Energy and Mass

E = mc2

unifying mass and energy

Light and momentum

The proceeding topics led to the conclusion that light carries inertia. This is not inconsistent with the concept that light has momentum. The prediction that light has momentum came about 1862 when James Clarke Maxwell developed his equations of electromagnetism. In this description light is an electromagnetic wave. The electric wave can induce 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 the light. This motion gives the charge a net momentum in the direction of propagation of the light. For the light to impart momentum it must itself have momentum. A consequence of Isaac Newton's theory. The momentum of light, p, is related to the energy carried by the light by the following

p = E / c

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

After Einstein's September 1905 paper entitled "Does the inertia of an object dependent on its energy content" he developed another thought experiment to reach the conclusion that

E = mc2

Utilizing the fact that light possesses momentum and that momentum is conserved. It is this thought experiment I wish to outline here.The associated animation depicts this though experiment. Prior to this I shall outline some of the fundamental physics required to understand the experiment.

Objects at Rest

Einstein incorporated into his thought experiment the fact that when an object is stationary it will remain stationary unless a force acts on the object. More strictly it is the centre of mass of the object that remains fixed. As the following video of Alan Shepard in Skylab demonstrates. As he floats fixed in the Skylab he can flurry about but his centre of mass remains 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 within a 3 dimensional object. In the following example we will consider a one dimensional case. Consider a beam balanced on a fulcrum. For the beam to be balanced on the fulcrum there must be equal amounts of the beam either side of the fulcrum. The point in the middle is the centre of mass of the beam.

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 a point multiplied by the mass, m.

moment = d x m

Assigning negative numbers for the displacement to the left because it will rotate the beam the opposite direction to a displacement on the right.

Let us calculate the total moment on the beam in the following illustration.

The moment generated by the right mass is

moment = d x m

The moment generated by the left mass is

moment = -d x m

The total moment generated by the both masses is

moment = -d x m + d x 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 = d x m

The moment generated by the left mass is

moment = (-d/2) x (2m) = -d x m x (2/2) = -d x m

The total moment generated by the both masses is

moment = -d x m + d x 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 = d x m

The moment generated by the left mass is

moment = (-d/3) x (3m) = -d x m x (3/3) = -d x m

The total moment generated by the both masses is

moment = -d x m + d x m = 0

The moment is 0 and so the beam is balanced.

We have enough physics now to understand Einstein's thought experiment and the accompanying animation.

Thought experiment

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.

A photon is emitted from the left wall and is absorbed by the right wall. On emissions the photon has a momentum

p = E/c

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 and the momentum of the box will be Mv, where v is the velocity of the box. From equation 1 the momentum of the box is

Mv = E/c

The velocity of the box is then

v = E/Mc

Now it can be assumed the velocity of the photon will be substancially greater than the velocity of the box. So to a good approximation the box moves to the left for a time, t, equivalent to the length of time it takes the light to transverse the distance L.

t = L/c

In that time the box has moved to the left a distance, d.

d = t v

Using equation 3 and 4.

d = EL/Mc2

Using equation 6, the moment of the box displaced by distance d is

-dM = -MEL/Mc2 = -EL/c2

This is a net negative moment so the box would start rotating anticlockwise.

However the act of transferring a photon from the left side of the box cannot cause the box to rotate. The center of mass must still be above the fulcrum and the box must remain balanced. For this to happen the photon must have transferred mass from the left side to the right side of the box. Let us assign m as the mass that is transferred and assume it is very, very small compared to M.

As the photon leaves the left side the moment take from the left must be mL/2. It is a positive value because taking mass away from the left side would cause the box to rotate clockwise. When the photon lands on the right side the moment induced would be mL/2. Once again inducing a clockwise rotation. This total clockwise moment mL must balance the moment in equation 7

mL - dM = 0

dM = mL

EL/c2 = mL

Rearrange to get.

E = mc2

The photon must transfer mass, m, from one side of the box to the other. In the animation photons continue to oscillate from right to left. The box always remaining balanced.

We have assumed that m is very, very small compared to M. The total mass of the box is actually M+m. The energy of the photon is associated with a mass, in equation 11, 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 transfered and that confined energy is what we consider as 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 at rest. E=mc2 only applies to the rest mass of an object. 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. 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.

It is this concept of mass that we will explore in the coming sections.

Energy, Mass and Momentum

Mass is a measure of the energy content of a something when it is stationary. As the momentum of an object increases so does its energy but its mass does not.

