Relativity Practice

The two postulates are:

1. The laws of physics are the same in all inertial frames.

2. The speed of light in vacuum is the same for all inertial observers.

These are the starting point for special relativity.

Use

Substitute $v = 0.6c$:

Compute the light-travel distance for the time separation:

Since

the interval is timelike.

First find $\gamma$ from $v = 0.6c$:

Then use time dilation:

For $v = 0.8c$,

Use length contraction:

If the events are simultaneous in the first frame, then $\Delta t = 0$ and

With $\gamma = 1.25$, $v = 0.6c$, and $\Delta x = 12\ \text{m}$:

Since $c = 3.00\times 10^8\ \text{m/s}$,

Use relativistic velocity addition:

Substitute $u' = 0.5c$ and $v = 0.5c$:

Use

So

That time is called the proper time.

It is the time measured in the rest frame of the clock.

For light, the relation is

So the momentum is

For $v = 0.6c$, we have $\gamma = 1.25$.

Use the Lorentz transformations:

and

First compute $vt$:

So

Next,

Therefore,

For $v = 0.8c$,

The proper time is

The Earth-frame distance is

So the ship experiences $6\ \mu\text{s}$, and it travels $2400\ \text{m}$ in Earth frame.

Compute the light-travel distance for the time separation:

Since

the interval is spacelike. A light signal cannot connect the events, so there is no causal connection at or below light speed.

For $v = 0.6c$, the Lorentz factor is

Use

So

That is

Use the invariant energy-momentum relation:

Substitute $p = \frac{3}{4}mc$:

Factor out $m^2c^4$:

So

For $v = 0.8c$, we have

So the lab-frame lifetime is

The distance traveled is

Since $880\ \text{m} > 500\ \text{m}$, the particle reaches the detector before decaying.

Use the simultaneity formula with $\Delta t = 0$:

For $v = 0.6c$, $\gamma = 1.25$. Then

Since $c = 3.00\times 10^8\ \text{m/s}$,

So the flashes are not simultaneous in the train frame.

For a photon,

so energy is proportional to frequency.

The frequency changed from $6.0\times 10^{14}$ to $5.4\times 10^{14}$, which is a factor of

So the energy dropped to $90\%$ of its original value, a decrease of $10\%$.

General relativity is essential, because the satellite and the ground clock are in different gravitational potentials.

Special relativity handles motion in inertial frames, but it does not account for gravitational time dilation. GPS needs both motion and gravity, so the gravitational effect requires general relativity.

For $v = 0.6c$, the Lorentz factor is

The lab-frame lifetime is

The distance traveled is

So the muon travels about $495\ \text{m}$, which is just short of $500\ \text{m}$. It does not quite reach the ground.

The time coordinate transforms as

If the two strikes occur at different positions, the $vx/c^2$ term is different for each event. That means equal $t$ in Earth frame does not imply equal $t'$ in the train frame.

So simultaneity depends on the frame.

For each particle moving at $0.6c$, $\gamma = 1.25$, so each has energy

The total energy before the collision is

Because the particles have equal and opposite momenta, the total momentum is zero. After they stick together, the composite object is at rest, so its rest energy is

Therefore,

For a photon,

so energy changes by the same percentage as frequency.

If the frequency drops by $12\%$, the energy also drops by $12\%$.

With $\Delta t = 0$,

Take magnitudes:

Substitute $|\Delta t'| = 80\ \text{ns}$ and $\Delta x = 24\ \text{m}$:

So

which simplifies to

where $\beta = v/c$.

Thus

Square both sides:

so

Therefore,

First compute the light-travel distance:

Since $3\ \text{m} > 1.8\ \text{m}$, the interval is timelike.

The invariant interval is

For a timelike interval,

So the proper time is $8\ \text{ns}$.

Each particle has $\gamma = 1.25$, so each carries energy

The total energy is

Because the momenta cancel, the final object is at rest. Its rest energy is therefore

so

That is larger than $2m$ because some kinetic energy has been converted into rest mass.

General relativity says mass-energy curves spacetime, and light follows the curved geometry of spacetime.

So light is not being pulled by a Newtonian force in the usual way. Instead, its path is a geodesic in curved spacetime.

The equivalence principle is the key idea behind this: locally, gravity and acceleration have the same physical effects, so even light is affected by the geometry associated with gravity.