# Relativity (mostly special)

Absolute, true, and mathematical time, of itself and from its own nature, flows equably without relation to anything external.

-Isaac Newton

It came to me that time was suspect!

-Albert Einstein

Outline

1. Galilean transformations of velocity
2. Invariance of physical laws
3. Breakdown of Galilean tranformations
4. Einstein's resolution
5. Time dilation
6. Length contraction
7. Experimental evidence?

The physical laws we're familiar with (e.g. $\myv F = m\myv a$) remain unchanged (are invariant)...

• in different locations,
• at different times,
• for different orientations,

• at different speeds. [as long as acceleration=0].

## Galilean transformations

Galileo was already familiar with this way of transforming from one moving reference frame to another...

We'll have Barack (frame ${\cal b}$) standing on the platform of a railway station, while Michelle (frame ${\cal m}$) is in a train car moving away with a speed $V$ in the $x$-direction. The origins coincide at $t=0$. $$x_{\cal m} =x_{\cal b} - Vt$$ $$y_{\cal m} = y_{\cal b}$$ $$z_{\cal m} = z_{\cal b}$$ $$t_{\cal m}=t_{\cal b}.$$

When it's clear from context, we shall frequently drop the ${\cal b}$ subscript from the platform quantities.

If an object on the platform is moving with a velocity $v_{x {\cal b}}$, then its speed with respect to the train is (take $\del/\del t$ of the relationship for the $x$ direction): $$v_{x {\cal m}} = v_{x {\cal b}} - V.$$

## Galilean transformations

Consider two billiard balls with equal mass moving towards each other, then we know that they'll rebound off each other with the same speed.

Together with Galilean relativity, (at 5 f/s to each ball) we can figure out what happens if one ball is incident on a stationary one...

OK, OK, this is something you can also figure out from energy/momentum conservation.

Here is a collision that is a bit harder to find, even if you know a bunch of conservation laws.

A light ball will rebound off a much heavier one with practically the same speed as it came in with.

But what if the heavy ball is moving towards a much lighter ball. What will the outcome of the collision be?

### Invariance of physical laws under transformation

Imagine that Michelle undertakes to accelerate some object in her $x$-direction with a calibrated spring. She'd see: $$F=m \frac{d^2 x_{ {\cal m}}}{dt^2} = m \frac{d^2}{dt^2}(x_{\cal b}-Vt) = m \frac{d^2 x_{ {\cal b}}}{dt^2}.$$

If she accelerates something in the $y$- direction, it's even easier to see: $$F=m \frac{d^2 y_{\cal m}}{dt^2} = m \frac{d^2 y_{b}}{dt^2} .$$

So the force law takes the same (invariant form) $\myv F= m \myv a$ in both frames.

Galileo used the argument that no experiment could tell whether the earth was still (and the sun moving) or the earth was moving (and the sun was still) in support of Kepler's model of a heliocentric solar system.

### Breakdown of Galilean transformations

Two results from electromagnetism do not 'fit' with the Galilean transformations at all:

1. Speed of light: The four Maxwell equations predict electromagnetic waves (light) that move at a speed $c$ in all reference frames!Not at $c$ relative to the velocity of the flashlight that I just turned on. Just $c$! How can that be?

2. Forces between two charges: The situation above might be hard to detect unless things are moving close to the speed of light. But here's a "slower" problem. Mount two charges on a train, and consider the forces between them in two different frames...

Michelle says:
"They're standing still, so there's an electric force between them, and that's it."

Barack says:
"Yes, honey, but you're moving, so each charge is generating a magnetic field. Since there's a magnetic field, there's also a magnetic force".

As pictured, there's also a torque. Trouton and Noble tried and tried to detect this torque (ca. 1902) but in vain.

### Einstein's resolution

Lorentz had come up with an alternative set of transformations to Galileo's which worked for E-M fields, but with no justification.

Einstein proposed these two Ansätze (postulates) as starting points from which to justify Lorentz's math ("Zur Elektrodynamik bewegter Körper")

1. The laws of physics [Einstein said Mechanics and Electro-Magnetism] are the same in all inertial reference frames*: absolute uniform motion cannot be detected.
2. The speed of light $c$ is the same no matter the motion of the source.

