Minimum energy of a photon to produce ##\pi^+##

[Frostman]

Homework Statement

A photon hits a proton stationary in the laboratory system and gives rise to the process
$$\gamma + p \rightarrow n + \pi^+$$
Assuming that the masses of the proton and the neutron are approximately equal, denoted by $m$, and that the mass of the pion is $m_\pi$:
1- Determine the minimum energy that the photon must have for the process to take place.
2- At this energy, calculate the average life of the $\pi^+$ in the reference system
ment of the laboratory, knowing that the mean life at rest is $\tau_0$.

Relevant Equations

Conservation of the four-momentum

I have a doubt about the first request:

Determine the minimum energy that the photon must have for the process to take place.

I suppose to find the minimum energy of $\gamma$ in the situation where $p$ is stationary, there is no reason to say that the proton is stationary if I were to calculate it in the CM, right?. So I have to consider che LAB-frame to find $E_\gamma$

In the LAB-frame so I have these four-momentums:

$p^\mu_\gamma=E_\gamma(1,0,0,1)$
$p^\mu_p=(m, 0, 0, 0)$
$p^\mu_n=(E_n, 0, 0, \sqrt{E_n^2-m^2})$
$p^\mu_\pi=(E_\pi, 0, 0, \sqrt{E_\pi^2 - m_\pi^2})$

In this case I have 3 unknowns: $E_\gamma$ (what the problem wants) and $E_n$, $E_\pi$, with 2 equations (energy and $p_z$).

I would like to understand if I first understood the request and in the case it was correct, to understand if the consideration I made is correct.

 

[TSny]

I suppose to find the minimum energy of $\gamma$ in the situation where $p$ is stationary, there is no reason to say that the proton is stationary if I were to calculate it in the CM, right? So I have to consider the LAB-frame to find $E_\gamma$.

Right, the proton is not at rest in the CM frame. But it is very helpful to think about the final state of the neutron and the pion in the CM frame for the case where the photon has the minimum possible energy.

There is a slick way to solve this type of problem using the idea that the "length" of a 4-vector has the same value in any frame of reference. If $E_{\rm tot}$ and $\vec P_{\rm tot}$ are the total energy and momentum in some frame of reference, how would you express the square of the length of the total energy-momenum 4-vector in terms of $E_{\rm tot}$ and $P_{\rm tot}$?

 

[Steve4Physics]

I suppose to find the minimum energy of $\gamma$ in the situation where $p$ is stationary, there is no reason to say that the proton is stationary if I were to calculate it in the CM, right?.

[EDIT. Ha! TSny beat me to it!]

The minimum $\gamma$ energy correponds to the neutron and pion being created stationary in the CM frame.

What can you say about the length of the initial energy-momentum 4-vector in the Lab frame and the length of the final energy-momentum 4-vector in the CM frame?

 

[Frostman]

Thank you so much guys!
I consider the initial state in the LAB frame:

$p^\mu_\gamma=E_\gamma(1,0,0,1)$
$p^\mu_p=(m, 0, 0, 0)$

And the final state in the CM frame:

$p^\mu_n=(m,0,0,0)$
$p^\mu_\pi=(m_\pi, 0, 0, 0)$

Recalling that the norm according to Minkowski is a Lorentz-invariant quantity I can write:

$(p^\mu_\gamma+p^\mu_p)^2=(p^\mu_n+p^\mu_\pi)^2$
$p^2_\gamma+p^2_p+2p^\mu_\gamma p_{\mu p}=p^2_n+p^2_\pi+2p^\mu_n p_{\mu \pi}$
$0+m^2+2E_\gamma m=m^2+m_\pi^2+2mm_\pi$

And finally I get:

$E_\gamma=m_\pi + \frac{m_\pi^2}{2m}$

From this I can calculate the average life of $\pi^+$ in the LAB-frame. In this case, we have a time dilation:

$\tau= \gamma \tau_0= \frac{E_\gamma}{m_\pi} \tau_0 = (1+\frac{m_\pi}{2m})\tau_0$

Is it correct?

 

[Steve4Physics]

$E_\gamma=m_\pi + \frac{m_\pi^2}{2m}$

From this I can calculate the average life of $\pi^+$ in the LAB-frame. In this case, we have a time dilation:

$\tau= \gamma \tau_0= \frac{E_\gamma}{m_\pi} \tau_0 = (1+\frac{m_\pi}{2m})\tau_0$

Is it correct?

$E_\gamma=m_\pi + \frac{m_\pi^2}{2m}$ looks correct but I can't see how you got $\gamma = \frac{E_\gamma}{m_\pi}$ when finding $\tau$.

Remember in the Lab' frame, we see the neutron and pion moving together so they are like a single particle of rest mass $m + m_\pi$.

If you consider the intial total energy and final total energy, both measured in the Lab' frame, you can get an equation from which $\gamma$ can be found.

