### Energy and Mass

### E = mc^{2}

*unifying mass and energy*

## Introduction

In this section we shall consider mass, momentum and energy as described by the theory of relativity.

In September of 1905 Einstein published a paper entitled

"Does the inertia of an object dependent on its energy content?"

The paper concluded that

### E = mc^{2}1

Where m is the mass of an object, while at rest, and E is the energy contained in the mass. This is an amazing unification and may change your concept of mass. Einstein was so surprised and unsure about this relationship he entitled his paper as a question.

After 1905 Einstein was to develop a simpler derivation of this equation through an elegant thought experiment. It is that second derivation we shall explore in the following sections. Ultimately we shall conclude that mass is confined energy.

Firstly we need to cover some fundamental concepts. We shall explore momentum and inertia before deriving the famous equation.

## Light and Momentum

The momentum of light directly led Einstein to a deeper understanding of mass and energy. In about 1873 James Clarke Maxwell's predicted that light has momentum, as derived from his equations of electromagnetism. In this description light is a propagating electric wave with a transverse magnetic component. The electric wave induces a free charge to oscillate transverse to the direction of propagation of the wave. The charge also oscillates through the magnetic component in the wave which induces the charge to move in the direction of propagation. This motion gives the charge a net momentum in the direction of motion of the light. For the light to impart momentum the light itself can be thought of as having momentum. Compton showed in the early 1920s that light having momentum was consistent with a particle description of light. A particle of light is called a photon. Momentum is conserved when a photon encounters a charged particle. The momentum of light, p, is related to the energy carried by the light by the following equation.

### p = E / c

E is the energy of a photon and c is the velocity of light.

Though the momentum of a photon is less familiar than the momentum of a massive object, it is actually more fundamental. A photon always has momentum, it is a fundamental property of a photon, whereas a massive object can have zero momentum by having zero velocity.

As stated above the mass of an object is measured while at rest. It is then clear that a photon is defined as not having mass, as it cannot be at rest. If a photon collides with an object and absorbed there is nothing left, no mass to measure. So a photon has momentum but no mass.

## Objects at Rest

Einstein incorporated into his thought experiment the fact that when an object is stationary it will remain stationary unless acted upon by a force. More strictly, it is the centre of mass of the object that remains fixed, even if parts of the object are in motion. As demonstrated by the following video of a NASA astronaut in the space station Skylab. As he floated fixed in Skylab he flurried his arms and legs about, but his centre of mass remained in the same position. He finally escapes his confinement by being pulled to freedom by a fellow crew member.

## Centre of Mass

Firstly, let us consider the centre of mass of an object. It is a point often associated with a 3 dimensional object and so the point has a location in 3 dimensional space. In the following example we will consider a one dimensional case. Consider a uniform beam balanced on a fulcrum. For the beam to be balanced on the fulcrum there must be an equal amount of the beam on either side of the fulcrum. The position of the fulcrum is the centre of mass of the beam. We can only find the location of the centre of mass along the beam and so we are finding the location of the centre of mass in one dimension.

If we add more mass to the beam to maintain a balance then the distribution of mass must follow a strict rule. The total "moment" of the beam must be zero. The moment is a measure of the torque about an axis. The moment is expressed as the displacement, d, of a mass from the fulcrum multiplied by the mass, m.

### moment = d m

Let us assign negative values for the displacement to the left. A displacement to the left will induce an anticlockwise rotation while a displacement to the right will induce a clockwise rotation.

Let us calculate the total moment on the beam induced by two weights.

In the above illustration

### a = -b

The moment generated by the right mass is

### moment = b m

The moment generated by the left mass is

### moment = a m

The total moment generated by the both masses is

### total moment = a m + b m = 0

With a = -b the moment is 0 and so the beam is balanced.

Consider the next illustration.

The moment generated by the right mass is

### moment = b m

The moment generated by the left mass is

### moment = (a/2) (2m) = a m (2/2) = a m

The total moment generated by the both masses is

### total moment = a m + b m = 0

The moment is 0 and so the beam is balanced.

Consider the next illustration.

The moment generated by the right mass is

### moment = b m

The moment generated by the left mass is

### moment = (a/3) (3m) = a m (3/3) = a m

The total moment generated by the both masses is

### total moment = a m + b m = 0

The moment is 0 and so the beam is balanced.

#### Mass Transfer

In the following animation a block of mass, M, includes a much smaller mass, m. The block starts off balanced and is then displaced by a distance d. By moving the mass m from one end of the block to the other, over a distance L, balance can be regained. It is important to realise that the center of mass of the system at the start and at the finish is in the same position, above the fulcrum.