*An inertial reference frame is one with no acceleration: Newton's law $\myv F=m \myv a$ holds in any inertial reference frame, but not (strictly) in a rotating reference frame like earth, or in an accelerating car.

We shall now explore some of the strange consequences (and experimental evidence) of these assumptions:

$\Rightarrow$ Time cannot be the same for all observers, and neither can length, if we'd like to stick to Einstein's postulates.

### Time dilation

Imagine flashing a bulb at position $A'$ in the train, and timing it until its return after bouncing off a mirror a distance $D$ above.

Michelle sees the situation at the left and measures the time for a light flash to return to $A'$ as $$\Delta t_{\cal m} = \frac{2D}{c}.$$

Meanwhile, Barack measures a time interval $\Delta t$ for the emission-bounce-return to happen. We're going to assume [Einstein] that $c$ is the same for Barack as for Michelle.. Then using this triangle... $$(c \frac{\Delta t}{2})^2 = D^2 + (V \frac{\Delta t}{2})^2.$$

Solving for $\Delta t$: $$\Delta t = \frac{2D}{\sqrt{c^2-V^2}} = \frac{2D}{c} \frac{1}{\sqrt{1-V^2/c^2}}.$$

Using Michelle's time interval: $$\Delta t = \gamma \Delta t_{\cal m}$$ where: $$\gamma \equiv \frac{1}{\sqrt{1-V^2/c^2}} \geq 1.$$

Michelle can use the same clock to measure the starting pulse and the return.

But because the flash and return happen at different locations on the platform, Barack has to use two previously synchronized clocks to measure the time.

The time in the frame in which just one clock can be used is called the proper time.

The proper time is always shorter than the time between these two events in any other frame. Folks moving with respect to the light pulse see events taking longer--a phenomenon called time dilation.

### Length contraction

Imagine that in the experiment above, a measuring scale was fixed to the platform. The ruler has its proper length $L_0$ in the frame (${\cal b}$) in which it is at rest.

what length do folks on the train attribute to the ruler?

Michelle sez':

"It took a time $\Delta t_{\cal m}$ for something moving at a speed of $V$ to pass by. It must have had a length": $$L_{\cal m}=V\Delta t_{\cal m} = V \frac{\Delta t}{\gamma} = \frac{L_0}{\gamma}.$$

So everybody moving relative to the ruler measures the ruler to be somewhat shorter than its proper length. This is the length contraction.

Length contraction only happens in the direction of motion--not in a direction ($y$- or $z$-direction) perpendicular to the motion.

### Experimental evidence?

Muons ($\mu^-$) are unstable particles that decay in the lab according to $$N(t)=N_0e^{-t/\tau}$$ with $\tau=2 \times 10^{-6}$ sec in a laboratory (where the muon is at rest).

They are produced high in the atmosphere as a by-product of cosmic rays hitting the atmosphere:

high energy $p^+$ $\to\ \pi$ meson ($\tau=3\times 10^{-8}$ s) $\to\ \mu^-$ ($\tau=2\times 10^{-6}$ s.

Muons have approximately the same momentum as the original proton, but a much smaller mass (~1/9) so will be moving faster than the original proton. Let's say that a particular population of muons have a speed of 0.998$c$. Then without special relativity, a muon with the mean lifetime should travel just $0.998 (3 \times 10^8 \text{m/s})*2 \times 10^{-6} s = 600 m$.

Muons are not produced lower in the atmosphere: The cosmic protons are charged particles (89% protons), which interact strongly with matter. So no cosmic ray protons penetrate beyond the outer layers of the atmosphere. (Fortunately for life on earth: If you though UV radiation was bad for your skin...

So, if muons are produced several thousand meters up in the atmosphere, very few should reach sea-level.

But if special relativity is true, then the lifetime of a muon measured by an observer stationary on the surface of the earth should be longer by a factor $\gamma =\frac{1}{\sqrt{1-V^2/c^2}}$. We have $\gamma(0.998c) \approx 16$, so it should appear to have a longer lifetime of $\approx 30 \times 10^{-6}$s, during which it travels 9000 m.

Or, from the point of view of the muon, the distance 9000 m is contracted in its reference frame to just 600 m, which it can easily traverse in its $2\times10^{-6}$s lifetime.

Measurements show the concentrations of muons high in the atmosphere and close to sea level are of the same order of magnitude.

JSmith Photo