Momentum and energy in Center-of-Momentum frame

[LagrangeEuler]

Homework Statement

Proton of energy 76,4GeV colides with the proton with the proton that staying in the rest. How to find energy and momentum in center of momentum frame? And from that how to find the maximal number of protons and antiprotons that could appear from that.

Relevant Equations

Center of momentum frame is the one very sum over momentum of particles is equal to zero vector.
$$\sum_i \vec{p}_i=\vec{0}$$

To my mind because one particle has momentum $\vec{p}$ and the other one $\vec{0}$. It is for instance necessary to find reference frame where one momentum will be for instance $\frac{1}{2}\vec{p}$ and the momentum of other particle should be $-\frac{1}{2}\vec{p}$. So it is necessary to find fram that is moving with some velocity left according to lab system. However it is also possible to get that if momentum of one particle $\frac{3}{4}\vec{p}$ and $-\frac{1}{4}\vec{p}$. Energies in that two cases are different. So I am confused with that...

 

[PeroK]

Homework Statement:: Proton of energy 76,4GeV colides with the proton with the proton that staying in the rest. How to find energy and momentum in center of momentum frame? And from that how to find the maximal number of protons and antiprotons that could appear from that.
Relevant Equations:: Center of momentum frame is the one very sum over momentum of particles is equal to zero vector.
$$\sum_i \vec{p}_i=\vec{0}$$

To my mind because one particle has momentum $\vec{p}$ and the other one $\vec{0}$. It is for instance necessary to find reference frame where one momentum will be for instance $\frac{1}{2}\vec{p}$ and the momentum of other particle should be $-\frac{1}{2}\vec{p}$. So it is necessary to find fram that is moving with some velocity left according to lab system. However it is also possible to get that if momentum of one particle $\frac{3}{4}\vec{p}$ and $-\frac{1}{4}\vec{p}$. Energies in that two cases are different. So I am confused with that...

Welcome back! Yes, it's not as easy as in classical physics.

Here's a trick: consider the energy-momentum transformation of the system. Note that the total energy-momentum of a system, being the sum of four-vectors, is itself a four-vector. So, transforms according to the Lorentz transformation.

What happens if you transform the total momentum-energy of the lab frame to the momentum of the CoM frame?

 

[PAllen]

I answered this in the other thread. I'll repeat here:

Compute the invariant mass, as energy (i.e, times c squared if not using c=1) of the system in the lab frame. This equals total energy in the COM frame. Then, the original two protons each simply have half this energy in the COM frame. Get the momentum of each in the COM frame from this energy and their rest energy.

Note, this is why the whole problem becomes trivial using invariant mass or energy. The invariant energy of the system, computed in any frame, divided by the rest energy, is the maximum number of nucleons you can produce. This number, is, of course, frame invariant.

 

[PeroK]

I answered this in the other thread. I'll repeat here:

Compute the invariant mass, as energy (i.e, times c squared if not using c=1) of the system in the lab frame. This equals total energy in the COM frame. Then, the original two protons each simply have half this energy in the COM frame. Get the momentum of each in the COM frame from this energy and their rest energy.

Note, this is why the whole problem becomes trivial using invariant mass or energy. The invariant energy of the system, computed in any frame, divided by the rest mass, is the maximum number of nucleons you can produce. This number, is, of course, frame invariant.

It's also pretty nifty to transform momentum; especially if the particles have different masses.

 

[PeroK]

Homework Statement:: Proton of energy 76,4GeV colides with the proton with the proton that staying in the rest. How to find energy and momentum in center of momentum frame? And from that how to find the maximal number of protons and antiprotons that could appear from that.
Relevant Equations:: Center of momentum frame is the one very sum over momentum of particles is equal to zero vector.
$$\sum_i \vec{p}_i=\vec{0}$$

To my mind because one particle has momentum $\vec{p}$ and the other one $\vec{0}$. It is for instance necessary to find reference frame where one momentum will be for instance $\frac{1}{2}\vec{p}$ and the momentum of other particle should be $-\frac{1}{2}\vec{p}$. So it is necessary to find fram that is moving with some velocity left according to lab system. However it is also possible to get that if momentum of one particle $\frac{3}{4}\vec{p}$ and $-\frac{1}{4}\vec{p}$. Energies in that two cases are different. So I am confused with that...

Let me show you the quick way to do these problems. This uses the "invariant mass" of a system of particles. This doesn't seem to be taught much, despite being by far the simplest method.

I'll use $c = 1$ units.

For any system of particles (with mass $m_k$, energy $E_k$, and momentum $\vec p_k$, we define the total energy $E$, total momentum $\vec P$ (magnitude $P$) and invariant mass $M$ as:
$$E = \sum E_k, \ \ \vec P = \sum \vec p_k, \ \ M^2 = E^2 - P^2$$
Note that $(E, \vec P)$ being the sum of four-vectors is itself a four vector, with $M$ its invariant quantity. This means that $E, \vec P$ transform according to the usual energy-momentum transformation.

