$\beta^-$ decay: expression for maximum electron energy

Question

I'm having some trouble finding an expression for the maximum electron energy in $\beta^-$ decay. In the frame where the neutron is initially at rest, conservation of momentum reads:

$$\vec{p}_p+\vec{p}_e+\vec{p}_\bar{\nu}=0$$

and energy conservation yields the following for the electron's energy:

$$E_e=\sqrt{m_e^4+{\vec{p}_e}^2}=m_n^2-\left(\sqrt{m_\bar{\nu}^4+{\vec{p}_\bar{\nu}}^2}\right)-\left(\sqrt{m_p^4+{\vec{p}_p}^2}\right)$$

How do I know how to choose $\vec{p}_\bar{\nu}$ and $\vec{p}_p$, such that $E_e$ is maximum? Or how can I calculate the maximum? I read in another post that "because of the mass scales the beta and the neutrino take the bulk of the energy". How can I see that from the formulas? Is that the key to the problem?

Answer 1

The way to approach the problem initially is to consider what you know about the reaction $$ n \longrightarrow \mathrm{p}^+ + \mathrm{e}^- + \bar{\nu}_e \,.$$

Because of the relativity principle we can consider the reaction in the rest-frame of the neutron without loss of generality and we know that both (three-)momentum and energy are conserved. $$\begin{align*} 0 &= p_e + p_\nu + p_p \\ m_n &= E_e + E_\nu + E_p \, \end{align*}$$ which is two equations with six unknowns, but we also have three relationships between the energy and momentum of the particles so, the system is only under-constrained by one degree of freedom. One assumption will solve the problem for us, it just has to be the right assumption for getting the electron the maximum energy.

It should be obvious that maximum energy is linked to maximum momentum, so we expect the proton and neutrino momenta to be co-linear. So let's give one all the recoil momentum and leave the other at rest. How does it shake out?

Leaving the nucleus at rest the two light particle recoil from each other, and the neutrino can be treated as ultra-relativistic, meaning the kinetic energy is $$ T_\nu = E_\nu = \left|p_e\right| \,. \tag{1} $$

If, instead we leave the neutrino at rest, the very heavy proton recoils at non-relativistic speed (check this later) so that $$ T_p = \frac{p_e^2}{2m_p} \,. \tag{2}$$

In these "one at rest" scenarios the more energy the other moving particle has the less the electron gets, so we need to figure which one of (1) and (2) is smallest and go with that scenario.

Two ways to proceed. The safe way is to follow each scenario all the way through, the other is to look at the scale of the quantities and try to guess. Let's start by guessing. Both have the same dimensionality and both have (at least) one factor of $p_e$. So if we know the size of the coefficients of $p_e$ we know the answer. $$\begin{align*} T_\nu &= (1) p_e \tag{1b} \\ T_p &= \left( \frac{p_e}{2 m_p} \right) p_e \tag{2b} \,, \end{align*}$$ but the available energy is of order $m_n - m_p$ which is much less that $m_p$, so $$ \frac{p_e}{2m_p} \ll 1 \,,$$ and our choice is obvious. Scenario (2) (with the neutrino at rest) results in the most energy for the electron.

Going back to our system of equations we get $$\begin{align*} 0 &= p_e - 0 - p_p \\ m_n &= E_e + m_\nu + E_p \, \end{align*}$$ and applying the non-relativisitc approximation to the proton $$\begin{align*} p_p &= p_e \\ m_n &= E_e + m_\nu + m_p + \frac{p_p^2}{2m_p} p_p^2 \\ m_n - m_p &\approx E_e + 0 + \frac{p_e^2}{2m_p} \\ &= E_e + \frac{E_e^2 - m_e^2}{2m_p} \,, \end{align*}$$ and the fraction on the RHS is small leading to $E_e \approx m_n - m_p$. So the electron ends up with essentially all the energy. This situation is even more extreme in the beta decay of a composite nucleus because the mass ratio is even larger.

