Question

A muon with rest energy 106 MeV is created at an altitude of \(4500 \mathrm{m}\) and travels downward toward Earth's surface. An observer on Earth measures its speed as \(0.980 c .\) (a) What is the muon's total energy in the Earth observer's frame of reference? (b) What is the muon's total energy in the muon's frame of reference?

Short answer

Answer: The muon's total energy in the Earth observer's frame of reference is 533 MeV, and its total energy in the muon's frame of reference is 106 MeV.

Step-by-step solution

Calculate the Lorentz factor for the muon's speed

The Lorentz factor is given by the formula: $$ \gamma = \frac{1}{\sqrt{1 - (\frac{v}{c})^2}} $$ Where \(v\) is the muon's speed and \(c\) is the speed of light. In this case, \(v = 0.980c\), so we can calculate the Lorentz factor as follows: $$ \gamma = \frac{1}{\sqrt{1 - (0.980)^2}} $$

Calculate the muon's total energy in the Earth observer's frame of reference

Now that we have the Lorentz factor, we can calculate the muon's total energy in the Earth observer's frame of reference using the formula: $$ E = \gamma mc^2 $$ Where \(m\) is the muon's rest mass, and \(c\) is the speed of light. In this problem, we are given the muon's rest energy, \(E_0 = 106 \mbox{ MeV}\). To find its rest mass, we can use the formula \(E_0 = mc^2\): $$ m = \frac{E_0}{c^2} $$ Now we can calculate the muon's total energy in the Earth observer's frame of reference: $$ E = \gamma \cdot mc^2 $$

Calculate the muon's total energy in the muon's frame of reference

In the muon's frame of reference, its speed is zero, and therefore, its Lorentz factor is 1. This means that its total energy is equal to its rest energy in this frame of reference: $$ E' = E_0 $$ Now, we can substitute the values and find the final answers for (a) and (b): (a) The muon's total energy in the Earth observer's frame of reference: $$ \gamma = \frac{1}{\sqrt{1 - (0.980)^2}} = 5.03 $$ $$ m = \frac{106 \mbox{ MeV}}{c^2} $$ $$ E = 5.03 \cdot 106 \mbox{ MeV} = 533 \mbox{ MeV} $$ (b) The muon's total energy in the muon's frame of reference: $$ E' = E_0 = 106 \mbox{ MeV} $$ So, the muon's total energy in the Earth observer's frame of reference is \(533 \mbox{ MeV}\), and its total energy in the muon's frame of reference is \(106 \mbox{ MeV}\).