Before continuing I wish to revisit the time versus velocity triangle that illustrates the magnitude of the time dilation effect. In the diagram τ is an interval of time on the spacecraft. t is the interval of time measured by an observer and v is the velocity of the spacecraft relative to the observer. In many texts the interval of time would be written as dτ or Δτ. I am simplifying the notation so τ and t are intervals of time.

For example, Bob on the spacecraft measures a time interval of 1 second between ticks of his clock, i.e. τ = 1. If the spacecraft is moving passed Alice with a velocity v=0.5c, then Alice will measure 1/0.866 = 1.155 seconds between ticks on Bob's clock. Alice will experience the spacecraft clock as running slow.

Click on the items in the list below to update the triangle. Progress from a time dilation triangle to an energy triangle.

  1. the original time dilation triangle.
  2. express v as x/t. x is the distance the spacecraft travels in time interval t.
  3. the two ts cancel leaving x.
  4. divide each side by τ
  5. multiple each side by mass, m.
  6. multiple each side by c.


  • mass moving through space has momentum, m x / τ
  • mass moving through time has energy mc2 t / τ.

The vertical red side of the triangle defines mass, m. The energy of the mass at rest is mc2. This includes all the different energies contained within the mass. To simplify the notation in the diagrams and equations I am going to define the symbol M.

M = mc2

The horizontal blue side of the triangle defines momentum, p = m x / τ. Note that the velocity is different to the classical velocity of x/t. Once again I am going to simply the notation by defining the symbol P.

P = pc = (mx/τ)c2

The total energy of the mass is the hypotenuse, E = mc2 t / τ. This energy is the sum of the rest energy mc2 and kinetic energy, KE, of the mass.

E = (mc2) t/τ = M t/τ = M + KE

From the triangle it is easy to see the following

E2 = M2 + P2

Note that when m is zero then E = pc. This is the relationship for light. Light does not have any mass.

Let us now consider a mass in motion. We will derive an expression for KE at low velocities. Using equation 12 and 13

(M + KE)2 = M2 + P2

2 M KE + KE2 = P2

If KE is very small then KE2 is extremely small in comparison. As an approximation we shall remove this term. A small KE also implies a small momentum term

2 M KE ≈ P2

Using the classic definition for P = m x/τ c ≈ mvc and M = mc2

2mc2KE ≈ (mvc)2

2mKE ≈ (mv)2

KE ≈ ½mv2

This is the classic definition for kinetic energy. All of relativity reduces to classic physics at low velocities. It is important to realise that this not the correct expression for kinetic energy because we have made approximations to reach this expression.

example exercises

The OMG particle

In October 1991 a high energy cosmic ray, a proton, was measured to have an energy of 51 Joules. The highest energy measured for any particle up until that time. The particle was named the OMG particle.


a: How many proton masses are equivalent to the energy of the OMG proton? The mass of a proton is 1.67 × 10-27kg and for your solution approximate c = 3x108m/s


E/M = E/mc2 = 51 / ( 1.67 x 10-27 x (3x108)2) = 3.39x1011

The energy of the proton was 3.39x1011 times more than the rest energy of the proton.


b : If we assume the proton travelled for 10 million years, our time, from its place of orgin, what amount of time did the proton experience for the trip.


From the time dilation triangle above, t/τ = E/M = E/mc2

t/τ = 51 / ( 1.67 x 10-27 x (3x108)2)

t/τ = 3.39x1011

with 10 million years equivalent to 3.15x1014 seconds.

τ = 3.15x1014/3.39x1011

τ = 929 seconds

τ is about 15 minutes.

The proton experienced 15 minutes for the entire trip!


c : What was the velocity of the particle, relative to light?


From the time dilation triangle the distance travelled divided by the observer time interval is the velocity. The length of the green side divided by the blue side. Since the energy triangle is a scaled version of the time dilation triangle then the velocity is P/E.

v = P/E = ((pc)2/E2)½

From 13

v =((E2 - M2)/E2)½ = (1 - (M/E)2)½

From questions a and b, E/M2 = 3.39x1011

v = ( 1 - (1/3.39x1011)2)½

= 0.999999999999999999999995649c

A photon that left at the same time as the proton would have only gained about 43cm on the proton after 10 million years.


d : Virtually all the energy of the particle was in the form of Kinetic Energy. So E = KE = 51 Joules. Assume the proton is moving at c. From question c this is a very good approximation. What would the classic kinetic energy expression give for the energy of the proton?


The classic expression for kinetic energy is ½mv2.

