The kinetic energy of a particle that is moving can be calculated using its rest energy, which is the energy when the particle is at rest. Specifically, if a particle has a kinetic energy that is 3 times its rest energy, the kinetic energy can be calculated as:
Quick Answer
If a particle has a rest energy E0, and its kinetic energy is 3 times its rest energy, then its kinetic energy Ek is:
Ek = 3 x E0
Where E0 is the rest energy of the particle and Ek is the kinetic energy.
Explanation
Rest energy is a concept from special relativity that refers to the intrinsic energy of a particle when it is at rest. This rest energy, denoted by E0, remains constant regardless of the motion of the particle. It is related to the rest mass m0 of the particle by the famous equation:
E0 = m0c2
Where c is the speed of light in a vacuum.
Kinetic energy, denoted by Ek, is the energy a particle has due to its motion. For slow moving particles at non-relativistic speeds, the kinetic energy is given by the classical physics equation:
Ek = 1/2mv2
Where m is the relativistic mass and v is the velocity of the particle.
However, for particles moving at relativistic speeds near the speed of light, the kinetic energy must be calculated using relativity theory. In this case, the kinetic energy of a particle is related to its total energy E and rest energy E0 by:
Ek = E – E0
Where the total energy E is given by:
E2 = p2c2 + E02
Here, p is the relativistic momentum of the particle. This velocity-dependent momentum accounts for the relativistic increase in mass at high speeds.
So for a particle with rest energy E0 moving at a relativistic speed v, its kinetic energy is increased compared to the classical case. If the kinetic energy is 3 times the rest energy, then:
Ek = 3E0
This could occur for a particle moving at an extremely high speed, approaching the speed of light. The faster it moves, the greater its kinetic energy becomes relative to the rest energy.
Examples
Here are some examples to illustrate the kinetic energy of a relativistic particle:
Example 1
An electron has a rest energy of 0.511 MeV (million electron volts). If it is moving at a speed where its kinetic energy is 3 times its rest energy, what is the electron’s kinetic energy?
Rest energy E0 = 0.511 MeV
Ek = 3 x E0 = 3 x 0.511 MeV = 1.533 MeV
So the electron’s kinetic energy is 1.533 MeV.
Example 2
A proton has a rest energy of 938.3 MeV. If it has a kinetic energy of 5000 MeV, what is the ratio of its kinetic energy to rest energy?
Rest energy E0 = 938.3 MeV
Kinetic energy Ek = 5000 MeV
Ratio = Ek/E0 = 5000 MeV / 938.3 MeV = 5.32
So the proton’s kinetic energy is 5.32 times its rest energy.
Kinetic Energy Equation Derivation
The kinetic energy equation relating the kinetic energy to the rest energy can be derived from the relativistic energy equations as follows:
Starting with:
E2 = p2c2 + E02
Where E is total energy, p is relativistic momentum, c is speed of light, and E0 is rest energy.
Also, from relativity:
E = γE0
Where γ is the Lorentz factor:
γ = 1/√(1 – v2/c2)
with v being the velocity. Substituting this into the first equation gives:
(γE0)2 = p2c2 + E02
γ2E02 = p2c2 + E02
Solving this for the momentum p:
p = γE0v/c
Where we used the relativistic momentum equation p = γmv
Substituting the momentum back into the energy equation:
(γE0)2 = (γE0v/c)2c2 + E02
γ2E02 = γ2E02v2/c2 + E02
Cancelling terms and rearranging gives:
γE0 – E0 = (γ – 1)E0
Since γE0 is the total energy E, and E0 is the rest energy, this shows:
E – E0 = (γ – 1)E0
But E – E0 is just the kinetic energy Ek, so:
Ek = (γ – 1)E0
This gives the final relationship between kinetic energy and rest energy for a relativistic particle.
When Kinetic Energy Is Much Greater than Rest Energy
In cases where the kinetic energy is much greater than the rest energy, some approximations can be made to the kinetic energy equation.
For example, if Ek >> E0, then γ >> 1. This is because γ gets very large as the velocity v approaches c.
In this case, γ – 1 ≈ γ, since 1 is negligible compared to γ.
So the kinetic energy equation can be approximated as:
Ek ≈ γE0
For a case where Ek = 3E0, then:
3E0 ≈ γE0
γ ≈ 3
So if the kinetic energy is 3 times rest energy, the Lorentz factor gamma is approximately 3.
