Massless Particles and the algebra involved

[forrealfyziks]

Homework Statement

The positive pion decays into a muon and a neutrino. The pion has rest mass m=140 MeV/c^2, the muon has m=106 MeV/c^2 while the neutrino is about m=0 (assume it is). Assuming the original pion was at rest, use conservation of energy and momentum to show that the speed of the muon is given by

U/c = [(m_$$\pi$$/m_$$\mu$$)² - 1] / [(m_$$\pi$$/m_$$\mu$$)² + 1]

Homework Equations

For massless particles, E=pc u=c $$\beta$$=1
p=$$\gamma$$mu E=$$\gamma$$mc²

The Attempt at a Solution

(Note $$\gamma$$ and u are for mu, since it's the only particle using them)
Alright, I know that P_$$\pi$$=0 and P_$$\mu$$+P_$$\nu$$=0. I found that P_$$\nu$$=-$$\gamma$$m_$$\mu$$u_$$\mu$$
I replace p in m_$$\pi$$c²=P_$$\nu$$c + $$\gamma$$m_$$\mu$$c² and reduced it down and got
m_$$\pi$$c=$$\gamma$$m_$$\mu$$(c-u)
I then unraveled the gamma, moved some stuff around, and squared both sides and got (m_$$\pi$$/m_$$\mu$$)²=(c-v)²/(1-v²/c²)
I don't know if this is the right path, but I have tried many different methods from this point and nothing seems to get any closer to what I need. Thanks for any help.

 

[gabbagabbahey]

m_$$\pi$$c=$$\gamma$$m_$$\mu$$(c-u)
I then unraveled the gamma, moved some stuff around, and squared both sides and got (m_$$\pi$$/m_$$\mu$$)²=(c-v)²/(1-v²/c²)
I don't know if this is the right path, but I have tried many different methods from this point and nothing seems to get any closer to what I need. Thanks for any help.

You seem to have dropped a factor of $c^2$;

$$m_{\pi}c=\gamma m_{\mu}(c-u)=\left(1-\frac{u^2}{c^2}\right)^{-1/2}m_{\mu}(c-u)$$

$$\implies m_{\pi}^2c^2= \left(1-\frac{u^2}{c^2}\right)^{-1}m_{\mu}^2(c-u)^2$$

$$\implies c^2-u^2=\left(\frac{m_{\mu}}{m_{\pi}}\right)^2(c-u)^2$$

From here, just expand $(u-c)^2$, group terms in powers of $u$ and solve via the quadratic formula (studying university physics is no excuse for forgetting high school algebra!)

 

[Count Iblis]

I think you started out with way too complicated algebra, by considering the 3-mometum seperately. Also, you should use c = 1 units to avoid making mistakes with factors of c.

The simple way to soleve these sorts of problems is to use the energy momentum relation in the form:

p^2 = m^2

where p is the four-momentum, the square denotes the Lorentz inner product of the momentum with itself and m is the mass. By bringing the four-momentum of the particle in which you are not interested in to one side and squaring, you eliminate its momentum and energy in one go. So, in this case, you would proceed like:

p_neut = p_pi - p_mu

0 = m_pi^2 + m_mu^2 - 2 m_pi E_mu

Then solving for E_mu and dividing by m_mu gives you the gamma factor and you're done.

 

[forrealfyziks]

eh I was up all night doing this so it was no doubt the fatigue, but I got to that point once and gave up(talking to gabba). I haven't been taught the four-momentum process so I can't use it unfortunately.

Thanks you!

 

[paranormal]

First of all, it is important to note that the equation provided in the Homework Statement is incorrect. The correct equation for the speed of the muon is given by:

u/c = [(m_pi/m_mu)^2 - 1] / [(m_pi/m_mu)^2 + 1]

Now, let's look at the conservation of energy and momentum equations for this decay:

Conservation of energy: E_pi = E_mu + E_nu

Conservation of momentum: p_pi = p_mu + p_nu

Since the pion is initially at rest, its energy is simply its rest energy, E_pi = m_pi*c^2. The neutrino is assumed to be massless, so its energy is also simply its momentum, E_nu = p_nu = c*p_nu = c*p_mu.

Substituting these into the conservation of energy equation, we get:

m_pi*c^2 = E_mu + c*p_mu

Now, we can use the equation for the energy of a relativistic particle, E = \gamma*m*c^2, to write the energy of the muon as:

E_mu = \gamma_mu*m_mu*c^2

Substituting this into our conservation of energy equation, we get:

m_pi*c^2 = \gamma_mu*m_mu*c^2 + c*p_mu

Dividing both sides by c, we get:

m_pi = \gamma_mu*m_mu + p_mu

Now, we can use the equation for the momentum of a relativistic particle, p = \gamma*m*v, to write the momentum of the muon as:

p_mu = \gamma_mu*m_mu*v_mu

Substituting this into our conservation of momentum equation, we get:

m_pi = \gamma_mu*m_mu + \gamma_mu*m_mu*v_mu

Dividing both sides by \gamma_mu*m_mu, we get:

1 = 1 + v_mu

Solving for v_mu, we get:

v_mu = 0

Therefore, the speed of the muon is given by u/c = 0, which simplifies to u/c = 0. This is consistent with the fact that the muon is a massive particle and cannot travel at the speed of light.