Question

A particle is moving with a velocity \( 0.8c \) (where \( c \) is the speed of light) in an inertial frame \( S_1 \). Frame \( S_2 \) is moving with a velocity \( 0.8c \) with respect to \( S_1 \). Let \( E_1 \) and \( E_2 \) be the respective energies of the particle in the two frames. Then, \( \frac{E_2}{E_1} \) is ............ (Round off to two decimal places).

Hint:

For relativistic transformations, always use the Lorentz factor \( \gamma \) and the relativistic velocity addition formula when dealing with moving frames.

Answer 1

Correct Answer: 1.64

Step 1: Understanding the relativistic energy transformation.
The energy of a particle in relativity is given by the formula \[ E = \gamma m c^2, \] where \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \) is the Lorentz factor, and \( v \) is the velocity of the particle.
Step 2: Energy in the initial frame \( S_1 \).
For the frame \( S_1 \), the particle is moving with a velocity \( 0.8c \), so the Lorentz factor \( \gamma_1 \) for frame \( S_1 \) is \[ \gamma_1 = \frac{1}{\sqrt{1 - \frac{(0.8c)^2}{c^2}}} = \frac{1}{\sqrt{1 - 0.64}} = \frac{1}{\sqrt{0.36}} = \frac{1}{0.6} = 1.6667. \] Thus, the energy \( E_1 \) in frame \( S_1 \) is \[ E_1 = \gamma_1 m c^2 = \frac{5}{3} m c^2. \] Step 3: Energy in the frame \( S_2 \).
In frame \( S_2 \), the particle is moving with a velocity \( 0.8c \) relative to \( S_1 \), and \( S_2 \) itself is moving at \( 0.8c \) with respect to \( S_1 \). The relative velocity between the particle and the new frame \( S_2 \) is given by the relativistic velocity addition formula: \[ v_{\text{relative}} = \frac{v - v_{\text{frame}}}{1 - \frac{v v_{\text{frame}}}{c^2}} = \frac{0.8c - 0.8c}{1 - \frac{(0.8c)(0.8c)}{c^2}} = \frac{0}{1 - 0.64} = \frac{0}{0.36} = 0. \] Thus, \( v_{\text{relative}} = 0 \), so the energy in the frame \( S_2 \) would be the same as in the original frame. Hence, \[ E_2 = E_1 = \frac{5}{3} m c^2. \] Step 4: Final Answer.
The ratio \( \frac{E_2}{E_1} = 1.00 \).