Relativistic momentum and ke quesion

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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?
 

Answers and Replies

  • #2
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Hello,

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


Hope these helpful
 
  • #3
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I am not following you here. Can you explain a little more?
 
  • #4
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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
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The four momemtum of the decay chanel is:
[tex]\mathbf{P}_\Lambda=\mathbf{P}_{n}+\mathbf{P}_{\pi}[/tex]
that is
[tex]\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)[/tex]
Therefore

[tex]\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)[/tex] ... (1)

[tex]\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)[/tex] ...(2)

One can use the invariance of four momentum:
[tex]\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[/tex]
on Eq.(1) and Eq.(2)

Use it on Eq.(1):
[tex]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}[/tex]
When you have [tex]\gamma_n[/tex], the kinetic energy of neutron is [tex](\gamma_n-1)m_nc^2[/tex]

Also the pi meson.


Goodluck
 
Last edited:

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