Question: Verify that the hypothetical proton decay scheme in Eq. 44-14 does not violate the conservation law of

(a) charge,

(b) energy, and

(c) linear momentum.

(d) How about angular momentum?

The proton decay process is

$\mathrm{p}\to {\mathrm{e}}^{+}+{\mathrm{\nu }}_{\mathrm{e}}.....1$

The charges of proton, electron and neutrino are

$$\begin{gathered} {\mathbf{Q}}_{\mathbf{p}}\mathbf{=}\mathbf{+}\mathbf{1} \\ {\mathbf{Q}}_{{\mathbf{e}}^{\mathbf{+}}}\mathbf{=}\mathbf{+}\mathbf{1}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{1} \\ {\mathbf{Q}}_{{\mathbf{\nu }}_{\mathbf{e}}}\mathbf{=}\mathbf{0} \end{gathered}$$

The rest energies of proton, electron and neutrino as third set of equation (III) below:

$$\begin{gathered} {\mathbf{E}}_{\mathbf{0}\mathbf{p}}\mathbf{=}\mathbf{938}\mathbf{.}\mathbf{3}\mathbf{MeV} \\ {\mathbf{E}}_{\mathbf{0}{\mathbf{e}}^{\mathbf{+}}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{511}\mathbf{MeV} \\ {\mathbf{E}}_{\mathbf{0}{\mathbf{\nu }}_{\mathbf{e}}}\mathbf{=}\mathbf{0} \end{gathered}$$

The spins of the proton, electron and neutrino as the equation (IV).

$$\begin{gathered} {\mathbf{S}}_{\mathbf{p}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}} \\ {\mathbf{S}}_{{\mathbf{e}}^{\mathbf{+}}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}} \\ {\mathbf{S}}_{{\mathbf{\nu }}_{\mathbf{e}}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}} \end{gathered}$$

(a)

Use equation (II) to calculate the change in charge in the interaction in equation (I)

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{p}}-{\mathrm{Q}}_{{\mathrm{e}}^{+}}-{\mathrm{Q}}_{{\mathrm{\nu }}_{\mathrm{e}}}=+1-1-0 \\ =0 \end{gathered}$$

Thus, charge is conserved.

(b)

Use equation (III) to calculate the difference in rest energies of the initial and final states in the interaction in equation (I)

$$\begin{gathered} {\mathrm{E}}_{0\mathrm{p}}-{\mathrm{E}}_{0{\mathrm{e}}^{+}}-{\mathrm{E}}_{0{\mathrm{\nu }}_{\mathrm{e}}}=938.3\text{MeV -}0.511\text{MeV}-0 \\ =982.989\text{MeV} \end{gathered}$$

Thus

${\mathrm{E}}_{0\mathrm{p}}>{\mathrm{E}}_{0{\mathrm{e}}^{+}}+{\mathrm{E}}_{0{\mathrm{\nu }}_{\mathrm{e}}}$

Hence, charge is conserved.

(c)

Since a proton at rest decays to two states, electron and neutrino, as long as the electron and neutrino move in opposite direction with equal magnitude of linear momentum, the linear momentum will be conserved.

(d)

Use equation (IV) to calculate the change in spin in the interaction in equation (I)

$$\begin{gathered} {\mathrm{S}}_{\mathrm{p}}-{\mathrm{S}}_{{\mathrm{e}}^{+}}-{\mathrm{S}}_{{\mathrm{\nu }}_{\mathrm{e}}}=\frac{1}{2}-\frac{1}{2}-\frac{1}{2} \\ =-\frac{1}{2} \end{gathered}$$

Thus, angular momentum is not conserved.