# Homework Help: Decay of pion on electron and antineutron. Energy of electron?

1. Nov 9, 2013

### skrat

1. The problem statement, all variables and given/known data
Negative pion with kinetic energy 100 Mev decays to electron and antineutron. What is the kinetic energy of electron , which will move in the same direction as pion did?

2. Relevant equations
$p^{\mu }=(\frac{E}{c},\vec{p})$

$E=T+mc^2=\sqrt{p^2c^2+m^2c^4}$

3. The attempt at a solution

Guys, I need lots of help here... I have got lots of similar problems in my book here but I can't finish a single one of them - which worries me just a bit. Ok, not true... I am really worried and frustrated and will hit somebody and ... -.- you get it.

At the beginning: $p_{0}^{\mu }=(\frac{T_\pi +m_\pi c^2}{c},p_\pi ,0 )$,

after pion decays: $p_{1}^{\mu }=(\frac{T_e +m_e c^2+E}{c},p_e+\frac{E}{c}cos\varphi ,\frac{E}{c}sin\varphi )$

for any direction of antineutron (E is the energy of antineutron). BUT since $p_{0}^{\mu }=p_{1}^{\mu }$ than also $\frac{E}{c}sin\varphi =0$, therefore $\varphi = 0$.

This means:

$p_{0}^{\mu }=(\frac{T_\pi +m_\pi c^2}{c},p_\pi ,0 )$ and

$p_{1}^{\mu }=(\frac{T_e +m_e c^2+E}{c},p_e+\frac{E}{c},0)$

Now I haven't got a single clue what to do! I tried to equate each component of $p^\mu$ but that leads me to some horrible and probably never ending calculations.

I also tried to use invariance of scalar product:

$(T_\pi +m_\pi c^2)^2-p_{\pi }^{2}c^2=(T_e+m_ec^2+E)^2-(cp_e+E)^2$

which obviously leads nowhere since I have to get rid of E and this just gives me all sorts of things multiplied by E...

Than I also tried to got to barycentric coordinate system where $c^2(\Sigma _{i}\vec{p_i})^2=0$ for $\vec{p_i}$ after the decay. Than:

$(T_\pi +m_\pi c^2)^2-p_{\pi }^{2}c^2=(T_e+m_ec^2+E)^2$

Which in my opinion also leads me nowhere, since there is no way to get rid of E...

Really guys, I am dealing with this for two days now. I beg you, teach me, help me understand so I can solve the other 25490785545 problems.

Thanks!

Last edited: Nov 9, 2013
2. Nov 9, 2013

### fzero

You should replace $T+mc^2$ by $\sqrt{(pc)^2 + (mc^2)^2}$. Then equating the 4-momenta leads to a pair of equations that determines $p_e,E$ in terms of $p_\pi$ and the masses. One equation lets you write $E$ as a linear function of $p_e,p_\pi$. The other equation can be squared in a certain way to get a quadratic equation for $p_e$. One of the roots should be the solution that you want. It will be a little messy, but if you're careful you should be able to get an expression that you can plug numbers into.

Also, the decay product in this decay is not the antineutron, it is the electron antineutrino. The antineutron is too massive to appear in the final state.

3. Nov 10, 2013

### skrat

So, there isn't like a recipe what to use:

-equating the components,
-calculating the invariance,
- ...

Everything works. ?

4. Nov 10, 2013

### vela

Staff Emeritus
I find it best to stick to using total energy, momentum, and mass whenever possible. In this case, you'd have the four-vectors
\begin{align*}
p_\pi^\mu &= (E_\pi, p_\pi, 0, 0) \\
p_e^\mu &= (E_e, p_e, 0, 0) \\
p_\bar\nu &= (p_\bar\nu, -p_\bar\nu, 0, 0)
\end{align*} where I've set $c=1$. Conservation of energy and momentum requires
$$p_\pi^\mu = p_e^\mu + p_\bar\nu^\mu.$$ Quite often, by simply rearranging and squaring, you can solve for some quantity very easily. For example, we can write
$$p_\pi^\mu - p_\bar\nu^\mu = p_e^\mu.$$ If you now square this equation, you get
$$p_\pi^2 + p_\bar\nu^2 - 2 p_\pi\cdot p_\bar\nu = p_e^2.$$ Using the expressions above for the four-vectors and doing a little bit of algebra, you can show that
$$p_\bar\nu = \frac{m_\pi^2-m_e^2}{2(E_\pi+p_\pi)}.$$ Once you have that, you can easily calculate the electron's energy from conservation of energy.

