Mass energy equivalence.
By the Work-Energy Theorem:
W = KE
Where W is work done,KE is kinetic energy,
In this case, we will consider PE = 0
Where PE is potential energy
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
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
m=
The relativistic mass is:
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(
0
v
m0 v
v
1- 2
c
2
) = m0 ò vd(
0
v
v
1- 2
c
2
)
Applying the Quotient Rule:
(keep in mind v is a variable and c is a constant)
v2
1
v 2 - 12 1
(dv) 1 - 2 - v( )(1 - 2 ) (- 2 )(2vdv)
v
c
2
c
c
d(
)=
2
2
v
v
(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
v
and denominator of
1-
v
c2
2
by
v2
(1 - 2 )
c
v2
(1 - 2 )
c
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
3
2 2
(c 2 - v )
3
c
v
E = m0 ò
0
c 3v
3
2 2
E = m0 c
dv
(c 2 - v )
v
3
dv
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 (
c
2
E = mo c (
2
c -v
2
1
2
- 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
Recall the relativistic mass:
Substituting this yields:
or
2
m=
m0
v2
(1 - 2 )
c
E = c2 (m - m0 )
E = (m - mo )c2