Assuming that m is very much smaller than M show that the balanced state can be approximated by the expression

### M d + m L = 0 2

By regaining a balance the centre of mass is once again directly over the fulcrum.

This result will be useful to us later.

For the initial unbalanced system the moment is

### M d

d is a negative value and so there is a negative moment.

Then m is taken from the left side at a displacement of - L/2 + d. This subtracts a moment from the left side. However because the left displacement is negative the removal of m from the left adds a positive moment.

### M d - m(-L/2 + d)

The mass m is then placed on the right at a displacement of L/2 + d to create a balanced system.

### M d - m(-L/2 + d) + m(L/2 + d) = 0

Reduce

### M d + m L = 0

This is the same as equation 2.

## The Thought experiment

In the following thought experiment Einstein assumed Newtonian dynamics are fine for low velocities and energies.

Imagine a box sitting on a fulcrum. The box is balanced on the fulcrum and is stationary. The fulcrum must sit directly below the centre of mass for the box to be balanced. The box has a mass of M and a length of L, across the interior walls.

A photon is emitted from the left wall and is absorbed by the right wall. As seen in the animation this induces the box to move to the left.

On emission the photon has a momentum

### p = E/c 3

and will impart an equivalent momentum in the opposite direction to the box. The box will move to the left with a momentum E/c. Since the mass of the box is M, the momentum of the box will be Mv, where v is the velocity of the box. From equation 3 the momentum of the box is

### - M v = E/c4

v is of opposite sign to c.

However, a closed system like the box cannot induce a movement of its centre of mass. The box must remain balanced. To maintain a balance there must be a transfer of mass by the photon from the left to right side of the box. Let us call the amount of this mass m.

#### Energy Mass Equivalence

Show that the mass transferred, m, is related to the energy, E, of the photon by

### E = mc^{2}

Use the result of the exercise called Mass Transfer, and assume, to a very good approximation, that the length of time, t, for the light to transverse the box is L/c.

*hint*: multiply both sides of equation 4 by t.

From equation 4. The momentum conservation equation.

### - M v = E / c

To equate this with equation 2, multiply both sides by t

### - M v t = E t / c

With d = vt and t = L/c

### - M d = E L / c^{2}

Mass has been transferred as in the Mass Transfer exercise . So from equation 2

### M d + m L = 0

### -M d = m L

then from the above two equations that equal -M d.

### E L / c^{2} = m L

### E / c^{2} = m

### E = m c^{2}

The confined photon with energy E has transferred a mass, m, from the left to right side of the box.

## Mass

The photon must transfer mass, m, from one side of the box to the other. In reality many photons would be traversing the box in every direction , transferring mass, but the box will always remaining balanced.

We have assumed that m is very, very small compared to M. The energy of the photon is associated with a mass, in equation 1, that contributes to the total mass of the box. This mass m is not in the form of a particle like an electron, proton etc. No electron or proton is transferred across the box. Energy was transferred and that energy confined to the box is what we consider as mass.

A free photon is not confined and has no mass.

The concept of mass has evolved since Einstein first derived his famous equation. Ever since about 1992 mass has been considered to be the confined energy content of an object at rest. The energy trapped within a system at rest is a measure of its mass. The concept of relativitistic mass, the concept that mass increases with velocity, has gone out of favour. There is only one mass associated with an object and that is the mass of the object at rest. E=mc^{2} only applies to the rest mass of an object. In all equations in relativity m is only the mass of an object at rest. In the example above the photon is confined to the box and so its energy contributes to the mass of the box. As we shall see all energy in the box contributes to the mass of the box when its centre of mass is stationary. The mass of the box is a measure of the energy content of the box in a reference frame in which it is at rest. It could be measured by weighing the box. If the energy content of the box is constant the weight of the box is constant.

Using this concept of mass it is conceivable that the mass of an object could be principally made up of confined photons.

Confined energy is the concept of mass that we will explore in the coming sections.

#### Energy of Mass

Consider a mass of 1kg. What does Einstein's equation say about its energy content?

### E = m c^{2}

### c = 299791458 m/s

then

### c^{2} = 299792458 * 299792458

### c^{2} = 89,875,517,873,681,764

This is a very large number. Substitute into Einstein's equation, with m=1.

### E = m c^{2}

### E = 1 * 89,875,517,873,681,764

### E = 89,875,517,873,681,764 Joules

1kg of mass could supply all the energy needs of Australia for 8 days. That is domestic, industrial and all transport. If only we had a way to convert it to useful energy we would easily satisfy all our energy needs.