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Energy and Momentum of a Particle
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The four-vector formalism is both powerful and elegant and is used
to derive the important equation that relates the physical quantities
energy, mass and momentum in Special Relativity.
The 4-velocity is the rate of change of both time and space coordinates
with respect to the proper time of the object. The 4-velocity is a
tangent vector to the world line. Thus,
4-vector of velocity, uμ = dxμ/dτ
Time component:
x0 = ct = cγτ
dx0/dτ = cγ
Spacial components:
dτ2 = dt2 - dx2/c2
Rewrite
dτ = √{(dt2 - dx2/c2)/dt2}dt
= √(1 - v2/c2)dt
dt/dτ = 1/√(1 - v2/c2)
dx/dτ = (dx/dt)(dt/dτ) = vx/√(1 - v2/c2) same for y and z
The complete 4-vector is written notationally in terms of the time
and space components as:
dxμ/dτ = {c/√(1 - v2/c2), vx/√(1 - v2/c2)
Alternatively, and generalizing to all directions
uμ = {γc, γv}
Likewise we can identify the 4-momentum as
pμ = {γmc, γmv}
Note: The relativistic mass is defined as γm where m is the invariant/rest/proper
mass. Some physicists reject this as not really meaningful, however.
Einstein recognized p0 when multiplied by c as the KE. Thus
E = mc2/√(1 - v2/c2)
For small v/c expand as binomial.
mc2(1 - v2/c2)-1/2 = (1 + v2/2c2)mc2
E = mc2 + mv2/2
Which is agreement with Newton (rest energy + KE).
Relationship between Energy and Momentum
----------------------------------------
E2 = m2c4/(1 - v2/c2)
Put c = 1 to simplify math.
E2 = m2/(1 - v2)
p2 = m2v2/(1 - v2)
E2 - p2 = m2
Add back in c to make it dimensionally correct.
E2 - p2c2 = m2c4
where m (the rest or invariant mass) is the mass of a body that is isolated (free) and
at rest relative to the observer.
When p = 0, E = mc2
For massless particle m = 0
E = √p2c2 = pc