Special Relativity

They expected an "aether wind" to shift the interference fringes as the apparatus rotated, but no shift was ever seen. Light's speed is independent of direction and of Earth's motion, so the luminiferous aether does not exist.

The speed of light in vacuum is $c$ for every observer, regardless of the motion of the source or observer. Velocities do not simply add for light, so both measure $c$, not $1.9c$.

An inertial frame moves at constant velocity (zero acceleration). Turning a corner, firing engines, spinning, and starting or stopping all involve acceleration, so they are non-inertial. Special relativity applies to inertial frames.

$t = \dfrac{t_0}{\sqrt{1 - v^2/c^2}} = \dfrac{1.0}{\sqrt{1 - 0.36}} = \dfrac{1.0}{0.8} = 1.25\text{ h}$. The moving clock runs slow, so Earth sees more time pass.

Simultaneity is relative. The passenger moves toward the light from the front and away from the light from the back; since both flashes travel at $c$ in the train frame too, the front flash arrives first, so the passenger concludes the front was struck first. Events simultaneous in one inertial frame need not be in another.

Proper time $t_0$ is measured by one clock present at both events (carried from one to the other). It is the shortest interval; every other observer, for whom that clock is moving, measures a longer, dilated time.

$l = l_0\sqrt{1 - v^2/c^2} = 100\sqrt{1 - 0.64} = 100\times0.6 = 60\text{ m}$. A moving object is contracted along its direction of motion.

$E = mc^2 = (2.0)(3.0\times10^8)^2 = (2.0)(9.0\times10^{16}) = 1.8\times10^{17}\text{ J}$.

From the ground the pulse traces a longer zig-zag between the moving mirrors. Because $c$ is the same in every frame, covering the longer path takes more time, so the moving clock ticks slower. This is the geometric origin of time dilation.

$\gamma = \dfrac{1}{\sqrt{1 - 0.6^2}} = \dfrac{1}{\sqrt{0.64}} = \dfrac{1}{0.8} = 1.25$.

Length contraction acts only parallel to the velocity. Dimensions perpendicular to the motion are unaffected, so the cube is seen as a shortened box (its volume therefore decreases).

At about $0.98c$ the Lorentz factor is roughly $5$, so the dilated lifetime is about $11\ \mu\text{s}$, long enough to cross $10\text{ km}$. Detecting muons at sea level is direct evidence of time dilation. (In the muon's own frame the distance is length-contracted instead.)

The astronaut holds the clock, so $5.0$ years is proper time. Earth measures $t = \gamma t_0 = 1.25\times5.0 = 6.25$ years.

Earth time $t = \dfrac{d}{v} = \dfrac{6.0}{0.6} = 10$ years. The ship measures proper time $t_0 = t\sqrt{1 - v^2/c^2} = 10\times0.8 = 8.0$ years.

In the muon's frame its lifetime is the short proper time, but the atmosphere rushes past so the $10\text{ km}$ is length-contracted to a few hundred metres, crossed within $2.2\ \mu\text{s}$. Earth explains the survival by time dilation, the muon by length contraction, and both frames agree it arrives.

Mass defect $\Delta m = (6.693 - 6.645)\times10^{-27} = 4.8\times10^{-29}\text{ kg}$. Then $E = (\Delta m)c^2 = (4.8\times10^{-29})(9.0\times10^{16}) = 4.3\times10^{-12}\text{ J}$. Using $\Delta m\,c$ instead of $\Delta m\,c^2$ gives the trap $1.4\times10^{-20}$.

Energy in a year: $E = Pt = (1.0\times10^9)(3.15\times10^7) = 3.15\times10^{16}\text{ J}$. Then $m = \dfrac{E}{c^2} = \dfrac{3.15\times10^{16}}{9.0\times10^{16}} = 0.35\text{ kg}$. About a third of a kilogram powers a city for a year. (Dividing by $c$ instead of $c^2$ gives the trap $1.05\times10^8$.)

$\gamma = 2$ means $\sqrt{1 - v^2/c^2} = 0.5$, so $1 - v^2/c^2 = 0.25$, giving $v^2/c^2 = 0.75$ and $v = \sqrt{0.75}\,c \approx 0.87c$.

As $v \to c$, $\gamma \to \infty$, so the kinetic energy $K = mc^2(\gamma - 1)$ also diverges. Reaching $c$ would need infinite energy, which is impossible, so only massless particles (like photons) travel at $c$.

Although each sees the other's clock run slow during the trip, the situation is not symmetric: B turns around (accelerates and changes inertial frames) while A remains inertial the whole time. The consistent result is that the traveller B logs less proper time and returns younger.

The experiment set out to detect the aether wind and found nothing. That null result undermined the aether model and was later explained by the constancy of $c$, a cornerstone of Einstein's relativity, showing how a "failed" search can transform physics.

This is the relativity of simultaneity: events separated in space that are simultaneous in one frame are not simultaneous in another. For events with no causal link, even their time order can differ between observers.

In the ship's frame the distance contracts to $l = 8.0\times0.6 = 4.8$ ly, covered at $0.8c$ in $\dfrac{4.8}{0.8} = 6.0$ years, exactly the crew's clock reading. Earth uses time dilation, the crew use length contraction, and both give $6.0$ years on the ship's clock.

$\gamma = 5$ means $\sqrt{1 - v^2/c^2} = 0.2$, so $1 - v^2/c^2 = 0.04$, giving $v^2/c^2 = 0.96$ and $v = \sqrt{0.96}\,c \approx 0.98c$.

First $\gamma = \dfrac{1}{\sqrt{1 - 0.8^2}} = \dfrac{1}{0.6} = 1.667$. Then $K = mc^2(\gamma - 1) = 500(0.667) \approx 330\text{ MeV}$. The total energy $\gamma mc^2 \approx 833\text{ MeV}$ is the rest energy plus this KE.