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# 4: Einstein Relativity

In the 19th century it was discovered that the Maxwell Equations describing electric and magnetic fields, a grand synthesis of the results of many different experiments, unlike Newton's laws of motion, are not consistent with Galilean relativity. A priori, the solution was not clear. One possible reason for this inconsistency, taken seriously at the time, was that the principle of relativity is wrong; i.e., there actually is an absolute rest frame, and our motion could be detected with respect to it with the appropriate experiment. Indeed, there was a significant experimental program to detect our motion with respect to absolute rest defined by a medium called "the ether."

Another possibility, is that while the principle of relativity holds, its specific implementation as Galilean relativity does not. As you know, because you have studied special relativity, this is indeed the correct solution to the puzzle of the Maxwell Equations lack of invariance under a Galilean transformation.

It turns out that the "Galilean Boost" can be generalized to a "Lorentz Boost" that is also consistent with the principle of relativity. The primed and unprimed coordinate systems constructed as before, under a Lorentz boost are related as:

\begin{aligned} t' & = \gamma (t-vx/c^2) \\ x' & = \gamma (x - vt), \\ y' & = y,\ {\rm and} \\ z' & = z. \end{aligned}

where $$\gamma \equiv 1/\sqrt{1-v^2/c^2}$$. In the limit that $$c \rightarrow \infty$$ this reduces to the Galilean boost. As can be easily shown (see the homework problem) the reverse transformation is the same rule with $$v \rightarrow -v$$. Most importantly, the Maxwell equations are invariant under this transformation.

One of the more spectacular consequences of the Maxwell Equations is that one of their solutions is waves traveling at the speed of light. If the Maxwell equations are correct in all inertial frames, then this implies that these waves will be moving at the speed of light in all inertial frames. To your Galilean intuition this is quite startling as it violates the simple rule for addition of velocities you derived in the previous chapter.

The result can be easily demonstrated from the Lorentz transformation above. Here we sketch out the process, and you can fill in the details by performing the exercise that follows. 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

$\frac{dx'}{dt'} = \frac{dx'}{d\lambda}\frac{d\lambda}{dt'} = \frac{dx'}{d\lambda}\left(\frac{dt'}{d\lambda}\right)^{-1} = c.$

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. This result is consistent with our claim that the Maxwell equations are invariant under the Lorentz transformation, since a consequence of the Maxwell Equations is that electromagnetic waves travel at speed $$c$$.

Box $$\PageIndex{1}$$

Exercise: 4.1.1: Fill in the steps in the above derivation.

\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*}

Unlike rotational coordinate transformations that preserve spatial distances between pairs of points, a Lorentz transformation does not. The spatial separation between $$(x,t)$$ and $$(x+dx,t)$$ is $$dx$$. The spatial separation between these points in the prime frame is $$\gamma dx$$, as one can see from the transformation rule. How can length depend on reference frame? Key to resolving this apparent paradox is the fact that in the primed frame the two events are not simultaneous. We won't go through sorting out these apparent paradoxes here.

We will, however, introduce a quantity that, unlike spatial length, is invariant under Lorentz transformations. For Cartesian spatial coordinates, the square of the invariant distance between event $$(t,x,y,z)$$ and event $$(t+dt,x+dx, y+dy, z+dz)$$ is given by

$ds^2 = -c^2 dt^2 + dx^2 + dy^2 + dz^2. \label{eqn:invdist}$

This quantity has the following two-part physical interpretation:

1. For $$ds^2 > 0$$, $$\sqrt{ds^2}$$ is the length of a ruler that connects the two events and is at rest in the frame in which the two events are simultaneous.
2. For $$ds^2 < 0$$, $$\sqrt{-ds^2}$$ is the time elapsed on a clock that moves between the two events with no acceleration.

Why is this quantity invariant under boosts? That's a deep question, and I'm not sure we have the fullest possible answer yet sorted out. We do know that the Maxwell equations are a synthesis from experiments, their form is invariant under a Lorentz transformation, and the Lorentz transformation preserves the invariant distance.

Box $$\PageIndex{2}$$

Exercise 4.2.1: Show that the invariant distance is indeed invariant under a Lorentz transformation. For specificity, take it to be the transformation appropriate for a boost in the $$+x$$ direction with speed $$v$$. 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.

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*}

Rather than the Lorentz transformation itself, the key thing to take away from this chapter is the definition of the invariant distance. We will be using it for the rest of the course, generalized to spacetimes with "curvature." Before doing so, we give some exercises here in which you get to make use of the invariant distance to solve problems in the more familiar context of a flat spacetime, the so-called Minkowski space you are familiar with from special relativity. A Minkowski space is simply a spacetime that can be labeled with $$t,x,y,z$$ such that the invariant distance is given by Eq. \ref{eqn:invdist}.

In Minkowski space, as one of the homework problems asks you to show, a finite (as opposed to infinitesmial) version of the invariant distance equation is also true:

$(\Delta s)^2 = -c^2 (\Delta t)^2 + (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2$

for trajectories that are straight lines, with $$\Delta s \equiv \int d\lambda \frac{ds}{d\lambda}$$ also invariant under Lorentz transformations.

Box $$\PageIndex{3}$$

Exercise 4.3.1: 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 that here we have used these facts: 1) 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 2) the invariant distance is invariant (the same in both coordinate systems). 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.

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*}

# HOMEWORK Problems

Problem $$\PageIndex{1}$$

Show, by solving for $$x$$ and $$t$$ that the inverse Lorentz transformation is the same as the forward transformation but with $$v \rightarrow -v$$. Explain what this has to do with the principle of relativity.

Problem $$\PageIndex{2}$$

Show that for straight paths in spacetime, that $$(\Delta s)^2 = -c^2 (\Delta t)^2 + (\Delta x)^2$$ follows from $$ds^2 = -c^2 dt^2 + dx^2$$. Hint: all straight paths in spacetime (at least the flat spacetime of special relativity we are studying now) can be parametrized via: $$t-t_0=\lambda, x=x_0 +v\lambda$$.

Problem $$\PageIndex{3}$$

Events A and B occur 10 meters and 100 ns apart in time in frame 1. If they occur 95 ns apart in frame 2, what must their spatial separation be in frame 2?

Problem $$\PageIndex{4}$$

Derive the phenomenon of time dilation. Consider the path taken by a clock from point 1 to point 2 in two different coordinate systems, a primed one in which the clock is at rest, and an uprimed one in which the clock is moving at constant speed $$v$$. Use the invariance of the invariant distance to show that the time elapsed on the clock is less than $$t_2 - t_1$$. [Yes, this is basically the same as one of the exercises.]