Question

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?

Short answer

(a) It is verified that the proton decay scheme does not violate charge conservation.

(b) It is verified that the proton decay scheme does not violate energy conservation.

(c) It is verified that the proton decay scheme does not violate linear momentum conservation.

(d) It is found that the proton decay scheme violates angular momentum conservation.

Step-by-step solution

Given data

The proton decay process is

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

Quantum numbers and rest energy of proton, electron and neutrino

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).

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

Determine the change in charge in proton decay

(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.

Determine whether proton decay conserves energy

(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.

Determine whether proton decay conserves linear momentum

(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.

Determine the change in spin in proton decay

(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.