If you can get a copy of Griffith's Introduction to Elementary Particles, try looking over the chapter on special relativity. He shows how to do these types of problems efficiently. As you've found out, if you don't choose a good method, you can get completely bogged down in algebra.

5. Nov 10, 2013

### skrat

Firstly, thanks for all the help so far.

But, what do I do here:
What is the minimum amount of energy of photon that hits proton so $\pi ^+$, $\pi ^-$ and $\pi ^0$ can be "born" (that would be direct translation from my language, I don't know the correct english word for this...).

What I don't know is what is my system after the photon hits proton....

Before:

$p_{0}^{\mu }=(\frac{E_\gamma +E_{e^{+}}}{c},\frac{E_\gamma }{c})$

after:

Now what? For sure these three $\pi ^+$, $\pi ^-$ and $\pi ^0$ , all of them with rest energy but what happens to $e^+$ and $\gamma$...? Something has to take the momentum $\frac{E_\gamma }{c}$ from photon....

6. Nov 10, 2013

### skrat

Similar question goes for this one:
A proton hits another proton and starts a reaction: $p+p\rightarrow p+n+\pi ^{+}$ (p=proton, n=neutron).

Minimum kinetic energy should than be:

$E_p+m_pc^2=m_pc^2+m_nc^2+m_{\pi ^+}c^2$

$T_p+2m_pc^2=m_pc^2+m_nc^2+m_{\pi ^+}c^2$

so

$T_p=m_nc^2+m_{\pi ^+}c^2-m_pc^2= 141,6 Mev$ instead of 292 MeV (the "official" result...)

OMG -.-

7. Nov 10, 2013

### vela

Staff Emeritus
You should open new threads for new problems. Were you able to work out the original problem?

8. Nov 11, 2013

### fzero

I agree with vela that you should start a new thread for the new problems, but in the interest of trying to say something useful, I will reply anyway. If you really want to figure out what's going on here, we can use the quark content. The proton is $uud$, while the neutron is $udd$. The pions are $\pi^+ = u \bar{d}$, $\pi^- = \bar{u}d$, while $\pi^0$ is a linear combination of $u\bar{u}$ and $d\bar{d}$. In the vicinity of the proton, the photon can pair create $u\bar{u}$ or $d\bar{d}$. So as intermediate states, we can have

$$\gamma + p = u\bar{u} + uud ~~\mathrm{or} ~~d\bar{d} + uud.$$

We can assemble the linear combination of these states into $\pi^0+ p$, so that is one allowed final state. We can also assemble the first state into

$$u\bar{u} + uud = \bar{u}d + uuu = \pi^- + \Delta^{++}.$$

The $\Delta^{++}$ is very unstable into decay to $p + \pi^+$, but for the threshold pion production calculation would ignore this and just consider the first step of $\pi^-$ production.

The second state above can be assembled into

$$d\bar{d} + uud = u\bar{d} + udd = \pi^+ + n.$$

So we have 3 channels to consider:

$$\gamma + p \rightarrow \pi^0 + p ,~~~ \pi^- + \Delta^{++}, ~~~ \pi^+ + n.$$

By considering energy and momentum conservation for each channel, we can determine the minimum photon energy needed.

You are ignoring momentum conservation. Since in the lab frame, there is one proton moving with respect to the second proton at rest, the system has net momentum in the lab frame. It is therefore impossible for all of the final states to be at rest in the lab frame. The easiest way to solve the problem is to work in the center of mass frame.