4-Momentum

We have found that velocities in relativity add in strange ways. We still want momentum to be conserved in some form, but the classical formulation won’t work.

TODO: derivation (lec 6)

We find a 3-momentum that is conserved as $\tv p = \gamma(u) m \tv u$. Intuitively, this works because when we change reference frames, velocity addition gives us a $\gamma$ in the denominator, which this accounts for.

What should the 0 component be?

Doing some integrals (TODO) we find that $K = (\gamma(u) - 1) mc^2$. We define the total energy to be $E = \gamma(u)mc^2 = K + mc^2$.

This total energy $E$ respects conservation laws. We find that setting $\fv p^0 = E/c$ creates a geometric object that follows the Lorentz transform.

As we show know, the $0$ component of 4-velocity in the rest frame of the object is $c$. This means that in that frame, $-p^\alpha v_\alpha = E$. Since these are 4-vectors and this is an inner product, the result is a scalar — a Lorentz invariant. If we know $\fv p$ and $\fv v$ in any frame, we can find the energy.

Equivalence of mass and energy

Consider two masses colliding and forming one mass at rest.

The momentum conservation equation in the center of mass frame gives us

Classically we would expect $M = 2m$. We recover this in the classical limit, but at high speeds, we see the resulting object actually gains some mass. The object at rest after the collision is warmer than the two initial objects, since their kinetic energy was converted to thermal energy. Because of the mass-energy equivalence, this warmer object is more massive than before. As it cools down, it will also lose mass.

We see this effect in reality in e.g. a pion decaying into a muon and a neutrino.