Finding Maximum and Minimum Electron Energy in Muon Decay

[Tolya]

[SOLVED] myuon decay

First, sorry for my English. I'm not very well in it... Please, try to understand.
I wrote this problem in "Introductory Physics", but some man sayed its not "introductory", so I decided to post it in "Advenced Physcis".
The problem.
We have reaction: $$\mu \rightarrow e + \nu + \tilde{\nu}$$
We know energy of myuon - E.
Question: Find the maximum and the minimum energy of electron.

My attept:

Conservation of energy: $$E = E_e+E_{\nu}+E_{\tilde{\nu}}$$ (1)
Conservation of impulse: $$\vec{p} = \vec{p_e}+\vec{p_{\nu}}+\vec{p_{\tilde{\nu}}}}$$ (2)

The mass of the rest of neutrino and antineutrino is 0. So, $$E_{\nu}=p_{\nu}c$$
$$E_{\tilde{\nu}}=p_{\tilde{\nu}}c$$ and from the first equation:

$$E_e = E-c(p_{\nu}+p_{\tilde{\nu}})$$

Therefore, we must find the minumum and the maximum value of $$(p_{\nu}+p_{\tilde{\nu}})$$ Then, the minimum value of this expression gives us the maximum value of $$E_e$$ and the maximum value gives the minimum of energy. Am I right in this statement?

From (2):
$$\vec{p_{\nu}}+\vec{p_{\tilde{\nu}}}} = \vec{p} - \vec{p_e}$$
Also we know that: $$p= \sqrt{\frac{E^2}{c^2}-m^2c^2}$$
and: $$p_e= \sqrt{\frac{{E_e}^2}{c^2}-{m_e}^2c^2}$$

How can I find the minumum and the maximum value of $$(p_{\nu}+p_{\tilde{\nu}})$$ with the help of all I wrote here? :)

I also have an assumption that we can find half of the answer simply in the following way:
$$E_e$$ reaches its maximum when impulses of neutrino and antineutrino have opposite directions and equals in absolute. (It's easy to understand this fact because in this case impulses of neutrino and antineutrino compensate each other and the value $$p_e$$ reaches its maximum, so does $$E_e$$). Then, almost easy:
$$p=\sqrt{\frac{E^2}{c^2}-m^2c^2}=p_e=\sqrt{\frac{{E_e}^2}{c^2}-{m_e}^2c^2}$$
And we have: $$E_e^{max}=\sqrt{E^2+c^4({m_e}^2-m^2)}$$ Am I right? How can I find the minimum value of $$E_e$$?

Please, help.

 

[Meir Achuz]

EMax is when the two neurinos go backward.
Emin is when the two neurinos go forward.
Just write the energy and momentum equations.
Then add or subtract the equations to eliminate the neutrinos.

 

[Tolya]

I'm sorry, what do you mean backward or forward?
Please, read my calculations.
Emax is when impulses of neutrino and antineutrino have opposite directions and equals in absolute! Isn't it?

 

[arivero]

Let me see. Assume the muon moves along x. What we have is
$$E^2= m_\mu^2 c^4 + p_ x^2 c^2$$
$$e_0^2= m_e^2 c^4 + e_ x^2 c^2 + e_ y^2 c^2 +e_ z^2 c^2$$
$$r_0^2= r_ x^2 c^2 + r_ y^2 c^2 +r_ z^2 c^2$$
$$s_0^2= s_ x^2 c^2 + s_ y^2 c^2 +s_ z^2 c^2$$

with
$$E=e_0+r_0+s_0$$
$$p_i=e_i+r_i+s_i$$

Meir Achuz suggests, it seems to me, to consider first the solutions where the neutrinos are along the direction x too. This simplifies the issue to

$$E^2= m_\mu^2 c^4 + p_ x^2 c^2$$
$$e_0^2= m_e^2 c^4 + e_ x^2 c^2$$
$$r_0^2= r_ x^2 c^2$$
$$s_0^2= s_ x^2 c^2$$And then each of two last equations have two solutions, + and -, which are the forward and backward he refers to, I guess. His point surely is that the equation for the energy of the electron is then

$$e_0^2= m_e^2 c^4 + (p_x-(\pm r_0) - (\pm s_0) )^2 c^2$$
or, only with energies:
$$e_0^2= m_e^2 c^4 + ( +\sqrt {E^2 - m_\mu^2 c^4 \over c^2 }-(\pm r_0) - (\pm s_0) )^2 c^2$$

so at the end if you put the two neutrinos going "forward" along the path of the incoming muon you get the minimum of energy for the electron, and if you choose the other two signs you get the maximum

