Derivation of E=mc2
Yum-Yum Physics!
By the Work-Energy Theorem:
W = KE
In this case, we will consider PE = 0 so:
W = KE = E
From the definition of work:
W = F(Dx)
E = F(Dx)
x
Therefore:
E = ò F dx
0
From Newton’s original statement of F=ma:
dm
dv
F=
v+m
dt
dt
d
F = (mv)
dt
This, along with the previous equation yields:
x
E = ò F dx
0
x
d
E = ò (mv)dx
dt
0
We have dt and dx in the equation.
Let’s write it all in terms of dt:
Dx dx
v=
=
Dt dt
dx = vdt
When the variable changes from dx to dt,
we must also change the bounds of the integral
x
t
d
d
E = ò (mv)dx = ò (mv)v dt
dt
dt
0
o
We can eliminate the dt’s so:
Notice that the dt changed to d(mv) so the bounds also changed
For velocities approaching c, the mass increases
The relativistic mass is: m =
mv
E=
mv
ò vd(mv) = ò vd(
0
0
m0
v2
1- 2
c
m0 v
v
1- 2
c
2
)
Since m0 is a constant, it can factored out of the
integral:
mv
E=
ò vd(
v
m0 v
) = m0 ò vd(
v
)
v
v
0
1- 2
1- 2
c
c
Applying the Quotient Rule:
(keep in mind v is a variable and c is a constant)
0
2
2
v2
1
v 2 - 12 1
(dv) 1 - 2 - v( )(1 - 2 ) (- 2 )(2vdv)
v
c
2
c
c
d(
)=
2
2
v
v
Yikes!
(1 - 2 )
1- 2
c
c
Combining these expressions for Energy yields:
v
E = m0 ò vd(
0
v
v
1- 2
c
2
)
v2
1
v 2 - 12 1
(- 2 )(2vdv)
v (dv) 1 - 2 - v( )(1 - 2 )
c
2
c
c
E = m0 ò v
2
v
0
(1 - 2 )
c
v3
v
2
v
c
E = m0 ò (
+
)dv
3
2
2
v 2
v
0
1 - 2 (1 - 2 )
c
c
To get a common denominator, we will multiply
the numerator
2
v
v
and denominator of
v
1- 2
c
2
by
(1 -
)
c
v2
(1 - 2 )
c
2
v3
v
2
v
c
E = m0 ò (
+
)dv
2 3
2
v 2
v
0
1 - 2 (1 - 2 )
c
c
v2
v3
v (1 )v
2
2
c
c
E = m0 ò [
+
]dv
2 3
2 3
v 2
v 2
0
(1 - 2 )
(1 - 2 )
c
c
Simplifying:
v2
v3
v (1 )v
2
2
c
c
E = m0 ò [
+
]dv
2 3
2 3
v 2
v 2
0
(1 - 2 )
(1 - 2 )
c
c
v3 v3
v v+ 2
2
c
c dv
E = m0 ò
2 3
v
0
(1 - 2 ) 2
c
v
v
E = m0 ò
dv
2 3
v 2
0
(1 - 2 )
c
v
v
E = m0 ò
dv
2 3
v 2
0
(1 - 2 )
c
v
v
E = m0 ò 2
dv
2 3
c
v 2
0
( 2 - 2)
c
c
v
v
E = m0 ò 2
dv
2 3
c -v 2
0
( 2 )
c
v
E = m0 ò
0
v
3
2 2
(c 2 - v )
c3
dv
v
E = m0 ò
0
v
0
c 3v
3
2 2
E = m0 c
dv
(c 2 - v )
v
3
dv
(c - v )
c3
2
v
E = m0 ò
3
2 2
ò
0
v
3
2 2
(c - v )
2
dv
To evaluate the integral we make the substitution:
u = c2 - v2
Therefore:
du
= -2vdv
dv
1
- du = vdv
2
v
vdv
3
E = m0 c ò
3
0
(c 2 - v 2 ) 2
u(v)
E = m0 c 3
ò
u(v= 0)
1
- du
2
u
3
2
1
- du
2
u(v)
ò
E = m0 c 3
u
u(v= 0)
m0 c
E=2
2
2
3 c -v
ò
du
u
c2
-
3
2
3
2
1
2
m0 c u c2 - v2
E=(
|c2 )
1
2 2
3
E = mo c (
1
3
c -v
2
2
-
1
c
2
)
1
E = mo c (
3
c2 - v2
1
-
c2
)
1
1
E = mo c (
- )
c2 - v2 c
3
E = mo c 3 (
c
c c -v
2
E = mo c 3 (
-
2
c2 - v2
c c -v
2
c - c2 - v2
c c -v
2
2
)
2
)
E = mo c 3 (
E = mo c 2 (
E = mo c 2 (
c - c2 - v2
c c2 - v2
c - c2 - v2
c2 - v2
c
c2 - v2
E = mo c (
2
c
c2 - v2
)
)
c2 - v2
c2 - v2
- 1)
)
E = mo c (
2
E = mo c (
2
c
c2 - v2
1
- 1)
- 1)
1 2
c - v2
c
1
2
E = mo c (
- 1)
1 2
( 2 )(c - v 2 )
c
1
2
E = mo c (
- 1)
2
v
(1 - 2 )
c
E = mo c (
2
1
- 1)
v
(1 - 2 )
c
m0
2
E=c (
- m0 )
2
v
(1 - 2 )
c
2
Recall the relativistic mass: m =
m0
v2
(1 - 2 )
c
Substituting this yields: E = c2 (m - m0 )
or
E = (m - mo )c2
Make
everything
as simple as
possible,
but not
simpler!
Second Derivation
Consider a box with a light source of
negligible mass at the left end and a black
absorber of negligible mass at the other
end. The total mass of the box, source,
and absorber is M. a Photon is emitted
from the left. It strikes the absorber and
is absorbed.
By conservation of momentum,
when the photon leaves the
source, and travels right, the box
must move to the left to maintain
a total momentum of 0 in the
system. We will use two methods
to calculate how far the box
moves, and equate the results.
The energy of a photon is given by:
E = hf =
hc
l
From de Broglie’s Equation:
h
h
l= =
p mv
Thus:
hc
E=
h
mv
E = cmv
E
= mv = p
c
By the Law of Conservation of Momentum, the momentum
of the photon equals the momentum of the box
E
p = MV =
c
E
V=
Mc
From its definition, the velocity of the photon is given
by the displacement divided by the time
L
c=
t
L
t=
c
E
The velocity of the box is: V =
Mc
L
The amount of time that the box moves is: t =
c
The distance the box moves is:
d = Vt
E L
d=
Mc c
EL
d=
Mc 2
From the mass-energy equivalence, the photon has
mass since it has energy. Let this mass be m (m<<M).
Since there is no external force on the system, the
center of mass must remain in the same place.
Therefore the moments of the box and photon must
have the same value:
Md = mL
mL
d=
M
Equating the two expressions for d yields:
mL EL
d=
=
M
Mc 2
E
m= 2
c
E = mc2