Now, let's consider the special case of one-dimensional collisions. Typically this will be two particles colliding to form a system of several particles. One particle may be at rest, both may be moving in the lab frame and they may have different masses. If we transform the total momentum to another frame we get:
$$P' = \gamma(P - vE)$$
For the CoM (centre of momentum frame) we want $P' = 0$, hence:
$$v_{CoM} = \frac P E$$
And, the gamma factor of the CoM frame is:
$$\gamma_{CoM} = \frac E M$$
If the particles collide to create a single particle, then that particle must have mass $M$. You can see this by considering the invariant mass in the CoM frame, which is just the rest energy of the particle. And this applies where the collision creates several particles all at rest in the CoM frame - which equates to being at the threshold energy. The sum of the masses must be $M$.

To see how this works, let's go back to your previous problem, where the incident proton had energy of $E_1 = 76.4GeV$. The total energy and momentum of the system are:
$$E = E_1 + m, \ \ P = p_1, \ \ M^2 = E^2 - P^2 = 2E_1m + 2m^2$$
We want to know how many multiples of $m$ we can get from $M$:
$$n = \frac M m = \sqrt{2(\frac {E_1}{m} + 1)} = 12.8$$

 

[robphy]

An even simpler method comes from drawing an energy-momentum diagram and doing some trigonometry.
This is equivalent to @PeroK 's calculation.

If you can add vectors and do some trig for a free-body diagram, you can do this:

We add two proton 4-momentum vectors $\tilde p_1$ and $\tilde p_2$,
one at rest and the other with nonzero velocity in this frame. Each has magnitude equal to the proton-rest-mass $m_p$.
The sum of these 4-momenta is the 4-momentum of the system, which has magnitude $m_{inv}$, called the invariant mass.
(By the way, this is an isosceles triangle in energy-momentum space.)

Using rapidity, where $v=\tanh\theta$ so that $\gamma=\cosh\theta$ and $\gamma v =\sinh\theta$ ...

In this lab frame, the "high-energy proton" has energy $E_2=m_p\cosh\theta = \gamma m_p$ and spatial momentum $p_2=m_p\sinh\theta =\gamma m_p v$.

So, the magnitude of $\tilde p_1 + \tilde p_2$ is
$$\begin{align*} m_{inv}=|\tilde p_1 + \tilde p_2| &=\sqrt{ (\tilde p_1 +\tilde p_2) \cdot (\tilde p_1 +\tilde p_2) } \\ &=\sqrt{ (E_1 + E_2)^2 - (p_1 + p_2)^2 } \\ &=\sqrt{ (m _p + m_p\cosh\theta)^2 - (0 + m_p\sinh\theta)^2 } \\ &=m_p\sqrt{ (1 + \cosh\theta)^2 - (0 + \sinh\theta)^2 } \\ \end{align*}$$

With $m_p=0.938$ and $E_2=76.4$,
we have
$\cosh\theta =\displaystyle \frac{adj}{hyp}=\frac{ 76.4}{0.938}$.

• You can solve for $\theta=\texttt{arccosh(76.4/0.938)}=5.09...$
  so that $\sinh\theta=\texttt{sinh(arccosh(76.4/0.938))}= 81.44...$
  
  Thus, $m_{inv}=m_p \texttt{ sqrt((1+(76.4/0.938))^2-sinh(arccosh(76.4/0.938))^2)}=(12.84...)m_p$

or

• You can use some identities:
  $\cosh^2\theta-\sinh^2\theta = 1$ implies
  $\sinh^2\theta=\cosh^2\theta - 1$.
  So,
  $\begin{align*} m_{inv} &=m_p\sqrt{ (1 + \cosh\theta)^2 - (\cosh^2\theta - 1)}\\ &=m_p\sqrt{ 2+ 2\cosh\theta}\\ &=m_p\sqrt{ 2\left(1+ \frac{ 76.4}{0.938}\right) }\\ &=m_p\texttt{ sqrt(2(1+(76.4/0.938)))}=(12.84..)m_p \end{align*}$

No need to boost into the center-of-momentum frame.
When analyzing a free-body diagram, do you generally rotate your axes along the net-force to compute its magnitude?

UPDATE:

• You can also use the law of cosines.
  $\begin{align*} \tilde P&= \tilde p_1 + \tilde p_2\\ \tilde P^2&= \tilde p_1{}^2 + \tilde p_2{}^2 + 2\tilde p_1\cdot \tilde p_2 \\ m_{inv}^2 &= m_p^2 + m_p^2 + 2m_p m_p\cosh\theta \qquad \mbox{we use the external angle}\\ &= m_p^2\ 2(1+ \cosh\theta) \\ \end{align*}$

 

[PAllen]

Here is my take on the most direct approach, that I wrote down in a few lines on seeing the original problem.

Reasoning: The invariant mass (as energy) is what is available to make particles because it is the total energy in the COM frame; being in the COM frame, momentum is conserved converting all available energy to particles. The maximum number of particles created is obviously frame independent, so just computing invariant energy and allocating to particles is sufficient to solve any similar problem.

I choose to use E for the total energy in the lab frame, P for the momentum of the accelerated proton (which is also the total momentum in the lab frame), and $E_0$ for the rest energy of a nucleon. Then, the invariant energy $(E_i)$ is simply given by:
$$E_i^2 = E^2 - P^2 = E^2 - ((E-E_0)^2 - E_0^2) = 2E E_0$$
So the maximum number of nucleons that could be produced is simply:
$$\sqrt {2 E/E_0 }$$
This compares to the case of colliding nucleons with equal energies, giving simply $E/E_0$. Thus the huge advantage of colliders rather than stationary targets.