Derivation of E=mc2

Yum-Yum Physics!

By the Work-Energy Theorem:

W = KEIn this case, we will consider PE = 0 so:

W = KE = EFrom the definition of work:

W = F(!x)E = F(!x)

Therefore: E = Fdx0

x

!

From Newton’s original statement of F=ma:

F =dmdt

v + m dvdt

F =ddt(mv)

E = Fdx0

x

!This, along with the previous equation yields:

E =ddt(mv)dx

0

x

!

We have dt and dx in the equation. Let’s write it all in terms of dt:

v =!x!t

=dxdt

dx = vdt

When the variable changes from dx to dt, we must also change the bounds of the integral

E =ddt(mv)dx = d

dt(mv)vdt

o

t

!0

x

!

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 =m0

1! v2

c2

E = vd(mv)0

mv

! = vd(0

mv

!m0v

1" v2

c2

)

Since m0 is a constant, it can factored out of the integral:

Applying the Quotient Rule: (keep in mind v is a variable and c is a constant)

d( v

1! v2

c2

) =(dv) 1! v

2

c2! v(1

2)(1! v

2

c2)!12 (! 1

c2)(2vdv)

(1! v2

c2)

E = vd( m0v

1! v2

c2

) = m00

mv

" vd( v

1! v2

c2

)0

v

"

Yikes!

Combining these expressions for Energy yields:

E = m0 v0

v

!(dv) 1" v

2

c2" v(1

2)(1" v

2

c2)"12 (" 1

c2)(2vdv)

(1" v2

c2)

E = m0 ( v

1! v2

c2

+

v3

c2

(1! v2

c2)32

)dv0

v

"

E = m0 vd( v

1! v2

c2

)0

v

"

To get a common denominator, we will multiply the numerator and denominator of by

v

1! v2

c2

(1! v2

c2)

(1! v2

c2)

E = m0 [(1! v

2

c2)v

(1! v2

c2)32

+

v3

c2

(1! v2

c2)32

]dv0

v

"

E = m0 ( v

1! v2

c2

+

v3

c2

(1! v2

c2)32

)dv0

v

"

Simplifying:

E = m0 [(1! v

2

c2)v

(1! v2

c2)32

+

v3

c2

(1! v2

c2)32

]dv0

v

"

E = m0

v ! v3

c2+ v

3

c2

(1! v2

c2)320

v

" dv

E = m0v

(1! v2

c2)320

v

" dv

E = m0v

(1! v2

c2)320

v

" dv

E = m0v

(c2

c2! v

2

c2)320

v

" dv

E = m0v

(c2 ! v2

c2)320

v

" dv

E = m0v

(c2 ! v2 )32

c30

v

" dv

E = m0v

(c2 ! v2 )32

c30

v

" dv

E = m0c3v

(c2 ! v2 )320

v

" dv

E = m0c3 v

(c2 ! v2 )320

v

" dv

To evaluate the integral we make the substitution:

u = c2 ! v2

dudv

= !2vdv

!12du = vdv

Therefore: E = m0c3 vdv

(c2 ! v2 )320

v

"

E = m0c3

! 12du

u32u(v=0)

u(v)

"

E = m0c3

! 12du

u32u(v=0)

u(v)

"

E = !m0c

3

2du

u32c2

c2 !v2

"

E = !m0c

3

2(u

!12

! 12

|c2c2 !v2 )

E = moc3( 1

c2 ! v2!1c2)

E = moc3( 1

c2 ! v2!1c2)

E = moc3( 1

c2 ! v2!1c)

E = moc3( cc c2 ! v2

!c2 ! v2

c c2 ! v2)

E = moc3(c ! c2 ! v2

c c2 ! v2)

E = moc3(c ! c2 ! v2

c c2 ! v2)

E = moc2 (c ! c2 ! v2

c2 ! v2)

E = moc2 ( c

c2 ! v2!

c2 ! v2

c2 ! v2)

E = moc2 ( c

c2 ! v2!1)

E = moc2 ( c

c2 ! v2!1)

E = moc2 ( 11c

c2 ! v2!1)

E = moc2 ( 1

( 1c2)(c2 ! v2 )

!1)

E = moc2 ( 1

(1! v2

c2)!1)

E = moc2 ( 1

(1! v2

c2)!1)

E = c2 ( m0

(1! v2

c2)! m0 )

Recall the relativistic mass: m =m0

(1! v2

c2)

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:

From de Broglie’s Equation:

! =hp=

hmv

E = hf = hc!

Thus: E =hchmv

E = cmvEc= mv = p

By the Law of Conservation of Momentum, the momentum of the photon equals the momentum of the box

p = MV =Ec

V =EMc

From its definition, the velocity of the photon is given by the displacement divided by the time

c = Lt

t = Lc

The velocity of the box is:

The amount of time that the box moves is:

V =EMc

t = Lc

The distance the box moves is:

d = Vt

d =EMc

Lc

d =ELMc2

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

d =mLM

Equating the two expressions for d yields:

d =mLM

=ELMc2

m =Ec2

E = mc2