## A System of Particles

As we have seen the mass of an object is a measure of its energy content when stationary. When an object is considered stationary it is the centre of mass that is stationary. An object is, more often than not, a system of particles confined to a volume. For example it could be a block of wood, or a metal container full of air, both have mass which could be measured. Both objects are full of atoms. When we weight the object we are weighting the mass contribution 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 kinetic 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.

Previously, from the energy triangle, the kinetic energy, KE, does not change the mass of a particle but is a part of the overall energy of a particle in motion. Does kinetic energy contribute to mass, or does it not?

The key word in the highlighted sentence above is the plural of atom, atoms. The sentence is referring to many atoms. If from the system of atoms you can construct a stationary centre of mass then all the energies of the atoms contribute to the mass of the system of particles. Just like confined photons contribute to the mass of an object, but one single photon cannot have a stationary centre of mass, and so a single photon does not have mass.

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.

Lets look at this concept a little more closely. For one atom the kinetic energy does not contribute to the mass, but for a system of atoms the kinetic energy of the random motions 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 kinetic energy in the object. The centre of mass of the object is stationary. 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 would be moving. The whole object would be in motion, then 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 an 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 of the object. The wonderful thing about relativity is these two concepts are unified into the one relationship.

### E^{2} = M^{2} + P^{2}

## Conservation rules

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

Consider a radioactive radon-198 atom, initially at rest, that undergoes alpha decay. The masses of the atoms involved are shown in atomic mass units (u).

radon-198 | → | polonium-194 | + | helium-4 |

197.999u | 193.988 u | 4.00260 u |

The total energies and momenta on each side of the equation are equal. The momentum of the radon atom is zero so the total momentum of the right side is zero. The energy of the radon is measured by its mass, which is the total energy available to the right side of the equation.

We shall look at the energy triangles to illustrate the energy components of this decay. The left side of the illustration depicts the energy triangle for the radon at rest before the decay. To the right at the two energy triangles for the polonium and helium atoms. The scale of the triangles have been altered for illustrative purposes.

The energy on the left of the equation, or before decay, is simply the energy of the particle at rest. So

### M_{r} = E_{r} = m_{r}c^{2}

### P_{r} = 0

E_{r} must equal the magnitude of the total energy of the polonium, E_{p} and helium, E_{h} added together.

### M_{r} = E_{r} = E_{p} + E_{h}1

However the mass energy of the polonium, M_{p}, and the helium, M_{h} do not add up to to the original energy E_{r}, as can be seen in the illustration. The difference

### M_{r} - (M_{p} + M_{h} )2

is called the *mass defect*.

It can be easily shown that the mass defect is equal to the kinetic energy on the right side of the equation. From equation 1.

### M_{r} = E_{p} + E_{h}

Refer to the energy triangles.

### M_{r} = (M_{p} + KE_{p}) + (M_{h}+ KE_{h})

### M_{r} = (M_{p} + M_{h}) + (KE_{p} + KE_{h})

### M_{r} - (M_{p} + M_{h}) = (KE_{p} + KE_{h})

Referring to equation 2.

### mass defect = (KE_{p} + KE_{h})

Referring to the energy triangle in the previous section it can be seen that the mass defect gives the total kinetic energy after the decay. The particles must be in motion. The momenta, P_{p} and P_{h} must be equal and opposite to give a total momentum of zero.

It can also be seen that the kinetic energy of the helium will be much greater than the polonium.

## example exercises

If the kinetic energy of the polonium atom produced is 2.55 × 10^{−14} J.
By considering mass defect, calculate the kinetic energy of the alpha particle, and
explain why it is significantly greater than that of the polonium atom.

The energy on both sides of the equation must be equal.

On the left side all the energy is in the form of mass.

On the right the energy is divided between two particles.

We will see that the total mass on the right is less than the mass on the left.

Each particle on the right must have some momentum so the total energies are equal on both sides of the equation.

The mass defect is the difference in the mass between the left and right side. Because the particle is at rest in the beginning the mass defect is the amount of energy available to kinetic energy on the right.

### Mass defect = 197.99 - 193.988 - 4.00260

### Mass defect = 0.0084u

use E=mc^{2} to convert mass to energy.

### KE = 0.0084 * 1.661x10^{-27} * c^{2} = 1.23x10^{-12}J

This is very much larger than 2.55x10^{-14}.

Most of the kinetic energy goes into the Helium to conserve momentum.

#### Photon Emission

In this question we have a spaceship of mass M_{1} cruising to the left with total energy E_{1} and momentum P=pc. It emits an energetic photon of energy E_{g} 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.

