Calculate the speed of a particle (a) whose kinetic energy is equal to twice its rest energy and (b) whose total energy is equal to twice its rest energy.

To tackle the problem of calculating the speed of a particle under two different conditions—one where its kinetic energy equals twice its rest energy, and the other where its total energy equals twice its rest energy—we need to delve into some fundamental concepts of relativistic physics.

Understanding Rest Energy and Kinetic Energy

First, let's establish the relationship between rest energy, kinetic energy, and total energy. The rest energy (\(E_0\)) of a particle is given by the equation:

E_0 = m_0 c^2

where \(m_0\) is the rest mass of the particle and \(c\) is the speed of light. The total energy (\(E\)) of a particle moving at relativistic speeds is expressed as:

E = \gamma m_0 c^2

Here, \(\gamma\) (the Lorentz factor) is defined as:

\(\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\)

where \(v\) is the velocity of the particle.

Case (a): Kinetic Energy Equals Twice Rest Energy

In this scenario, we have:

Kinetic Energy (KE) = 2 E_0

Using the definition of kinetic energy in relativistic terms:

KE = E - E_0 = \gamma m_0 c^2 - m_0 c^2

This can be simplified to:

KE = (\gamma - 1) m_0 c^2

Setting this equal to \(2 E_0\), we get:

(\gamma - 1) m_0 c^2 = 2 m_0 c^2

Dividing both sides by \(m_0 c^2\), we find:

\(\gamma - 1 = 2\)

Thus, \(\gamma = 3\). Now, we can relate \(\gamma\) back to the velocity:

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

Squaring both sides gives:

9 = \frac{1}{1 - \frac{v^2}{c^2}}

Rearranging leads us to:

1 - \frac{v^2}{c^2} = \frac{1}{9}

Thus:

\(\frac{v^2}{c^2} = \frac{8}{9}\)

Taking the square root provides:

v = c \sqrt{\frac{8}{9}} \approx 0.943c

Case (b): Total Energy Equals Twice Rest Energy

Now, let's analyze the second case where the total energy is twice the rest energy:

Total Energy (E) = 2 E_0

Substituting the total energy expression yields:

\(\gamma m_0 c^2 = 2 m_0 c^2\)

Dividing by \(m_0 c^2\) gives:

\(\gamma = 2\)

Relating this back to velocity, we have:

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

Squaring both sides results in:

4 = \frac{1}{1 - \frac{v^2}{c^2}}

Rearranging leads us to:

1 - \frac{v^2}{c^2} = \frac{1}{4}

This implies:

\(\frac{v^2}{c^2} = \frac{3}{4}\)

Therefore:

v = c \sqrt{\frac{3}{4}} \approx 0.866c

Summary of Findings

To summarize:

• For a particle whose kinetic energy is twice its rest energy, the speed is approximately 0.943c.
• For a particle whose total energy is twice its rest energy, the speed is approximately 0.866c.

These calculations highlight the intriguing interplay between energy and velocity in the realm of relativistic physics. Understanding these relationships allows us to explore the nature of particles moving at high speeds, which is crucial in fields like astrophysics and particle physics.