## Energy, Mass and Momentum

We will now consider the relationship between energy, mass and momentum. Relativity unified all these concepts together. However I wish also to explore a deeper unification with space and time.

#### Space, Time, Energy, Mass and Momentum

I want you to consider the characteristics of Space, and the characteristics of Time, and how we move through them. These characteristics do not necessarily have to be related to relativity but can be related purely to your current understanding.

Now consider the characteristics of Energy and the characteristics of Momentum.

Of Energy and Momentum which is more Space like and which is more Time like?

Mass is a measure of the energy content of something when its centre of mass is stationary. An object has energy when stationary but no momentum. As the object moves the momentum of the object increases so does its energy, but its mass does not increase. This increase in energy is called the Kinteic Energy.

Before continuing I wish to revisit the time versus velocity triangle that illustrates the magnitude of the time dilation effect. In this discussion I will use τ as an interval of time on the spacecraft. For example the interval between two ticks of a clock. t is the interval of time between the spacecraft ticks as measured by an observer outside the spacecraft, 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 a 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

### ((1c)^{2} + (0.5c)^{2})^{½} = 1.12c

Alice measures t = 1.12 seconds between ticks on Bob's clock. Alice will experience Bob's spacecraft clock as running slower than hers.

Another way of expressing this is Alice experiences more time than Bob. Alice experiences 1.12 seconds for every 1 second of time experienced by Bob.

Below is an interactive graphic illustrating how Space and Time are related to Energy and Momentum by simply rescaling the time dilation triangle.

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

- the original time dilation triangle.
- express v as x/t. x is the distance the spacecraft travels in time interval t.
- the two ts cancel leaving x.
- divide each side by τ
- multiply each side by mass, m.
- multiply each side by c.

Fundamentally

- mass moving through time has energy mc
^{2}t / τ. - think of an object travelling in time like a momentum restricted to one direction, always going forward.
- traditionally γ = t/τ so the total energy of an object is mc
^{2}γ - mass moving through space has momentum, m x / τ

So movement of an object through time gives it energy. The more it moves through time the more energy it possesses, mc^{2} t / τ. Rest energy mc^{2} multiplies by t / τ. Where t / τ is always equal to 1 or greater. In the example above the observer, Alice, measures the spacecraft as moving through 1.12 seconds of time for each second of spacecraft time. Alice measures the spacecraft as having 1.12 times more energy than Bob on the spacecraft. The extra energy is in the form of kinetic energy.

A spacecraft is travelling with a velocity of 0.866c. What is the energy of the spacecraft in comparison to its rest mass?

At 0.866c time is travelling half the rate on board the spacecraft, i.e. t/τ = 2. Therefore the energy of the spacecraft is twice the energy of the rest mass energy.

So for every kilogram of spacecraft mass a kilogram of energy needs to be given to the spacecraft to get it to 0.866c. As previously calculated 1kg of mass at rest is equivalent to the energy consumption of Australia over 8 days. So to get the spacecraft to 0.866c requires the energy consumption of Australia over 8 days for every 1kg of spacecraft mass.

## simplify symbols

We will now simplify the symbols used in the triangle. The vertical orange side of the triangle defines mass, m. The energy of the mass at rest is mc^{2}. 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 = mc^{2}

The total energy, E, of the mass is the hypotenuse of the triangle,

### E = mc^{2} t / τ = M γ

This energy is the sum of the rest energy mc^{2} and kinetic energy, KE, of the mass.

### E = M + KE1

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 simplify the notation by defining the symbol P.

### P = pc = (mx/τ)c = E v

v is now the classical velocity relative to the speed of light, (x/t)/c. v is a velocity divided c and so does not have any units. v is then P/E.

From now on I will use the symbol v for x/t. So v = v/c.

From the triangle it is easy to see the following

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

Note that when m is zero

### P = E

This is the relationship for light.

## Classical Kinetic Energy

Let us now consider a mass in motion , with velocity v. Which is x/t, measured in metres per second. We will derive the classical expression for KE.

### KE = ½mv^{2}

The classical expression only applies when v, x/t in metres per second, is very small compared to c. Using equation 1 and 2

###
(M + KE)^{2} = M^{2} + P^{2}

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

In our daily existence the kinetic energy of an object is very small compared to the energy content of an object. With KE very small compare to M then the term KE^{2} is very small in comparison to the M KE term. As an approximation we shall remove this term.

