# Relativistic momentum and ke quesion

## Homework Statement

A Lambda0 (L0) hyperon at rest decays into a neutron and what? (a) Find the total kinetic energy of the decay products. (b) What fraction of the total kinetic energy is carried off by each particle?

I have both the decay process and the total kinetic energy.
L0 -> n+ p°
a) 41 MeV. This was solved by finding the rest energy of L0 and subtracting from it the total rest energy of the n+ p°.
As for b, I am somewhat lost. I know that the initial momentum of the system is 0; therefore, the final momentum must be 0. So P1=P2. I also know that KE1 + KE2 = 41MeV. The problem comes because both particles are relativistic, and using them in a system of equations is a nightmare. Any suggestions?

Hello,

For the second question, you can use the invariance of four-momentum and obtain simple relations.
For example,
$$\mathbf{P}_\Lambda=\mathbf{P}_{n}+\mathbf{P}_{\pi}$$ here,
it can be
$$\mathbf{P}_\Lambda-\mathbf{P}_{n}=\mathbf{P}_{\pi}$$
and one can take inner pruoduct of each side respectively.
In the example, one can obtain $$\gamma_{n}$$.

I am not following you here. Can you explain a little more?

If I took the equation E=(p^2c^2+m^2c^4)^(1/2) and subtracted mc^2 from it, I would obtain the kinetic energy of the particle correct? I played with the equation v=c^2p/E and got to (p^2)=(m^2c^2v^2)/(c^2-v^2). I placed that in [(p^2c^2+m^2c^4)^(1/2)-mc^2] to get the ke of one patricle. The ke of the other particle is the excat same equation except the mass value is different.

The four momemtum of the decay chanel is:
$$\mathbf{P}_\Lambda=\mathbf{P}_{n}+\mathbf{P}_{\pi}$$
that is
$$\left(\begin{array}{c}m_{\Lambda}c^2/c\\0\end{array}\right)=\left(\begin{array}{c}\gamma_nm_{n}c^2/c\\\gamma_nm_n\vec{v}_n\end{array}\right)+\left(\begin{array}{c}\gamma_\pi m_{\pi}c^2/c\\\gamma_\pi m_\pi \vec{v}_\pi\end{array}\right)$$
Therefore

$$\left(\begin{array}{c}m_{\Lambda}c^2/c\\0\end{array}\right)-\left(\begin{array}{c}\gamma_nm_{n}c^2/c\\\gamma_nm_n\vec{v}_n\end{array}\right)=\left(\begin{array}{c}\gamma_\pi m_{\pi}c^2/c\\\gamma_\pi m_\pi \vec{v}_\pi\end{array}\right)$$ ... (1)

$$\left(\begin{array}{c}m_{\Lambda}c^2/c\\0\end{array}\right)-\left(\begin{array}{c}\gamma_\pi m_{\pi}c^2/c\\\gamma_\pi m_\pi v\vec{}_\pi\end{array}\right)=\left(\begin{array}{c}\gamma_nm_{n}c^2/c\\\gamma_nm_n\vec{v}_n\end{array}\right)$$ ...(2)

One can use the invariance of four momentum:
$$\left(\begin{array}{c}E/c\\\vec{p}\end{array}\right)\cdot\left(\begin{array}{c}E/c\\\vec{p}\end{array}\right)=E^2/c^2-\vec{p}\cdot\vec{p}=m^2c^2$$
on Eq.(1) and Eq.(2)

Use it on Eq.(1):
$$m^2_{\Lambda}c^2+m^2_nc^2-2m_{\Lambda}\gamma_nm_nc^2=m^2_{\pi}c^2\quad\Rightarrow\quad\gamma_{n}=\frac{m^2_\Lambda+m^2_n-m^2_\pi}{2m_\Lambda m_n}$$
When you have $$\gamma_n$$, the kinetic energy of neutron is $$(\gamma_n-1)m_nc^2$$

Also the pi meson.

Goodluck

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