While I think of it, a couple of points for information which I hope are useful:
- the plural of 'momentum' is usually 'momenta' ('momentums' is sometimes used but usually not in a physics context);
- the usual convention for simple 4-vector problems is to consider motion to be in the x-direction (not the z-direction). So, for example, rather than:
$E_\gamma(1,0,0,1)$
we would use:
$E_\gamma(1,1,0,0)$
and for neatness/brevity some authors would just write it as $E_\gamma(1,1)$.

 

[Steve4Physics]

$E_\gamma=m_\pi + \frac{m_\pi^2}{2m}$

From this I can calculate the average life of $\pi^+$ in the LAB-frame. In this case, we have a time dilation:

$\tau= \gamma \tau_0= \frac{E_\gamma}{m_\pi} \tau_0 = (1+\frac{m_\pi}{2m})\tau_0$

Is it correct?

$E_\gamma=m_\pi + \frac{m_\pi^2}{2m}$ looks correct but I can't see how you got $\gamma = \frac{E_\gamma}{m_\pi}$ when finding $\tau$.

Remember in the Lab' frame, we see the neutron and pion moving together so they are like a single particle of rest mass $m + m_\pi$.

If you consider the intial total energy and final total energy, both measured in the Lab' frame, you can get an equation from which $\gamma$ can be found.

While I think of it, a couple of points for information which I hope are useful:
- the plural of 'momentum' is usually 'momenta' ('momentums' is sometimes used but usually not in a physics context);
- the usual convention for simple 4-vector problems is to consider motion to be in the x-direction (not the z-direction). So, for example, rather than:
$E_\gamma(1,0,0,1)$
we would use:
$E_\gamma(1,1,0,0)$
and for neatness/brevity some authors would just write it as $E_\gamma(1,1)$.
[EDIT: My preference would be to write $(E_\gamma, E_\gamma)$.]

 

[Frostman]

$E_\gamma=m_\pi + \frac{m_\pi^2}{2m}$ looks correct but I can't see how you got $\gamma = \frac{E_\gamma}{m_\pi}$ when finding $\tau$.

Remember in the Lab' frame, we see the neutron and pion moving together so they are like a single particle of rest mass $m + m_\pi$.

If you consider the intial total energy and final total energy, both measured in the Lab' frame, you can get an equation from which $\gamma$ can be found.

Yes, I can see it.

So the initial state in LAB frame is:

$p^\mu_\gamma = (m_\pi+\frac{m_\pi^2}{2m}, 0, 0, m_\pi+\frac{m_\pi^2}{2m})$
$p^\mu_p = (m, 0, 0, 0)$

While the final state in LAB frame is:

$p^\mu = (\sqrt{|p|^2+(m_\pi+m)^2}, 0, 0, p)$

I suppose now that I have to work with $p = \gamma (m_\pi + m) v=\frac 1 {\sqrt{1-v^2}}(m_\pi + m)v$

And find $v$ from:

$m_\pi+\frac{m_\pi^2}{2m} = \frac 1 {\sqrt{1-v^2}}(m_\pi + m)v$

While I think of it, a couple of points for information which I hope are useful:
- the plural of 'momentum' is usually 'momenta' ('momentums' is sometimes used but usually not in a physics context);
- the usual convention for simple 4-vector problems is to consider motion to be in the x-direction (not the z-direction). So, for example, rather than:
$E_\gamma(1,0,0,1)$
we would use:
$E_\gamma(1,1,0,0)$
and for neatness/brevity some authors would just write it as $E_\gamma(1,1)$.
[EDIT: My preference would be to write $(E_\gamma, E_\gamma)$.]

Okay! Thank you for these conventions! I will apply them next time!

 

[Steve4Physics]

I suppose now that I have to work with $p = \gamma (m_\pi + m) v=\frac 1 {\sqrt{1-v^2}}(m_\pi + m)v$

I think it's easier than that. Consider simple conservation of energy in the Lab' frame.

Initial total energy = $m + E_\gamma = m + m_\pi+\frac{m_\pi^2}{2m}$

The neutron and pion are moving together like a single particle of mass $m+m_\pi$ so their total energy in the Lab' frame is $\gamma(m+m_\pi)$ giving

$m + m_\pi+\frac{m_\pi^2}{2m}$ = $\gamma(m+m_\pi)$

which gives $\gamma$

 

[Frostman]

I think it's easier than that. Consider simple conservation of energy in the Lab' frame.

Initial total energy = $m + E_\gamma = m + m_\pi+\frac{m_\pi^2}{2m}$

The neutron and pion are moving together like a single particle of mass $m+m_\pi$ so their total energy in the Lab' frame is $\gamma(m+m_\pi)$ giving

$m + m_\pi+\frac{m_\pi^2}{2m}$ = $\gamma(m+m_\pi)$

which gives $\gamma$

Absolutely yes. Thank you very much for helping! Have a good day! :)