Now check the "non-relativistic" assumption for the proton. The kinetic energy available is approximately $$\begin{align*} T_p &= \frac{p_e^2}{2m_p} \\ &= \frac{\left(m_n - m_p\right)^2 - m_e^2}{2m_p} \\ &\approx \frac{\left(939.6 - 938.2 \right)^2 - (.5)^2}{2 (938.2)} \tag{masses in MeV} \\ &= 9.11 \times 10^{-4} \,\mathrm{MeV} \,. \end{align*}$$ Kinetic energy on order of one part in $10^7$ of mass is safely non-relativistic.

Finally, consider that the given the tiny mass of the neutrino the "neutrino at rest" scenario is essentially what was expected of beta decay as a two-body problem, Meaning that the energy we found for the electron is approximately the end-point energy of the spectrum (yeah, we did it right). You should also notice that with the neutrino momentum vanishingly small the phase space for this outcome is likewise vanishing which is part of what makes decay end-point measurements hard.

Answer 2

finding an expression for the maximum electron energy in beta minus decay

Considering any general three-body decay, $$ A \rightarrow B + C + D, $$ where $A$ shall have been initially a member of a suitable inertial system $\mathcal A$, and asking to maximize the energy (as well as the speed, and the momentum) of decay product $B$ with respect to the members of inertial system $\mathcal A$,
we can consider $B$ recoiling against the system constituted of decay products $C$ and $D$: $$ A \rightarrow B + (C~D). $$

Accordingly, $$ E_{\mathcal A}[~A~] = E_{\mathcal A}[~B~] + E_{\mathcal A}[~(C~D)~],$$ $$ \small{ m[~A~]~c^2 = \sqrt{ (m[~B~]~c^2)^2 + (\| \vec p_{\mathcal A}[~B~] \|~c)^2 } + \sqrt{ (m[~(C~D)~]~c^2)^2 + (\| \vec p_{\mathcal A}[~(C~D)~] \|~c)^2 } }, $$ and $$ \| \vec p_{\mathcal A}[~B~] \|^2 = \| \vec p_{\mathcal A}[~(C~D)~] \|^2 = $$ $$\Tiny{ c^2~\frac{m[~A~]^4 + m[~B~]^4 + m[~(C~D)~]^4 - 2~m[~A~]^2~m[~B~]^2 - 2~m[~A~]^2~m[~(C~D)~]^2 - 2~m[~B~]^2~m[~(C~D)~]^2}{4~m[~A~]^2} }.$$

Since obviously the mass of the decaying particle is larger than the invariant mass of any systems of decay products, therefore $$m[~(C~D)~]^2 \lt 2~m[~A~]^2 + 2~m[~B~]^2.$$ Consequently the maximum momentum (norm) is attained if the invariant mass $m[~(C~D)~]$ has its minimum value; namely just the plain sum of the constituent masses, $$m_{\text{min}}[~(C~D)~] = m[~C~] + m[~D~],$$ which holds in the case that decay products $C$ and $D$ don't move with respect to each other at all.

Therefore $$\Tiny { \| \vec p_{\mathcal A}[~B~] \|^2 \le c^2~\frac{m[~A~]^4 + m[~B~]^4 + (m[~C~] + m[~D~])^4 - 2~m[~A~]^2~m[~B~]^2 - 2~m[~A~]^2~(m[~C~] + m[~D~])^2 - 2~m[~B~]^2~(m[~C~] + m[~D~])^2}{4~m[~A~]^2} },$$

and correspondingly

$$\Tiny { E_{\mathcal A}[~B~] \le \sqrt{ (m[~B~]~c^2)^2 + c^4~\frac{m[~A~]^4 + m[~B~]^4 + (m[~C~] + m[~D~])^4 - 2~m[~A~]^2~m[~B~]^2 - 2~m[~A~]^2~(m[~C~] + m[~D~])^2 - 2~m[~B~]^2~(m[~C~] + m[~D~])^2}{4~m[~A~]^2} } }.$$

