Section 1.3 - Mass-Energy Equivalence

The most famous equation of all time

Just as Einstein unified space and time with special relativity, he also used it to unify mass and energy: hence the mass-energy equivalence equation, $E=mc^2$

A thought experiment demonstrating $E=mc^2$

Note that this is just a simple demonstration taken from this video. The complete proof is more intense. 

Suppose we have a radioactive cat floating freely in space that emits a flash of energy, $E$. The law of conservation of energy says the cat must've lost energy $E$, but since the energy was emitted symmetrically in all directions, the cat's velocity won't change

Now let's assume we zoom past the cat in a spaceship during the experiment. Intuitively we think that the cat is still and we're moving, but remember all motion is relative, so we'll define ourselves and stationary in our spaceship and the cat as moving

But since one of us is moving, time is different according to special relativity, so we'll measure different frequencies for the light flash, and by extension, different energies (recall the Planck-Einstein equation, $E = hf$, which states that the energy of a photon is equal to its frequency * Planck's constant)

This is known as the relativistic Doppler effect, similar to the classical Doppler effect we experience with sound but adjusted to account for relativity
$E_{observed} = E_{emitted} \cdot (1+\frac{v^2}{2c^2})$

Now let's consider two cases:

  1. We take off at velocity $v$, meaning the cat gains kinetic energy, $KE_1$. Then the cat flashes, so it loses energy, $E \cdot (1+\frac{v^2}{2c^2})$
  2. Cat flashes first before anyone starts moving, so it loses energy $E$, then we take off at velocity $v$, meaning the cat gains kinetic energy, $KE_2$

Obviously the two must be equal because we never influenced the cat, giving us:

$-E + KE_2 = KE_1 - E \cdot (1+\frac{v^2}{2c^2})$

From here, simple algebra will give us the final formula:

$E \cdot \frac{v^2}{2c^2} + KE_2 = KE_1$

Substitute the formula for kinetic energy, $KE = \frac{1}{2}mv^2$

$E \cdot \frac{v^2}{2c^2} + \frac{1}{2}m_2v_2^2 = \frac{1}{2}m_1v_1^2$

We know the velocity of the cat must be the same on both sides, so $v = v_1 = v_2$. Therefore it must be the mass of the cat that changed. Divide both sides by $\frac{1}{2}v^2$ to cancel everything out, then subtract $m_2$ from both sides

$E = mc^2$

Equating mass and energy resolves many apparent violations of the law of conservation of mass and the law of conservation of energy