Or it seems. It is not yet a complete answer because you need to proof the existence of the solutions for each choosing of signs, and also to study the solutions where the momentum of the neutrinos and/or the electron along y or z are not null.

(Ah, note that there are two reference frames in this kind of problems: the Lab reference frame, and the Center of Mass reference frame. Sometimes it is better first to boost the system from Lab to Center of Mass, and in this particular case you gain a good simplification, you know that in CM the initial $\vec p$ is zero and the energy is the rest mass of the muon, about 105 MeV. Problem is, that Energy is not a Lorentz invariant, so I am not sure if the conclusions are he same)

 

[arivero]

On other hand, your argumentation

maximum of energy in electron = maximum of momentum in electron = minimum of the sum of momenta in neutrino = opposite directions for neutrinos.

seems also compelling, and contradicts the claim of Meir Achuz. But it is wrong, because you can always increase the momentum of a particle by increasing the momentum of another one going exactly in the opposite direction. So it is not true that "maximum of ... = minimum of... " for momentum. Of course it is true for Energy, but we can not put the energy of the neutrinos equal zero; it is not a solution of the equations.

You can try to argue (with the caveat above) that in the rest frame the equations are
$$E^2= m_\mu^2 c^4$$
$$e_0^2= m_e^2 c^4 + e_ x^2 c^2$$
$$r_0^2= r_ x^2 c^2$$
$$s_0^2= s_ x^2 c^2$$

and so you can consider the energy given as
$$e_0^2= m_e^2 c^4 + (-(\pm r_0) - (\pm s_0))^2 c^2$$
and well, again it seems NOT to favor the idea of having opposite directions for neutrinos

In any case, this equation must be compatible with
$$e_0 + r_0 + s_0 = m_\mu c^2$$

is it?

 

[Tolya]

arivero, thank you for almost complete answer. You showed my mistake very clearly. But it still bother me, how can I prove, that your solution is right, when we say nothing about momentum of the neutrinos and/or the electron along y or z? When we simplify our equations, don't we make a mistake?

 

[arivero]

arivero, thank you for almost complete answer. You showed my mistake very clearly. But it still bother me, how can I prove, that your solution is right, when we say nothing about momentum of the neutrinos and/or the electron along y or z? When we simplify our equations, don't we make a mistake?

I am also of your same opinion, we need to consider the "orthogonal" direction. Let me to stay still in the CM reference frame (we can boost to Lab Frame later). The problem is that we have a lot of variables and no so many equations. Let e the variables for the electron, as above, and r, s the ones of the neutrinos.

The way I see to do some use of the opposite directions for neutrinos (in this reference frame) is to label colectively all of them (y and z) as $\vec r_\perp$. So the equations are
$$\vec r_\perp=-\vec s_\perp$$
$$e_x + r_x + s_x =0$$
$$e_0 + r_0 + s_0 = m_\mu c^2$$
$$e_0^2 = m _e^2 c^4 + e_x ^2 c^2$$
$$r_0^2 = + (r_x^2 + r_\perp^2) c^2$$
$$s_0^2 = + (s_x^2 + s_\perp^2) c^2$$

IE, we have six equations and eight variables. With the first equation, which implies $r_\perp^2=s_\perp^2$, we can reduce trivially to five equations with seven variables. It means that we must to choose two variables as independent, and to calculate all the other five as function of them. Note that the squares complicate the process of solving, but they also work in our favour because imaginary solutions are forbidden, then not all the plane of the independent variables is available.

