Solving Muon Momentum Decay: Maximizing Expression & Intuition

[Aleolomorfo]

Homework Statement

Consider the decay $K \rightarrow \mu + \nu$. Find the maximum $\mu$ momentum in a frame in which the $K$ momentum is 5 GeV

Relevant Equations

$P = (E, p_x, p_y, p_z)$
$P^2 = m^2$

My first idea was to find an expression of the muon momentum as a function of the angle and then maximaze the expression. But my attempts were not succesful. I report here my attempt.

Set up (referring to the attached file "bettini.pdf"):
$$ p = (\sqrt{p_k^2+m_k^2},0,0,p_k)$$
$$ k_1 = (\sqrt{p_{\mu}^2+m_{\mu}^2},0,p_{\mu}\sin{\theta}, p_{\mu}\cos{\theta})$$
$$ k_2 = (p_{\nu},0,-p_{\nu}\sin{\phi}, p_{\nu}\cos{\phi})$$

My attempt:

$$ p = k_1 + k_2 $$
$$ k_2 = p - k_1$$

Squaring

$$ k_2^2 = p^2 + k_1^2 -2pk_1$$
$$ 0 = m_k^2 + m_{\mu}^2 - 2\sqrt{(p_k^2+m_k^2)(p_{\mu}^2+m_{\mu}^2)} + 2p_kp_{\mu}\cos{\theta}$$

Continuing from here the computation to obtain $p_{\mu}$ as a function of $\cos{\theta}$ is quite demanding.

My second idea was to think intuitively that the case in which the muon momentum is maximum is when it flies forward and the neutrino backward. And in this particular case the computation is easier.

To sum up, my two questions are: Is there a way to perform my first idea? Is my second idea right?

Thanks in advance!

 

[vela]

What I'd suggest is solving the problem in the rest frame of the kaon and then transforming back to the lab frame.

 

[Aleolomorfo]

What I'd suggest is solving the problem in the rest frame of the kaon and then transforming back to the lab frame.

I have attached the file with the set up.

$$ p = (m_k, 0, 0, 0)$$
$$ k_1 = (E_{\nu},0,0, E_{\nu})$$
$$ k_2 = (E_{\mu}, 0,0, -\sqrt{E_{\mu}^2-m_{\mu}^2})$$

Using the conservation of momentum and energy I get:

$$E_{\mu} = \frac{m_k^2+m_{\mu}^2}{2m_k} = 258.15 MeV $$
$$ p_{\mu} = \sqrt{E_{\mu}^2-m_{\mu}^2} = 235,53 MeV $$

Then I make the assumption that the muon has the maximum energy when it flies forward in the LAB frame:

$$\beta = \frac{p_k}{\sqrt{p_k^2+m_k^2}} = 1 $$
$$ \gamma = \frac{\sqrt{p_k^2+m_k^2}}{m_k} = 10$$

Using Lorentz transformation:

$$ p_{\mu}^{lab, max} = \gamma(p_{\mu}^{CM} + \beta E_{\mu}^{CM}) $$

Then substituting the numerical value I get the answer. Is it right?

 

[vela]

Looks good.

 

[Aleolomorfo]

Looks good.

Thank you very much!