Relativistic Energy

Relativistic Kinetic Energy

We know that an object in motion relative to an observer has energy E = (γm₀)c² and its rest energy is E₀ = m₀c²
- The kinetic store is the energy store associated with motion
- If an object in motion has a total energy E which is greater than its rest mass E₀, the additional energy must be in the kinetic store

$E_{k} = m c^{2} - m_{0} c^{2}$

Where:
- E_k is energy in the kinetic store
- m is relativistic mass
- m₀ is the object's rest mass
- c is the speed of light
This can be rearranged for an expression for the total energy, E, of a particle:

$E = m c^{2}$

$E = m_{0} c^{2} + E_{k}$

The total energy of an object increases at a rapidly increasing rate as it's speed approaches the speed of light

Graph showing how total energy increases as an object's speed approaches the speed of light

12-3-9-energy-vs-speed

As with mass on the previous page, the asymptote on this graph shows that an infinite amount of energy is needed to make an object with mass reach the speed of light

Worked example

A spacecraft with a proper mass of 450 kg accelerates to a speed of 0.7c. Calculate the difference in its relativistic kinetic energy and its kinetic energy calculated using Newtonian physics.

Answer:

Step 1: List the known quantities:

Rest mass, m₀ = 450 kg
Speed, v = 0.7c
Speed of light, c = 3.0 × 10⁸ ms^-1

Step 2: List the relevant equations:

Newtonian kinetic energy, E_kn = ½m₀v²
Relativistic kinetic energy, E_kr = γm₀c² − m₀c²

Step 3: Calculate the Newtonian kinetic energy:

Substitute the known quantities:

$E_{k n} = \frac{1}{2} \times 450 \times {(0.7 \times (3.0 \times 10^{8}))}^{2} = 9.9 \times 10^{18} J$

Step 4: Calculate the gamma term:

Substitute the speed into the gamma term:

$γ = \frac{1}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$

$γ = \frac{1}{\sqrt{1 - \frac{{(0.7 c)}^{2}}{c^{2}}}} = \frac{1}{\sqrt{1 - 0 . 7^{2}}} = 1.40$

Step 5: Substitute the gamma term into the relativistic kinetic energy equation:

Substitute the rest mass, speed of light and speed as well:

$E_{k r} = γ m_{0} c^{2} - m_{0} c^{2} = m_{0} c^{2} (γ - 1)$

$E_{k r} = 450 \times {(3.0 \times 10^{8})}^{2} \times (1.40 - 1) = 1.6 \times 10^{19} J$

Step 6: Calculate the difference between the two kinetic energies:

Subtract the smaller Newtonian kinetic energy from the larger relativistic energy

$Δ E_{k} = (1.6 \times 10^{19}) - (9.9 \times 10^{18}) = 6.1 \times 10^{18} J$

As you can see, the relativistic kinetic energy is almost twice as large as the Newtonian kinetic energy!

Relativistic Energy (AQA A Level Physics)

Revision Note