Question

An extremely relativistic particle is one whose kinetic energy is much larger than its rest energy. Show that for an extremely relativistic particle $E \approx p c$.

Short answer

Answer: For an extremely relativistic particle, the total energy E is approximately equal to its momentum p times the speed of light c, i.e., E ≈ pc.

Step-by-step solution

Write down the relativistic energy-momentum relation

We have the energy-momentum relation for a relativistic particle as follows: \(E^2 = (mc^2)^2 + (pc)^2\), where E is the total energy, m is the rest mass, c is the speed of light, and p is the momentum of the particle.

Define the kinetic energy for an extremely relativistic particle

Given that kinetic energy (K) is much larger than the rest energy, we can write the total energy as \(E = mc^2 + K \approx K\).

Apply the approximation for an extremely relativistic particle

Since the kinetic energy is much larger than the rest energy, we have \(K \gg mc^2\). In other words, \((pc)^2 \gg (mc^2)^2\).

Rewrite the energy-momentum relation with the new approximation

From step 3, we can rewrite the energy-momentum relation as \(E^2 \approx (pc)^2\).

Find the expression for the total energy E

Taking the square root of both sides in step 4, we can write \(E \approx pc\). Thus, for an extremely relativistic particle, we have shown that its total energy E is approximately equal to its momentum p times the speed of light c.