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Chapter 26: Problem 30

Relative to the laboratory, a proton moves to the right with a speed of \(\frac{4}{5} c,\) while relative to the proton, an electron moves to the left with a speed of \(\frac{5}{7} c .\) What is the speed of the electron relative to the lab?

Short Answer

Expert verified
Answer: The speed of the electron relative to the lab frame is \(\frac{c}{5}\).

Step by step solution

01

Identify the given velocities

The given velocities in the problem are as follows: - Velocity of the proton relative to the lab frame: \(v_1 = \frac{4}{5}c\) - Velocity of the electron relative to the proton frame: \(v_2 = -\frac{5}{7}c\) (negative sign because it's moving to the left)
02

Use the relativistic velocity addition formula

Now, we will use the relativistic velocity addition formula mentioned above: \(v_{total} = \frac{v_1 + v_2}{1 + \frac{v_1 v_2}{c^2}}\) Substitute \(v_1 = \frac{4}{5}c\) and \(v_2 = -\frac{5}{7}c\) into the formula: \(v_{total} = \frac{\frac{4}{5}c - \frac{5}{7}c}{1 + \frac{(\frac{4}{5}c)(-\frac{5}{7}c)}{c^2}}\)
03

Simplify the expression

Now, we can simplify the expression to calculate the velocity of the electron relative to the lab: \(v_{total} = \frac{\frac{4}{5}c - \frac{5}{7}c}{1 - \frac{20}{35}}\) Calculate the numerator: \(Numerator = \frac{4}{5}c - \frac{5}{7}c = \frac{28c - 25c}{35} = \frac{3c}{35}\) Calculate the denominator: \(Denominator = 1 - \frac{20}{35} = \frac{15}{35}\) Finally, divide the numerator by the denominator: \(v_{total} = \frac{\frac{3c}{35}}{\frac{15}{35}} = \frac{3c}{15} = \frac{c}{5}\)
04

State the final result

The velocity of the electron relative to the lab frame is \(\frac{c}{5}\).

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Most popular questions from this chapter

The light-second is a unit of distance; 1 light-second is the distance that light travels in 1 second. (a) Find the conversion between light-seconds and meters: 1 lightsecond \(=? \mathrm{m} .\) (b) What is the speed of light in units of light-seconds per second?
Derive the energy-momentum relation $$E^{2}=E_{0}^{2}+(p c)^{2}$$ Start by squaring the definition of total energy \(\left(E=K+E_{0}\right)\) and then use the relativistic expressions for momentum and kinetic energy $[\text{Eqs}. (26-6) \text { and }(26-8)]$.
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