Now, specifying the decaying particle $A$ as neutron $n$, which shall have been initially a member of a suitable inertial system $\mathcal N$, decay product $B$ as electron $e^-$, decay product $C$ as proton $p$, and decay product $D$ as anti-electron-neutrino $\overline \nu$, an experimentally reasonable approximation seems

$$\small { E_{\mathcal N}[~e^-~] \lessapprox \sqrt{ (m[~e^-~]~c^2)^2 + c^4~\frac{m[~n~]^4 + m[~p~]^4 - 2~m[~n~]^2~m[~p~]^2 - 2~m[~e^-~]^2~m[~n~]^2 - 2~m[~e^-~]^2~m[~p~]^2}{4~m[~n~]^2} } }.$$

Inserting the tabulated mass values I find $$E_{\mathcal N}[~e^-~] \lessapprox 1.2927 \text{ MeV}.$$

Answer 3

The answer is the one last shown by user12262, but the expression can be further simplified and this is not only illuminating but one also ends with the energy of the B particle in the "two body" decay $ A \rightarrow B + (C+D) $ :

$$ { E_{\mathcal N}[B] \lessapprox c^2\frac{m[A]^2 + m[B]^2 - (m[C]+m[D])^2}{2m[A]} }$$

and this leads to the maximum kinetic energy $$ T_{max}^e = E_{max}^e - m[e] c^2 $$ for the electron/positron in beta decay

$$ { T_{\mathcal N}[e^-] \lessapprox c^2\frac{ ( m[n]- m[e] )^2 - m[p]^2 }{2m[n]} }$$

which in fact is very close ( also in most beta decays) to the Q-value of the reaction

$$ Q = ( m[n] - m[p] - m[e]) c^2 .$$

Answer 4

rather "shorter" answer would be going directly from conservation of 4-momentum and it's squares.

Let us denote 4-momenta of particles by N, P, k, k' for neutron, proton, electron and electron anti-neutrino in this respective order.

Total 4-momentum of an isolated system is conserved, thusly it holds $N - P - k - k' = 0$, or $(N - k) = (P + k')$.

Note that when either LHS or RHS are squared (i.e. contracted with metric) you get and invariant. A quantity which is conserved under Lorentzian transform is precisely the same in value in every inertial frame.

1] Let's square both LHS and RHS.

2] I am using $g_{ab} = diag(1, -1, -1, -1)$ and thus 4-momentum is normalized to $+m^2$

3] We wish to get, after squaring, on RHS just linear term in $E_e$ (electron energy). It is convenient to use neutron rest frame, where in particular $N = \left(E_n, \overrightarrow{0}\right) = \left(m_n, \overrightarrow{0}\right)$

$$RHS^2 = N^2 -2N.k + k^2 = m_n^2 + m_e^2 - 2m_n E_e$$ $$LHS^2 = P^2 + k^2 + 2P.k = m_p^2 + m_\nu^2 + 2P.k \ge m_p^2 + m_\nu^2 $$

Together ($LHS^2 = RHS^2$ due to invariance) that is

$$ -2m_n E_e \ge - m_n^2 - m_e^2 + m_p^2 + m_\nu^2 $$ $$ E_e \le \frac{m_n^2 - m_p^2 + m_e^2 - m_\nu^2}{2m_n} $$

Now to arrive at your result:

1] $m_e$ is orders of magnitude smaller than $m_n$ in the denominator, $m_\nu = 0$ for all practical purposes of this level of calculation. We neglect both squares of those masses.

2] $(m_n^2-m_p^2) = (m_n + m_p)(m_n - m_p) =: 2M(m_n - m_p)$, where I have introduced approximate mass of nucleons $M$ for sum of two nucleons (not for their difference). If you goof around which is fun you find this brings out around $1/1000$ error if I remember correctly.

$$ E_e \le \dots \approx \frac{2M(m_n - m_p)}{2m_n} \approx \frac{2M(m_n - m_p)}{2M} \approx \frac{2M(m_n - m_p)}{2M} = m_n - m_p$$

i.e. upper energy boundary for electron is about the difference of the proton-neutron mass.

Sidenote: If I wanted to extract information about instead $E_p$, I would have started from neutron rest frame and squared $(N-P)$ to get linear term in $E_p$. I would need to use also for LHS electron-neutrino COM frame to get $\overrightarrow{k'} = -\overrightarrow{k}$ and use it in inequality of the squares similarly as above.