Relativity: Momentum-Energy 4-Vector

1. Velocity 4-vector

We need to use the time derivative of the components of the 4-vector displacement. That is d(ct,x,y,z)/dt. We have with x = v t:

V^μ = d(ct, x)/dt = (c,v)
So
V^μV_μ = - c² + v²,
which is not invariant. To become invariant, we use instead the proper time derivative. We recall that the proper time is the time in the rest frame of the particle; that is t' that we will denote by τ.
Then, with t = γ t'= γ τ:

V^μ = d(ct, x)/dt' = γ d(ct, x)/dt = γ (c, u)
So, dotting it into itself gives:
V^μV_μ = γ² (- c² + v²)
= - c²; which is invariant.

γ = 1/[1 - β²]^1/2, where v is the velocity of the moving frame defined as x = vt from the rest frame.

V^μV_μ = - c²

The magnitude of the velocity 4-Vector is c, no matter the velocity v is.

Therefore:

V^μ = d(ct, x)/dt' = γ d(ct, x)/dt = γ (c, v)
is a velocity 4-Vector.

V^μ = γ (c, u) is a velocity 4-Vector

2. Relativistic energy

The relativistic energy E of the particle of mass at rest m_o, then of mass m = γm_o, when moving at v, is derived as:

m = γm_o

dm/dv = m_o dγ/dv

dγ/dv = d(1/[1 - β²]^1/2)/dv =
v/c² [1 - (v/c)²]^-3/2 = (v/c²) (γ³)

Therefore:
dm/dv = m_o (v/c²) (γ³) = m (v/c²) (γ²) =
m v/ (c² - v²)

The force acting on the particle is F = dP/dt
F = dP/dt = d(mv)/dt = m dv/dt + v dm/dt =

We have:
dv/dt = (dv/dm)(dm/dt) = [(c² - v²)/mv](dm/dt) = [(c² - v²)/mv](dm/dt)

So
F = [(c² - v²)/v](dm/dt) + v dm/dt = (c²/v)dm/dt


The energy of the particle as the received work:

dE = F dx = c²dm

Integrating gives:

E = c²m + constant.

If there is no mass then there is no energy. Hence Constant = 0; and we find again the equivalence mass-energy formula:
E = c²m

E = m c²

3. Momentum 4-Vector

According to the result from the velocity -4Vector, we define the momentum 4-Vector as the product of the velocity 4-Vector and the mass at rest m_o of the particle moving at v from the frame at rest:

P^μ = m_o V^μ = m_o γ (c, v)

It is still a 4-Vector because we have multiplied by a constant.

P^μ is a 4-Vector.

We write:
m = m_o γ , and then:

P^μ = m(c, v)

P^μ = m (c, v) = γ m_o(c,v)

According to the result for the energy: E = m c², we write:

P^μ = m (c, v) = (mc, mv) = (mc, p) = (mc²/c , p) = (E/c,p)

P^μ = (E/c,p)

P^μ = (E/c,p)

V^μV_μ = - c² is the velocity 4-Vector Lorentz-invariant.
P^μP_μ = - m_o²c² is the momentum 4-Vector Lorentz-invariant.

Therefore, the Lorentz-invariant of the moment becomes:

= - (E/c)² + p² = - m_o²c², or
- E² + p²c² = - m_o²c⁴

That is:

E² = p²c²+ m_o²c⁴

E² = p²c²+ m_o²c⁴