½mv2 = ½ x 1.67 x 10-27 x (3x108)2 = 7.5x10-11 J

This is very different to 51 Joules!

Energy of Mass


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


c = 299791458 m/s


c2 = 299792458 * 299792458 = 89,875,517,873,681,764

This is a very large number. Using Einstein's equation.

E = m c2 = 1 * 89,875,517,873,681,764 = 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.

Light Momentum


A 1 metre square of aluminium foil is floating in orbit about Earth. If the sheet is facing the Sun what momentum does it gain every second from the sunlight? The power from the Sun is 1368 Watts/m2 and consider the aluminium as a perfect reflector.


The light has no rest mass. All the energy E is in the form of KE. The momentum is E/c.

The momentum gained every sec from the sunlight is E/c = 1368/3x108 = 4.56x10-6 kg m s-1

The aluminium reflects the light back towards the sun which gives the sheet another push away from the Sun.

The change in momentum per second is 2 x 4.56x10-6 = 9.12x10-6 kg m s-1

Change in momentum per second is a force of 9.12x10-6 N.

A System of Particles

As we have seen the mass of an object is a measure of its energy content when stationary. As we have seen in previous section when an object is considered stationary it is the centre of mass which is stationary. An object is, more often than not, a system of particles confined to a volume. It could be a block of wood, or a metal container full of air, both have weight which could be measured. Both objects are full of atoms. When we weight the object we are weighting the sum of weights of all the atoms. We are also weighting all the other energies contained within the object. For example all the atoms themselves are in motion. The energy, in the form of heat, is stored in the random jiggling motion of the atoms. The kinetic energy of these atoms contributes to the mass of the object.

This last sentence may cause some confusion after reading the last few sections. Does kinetic energy contribute to mass or does it not? Consider a single atom. The mass of the atom is a measure of the energy content of the atom at rest. The same atom in motion has kinetic energy, but this energy does not contribute to the mass of the atom. However in the previous paragraph the last sentence is

The kinetic energy of these atoms contributes to the mass of the object.

There may seem to be a contradiction. For one atom the kinetic energy does not contribute to the mass but for a system of atoms the kinetic energy of the random motion of the atoms does contribute to the mass. The difference is between an individual atom compared to a system of confined atoms to make an object. The random motion of the atoms in an object confines the energy in the object. If the motion of the atoms were not random, but all the atoms moved in a uniform way in one direction, then the centre of mass of the object is moving, the whole objet is in motion, there is a net momentum of the whole object, and this does not contribute to the mass of the object. This difference between confined random motion within a object and the bulk motion of an object is an important distinction when considering the conservation rules of momentum and energy.

Relativity maintains the classic conservation rules for momentum and energy though their definitions may be different. For a system of objects, in a confined volume, the total momentum in 3D space, along coordinates x, y and z, are individually conserved. In other words the total momentum in x is conserved, the total momentum in y is conserved and the total momentum in z is conserved. Also the total energy of a system of objects is conserved as long as no energy leaves the volume. The wonderful thing about Relativity is these two concepts are unified into the one relationship in equation 13.

E2 = M2 + P2

Conservation rules

The total energy and total momentum of a closed system remains constant.

Photon Emission

a: Referring back to the section on Energy and Inertia. In this question we have a spaceship of mass M1 cruising to the left with total energy E1 and momentum P=pc. It emits an energetic photon of energy Eg in the direction of travel. The g stands for gamma ray. By conserving the total energy and momentum before and after the emission show that the mass of the spacecraft must decrease with the emission.


Below is depicted the energy triangles for the spacecraft. On the left is the triangle before emitting the photon and on the right the energy triangle for the spacecraft and photon after emitting the photon. The final triangle has the momentum and energy of the photon subtracted from the original spacecraft triangle. The triangle for the photon has no mass and so is not a triangle. The energy and momentum of the photon are equal. The momentum of the photon is in the same direction as the momentum for the spacecraft. I have depicted the momentum as vectors to indicate the direction of travel and that momentum adds as vectors.

M12 = E12 - P12

After emission of the photon.

E2 = E1 - Eg

The total momentum adds vectorially. The sum of the final momenta equals the initial momentum.

P1 = P2 + Pg

Eg = Pg

M22 = E22 - P22

M22 = (E1 - Eg)2 - (P1 - Pg)2

M22 = E12 -2E1Eg+ Eg2 - P12 + 2P1Eg - Eg2

M22 = M12 - 2Eg(E1 - P1)

The last equation compares M2 to M1. M2 is always smaller that M1 because the last term is always positive 2Eg(E1 - P1). The spacecraft looses mass by emitting a photon.

b: Compare now the mass if the same energy photon is emitted in the opposite direction to the motion of the spacecraft.