We can also estimate the velocity using:
γ = 1/√(1 – v2/c2)
With γ ≈ 3, this gives:
v ≈ 0.94c
So the particle has a velocity about 94% the speed of light in this case.
This demonstrates that ultra-relativistic particles with kinetic energies much greater than their rest energy have speeds very close to c.
Significance for High Energy Physics
The relationship between kinetic energy and rest energy is very important in high energy particle physics experiments that investigate the properties of elementary particles. Some key examples include:
- Particle accelerators like the Large Hadron Collider can accelerate protons and ions close to the speed of light, reaching kinetic energies over 1000 times their rest energy. This allows them to collide with enough energy to create massive unstable particles like the Higgs boson.
- Cosmic rays consist of charged particles from space, typically protons or atomic nuclei, reaching Earth with tremendous kinetic energies millions of times their rest energies. This enables them to penetrate deeply into the atmosphere.
- An electron volt (eV) is a common unit of energy in particle physics, and can represent either rest energy or kinetic energy. 1 eV is approximately the rest energy of an electron. Particle energies are often stated in keV (thousand eV), MeV (million eV), GeV (billion eV), etc.
- The annihilation of electron-positron pairs converts their combined rest energies fully into the kinetic energies of the resulting photons per E=mc2. These photon energies can be in the MeV range.
- Neutrinos are nearly massless particles that travel at almost the speed of light. Thus their kinetic energy dominates over their sub-eV rest mass.
Being able to relate the readily measurable kinetic energy to the intrinsic rest energy is key for determining particle masses and energies in high-energy physics experiments and astrophysical observations.
Energy-Momentum Relationship
We can also derive the connection between kinetic energy and momentum for a relativistic particle. Starting again with the relativistic energy equation:
E2 = p2c2 + m02c4
Where p is momentum, m0 is rest mass, c is speed of light, and E is total energy.
The rest energy E0 is related to the rest mass by:
E0 = m0c2
Substituting this into the equation gives:
E2 = p2c2 + E02
Again, kinetic energy Ek is given by:
Ek = E – E0
So the energy equation can be written:
(E0 + Ek)2 = p2c2 + E02
Expanding this out and rearranging:
Ek2 + 2EkE0 = p2c2
Dividing through by c2 gives:
(Ek2/c2) + (2EkE0/c2) = p2
But Ek2/c2 is just the relativistic expression for (momentum)2!
So:
p2 = (Ek2/c2) + (2EkE0/c2)
This shows that for a moving particle, momentum p is dependent on both kinetic energy Ek and rest energy E0.
In the non-relativistic case where Ek << E0, the E0 term dominates, so:
p ≈ (2EkE0/c2)1/2 = (2E0Ek/c2)1/2
Which is the familiar classical momentum equation p = mv.
But in the ultra-relativistic case Ek >> E0, the kinetic energy term dominates:
p ≈ (Ek2/c2)1/2
This illustrates how at very high energies, momentum is approximately proportional to the square root of the kinetic energy.
Tables of Example Values
Below are some tables with example values for the rest energy, kinetic energy, total energy, and momentum of a sample relativistic particle at different velocities.
Energies
Velocity (v/c) | Rest Energy (MeV) | Kinetic Energy (MeV) | Total Energy (MeV) |
---|---|---|---|
0 | 10 | 0 | 10 |
0.1 | 10 | 0.5 | 10.5 |
0.5 | 10 | 15 | 25 |
0.9 | 10 | 162 | 172 |
0.99 | 10 | 1410 | 1420 |
Momenta
Velocity (v/c) | Rest Energy (MeV) | Momentum (MeV/c) |
---|---|---|
0 | 10 | 0 |
0.1 | 10 | 1.58 |
0.5 | 10 | 15.81 |
0.9 | 10 | 162.3 |
0.99 | 10 | 1414 |
This demonstrates how at relativistic velocities, both kinetic energy and momentum increase dramatically compared to the rest energy and rest mass.
Conclusion
In summary, the kinetic energy Ek of a relativistic particle with rest energy E0 is related to its rest energy by:
Ek = (γ – 1)E0
Where γ is the Lorentz factor. If the kinetic energy is much greater than rest energy, γ ≈ Ek/E0.
So for a particle with kinetic energy 3 times its rest energy, Ek = 3E0 and γ ≈ 3. This occurs at speeds approaching the speed of light c.
The concepts of rest energy and variable relativistic mass are key to understanding the behavior of particles at velocities near c, which is essential for high energy particle physics experiments and astrophysics.