### M_{1}^{2} = E_{1}^{2} - P_{1}^{2}

After emission of the photon.

### E_{2} = E_{1} - E_{g}

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

### P_{1} = P_{2} + P_{g}

For a photon the energy and momentum are the same.

### E_{g} = P_{g}

For the second triangle

### M_{2}^{2} = E_{2}^{2} - P_{2}^{2}

Substitute from above

### M_{2}^{2} = (E_{1} - E_{g})^{2} - (P_{1} - P_{g})^{2}

### M_{2}^{2} = E_{1}^{2} -2E_{1}E_{g}+ E_{g}^{2} - P_{1}^{2} + 2P_{1}E_{g} - E_{g}^{2}

### M_{2}^{2} = M_{1}^{2} - 2E_{g}(E_{1} - P_{1})

The last equation compares M_{2} to M_{1}. M_{2} is always smaller that M_{1} because the last term is always positive 2E_{g}(E_{1} - P_{1}). 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.

### E_{2} = E_{1} - E_{g}

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

### P_{2} = P_{1} + P_{g}

### E_{g} = P_{g}

### M_{2}^{2} = E_{2}^{2} + P_{2}^{2}

### M_{2}^{2} = (E_{1} - E_{g})^{2} - (P_{1} + P_{g})^{2}

### M_{2}^{2} = E_{1}^{2} -2E_{1}E_{g}+ E_{g}^{2} - P_{1}^{2} - 2P_{1}E_{g} - E_{g}^{2}

### M_{2}^{2} = M_{1}^{2} - 2E_{g}(E_{1} + P_{1})

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

### 2E_{g}(E_{1} + P_{1}) ≥ 2E_{g}(E_{1} - P_{1})

c :If we measure the same photon energy in 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.

d : Show that the photon emitted in the direction of travel has undergone a Doppler shift. The energy of the Doppler shifted photon is E_{g}. The energy of the photon relative to the spacecraft is X. Show that

### E_{g} = X (1-v^{2})^{½}/(1-v)

Use the fact that v = P_{1}/E

_{1}and E

_{1}/M

_{1}= 1/(1 - v

^{2})

^{½}

Take the case in question a : where the initial momentum is zero, P_{1} = 0. The photon energy is X and initial energy of the spacecraft is the same as its mass. The initial energy E_{1} = M_{1}. With this in mind modify equation 14.

### M_{2}^{2} = M_{1}^{2} - 2 X M_{1}

so from equation 14 when the spacecraft has momentum P_{1.}

### M_{2}^{2} = M_{1}^{2} - 2E_{g}(E_{1} - P_{1})

so

### E_{g}(E_{1} - P_{1}) = X M_{1}

### E_{g} = X M_{1}/(E_{1} - P_{1})

divide top and bottom by E_{1}

### E_{g} = X (M_{1}/E_{1}) / (E_{1}/E_{1} - P_{1}/E_{1})

### E_{g} = X (1 - v^{2})^{½}/(1 - v)

#### Gamma Ray Absorption

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.

- E
_{g}is the energy of the gamma ray. - P = pc, p is the momentum of the gamma ray.
- E
_{n}is the energy of of the nucleus before the collision - M
_{a}= m_{a}c^{2}, m_{a}is the mass of the nucleus before the collision - M
_{b}= m_{b}c^{2}, m_{b}is the mass of the nucleus after the collision

Energy conservation requires the total energy remains the same before and after the collision.

### E_{g} + E_{n}

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 M_{b} = 1.01M_{a}.

After the collision

### (E_{g} + E_{n})^{2} = P^{2} + M_{b}^{2}

Substitute E_{n} = M_{a}, E_{g} = P, M_{b} = 1.01 M_{a}

### (E_{g}+ M_{a})^{2} =E_{g}^{2} + (1.01M)_{a}^{2}

### E_{g}^{2}+2M_{a}E_{g}+M_{a}^{2}=E_{g}^{2}+1.0201M_{a}^{2}

### 2M_{a}E_{g} +M_{a}^{2} = 1.0201M_{a}^{2}

### E_{g} = 0.0201M_{a}^{2}/2M_{a}

### E_{g} =0.01005M_{a}

The energy of the gamma ray needs to be greater than 0.01M_{a} 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

and

###
(m + KE)^{2} = m^{2} + p^{2}

###
(3m)^{2} = m^{2} + p^{2}

###
9m^{2} - m^{2} = p^{2}

### p = √8m

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

### p = √8m c

### 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 = 2.83x10^{8}m/s