It is pretty obvious that we can not choose both $e_0, e_x$ as independent. But I am not sure which is the best pair to choose (Of course there are some good ideas in the textbooks, but if we have come up to here without a textbook, we can try to follow a little bit on our own).

 

[Tolya]

Please, give me advice about the textbook, where I can find the answer or, at least, some ideas about this.

 

[arivero]

Please, give me advice about the textbook, where I can find the answer or, at least, some ideas about this.

There are some notes about kinematics online, for instance
http://pdg.lbl.gov/2007/reviews/kinemarpp.pdf

But they are really "reminder notes", this is, you are supposed to know how to work out the problem.

Second, the classical analysis of four legs is done in the so-called "Mandelstam variables". Google for it, or look for them in the index of any book in the library. Problem is, usually the analysis is done with two incoming legs and two outcoming, that is "scattering". You want to use one incoming leg and three outcoming legs. But is is a good idea to start looking for books; if they do not speak of Mandelstam, they will not to do a complete analysis of the relativistic problem. I'd bet first for Russian books, of the Landau series for instance.

If you can go even deeper, the next concept to look is called "Dalitz plot". Again google or browse indexes for it. It is arguably the best idea to represent whole sets of three body decays. But to understand it we need to introduce the "invariant mass" of a pair of particles. The idea is that not only $E^2-p^2$ is a relativistic invariant, but also the difference [itex](\sum_j E)^2- (\sum_j p)^2[/tex] is, for a sum along any set of particles.

This is called the "invariant mass squared" of a set of particles. Dalitz suggestion, mad as it can appear, is to choose as independent variables for the 3-body decays the "invariant mass" of two pairs. We have three different pairs, $W_{er},W_{es},W_{rs}$. And as r and s are the same kind of particle, we can really build here two kinds of Dalitz plots:

A symmetrical one, taking as independent variables
$$W_{er}^2 \equiv (e_0+r_0)^2 - (e_x + r_x)^2 - r_\perp^2$$
$$W_{es}^2 \equiv (e_0+s_0)^2 - (e_x + s_x)^2 - s_\perp^2$$Or and asymmetrical one, taking as independent variables
$$W_{er}^2 \equiv (e_0+r_0)^2 - (e_x + r_x)^2 - r_\perp^2$$
$$W_{rs}^2 \equiv (r_0+s_0)^2 - (r_x + s_x)^2$$

In any case, the problem now is well defined: to build the function, say, $$e_0(W_{er}^2,W_{es}^2)$$, to calculate its domain and its maxima and minima, and then to describe them back in terms of the original variables.

I am afraid that your teacher will become suspicious of your work if you come to present it in terms of exactly this plane of analysis. Dalitz got to use it in 1953, a lot years later than the original relativity theory (but still before Mandelstam!).

 

[Tolya]

Thank you, arivero.
The second book of the Landau series laying on my table (in Russian of course) :) But the only thing I found in this book is collision of two particles, when after boom we have the same two particles (and also I found decay of the particle when we have two particles in the end).
About "Mandelstam variables" and "Dalitz plot", and my teacher. :)
I think this problem have the simple answer (not very simple, but you understand what I'm trying to say), because in other case, my teacher didn't give it to me :) You see, I don't familiar with "Mandelstam variables" and "Dalitz plot". Moreover, my programm of the subject "Theoretical Physics" in this term doesn't include these things (I'm studying in MIPT, Moscow, Russia, but I'm originally from Chernihiv, Ukraine).
Thanks for http://pdg.lbl.gov/2007/reviews/kinemarpp.pdf

 

[arivero]

Thank you, arivero.
The second book of the Landau series laying on my table (in Russian of course) :) But the only thing I found in this book is collision of two particles, when after boom we have the same two particles

This is the one of Mandelstam. There is a channel named, I believe, the "t-channel", where the same kinematics is interpreted as one particle decaying into three (but in two steps, as in a decat via W-boson).

Anyway, the only point is that you need to choose a pair of variables, any (except of course energy and momentum of the electron, as one depends of the other), and then use all the equations to put the energy of the electron (e0 in my notation) as a function of this pair of variables BUT considering the domain of the function, in the sense that all e0, r0, s0 must be real and positive. My guess is that your teacher expects you to do some non-invariant choosing, as for instance the two variables for the momentum of one of the neutrinos.

