Homework Help: Relativistic energy and momentum question

1. Feb 11, 2010

doublemint

1. A lambda particle decays into a proton and pion and it is observed that the proton is left at rest.
a. what is the energy of the pion?
b. what is the energy of the original lambda? m of lambda = 1116MeV/c^2, m of proton = 938 MeV/c^2, and m of pion = 140 MeV/c^2

2. Relevant equations
Conservation of energy and momentum. E$$\lambda$$ = E $$\rho$$ + E$$\pi$$, P $$\lambda$$ = P$$\rho$$ + P$$\pi$$
Lorentz Transformation

3. The attempt at a solution
I tried breaking down conservation of energy and momentum so solve for what i need but i cant seem to get it right. I've tried using lorentz transformation with the energy and momentum but no lucky either.
Any help is appreciated!
Thanks

2. Feb 11, 2010

vela

Staff Emeritus
The energy and momentum of the proton are straightforward to write down.

Use $m^2 = E^2-p^2$.

There's no need for a Lorentz transformation because you only have one frame in this problem.

You can save yourself a bit of algebra work if you know how to work with four-vectors.

3. Feb 11, 2010

doublemint

So you mean m$$^{2}$$c$$^{2}$$=E$$^{2}$$ - p$$^{2}$$c$$^{2}$$ ?
so,
E$$\lambda$$ = E$$\rho$$ + E$$\pi$$
and since the proton is not moving, p = 0
the energy of the pion is,
E$$\pi$$ = E$$\lambda$$ - mc$$^{2}$$

i dont know anything about doing work in 4 vectors...

for part b, what do i need to do to get the energy of the lambda particle? i have no values for velocity.

4. Feb 11, 2010

vela

Staff Emeritus
OK, so you have, so far, these relations:

$$E_\lambda = m_pc^2+E_\pi$$
$$p_\lambda = p_\pi$$

Multiply the second by c, square both equations, and then subtract the second from the first. Hopefully, you'll see where to go from there.

5. Feb 11, 2010

doublemint

I see what you are doing! Thanks!
If i did it right, it simplifies to m$$_{\lambda}$$c$$^{2}$$ = m$$_{\pi}$$c$$^{2}$$ + m$$_{p}$$c$$^{2}$$ = E of the original lambda
where E = 1078 MeV
but i think i did something wrong though

Last edited: Feb 11, 2010
6. Feb 11, 2010

vela

Staff Emeritus
What happened to the cross term on the RHS? That's the one that'll allow you to solve for the energy of the pion. Also, the other terms should be of the form $(mc^2)^2$, not just $mc^2$.

7. Feb 11, 2010

doublemint

I found my mistake! I knew there was a problem with my answer.
So, from the energy of the pion which i got as 184.4MeV, I use that to solve for lambda which i got to be 1122MeV!
I hope that is right!

Thanks a bunch vela!