Question

Take the ratio of relativistic rest energy, (E = frac{γ m c²}{mc²}), to relativistic momentum, (p = γ mu), and show that in the limit that mass approaches zero, you find (E/p = c).

a) (E/p = 1)
b) (E/p = frac{1}{c})
c) (E/p = c)
d) (E/p = frac{c}{2})

Answer 1

The correct answer is c) E/p = c, which is shown by the ratio of relativistic rest energy to relativistic momentum; as mass approaches zero, the particle's velocity approaches the speed of light, simplifying the ratio to the speed of light itself.

Step-by-step explanation:

When taking the ratio of relativistic rest energy (E = γmc²) to relativistic momentum (p = γmu), the formula becomes E/p = (γmc²) / (γmu). Here, γ is the Lorentz factor, m is the mass of the particle, c is the speed of light, and u is the particle's velocity. We can simplify this ratio by canceling out the γ and m to get E/p = c²/u.

As the mass of the particle approaches zero, its velocity u approaches c, the speed of light. According to the principles of special relativity, massless particles, such as photons, move at the speed of light. In this limit, the previous equation becomes E/p = c²/c, which simplifies to E/p = c. This shows that for massless particles, the ratio of energy to momentum is equal to the speed of light, confirming option c) as the correct answer.