An invariant choosing can be an advantage when you are asked more sophisticated questions, for instance the _average_ of the energy carried by the electron.

 

[arivero]

Well, my choosing last night has been is to define $a_\perp \equiv r_\perp = -s_\perp$ and to check the domain of validity of $a_\perp(r_x,s_x)$. Pretty non covariant thing, but in any case we will boost to Lab Frame later.

Of course the implicit equation for a_\perp is
$$\sqrt{r_x^2 + a_\perp^2} + \sqrt {s_x^2+ a_\perp^2} + \sqrt{ (-(r_x+s_x))^2 + m_e^2} = m_\mu$$
where I am already using the traditional notation c=1.

To avoid to solve the whole exercise, I will analyse the limit case m_e=0 for the moment. Actually it is more difficult, I get some square corner instead of an hyperbole.

Anyway, it is clear that the frontier of the domain of vailidity will be at $a_^\perp=0$. This is
$$\sqrt{r_x^2 } + \sqrt {s_x^2} + \sqrt{ (-(r_x+s_x))^2 } = m_\mu$$

when solving we need to consider each quadrant separately, because of the signs of $r_x$ against $|r_x|$ etc.

The answer is a figure of six sides
$$(0, {m_\mu \over 2}) \to ( {m_\mu\over 2},0 ) \to ( {m_\mu\over 2}, -{m_\mu\over 2} ) \to ( 0, -{m_\mu\over 2} ) \to (- {m_\mu\over 2}, 0 ) \to ( -{m_\mu\over 2}, {m_\mu\over 2} ) \to ( 0, -{m_\mu\over 2} )$$

and the function is valid inside the polygon. Note that while solving we do a squaring, then introducing a spurious solution for outside of the polygon in the +- quadrants, but it is easy to argue against it. Even it is easy to do not notice it!

Now we ask where in the poligon (in the "colinear" boundary as well as inside) the energy of the "electron" gets into maximum or minimum. In this reference frame, the case where both neutrinos are opposite and orthogonal to the trajectory is the minimum, while the case where both are forward or both are backward can saturate the maximum,
$$E_{max}=\sqrt { (-(r_x+s_x))^2 } = \sqrt { (-({m_\mu\over 2}))^2 }$$

but it is not yet the answer because we need to recover the lab reference frame; in the CM reference frame the electron direction was arbitrary respect to the Lab->CM boost, so we can now consider any angle to do the reverse CM->Lab boost. I will come to this later if nobody works it out before. At this point, I suggest to contemplate the CM answer because it is already funny to see:
- all the three particles being massless, one of them can carry only at most one half of the rest energy of the decaying particle. Not full, not 1/3.
- The "colinear" solution where all the three particles are produced in a same straight line has some peculiarity in the sense that when the particle "e" saturates the energy, the particles "r" and "s" can still combine themselves to various configurations of momentum; but if the particle "e" does not saturate, then "r" and "s" are fixed too. Amusing? Not so rare: what happens is that if "e" does not saturate, either "r" or "s" are forced to saturate. Thus:
- In the "colinear" solution there is always a particle carrying away 1/2 of the initial rest energy.

 

[Tolya]

Thanks!
I'm reading, translating and trying to understand...

 

[arivero]

Thanks!
I'm reading, translating and trying to understand...

Take your time, and let me to know what you do not understand. Meanwhile, let me to explain a little bit of the boost process (Actually I am not sure if "boost" is the right word"). The muon, that in the CM frame has zero momentum, and energy
$$E_{CM}= m_\mu c^2$$
must be boosted (restored) to Lab Frame, with the initial energy E of your problem statement. The transformation is
$$E=E_{Lab}={1 \over \sqrt { 1- \beta^2}} m_\mu c^2$$
so that we can calculate from here $\beta$ and thus its velocity $v=\beta c$. Of course the muon has now a momentum
$$\vec P_{Lab}={1 \over \sqrt { 1- \beta^2}} m_\mu \vec v$$
so that
$$E_{Lab}^2-P_{Lab}^2= {1 \over { 1- \beta^2}} m_\mu^2 (c^4 -v^2 c^2) = {1 \over { 1- \beta^2}} m_\mu^2 (c^4 - (\beta c)^2 c^2)= {1 \over { 1- \beta^2}} m_\mu^2 c^4 (1- \beta^2) = (m_\mu c^2) ^2$$

