Finding Maximum and Minimum Electron Energy in Muon Decay

[arivero]

Now, I see that it is always possible to extract an electron with momentum \vec e = 0 because
<br /> E&#039;^2 - p&#039;^2 = (E-e_0)^2 - p^2 = E^2 - p^2 + e_0^2 - 2 E e_0 =<br /> m_\mu^2 + m_e^2 - 2 E m_ e > m_\mu^2 + m_e^2 - 2 m_\mu m_ e = <br /> (m_\mu - m_e)^2 &gt; 0<br />
so the answer is that Emin=m_e c^2

Uff, last step wrong :

<br /> m_\mu^2 + m_e^2 - 2 E m_ e < m_\mu^2 + m_e^2 - 2 m_\mu m_ e<br />
so this line only proves that if we produce an electron at rest, then E&#039;^2 - p&#039;^2 &lt; (m_\mu - m_e)^2

In fact I need 2 E m_ e &lt; m_\mu^2 + m_e^2 and it obviously works when E=m_\mu, the muon rest case I put out as counterexample. What surprises me is that above E &gt; (m_\mu^2 + m_e^2) / 2 m_e (about 10 GeV) then it seems as if we can not produce electrons at rest anymore. This is puzzling but surely right, reflecting that in CM decay there is a maximum speed for an electron: if this "CM max speed" is less than the speed of the muon, then the Lab Frame can produce rest electrons. Above this speed, all the electrons in the LabFrame have some momentum greater than zero.

Incidentally, note that any approximation m_e \to 0 blocks the access to this regime, so the origin of the disagreement with M.A. can be here.

 

[arivero]

I have to run to a class right now but it seems to me that using four-momentum conservation is a much more efficient way to tackle this type of problem. I will look at it tonight. Just a thought.

Hi, I hope you get some time to think about it, tonight or tomorrow. As the question was for a non invariant quantity (the Energy in laboratory system) I did not worry about four-momentum, but any hint in this way could be very very helpful.

perhaps the trick is to find a fast method to derive the inequality
(E, \vec p) . (e_0, \vec e) &lt; {m_\mu^2 + m_e^2 \over 2}
and then to maximize and minimize e_0

 

[arivero]

hmm,
(E, \vec p) . (e_0, \vec e) &lt; {m_\mu^2 + m_e^2 \over 2}
is
E e_0 - p e \cos \phi &lt; {m_\mu^2 + m_e^2 \over 2}
and in the extreme cases \cos \phi = \pm 1, as we have been told, the equality solves to

<br /> e_0= { E ( m_\mu^2 + m_e^2 ) \pm p (m_\mu^2 - m_e^2)<br /> \over 2 m_\mu^2} <br /> ={ m_\mu^2 (E \pm p )+ m_e^2 (E \mp p) <br /> \over 2 (E + p ) (E - p ) }<br /> ={ (E + p ) (E - p ) (E \pm p )+ m_e^2 (E \mp p) <br /> \over 2 (E + p ) (E - p ) }=<br /> { (E \pm p )^2+ m_e^2 <br /> \over 2 (E \pm p ) }<br /> <br />

Thus, if we assume the solutions come from the extreme cases, it is possible to obtain BOTH my solution and Meir Achuz's ones! Mine, e_0= m_e, comes when the "<" is not saturated, which happens when
E &lt; {m_\mu^2 + m_e^2 \over 2 m_e}
p &lt; { m_\mu^2 - m_e^2 \over 2 m_e}
Note that it is in this point when the energy calculated with M.A. formula is exactly m_e.

So in some sense it was a "complete" answer with only telling that in the case of getting under saturation, the minimum is m_e (and then the extreme case of the cosine does not need to apply). The problem was that it is not easy to tell, from the equality formula, when is the saturation happening, because the function is never smaller than m_e.

 

[physicsbhelp]

Can Anyone Help Me With My Problem, Its Labeled Inelastic Problem Help

 

[Meir Achuz]

Arivero is right. My formula in post #15
E_{min}=[(E_\mu-p_\mu)^2+m^2]/[2(E_\mu-p_\mu)]
is only correct for E_\mu > 10 GeV.
For energies less than that, my formula would give the energy of an electron going backwards in the laboratory. Emin should then be when the electron is at rest, so that Emin=m (mass of the electron) for these muon energies.

My formula for Emin comes from the case when both neutrinos go forward.
This results in E_e-p_e=E_\mu-p_\mu. Solving for E_e
gives the formula above for Emin. Solving instead for p_e gives
p_e=\frac{m^2-(E_\mu-p_\mu)^2}{2(E_\mu-p_\mu)}.
p_e=0 when the numerator vanishes, which occurs for
E_\mu=\frac{M_\mu^2+m^2}{2m_e} ~ 10 GeV.

I apologize for some of my earlier posts which did not recognize
that p_e, which is really the z component of p_e,
would not be negative for Emin.

 

[Tolya]

Thank you all.

 

[CarlB]

It amazes me how complicated even a very elementary problem in relativistic kinematics is. This is most of the reason I work in qubits instead of the full field theory.

 

[arivero]

It was surely my fault to complicate it :-( I went for a limit solution to avoid giving Tolya an explicit one, and the limit obscured the discussion. It is elementary if you have the right physical intuition, but complicated if you need to trade physics by math. On the positive side, the complication gave us a complete view of the panorama, and I am reassured about keeping the thread on decay rates going forward

Hmm I am also amazed that the label [SOLVED] does not close the thread. Well.