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Chapter 26: Problem 57

Starting with the energy-momentum relation \(E^{2}=E_{0}^{2}+\) \((p c)^{2}\) and the definition of total energy, show that $(p c)^{2}=K^{2}+2 K E_{0}[\mathrm{Eq} .(26-11)]$.

Short Answer

Expert verified
Question: Derive the relation \((pc)^2 = K^2 + 2KE_0\) using the energy-momentum relation \(E^2 = E_0^2 + (pc)^2\) and the definition of total energy. Answer: \((pc)^2 = K^2 + 2KE_0\)

Step by step solution

01

Express total energy in terms of kinetic energy and rest energy

Total energy (E) is the sum of kinetic energy (K) and rest energy (\(E_0\)) for a particle: $$ E = K + E_0 $$
02

Substitute the expression for total energy into the energy-momentum relation

We have the energy-momentum relation: $$ E^2 = E_0^2 + (pc)^2 $$ Substitute \(E = K + E_0\) into the relation: $$ (K + E_0)^2 = E_0^2 +(pc)^2 $$
03

Expand and simplify

Expand the left side of the equation: $$ (K^2 + 2KE_0 + E_0^2) = E_0^2 + (pc)^2 $$ Subtract \(E_0^2\) from both sides: $$ K^2 + 2KE_0 = (pc)^2 $$
04

Final result

We have shown that using the energy-momentum relation and the definition of total energy, we can derive the equation: $$ (pc)^2 = K^2 + 2KE_0 $$ This is the desired result.

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