The problem now is that we must to apply the same $\beta$ to the outgoing electron, but we need to account that its CM momentum $e_x$ can be produced with the direction x being at any angle $\theta$ respect to the direction of the incoming muon.

So, let my massless electron to have a CM momentum $e_x \cos \theta$ along the direction of incoming muon, which is the direction of the boost. In the massive case, the one for you to solve, we should use Lorentz transformation for the energy. In this case, massless, we can say that the momentum along the boost gets relativistic Doppler shift so it becomes
$${\sqrt{1 + \beta} \over \sqrt {1-\beta} } e_x \cos \theta$$
So that
$$e^2_{0,Lab}= (e_x \sin \theta)^2 c^2+ ( {\sqrt{1 + \beta} \over \sqrt {1-\beta} } e_x \cos \theta )^2 c^2=$$
$$=e^2_{0,CM} (\sin^2 \theta+ { {1 + \beta} \over {1-\beta} } \cos^2 \theta)$$
so that the maximum and minimum in CM can be still further amplified of decreased depending on theta (NOTE: I am still not sure of the signs for beta in the shift. Reference online: http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/vec4.html#c2).

Edited, or continued: in principle we expect to use ${ {1 + \beta} \over {1-\beta} }$ or the contrary ${ {1 - \beta} \over {1+\beta} }$ depending if this component of the electron momentum is sent along the muon path or backward against it. It is a bit diffuse, and it proves that it is better to apply straigthforwardly the Lorentz transformation for energy-momentum, but this is left to Tolya. Note that the orthogonal situation, $\theta = \pi/2$, obviously does not depend of [itex]\beta[/tex], so the sign can be exchanged between the blue and red shifting.

In any case, the maximum and minimum for this massless situation are clear: the minimum is zero (all the energy in the hands of the neutrinos, which is not possible in the massive situation); and on the other side, using
$$\beta= \sqrt{ 1-{m_\mu^2 c^4 \over E_{Lab}^2}}$$
we get that the maximum is
$$E_{MAX}= \sqrt {e^2_{0,CM} {{1 + \beta} \over {1-\beta} }} = ({ m_\mu c^2 \over 2}) \sqrt {{1 + \beta} \over {1-\beta} }= ({ m_\mu c^2 \over 2}) \sqrt {{1 +\sqrt{ 1-{m_\mu^2 c^4 \over E_{Lab}^2}}} \over {1-\sqrt{ 1-{m_\mu^2 c^4 \over E_{Lab}^2}}} }= ({ m_\mu c^2 \over 2}) \sqrt {{E_{Lab} + \sqrt{ E_{Lab}^2 -{m_\mu^2 c^4 }}} \over { {E_{Lab} -\sqrt{ {E_{Lab}^2-{m_\mu^2 c^4 }}} }$$

(note to self: the problem seems to blow if the mass of the muon is also taken towards zero)

EDIT, TO compare with M.A. answer:

$$=({ m_\mu c^2 \over 2}) \sqrt { E_{Lab} + \sqrt{p_\mu ^2 c^2 } \over E_{Lab} -\sqrt{ p_\mu ^2 c^2 } } = ({ m_\mu c^2 \over 2}) \sqrt { {E_\mu + p_\mu c } \over {E_\mu - p_\mu c} } } = ({ m_\mu c^2 \over 2}) { {E_\mu + p_\mu c } \over \sqrt {E_\mu^2 - p_\mu^2 c^2} } } =({ m_\mu c^2 \over 2}) { {E_\mu + p_\mu c } \over m_\mu c^2 } ={ 1 \over 2} ( {E_\mu + p_\mu c })$$
YES, it coincides with Meir Achuz' answer in the next post (for m->0).

