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)x2 + ...
when |x| < 1.
2. Total energy:
E = mc2 = γm0c2 =
m0c2/[1 - u2/c2]1/2 =
m0c2 [1 - u2/c2]- 1/2
Here, in our case, x = - u2/c2, n = - 1/2,
and v << c ( the velocity of the particle is very
smaller that c is). Therefore:
E = m0c2 [1 + (1/2)u2/c2 +
(3/8)u4/c4 + ... ]
= m0c2 + m0c2[(1/2)u2/c2 +
(3/8)u4/c4 + ... ] = E0 + KE,
with:
E0 = m0c2, the energy at rest, and
KE = m0c2[ (1/2)u2/c2 +
(3/8)u4/c4 + ... ]
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 = m0c2 (1/2)u2/c2
= (1/2) m0u2
KE = (1/2)m0u2
We find the well known results from Classical Mechanics.
E0 + KE
means that to move a particle, it's necessary at first to
have at least its energy at rest E0 = m0c2.
Even at rest, an object has energy, that is E0 = m0c2,
Mass is energy.
3. The relativistic linear momentum
The relativistic linear momentum is P = mu
m is the relativistic mass (= γm0)
and u is the velocity of the particle) m and u are
measured in the rest frame.
γ = 1/[1 - u2/c2]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 - u2/cγ = 1/[1 - u2/c2]1/2]1/2
1/ (γ2 - 1 ) = c2/γ2u2
E2 = m2c4 = P2c4/u2
= P2c2. c2/u2
c2/u2 = γ2/(γ2 - 1)
Then:
E2 = P2c2[(γ2 - 1 + 1)/γ2 - 1]
= P2c2[1 + 1/(γ2 - 1)]
P2c2/(γ2 - 1) = m02c4
It follows that:
E2 = P2c2 + m02c4
E2 = P2c2 + m02c4
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 = m0u (classic)
E2 = m02u2 + m02c4 =
m02c4[1 + u2/c2].
Therefore:
E = m0c2[1 + u2/2c2]
= m0c2 + (1/2)m0u2
For a photon:
E = P c = hν = hc/λ
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