Contents
Special Relativity
© The scientific sentence. 2010
 Total enegy of any particle
1. Preliminary
1. The Binomial series
(1 + x)^{n} can be written as
the Taylor series, that is:
(1 + x)^{n} = 1 + nx +(1/2)n(n  1)x^{2} + ...
when x < 1.
2. Total energy:
E = mc^{2} = γm_{0}c^{2} =
m_{0}c^{2}/[1  u^{2}/c^{2}]^{1/2} =
m_{0}c^{2} [1  u^{2}/c^{2}]^{ 1/2}
Here, in our case, x =  u^{2}/c^{2}, n =  1/2,
and v << c ( the velocity of the particle is very
smaller that c is). Therefore:
E = m_{0}c^{2} [1 + (1/2)u^{2}/c^{2} +
(3/8)u^{4}/c^{4} + ... ]
= m_{0}c^{2} + m_{0}c^{2}[(1/2)u^{2}/c^{2} +
(3/8)u^{4}/c^{4} + ... ] = E_{0} + KE,
with:
E_{0} = m_{0}c^{2}, the energy at rest, and
KE = m_{0}c^{2}[ (1/2)u^{2}/c^{2} +
(3/8)u^{4}/c^{4} + ... ]
the kinetic energy of the prticle.
The more we go on the terms of the series, the less these
terms significant become. We disregard them from the second term
for the linear momentum and from the third one regarding the
total energy, and write at the first order:
KE = m_{0}c^{2} (1/2)u^{2}/c^{2}
= (1/2) m_{0}u^{2}
KE = (1/2)m_{0}u^{2}
We find the well known results from Classical Mechanics.
E_{0} + KE
means that to move a particle, it's necessary at first to
have at least its energy at rest E_{0} = m_{0}c^{2}.
Even at rest, an object has energy, that is E_{0} = m_{0}c^{2},
Mass is energy.
3. The relativistic linear momentum
The relativistic linear momentum is P = mu
m is the relativistic mass (= γm_{0})
and u is the velocity of the particle) m and u are
measured in the rest frame.
γ = 1/[1  u^{2}/c^{2}]^{1/2}.
2.Total energy and the Linear momentum
We are interested here to find the relation between
the total energy E and the relativistic Linear
momentum P.
γ = 1/[1  u^{2}/cγ = 1/[1  u^{2}/c^{2}]^{1/2}]^{1/2}
1/ (γ^{2}  1 ) = c^{2}/γ^{2}u^{2}
E^{2} = m^{2}c^{4} = P^{2}c^{4}/u^{2}
= P^{2}c^{2}. c^{2}/u^{2}
c^{2}/u^{2} = γ^{2}/(γ^{2}  1)
Then:
E^{2} = P^{2}c^{2}[(γ^{2}  1 + 1)/γ^{2}  1]
= P^{2}c^{2}[1 + 1/(γ^{2}  1)]
P^{2}c^{2}/(γ^{2}  1) = m_{0}^{2}c^{4}
It follows that:
E^{2} = P^{2}c^{2} + m_{0}^{2}c^{4}
E^{2} = P^{2}c^{2} + m_{0}^{2}c^{4}
The case of photons
When the velocity of a particle approches the value of c,
γ becomes infinite and its related mass as well.
The mass becomes infinite does not mean that it weights
infinity!. It just mean that when a particle moves faster
and its speed approches c, its inertial mass becomes strong
enough to be stopped.
For a photon, the mass at rest in null, then: E = P c
For low speeds, v/c << 1, we have:
P = m_{0}u (classic)
E^{2} = m_{0}^{2}u^{2} + m_{0}^{2}c^{4} =
m_{0}^{2}c^{4}[1 + u^{2}/c^{2}].
Therefore:
E = m_{0}c^{2}[1 + u^{2}/2c^{2}]
= m_{0}c^{2} + (1/2)m_{0}u^{2}
For a photon:
E = P c = hν = hc/λ

