Quiz: On the reading
Energy
0
0
E=
u
0
c
1. At low velocities, the value E of the new
definition must match the classical definition.
E
2
K + mc
0
d)
E
c)
c
b)
γmc2 =
Similar to the definition of the relativistic
momentum we want to find a definition for the
energy E of an object that fulfills the following:
c
u
c
u
2. The total energy (ΣE) of an isolated system of
bodies must be conserved in all inertial frames.
E
a) 
E
Which graph best represents the total energy
of a particle (particle's mass m>0) as a
function of its velocity u, in special relativity?
u
0
Kinetic energy
The work done by a force F to move a particle
from position 1 to 2 along a path s is defined by:
F≡
2
W12 =
∫ F ⋅ ds = K
2
K=
− K1
1
ds=udt
d
= m ∫ ud (γ u) = γ mc 2 − mc 2
0
Integration by parts
!
d
∫ F ⋅ ds = ∫ dt (γ mu) ⋅ u dt = m ∫ dt dt (γ u)u)
γu
K1,2 being the particle's kinetic energy at positions
1 and 2, respectively (true only for frictionless
system!).
Using our prior definition for the relativistic force we
can now find the relativistic kinetic energy of the
particle.
dp d
= (γ m u)
dt dt
Relativistic kinetic energy
The relativistic kinetic energy K of a particle with a rest
mass m is:
K = γmc2 - mc2 = (γ-1)mc2
Note: This is very different from the classical K=
! K = (γ-1)mc2 ≈ (1 + ½v2/c2 – 1)mc2 = ½ mv2
Rest energy of a particle
E = γmc2 = K + mc2
In the particle's rest frame, its energy is its rest
energy, E0. What is the value of E0?
0
c2
mc2
(γ-1)mc2
½ mc2
We rewrite the equation for the relativistic
kinetic energy and define the total energy of a
particle as:
E = γmc2 = K + mc2
½mv2 .
For slow velocities the relativistic energy equation gives the
same value as the classical equation! Remember the
binomial approximation for γ: γ ≈ 1+ ½v2/c2
A:
B:
C:
D:
E:
Total energy
Note: This suggests a connection
between mass and energy!
This definition of the relativistic mass-energy E
fulfills our condition of conservation of total energy.
(Not proven here, but we shall see several
examples where this proves to be correct.)
Relation between Mass and Energy
m
v
-v
m
E2 = γmc2 = K + mc2
E1 = γmc2 = K + mc2
Total energy:
Etot = E1+E2 = 2K + 2mc2
Equivalence of Mass and Energy
m
v
-v
m
Conservation of the total energy requires that the final
1 is the same as the energy Etot, before
energy Etot,final
2
2
thetot
collision. final
Therefore:
initial
E =γ
Etot,final =
Mc = 2K
Mc2 ≡
2K +
Example:
Rest energy of an object with 1kg
+ 2mc
2mc2 =
Etot,initial
We find that the total mass M of the final system is larger
than the sum of the masses of the two parts! M>2m.
E0 = mc2 = (1 kg)·(3·108 m/s )2 = 9·1016 J
9·1016 J = 2.5·1010 kWh = 2.9 GW · 1 year
This is a very large amount of energy! (Equivalent to the
yearly output of ~3 very large nuclear reactors.)
Potential energy inside an object contributes to its mass!!!
Enough to power all the homes in Colorado for a year!
E0 = mc2
OK, so what could we do with it?
What Dr. Evil didn't know
(fortunately)
Review: Relativistic mechanics
Relativistic momentum:
p≡m
Relativistic force:
F≡
Total energy of a
particle with mass ‘m’:
dr
dt propper
= γ mu
dp d
= (γ m u )
dt dt
Etot = γmc2 = K + mc2
We found that these definitions fulfill the momentum and
energy conservation laws. And for u<<c the definitions
for p, F, and K match the classical definitions. But we
found that funny stuff happens to the proper mass ‘m’.
E=mc2 ! Convert mass to energy?
Atomic cores are built from neutrons and protons. There
are very strong attractive forces between them. The
potential energy associated with the force keeping them
together in the core is called the binding energy EB.
We now know that the total rest energy of the particle
equals the sum of the rest energy of all constituents minus
the total binding energy EB:
Mc2 = Σ(mi c2) – EB
Or in terms of Mass per nucleon
Mc2 = Σ(mi c2) – EB