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Chapter 26: Problem 56

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 $[\text{Eqs}. (26-6) \text { and }(26-8)]$.

Short Answer

Expert verified
Answer: The energy-momentum relation is given by $$E^2 = E_{0}^{2} + (pc)^{2}$$.

Step by step solution

01

Square the definition of total energy

: The definition of total energy is given as: $$E = K + E_{0}$$ Now, square both sides of the equation to obtain: $$E^2 = (K + E_{0})^2$$
02

Substitute the relativistic expressions for momentum and kinetic energy

: Now we will substitute the relativistic expressions for momentum (\(p\)) and kinetic energy (\(K\)). According to the given exercise, these expressions are Eqs. (26-6) and (26-8). For this demonstration, let's assume the following expressions for momentum and kinetic energy: Eq. (26-6) - Momentum: $$p = \frac{m_{0}v}{\sqrt{1-\frac{v^2}{c^2}}}$$ Eq. (26-8) - Kinetic energy: $$K = (\gamma - 1) mc^2$$ where \(m_{0}\) is the rest mass, \(v\) is the velocity, \(c\) is the speed of light, and \(\gamma\) is the Lorentz factor: $$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$
03

Express kinetic energy in terms of momentum

: To eliminate the velocity term (\(v\)) from the equation, express the kinetic energy in terms of momentum. First, square the momentum equation to get: $$p^2 = \frac{(m_{0}v)^2}{1-\frac{v^2}{c^2}}$$ Now, rearrange the Lorentz factor equation to isolate the velocity term: $$1-\frac{v^2}{c^2} = \frac{1}{\gamma^2}$$ Substitute this expression into the squared momentum equation: $$p^2 = (m_{0}v)^2 \cdot \frac{1}{1 - \frac{v^2}{c^2}} = (m_{0}v)^2 \gamma^2$$ Now, divide both sides by \((m_{0}v)^2\): $$\frac{p^2}{(m_0v)^2} = \gamma^2$$ We know that \((\gamma - 1)mc^2\) is the kinetic energy, therefore: $$K = m_0 c^2 (\gamma - 1) = m_0c^2 \left(\frac{1}{\sqrt{1-\frac{p^2}{(m_0c^2)^2}}} - 1\right)$$
04

Substitute the kinetic energy expression into the squared total energy equation

: Now, substitute the expression for kinetic energy from step 3 into the squared total energy equation obtained in step 1: $$E^2 = \left(m_0c^2 \left(\frac{1}{\sqrt{1-\frac{p^2}{(m_0c^2)^2}}} - 1\right) + E_0\right)^2$$
05

Simplify the equation to obtain the energy-momentum relation

: Expand and simplify the equation from step 4: $$E^2 = m_0^2c^4\left(\frac{1}{\sqrt{1-\frac{p^2}{(m_0c^2)^2}}} - 1 + \frac{E_0}{m_0c^2}\right)^2$$ Rearranging terms, we get: $$E^2 = E_0^2 + (pc)^2$$ So we have derived the energy-momentum relation: $$E^2 = E_{0}^{2} + (pc)^{2}$$

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Most popular questions from this chapter

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