex{1}$$

Imagine a particle traveling at the speed of light. Let's parameterize its path through spacetime with the independent variable $$\lambda$$ so that $$t = \lambda$$ and $$x(\lambda) = c\lambda$$. Then we have (by direct substitution into the Lorentz transformation) that $$t' = (\gamma/c)(c-v)\lambda$$ and $$x'=\gamma (c-v)\lambda$$. The speed of this particle in the primed frame is

\begin{equation*} \begin{aligned} \frac{dx'}{dt'} = \frac{dx'}{d\lambda}\frac{d\lambda}{dt'} = \frac{dx'}{d\lambda}\left(\frac{dt'}{d\lambda}\right)^{-1} = c. \end{aligned} \end{equation*}

Thus we see the Lorentz transformation tells us that a particle traveling at speed $$c$$ in one frame will be traveling at speed $$c$$ in another.

Exercise: 3.1.1: Fill in the steps in the above derivation.

Solution $$\PageIndex{1.1}$$

\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*}

Box $$\PageIndex{2}$$

Exercise 3.2.1: Show that the invariant distance is indeed invariant under a Lorentz transformation. For simplicity, take your two coordinate systems to be coincident at their origins (i.e. $$t=x=y=z=0$$ is the same point as $$t'=x'=y'=z'=0$$), use the origin as one point, and $$t=dt, x=dx, y=dy, z = dz$$ as the other.

Solution $$\PageIndex{2.1}$$

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*}

Box $$\PageIndex{3}$$

Exercise 3.3.1: Calculate the time that elapses on a clock traveling in a straight line at speed $$v$$ from $$x_1,t_1$$ to $$x_2, t_2$$. Do so in the following manner: 1) Draw the clock's path in two coordinate systems: $$x$$ vs. $$t$$ and $$x'$$ vs. $$t'$$ where the prime system is the one where the clock is at rest. 2) Calculate $$(\Delta s)^2$$ along the path from point 1 to point 2 in both coordinate systems, set them equal, and solve for $$t_2'-t_1'$$.

Note: Here we have used the fact that the time that elapses on the clock will be equal to the difference in time coordinates in the frame in which it is at rest, and the invariance of the invariant distance. We could also have just calculated $$(\Delta s)^2$$ in the unprimed frame and used our physical interpretation of $$\sqrt{-(\Delta s)^2}$$ (for $$(\Delta s)^2 < 0$$ ) as the time that elapses on a clock traveling from point 1 to point 2.

Solution $$\PageIndex{3.1}$$

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*}