Conservation Laws

Conservation in proton to neutron nuclear reactions

Let’s consider an example of elemental charge and baryon number and lepton conservation in a weak interaction process (where the weak nuclear force intervenes).

In nuclear physics, for a proton to convert into a neutron, it must lose its charge while maintaining its baryon number. The proton to neutron process is:

\[^1_1{p} \rightarrow ^1_0{n}\]

The proton must lose its positive charge. However, the conservation law states that the total charge before and after the conversion must be the same.

\[\text{Total charge beginning = Total charge ending}\]

In this case, the total charge of the proton is 1, while the total neutron charge is 0.

\[^{\qquad \quad 1}_{charge = 1}{p} \rightarrow ^{\qquad \quad 1}_{charge = 0}{n}\]

Protons are composed of two up particles and one down particle, with charges of +⅔ and -⅓. The neutron, by contrast, consists of two down particles and one up particle.

\[p = (udu) = \Big(\frac{2}{3} - \frac{1}{3} + \frac{2}{3} \Big) = 1\]

\[n = (dud) = \Big( -\frac{1}{3} +\frac{2}{3} -\frac{1}{3} \Big) = 0\]

The proton must, therefore, lose 3⁄3 of the charge. The particle that needs to change is one of the proton’s up quarks, so the charges cancel each other out: u → d = ⅔ → -⅓. The sum for the proton will be ⅔-⅓-⅓ = 0.

The only way to achieve this is by releasing a charge of 1. A particle must, therefore, be released. In this case, the proton emits a positron.

\[^{\qquad \quad 1}_{charge = 1}{p} \rightarrow ^{\qquad \quad 1}_{charge = 0}{n} + ^{\qquad \quad 0}_{charge= 1}{e}^+\]

Now the charges are balanced and conserved. Before the process, we had a charge of 1, and afterwards, we still have a charge of 1.

The positron has a baryon number of zero, while the proton and the neutron have a baryon number of one, thus conserving the baryon number.

\[\text{Baryon number beginning = Baryon number ending}\]

\[_{\text{Baryon number = 1}}{p} \rightarrow ^{\text{Baryon number = 1}}{n} + _{\text{Baryon number = 0}}{e}^+\]

The electron is a lepton. Every lepton has a lepton number, as shown in the following table.

In the formula for the proton to neutron reaction, there is an abnormality. The proton and the neutron each have a lepton number of 0, but the positron has a lepton number of -1.

\[(\text{lepton number} = 0)p \rightarrow (\text{lepton number }= 0)n + (\text{lepton number} = -1)e^+\]

To balance the lepton number, we need to add another particle with a lepton number of 1 and a charge of 0. The particle to be added is an electron neutrino v_ε

\(p \rightarrow n + e^+ + v_{\varepsilon}\)

This can be summarised as follows:

\(^1_1p \rightarrow ^1_0n + ^0_1e^+ +v_{\varepsilon}\)

In this reaction, the baryon number is conserved, the charge is conserved, and the lepton number is also conserved.

The proton to neutron reaction can also be summarised as the transformation of one of the proton’s up quarks into a down quark, a positron, and a neutrino.

\(u \rightarrow d + e^+ + v_{\varepsilon}\)

Here is the reaction as represented by a Feynman diagram:

The particle that acts as a mediator in this reaction is a plus W boson.

Conservation in energy and momentum exchanges

Particles conserve both their momentum and their energy in quantum mechanics, as was observed in early experiments using x-rays. American physicist Arthur Compton noticed these effects as a decrease in wavelength and a scattering of photons. He theorised that the decrease in wavelength and the scattering were caused by the material’s electrons. The collisions cause the photons to scatter at an angle and reduce their energy by giving it to the electrons, an effect that is known as the Compton effect. Set the following example.

The Compton effect is important as it provides information about the momentum of a photon, which is given as the Planck constant divided by the photon’s wavelength:

The Compton effect was important as it gave information about the momentum of a photon, where the momentum was given as the Planck constant divided by the photon’s wavelength as below.

\(\text{Photon momentum} = \frac{h}{\lambda}\)