What is the Threshold Energy of a Neutrino in the Centre of Mass Frame?

[avkr]

Homework Statement

Mass of neutrino = 0
Mass of proton/neutron = 1 GeV
Mass of tau = 2 GeV

Homework Equations

Energy of neutrino in lab = e
Energy of neutrino in CM = e0

The Attempt at a Solution

In the centre of mass frame, we will take the momentum of the neutron to be e0/c, c = speed of light.
CoE in CM: e0 + sqrt(m_n^2*c^4 + e0^2) = (m_p + m_tau)c^2
this gives e0 =1.33 GeV

therefore the velocity of CM in lab frame = (e0/c)/(m_v + m_n) = e0/(c*m_n) = 1.33c > c.

Cannot find where I'm doing the mistake!

 

[kuruman]

Cannot find where I'm doing the mistake!

In the energy expression on the right you omitted the $p^2c^2$ term for the two particles. The fact that the sum of the momenta is zero in the CM frame does not necessarily mean that each momentum is zero.

 

[avkr]

I have included e0 as energy of neutrino whereas sqrt(m_n^2 c^4 + e0^2) contains the terms rest mass of neutron (m_n^2 c^4) and e0^2 is the (pc)^2 term ( p = e0/c, therefore (pc)^2 = e0^2).

 

[George Jones]

View attachment 214254In the centre of mass frame, we will take the momentum of the neutron to be e0/c, c = speed of light.
CoE in CM: e0 + sqrt(m_n^2*c^4 + e0^2) = (m_p + m_tau)c^2

This is what I get.

therefore the velocity of CM in lab frame = (e0/c)/(m_v + m_n) = e0/(c*m_n) = 1.33c > c.

In the CoM frame, the spatial momentum of the neutron is not $m_N v$. What is it?

For the relative speed between the two frames, I get
$$\frac{v}{c} = \tanh \left( \sinh^{-1} \left( \frac{e_0}{m_N} \right) \right) ,$$

but I don't think that you are expected to express it like this.

 

[kuruman]

I have included e0 as energy of neutrino whereas sqrt(m_n^2 c^4 + e0^2) contains the terms rest mass of neutron (m_n^2 c^4) and e0^2 is the (pc)^2 term ( p = e0/c, therefore (pc)^2 = e0^2).

That's the left side of the equation and it is correct. In post #2 I am questioning the right side which should be $\sqrt{m_p^2c^4 +p_{p}^2c^2}+\sqrt{m_{\tau}^2c^4 +p_{\tau}^2c^2}$. You have not conserved momentum in the CM frame.

On Edit: Sorry I missed the word "minimum" for the energy. I withdraw my objections.

 

[George Jones]

In the energy expression on the right you omitted the $p^2c^2$ term for the two particles. The fact that the sum of the momenta is zero in the CM frame does not necessarily mean that each momentum is zero.

In relativistic collision problems, "threshold" usually means that the product particles are all at rest in the zero momentum frame.

 

[Orodruin]

That's the left side of the equation and it is correct. In post #2 I am questioning the right side which should be $\sqrt{m_p^2c^4 +p_{p}^2c^2}+\sqrt{m_{\tau}^2c^4 +p_{\tau}^2c^2}$. You have not conserved momentum in the CM frame.

In the CM frame, the total momentum is zero by definition. You therefore have the total 4-momentum $P = p_p + p_\tau = (E_{cm},0,0,0)$. It follows that
$$
s = P^2 = E_{cm}^2 = m_p^2 + m_\tau^2 + 2p_p \cdot p_\tau \geq m_p^2 + m_\tau^2 + 2m_p m_\tau = (m_p+m_\tau)^2.
$$
The threshold CM energy is given by the equality and corresponds to the proton and tau being at relative rest.

 

[Orodruin]

In the CoM frame, the spatial momentum of the neutron is not $m_N v$. What is it?

For the relative speed between the two frames, I get
$$\frac{v}{c} = \tanh \left( \sinh^{-1} \left( \frac{e_0}{m_N} \right) \right) ,$$

but I don't think that you are expected to express it like this.

You cannot express the relative velocity using only quantities from the CoM frame (and invariants like $m_N$).

To OP: You do not need to compute the relative speed between the CM and lab frames to solve this problem. Are you familiar with the 4-vector formalism? I would then suggest working in terms of invariants. You can compute the invariants in whatever frame as they will take the same value in all frames.

therefore the velocity of CM in lab frame = (e0/c)/(m_v + m_n) = e0/(c*m_n) = 1.33c > c.

You seem to be using $v/c = p/mc$. This is not correct. The correct expression is $v/c = pc/E$.

 

[George Jones]

You cannot express the relative velocity using only quantities from the CoM frame (and invariants like $m_N$).

The neutron is at rest in the lab frame, so its speed $v$ in the zero-momentum frame is the relative speed between the two frames. In the zero-momentum frame (with $c = 1$), zero total spatial momentum (with zero mass for the neutrino) gives
$$e_0 = m_N \gamma v = m_N \sinh w$$
with $v = \tanh w$.

 

[Orodruin]

Ok, fine. But not the easiest way to express the relative velocity between the CM and lab frames with the information given in the problem (note that $E_0$ is not given, $E_\nu$ in the lab frame is the relevant quantity).

Edit: Also, there is no need to compute the relative speed of the CM frame and the lab frame.

Edit 2: I guess you can do it by computing $E_0$ at threshold and using the relative velocity between the frames to find $E_\nu$, but this seems like the long way around when the computation using invariants is literally one line.

Edit 3: As expected, the two approaches give the same result. The big difference is the amount of algebra you have to do ...

 

[avkr]

The neutron is at rest in the lab frame, so its speed $v$ in the zero-momentum frame is the relative speed between the two frames. In the zero-momentum frame (with $c = 1$), zero total spatial momentum (with zero mass for the neutrino) gives
$$e_0 = m_N \gamma v = m_N \sinh w$$
with $v = \tanh w$.

Thanks for this!
Got my mistake.