# Relativistic momentum and ke quesion

1. May 22, 2007

### 6Stang7

1. The problem statement, all variables and given/known data
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?

2. May 22, 2007

### variation

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 $$\gamma_{n}$$.

3. May 23, 2007

### 6Stang7

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

4. May 23, 2007

### 6Stang7

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.

5. May 23, 2007

### variation

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

Last edited: May 23, 2007