 

[Meir Achuz]

"Question: Find the maximum and the minimum energy of electron."
I don't understand all these posts if you just want to answer this question.
For EMax, Using my post #2, the answer is
$$E=[(E_\mu+p_\mu)^2+m^2]/[2(E_\mu+p_\mu)]$$,
and for Emin, $$E=[(E_\mu-p_\mu)^2+m^2]/[2(E_\mu-p_\mu)]$$.

 

[arivero]

"Question: Find the maximum and the minimum energy of electron."
I don't understand all these posts if you just want to answer this question.
For EMax, Using my post #2, the answer is $$E=(E_\mu+p_\mu)/2$$,
and for Emin, $$E=(E_\mu-p_\mu)/2$$.

I want to *proof* the answer. I do not understand how do you derive this result. Particularly, your answer is also for the limit where $$m_e \to 0$$, isn't it? How do you derive the answer for $$m_e \neq 0$$?

 

[arivero]

"Question: Find the maximum and the minimum energy of electron."
I don't understand all these posts if you just want to answer this question.
For EMax, Using my post #2, the answer is
$$E=[(E_\mu+p_\mu)^2+m^2]/[2(E_\mu+p_\mu)]$$,
and for Emin, $$E=[(E_\mu-p_\mu)^2+m^2]/[2(E_\mu-p_\mu)]$$.

Sorry again, m is the mass of the electron or the one of the muon?

 

[Meir Achuz]

I'm sorry. that first post was a poor latex input that did have m_e=0.
In English, we say "prove the result."
The colinearity is seen by a physical argument that the electron momentum will be maximum when the angle of both neutrinos is 180 degrees, and min when all three are at 0 degrees.
That is how Dalitz derived the limits of his plot.

 

[arivero]

I was doing the $$m_e=0$$ case because I wanted to illustrate the process before to solve it (or someone other to give a solution).

The colinearity is seen by a physical argument that the electron momentum will be maximum when the angle of both neutrinos is 180 degrees, and min when all three are at 0 degrees.
That is how Dalitz derived the limits of his plot.

Sorry the English. In any case, I do not see the physical argument to dispose of the proof process. I can prove it if we go all the way along, and then the maximum and minimum are -perhaps- as you say now. By the way, now you see I found practically impossible to understand fully the "backward and forward" argument in your first post: you said "EMax is when the two neurinos go backward" and this is not exactly the same as saying "when the angle of both neutrinos is 180 degrees".

Oh wait, you mean the angle between the neutrinos and the electron? So I have some mistake somewhere.

 

[arivero]

Hmm, Meir Achuz , consider decay from a static muon (or CM frame). Then both your Emax and Emin are the same value.

On the other hand, I have edited/finished my post #14 and yes, the result for Emax coincides with mine, at least for the massless electron. And I think it coincides for the rest frame muon too, so it is Emin which is under suspect.

 

[Meir Achuz]

Hmm, Meir Achuz , consider decay from a static muon (or CM frame). Then both your Emax and Emin are the same value.

On the other hand, I have edited/finished my post #14 and yes, the result for Emax coincides with mine, at least for the massless electron. And I think it coincides for the rest frame muon too, so it is Emin which is under suspect.

In the muon rest system, EMax would follow if the electron were going forward (in the original direction of the muon) with the two neutinos going backward, and Emin for the opposite directions. The problem can also be soved this way, and then LT to the frame in which the muon is moving.

My calculation is in the lab system, for Emin:
$$E+p_\nu+p_{\overline\nu}=E_\mu$$
$$p+p_\nu+p_{\overline\nu}=p_\mu$$
Subtract the equations, and solve for E., using $$p=\sqrt{E^2-m^2}$$.

 

[arivero]

I
My calculation is in the lab system, for Emin:
$$E+p_\nu+p_{\overline\nu}=E_\mu$$
$$p+p_\nu+p_{\overline\nu}=p_\mu$$
Subtract the equations, and solve for E., using $$p=\sqrt{E^2-m^2}$$.

But I can not subtract as it is; the $p_\nu$ in the the first equation is actually $|p_\nu|$.

