Question
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}$$ [Eq. (26-11)].$
Step 1
The energy-momentum relation is given by: $E^2 = E_0^2 + (pc)^2$ (1) The definition of total energy is: $E = K + E_0$ (2) where $E$ is the total energy, $K$ is the kinetic energy, and $E_0$ is the rest energy. Show more…
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Starting with the energy-momentum relation $E^{2}=E_{0}^{2}+$ $(p c)^{2}$ and the definition of total energy, show that $(p c)^{2}=K^{2}+2 K E_{0}[\mathrm{Eq} .(26-11)]$.
Starting with the energy-momentum relation $E^{2}=E_{0}^{2}+(p c)^{2}$ and the definition of total energy show that $(p c)^{2}=K^{2}+2 K E_{0}[\mathrm{Eq} \cdot(26-23)]$
Derive the energy-momentum relation $$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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