particle physics - Why is the spectrum of the $beta$-decay continuous?

the spectrum of the Gamma and Alpha decays are both discrete, i.e. the $\alpha$-particles and the $\gamma$-rays take on only discrete values when emitted from a decaying nucleus.

Why is it then, that the $\beta^{\pm}$ can take on continuous values?

The main thing that distinguishes the beta decay from the other two is, that it is a three body problem, i.e. the nucleus does not only decay into an electron/positron, but also into a electron-neutrino/antineutrino. I don't see yet how this immediately implies that the spectrum of the electrons is continuous though.

The way I understand the two body decays is, that the initial nucleus spontaneously decays into the two bodies, i.e. a smaller nucleus and a gamma or alpha particle. As the energy levels in both the nuclei are quantized, only certain values for the energy of the photon and the Helium core are allowed, sine Energy and momentum need to stay conserved.

Does the third particle change things in the way that there basically are two things (i.e. the electron and the neutrino in the beta decay) that are not restricted by an inner energy level hierarchy in the way the nuclei are, thus allowing the energy given by the nuclei to be split arbitrarily (and continuously) between the electron and the neutrino?

If this is the wrong explanation, please correct me.

Answer

At first, consider two particles decay:

$A\rightarrow B + e^-$ Where A is initially rest. So $\vec{p_B}+\vec{p_{e^-}}=0$

now \begin{align} \frac{p_B^2}{2m_B}+\sqrt{p_{e^-}^2+m_{e^-}^2}&=E_{released} \\ \frac{p_{e^-}^2}{2m_B}+\sqrt{p_{e^-}^2+m_{e^-}^2}&=E_{released} \tag{1} \end{align}

see here you have uncoupled equation (equ.1) for $p_{e^-}$ .. So, solving above (equ.1) you will get a fixed $p_{e^-}$. Hence the energy($\sqrt{p_{e^-}^2+m_{e^-}^2}$) of the $\beta$ particle is always fixed in the two body $\beta$ decay. (you can find $p_{B}$ too using the momentum conservation formula)

At first, consider three particles decay:

$A\rightarrow B + e^-+\nu_e$ Where A is initially rest. So $\vec{p_B}+\vec{p_{e^-}}+\vec{\nu_e}=0$

so \begin{align} \frac{p_B^2}{2m_B}+\sqrt{p_{e^-}^2+m_{e^-}^2}+p_{\nu_e}&=E_{released} \\ \frac{(\vec{p_{e^-}}+\vec {p_{\nu_e}})^2}{2m_B}+\sqrt{p_{e^-}^2+m_{e^-}^2}+p_{\nu_e}&=E_{released} \tag{2} \end{align}

Now see in contrast to the two particles decay equation, here we have five unknowns $\big{(}p_{e^-},p_{\nu},p_{B},\theta\ (\rm the\ angle\ between\ \vec{p_{e^-}},\vec{p_{\nu}}),\phi\ (\rm the\ angle\ between\ \vec{p_{e^-}},\vec{p_{B}})\big)$ but four coupled equations *. so we can't solve them uniquely. That's also what happens physically. You will get different values of $p_{e^-},p_{\nu},p_{B},\theta,\phi$ satisfying the four coupled equations. Hence different $\beta$ particles will have different energy($\sqrt{p_{e^-}^2+m_{e^-}^2}$) maintaining the statistics of decay process. Hence the continuous spectra.

*The four coupled equations are equ.2 and three equations which we can get by taking Dot products of $\vec{p_{e^-}},\vec{p_{\nu}}\rm\ and\ \vec{p_{B}}$ with the momentum conservation($\vec{p_B}+\vec{p_{e^-}}+\vec{\nu_e}=0$ ) and remember they lie in a plane so two angles ($\theta\rm\ and\ \phi$) are sufficient.