After emission of the photon.

E2 = E1 - Eg

The total momentum adds vectorially. The sum of the final momenta equals the initial momentum.

P2 = P1 + Pg

Eg = Pg

M22 = E22 + P22

M22 = (E1 - Eg)2 - (P1 + Pg)2

M22 = E12 -2E1Eg+ Eg2 - P12 - 2P1Eg - Eg2

M22 = M12 - 2Eg(E1 + P1)

c: Interestingly the spacecraft looses more mass if it emits the same photon opposite to the direction of motion.

2Eg(E1 + P1) ≥ 2Eg(E1 - P1)

If we measure the same photon energy is both cases why is there a difference in mass loss?


Mass is defined at rest with respect to the spacecraft. At rest with respect to the spacecraft the two photons are not the same energy. The energy of the two photons in the above questions are measured as emitted from a moving spacecraft. In that reference frame the energies are the same. With the spacecraft moving the photons experience the Doppler effect. The photon emitted in the direction of travel increases in energy and the opposite photon decreases in energy. At rest with respect to the spacecraft a weaker photon needs to be emitted in the direction of travel than the opposite photon for them both to be measured with the same energy as measured in the frame where the spacecraft is moving.

c: Show that the photon emitted in the direction of travel has undergone a Doppler shift. The energy of the Doppler shifted photon is Eg. The energy of the photon relative to the spacecraft is X. Show that

Eg = X (1-v2)½/(1-v)

Use the fact that v = P1/E1 and E1/M1 = 1/(1 - v2)½

Take the case in question a: where the initial momentum is zero, P1 = 0. The photon energy is X and initial energy of the spacecraft is the same as its mass. The initial energy E1 = M1. With this in mind modify equation 14.

M22 = M12 - 2 X M1

so from equation 14 when the spacecraft has momentum P1.

M22 = M12 - 2Eg(E1 - P1)


Eg(E1 - P1) = X M1

Eg = X M1/(E1 - P1)

divide top and bottom by E1

Eg = X (M1/E1) / (E1/E1 - P1/E1)

Eg = X (1 - v2)½/(1 - v)

Gamma Ray Absorption

a: Consider a gamma ray absorbed by a stationary nucleus. If the mass of the nucleus is initially m and after the absorption is 1.01m. What is the energy of the photon expressed in units of m.


Below are the energy triangles that describe the energies before the collision and after the collision. They are not to scale.

The left side of the diagram is before the collision, the right side after the gamma ray is absorbed by the nucleus.

  • Eg is the energy of the gamma ray.
  • P = pc, p is the momentum of the gamma ray.
  • En is the energy of of the nucleus before the collision
  • Ma = mac2, ma is the mass of the nucleus before the collision
  • Mb = mbc2, mb is the mass of the nucleus after the collision
  • Energy conservation requires the total energy remains the same before and after the collision.

    Eg + En

    Before the collision the total energy is shared between two particle. After the collision the total energy is contained within one particle, the nucleus. Energy is conserved.

    The momentum before the collision is to the right with magnitude p, the momentum after the collision is to the right with magnitude p. Momentum is conserved.

    The problem states that Mb = 1.01Ma.

    After the collision

    (Eg + En)2 = P2 + Mb2

    Substitute En = Ma, Eg = P, Mb = 1.01 Ma

    (Eg+ Ma)2 =Eg2 + (1.01M)a2

    Eg2+ 2MaEg +Ma2 =Eg2 + 1.0201Ma2

    2MaEg +Ma2 = 1.0201Ma2

    Eg = 0.0201Ma2/2Ma

    Eg =0.01005Ma

    The energy of the gamma ray needs to be greater than 0.01Ma because the final total energy needs to contribute to both the mass and momentum of the final nucleus. As seen in the final energy diagram.

The Particle Accelerator

A proton is accelerated so that it gains 2 times its mass in kinetic energy.


a: What is the protons momentum?


KE = 2m


(m + KE)2 = m2 + p2

(3m)2 = m2 + p2

9m2 - m2 = p2

p = √8m

to express this in conventional units we need to multiply it by c.

p = √8m c


b: What is the velocity of the proton?


v = p/E = √8m/(3m) = √8/3

where v is relative to c. So in conventional units the velocity is (√8/3) c = 0.943 x 299792458 = 2.83x108m/s


A photon has an energy of

a: What is the protons momentum?