###
2 M KE ≈ P^{2}

A small KE also implies a small momentum term. Using the classic definition for P = m x/τ c ≈ m x/t c = mvc with v expressed in metres per second. Also M = mc^{2}

###
2mc^{2}KE ≈ (mvc)^{2}

### 2mKE ≈ (mv)^{2}

### KE ≈ ½mv^{2}

This is the classical expression for kinetic energy. All of relativity reduces to classical physics at low velocities. It is important to realise that this is not the correct expression for kinetic energy and is only an approximation for low velocities.

## 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 cosmic ray up until that time. The particle was named the OMG particle. ( the Oh My God particle )

a : How many proton masses are equivalent to the energy of the OMG proton? The mass of a proton is 1.67 × 10^{-27}kg and for your solution approximate c = 3x10^{8}m/s

### E/M = E/mc^{2}

### E/M = 51 / (1.67x10^{-27}x(3x10^{8})^{2})

### E/M = 3.39x10^{11}

The energy of the proton was 3.39x10^{11} 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 origin, what amount of time did the proton experience for the trip.

From the time dilation triangle above, t/τ = E/M, refer to question a:

### t/τ = E/M = 3.39x10^{11}

with 10 million years equivalent to 3.15x10^{14} seconds.

### τ = 3.15x10^{14}/3.39x10^{11}

### τ = 929 seconds

τ is about 15 minutes.

The proton experienced about 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

From equation 2.

###
v =((E^{2} - M^{2})/E^{2})^{½} = (1 - (M/E)^{2})^{½}

From questions a and b, E/M = 3.39x10^{11}

###
v = ( 1 - (1/3.39x10^{11})^{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. Approximately 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 ½mv^{2}.

###
½mv^{2} = ½ x 1.67x10^{-27}x(3x10^{8})^{2}

###
½mv^{2} = 7.5x10^{-11} Joules

This is very different to 51 Joules!

The classic expression for kinetic energy underestimates the true value of the kinetic energy. This is an extreme example where the particle is almost travelling at c.

#### Light Momentum

A 1 metre square of aluminium foil is floating in orbit about Earth.

a : If the sheet is facing the Sun what momentum does it gain every second from the Sunlight?

The power from the Sun at Earth is 1368 Watts/m^{2}. Consider the aluminium is a perfect reflector.

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

### p = E/c

The momentum gained every sec from the sunlight is

### E/c = 1368/3x10^{8}

### E/c = 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 * E/c = 2 x 4.56x10^{-6}

### 2 * E/c = 9.12x10^{-6} kg m s^{-2}

The change in momentum per second is a force of 9.12x10^{-6} N.

b : The mass of the sheet is 44g. What is the kinetic energy of the sheet after one week? Use the energy triangle to calculate the value for the kinetic energy. Also calculate the kinetic energy using the classic expression ½ mv^{2}. Compare the values.

### 1 week = 604,800 seconds

From part a: the change in momentum per seconds is 9.12x10^{-6}. After one week

### 604,800 * 9.12x10^{-6} = 5.516 kg m s^{-1}

Referring to the energy triangle P = pc

### P = pc = 5.516 * 3x10^{8}= 1.655x10^{9} J

### P = 1.655x10^{9} J

The energy of the mass is

### M = mc^{2} = 0.044 * (3x10^{8})^{2} = 3.96x10^{15}J

### M = 3.96x10^{15}J

The total energy of the sheet is

### E = (M^{2} + P^{2})^{½}

### E = ((3.96x10^{15})^{2} + (1.655x10^{9})^{2})^{½}

### E = 3.960000000000346x10^{15}

Kinetic energy as determined by relativity

### KE = E - M

### KE = 346 J

Classic Kinetic energy is ½mv^{2}

### v = P/E = 4.179x10^{-7}c

### v = 125.4 ms^{-1}

### classical KE = ½x0.044x(125.4)^{2} = 345.7J

### classical KE = 345.7J

With the sheet moving at 125ms^{-1} the classic kinetic energy gives a value approximately 0.1% less than the correct relativistic value.