Contents
Special Relativity
© The scientific sentence. 2010
 Relativity: MomentumEnergy 4Vector
1. Velocity 4vector
We need to use the time derivative of the components of the 4vector 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^{μ}V_{μ} =  c^{2} + v^{2}, 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^{μ}V_{μ} = γ^{2} ( c^{2} + v^{2})
=  c^{2}; which is invariant.
γ = 1/[1  β^{2}]^{1/2}, where v is the velocity
of the moving frame defined as x = vt from the rest frame.
V^{μ}V_{μ} =  c^{2}
The magnitude of the velocity 4Vector 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 4Vector.
V^{μ} = γ (c, u) is a velocity 4Vector
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  β^{2}]^{1/2})/dv =
v/c^{2} [1  (v/c)^{2}]^{3/2} =
(v/c^{2}) (γ^{3})
Therefore:
dm/dv = m_{o} (v/c^{2}) (γ^{3}) =
m (v/c^{2}) (γ^{2}) =
m v/ (c^{2}  v^{2})
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) =
[(c^{2}  v^{2})/mv](dm/dt) =
[(c^{2}  v^{2})/mv](dm/dt)
So
F = [(c^{2}  v^{2})/v](dm/dt) + v dm/dt =
(c^{2}/v)dm/dt
The energy of the particle as the received work:
dE = F dx = c^{2}dm
Integrating gives:
E = c^{2}m + constant.
If there is no mass then there is no energy. Hence
Constant = 0; and
we find again the equivalence massenergy formula:
E = c^{2}m
E = m c^{2}
3. Momentum 4Vector
According to the result from the velocity 4Vector, we define
the momentum 4Vector as the product of the velocity 4Vector 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 4Vector because we have multiplied by a constant.
P^{μ} is a 4Vector.
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 c^{2}, we
write:
P^{μ} = m (c, v) = (mc, mv) = (mc, p) =
(mc^{2}/c , p) = (E/c,p)
P^{μ} = (E/c,p)
P^{μ} = (E/c,p)
V^{μ}V_{μ} =  c^{2}
is the velocity 4Vector Lorentzinvariant.
P^{μ}P_{μ} =  m_{o}^{2}c^{2}
is the momentum 4Vector Lorentzinvariant.
Therefore, the Lorentzinvariant of the moment becomes:
=  (E/c)^{2} + p^{2} =  m_{o}^{2}c^{2}, or
 E^{2} + p^{2}c^{2} =  m_{o}^{2}c^{4}
That is:
E^{2} = p^{2}c^{2}+ m_{o}^{2}c^{4}
E^{2} = p^{2}c^{2}+ m_{o}^{2}c^{4}