What I get is
$$E-p+|p_\nu|-p_\nu+|p_{\overline\nu}|-p_{\overline\nu}=E_\mu-p_\mu$$

 

[Meir Achuz]

|p|=p in this case.

 

[arivero]

|p|=p in this case.

That is your guess. My point is that from the equation of colinearity I am not finding it. You argue it is a physical argument, but I doubt if you remember correctly the argument because it should imply that a muon at rest can only decay into an electron of a very particular energy.

Consider the equations for the muon at rest. Not a CM trick, but the real Lab reference system too, with a zero momentum muon.
$$E+p_\nu+p_{\overline\nu}=m_\mu$$
$$p+p_\nu+p_{\overline\nu}= 0$$

You say that Emax follows from
$p \ge 0, p_\nu< 0, p_{\overline\nu}< 0$
and then from the sum of first and second equation
$$E+p=m_\mu$$
With this, we set
$$E^2-m_e^2 = (m_\mu - E)^2$$
and then solve to
$$-m_e^2 = m_\mu^2 - 2 E m_\mu$$
and
$$E=(m_\mu^2 +m_e^2 )/ (2 m_\mu )$$

Ok. But you say also that Emin
follows from $p \ge 0, p_\nu\ge 0, p_{\overline\nu}\ge 0$
In this case, it implies, via the second equation, that
$$p=p_\nu=p_{\overline\nu}= 0$$
which is only possible if $m_e=m_\mu$
So I am pretty sure your memory is having a gap.

If still we insist on subtracting first minus second equation, we get
$$E-p=m_\mu$$ and from it the same answer
$$E=(m_\mu^2 +m_e^2 )/ (2 m_\mu )$$
and thus Emax=Emin, but also the contradiction
$$p=E-m_\mu={m_e^2-m_\mu^2 \over 2 m_\mu} <0$$

 

[Meir Achuz]

"Question: Find the maximum and the minimum energy of electron."

This is the question I gave a complete answer to.
Now, you can continue on your own.

 

[arivero]

"Question: Find the maximum and the minimum energy of electron."

This is the question I gave a complete answer to.
Now, you can continue on your own.

NO You got the maximum right, you got the minimum wrong. What kind of counterexample do you need?

 

[Tolya]

NO You got the maximum right, you got the minimum wrong.

Please, arivero, tell me about your minimum answer. What did you get?

 

[arivero]

Ok, M.A. is right in a point; that the proof is very complicated. Let me to think another, if only for the minimum.

First, let's assume that any massive particle is able to decay into two massless particles This seems a plausible physical assumption, but surely it can also be proved as an extra exercise.

With this, all we need is to ask that the energy and momentum after extracting the electron is still the one of a massive particle, of course with different mass. This is:

we start with a muon of energy E, momentum p, so that $E^2-p^2=m_\mu^2$.

we extract an electron of energy $e_0$, momentum $\vec e$, so that $e_0^2- |\vec e|^2=m_e^2$.

after extraction of the electron we have

$$E'=E-e_0$$
$$\vec p'= \vec p - \vec e$$

and all we need is to check that $E'^2 - p'^2 > 0$

Now, I see that it is always possible to extract an electron with momentum $\vec e$ = 0 because
$$E'^2 - p'^2 = (E-e_0)^2 - p^2 = E^2 - p^2 + e_0^2 - 2 E e_0 = m_\mu^2 + m_e^2 - 2 E m_ e > m_\mu^2 + m_e^2 - 2 m_\mu m_ e = (m_\mu - m_e)^2 > 0$$
so the answer is that $Emin=m_e c^2$

EDITED: this is wrong if E > 10 GeV, see later posts

 

[arivero]

For the maximum from this point of view, let's examine with more detail the decay of a "particle" into two massless particles. Let the particles to have momenta of absolute value $r, s$ with an angle $\theta$ between. Then the match asks for

$$E'= r + s > 0$$
$$p'= \sqrt{r^2+s^2 + 2 r s \cos \theta} > 0$$ (thus $r-s < p' < r+s = E'$)
$$m'^2=E'^2-p'^2=(r^2 + s^2 + 2 r s) - (r^2+s^2 + 2 r s \cos \theta)= 2 r s (1- \cos \theta) >0$$

so really it is enough to keep $E'^2 - p'^2 > 0$ as in the previous case and we can grant the production of the two massless particles.

Probably here is the physical argument that M.A. remembered imperfectly: note that 0 and 180 degrees are extremal cases. But the argument does not include the electron nor the original muon, it is only an angle between neutrinos.
In any case. We have still $(E-e_0)^2 - (\vec p - \vec e)^2 > 0$ and again all we need is to maximize $e_0$ subject to it, as this condition is enough for granting enough energy and momentum to the two neutrinos. We can rewrite it as
$$0< E^2+e_0^2-2 E e_0 - (p^2 + e^2 - 2 \vec p . \vec e) = m_\mu^2 + m_e^2 - 2 ( E e_0 -\vec p . \vec e)$$

 

[nrqed]

Ok, M.A. is right in a point; that the proof is very complicated. Let me to think another, if only for the minimum.

First, let's assume that any massive particle is able to decay into two massless particles This seems a plausible physical assumption, but surely it can also be proved as an extra exercise.

With this, all we need is to ask that the energy and momentum after extracting the electron is still the one of a massive particle, of course with different mass. This is:

we start with a muon of energy E, momentum p, so that $E^2-p^2=m_\mu^2$.

we extract an electron of energy $e_0$, momentum $\vec e$, so that $E^2- |\vec e|^2=m_e^2$.

after extraction of the electron we have

$$E'=E-e_0$$
$$\vec p'= \vec p - \vec e$$

and all we need is to check that $E'^2 - p'^2 > 0$

Now, I see that it is always possible to extract an electron with momentum $\vec e$ = 0 because
$$E'^2 - p'^2 = (E-e_0)^2 - p^2 = E^2 - p^2 + e_0^2 - 2 E e_0 = m_\mu^2 + m_e^2 - 2 E m_ e > m_\mu^2 + m_e^2 - 2 m_\mu m_ e = (m_\mu - m_e)^2 > 0$$
so the answer is that $Emin=m_e c^2$

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.

 

[arivero]

Now, I see that it is always possible to extract an electron with momentum $\vec e$ = 0 because
$$E'^2 - p'^2 = (E-e_0)^2 - p^2 = E^2 - p^2 + e_0^2 - 2 E e_0 = m_\mu^2 + m_e^2 - 2 E m_ e$$ > $$m_\mu^2 + m_e^2 - 2 m_\mu m_ e = (m_\mu - m_e)^2 > 0$$
so the answer is that $Emin=m_e c^2$

Uff, last step wrong :

$$m_\mu^2 + m_e^2 - 2 E m_ e$$ < $$m_\mu^2 + m_e^2 - 2 m_\mu m_ e$$
so this line only proves that if we produce an electron at rest, then $E'^2 - p'^2 < (m_\mu - m_e)^2$

In fact I need $2 E m_ e < m_\mu^2 + m_e^2$ and it obviously works when E=m_\mu, the muon rest case I put out as counterexample. What surprises me is that above $E > (m_\mu^2 + m_e^2) / 2 m_e$ (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
$$(E, \vec p) . (e_0, \vec e) < {m_\mu^2 + m_e^2 \over 2}$$
and then to maximize and minimize $e_0$

 

[arivero]

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

$$e_0= { E ( m_\mu^2 + m_e^2 ) \pm p (m_\mu^2 - m_e^2) \over 2 m_\mu^2} ={ m_\mu^2 (E \pm p )+ m_e^2 (E \mp p) \over 2 (E + p ) (E - p ) } ={ (E + p ) (E - p ) (E \pm p )+ m_e^2 (E \mp p) \over 2 (E + p ) (E - p ) }= { (E \pm p )^2+ m_e^2 \over 2 (E \pm p ) }$$

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
$$E < {m_\mu^2 + m_e^2 \over 2 m_e}$$
$$p < { m_\mu^2 - m_e^2 \over 2 m_e}$$
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.