For my physics channel, I need to calculate the (minimal possible) momentum with which our beam has to run in order to produce a pair of D mesons. Momentum conservation. Easy, right?

Well, yes, indeed it is. But I’m getting confused by the stuff every time. So, here, me, noting it down, for a future me to look at again. Because: Short-term memory.

Note: I want to produce two particles with an antiproton beam hitting a fixed proton target. But I start general and note all simplifications arising of my boundary conditions.

## Four-Momentum

Going over from a classic momentum \(\vec{p}\) to a four-momentum \(p^{\mu}\) (or \(\mathbf P\)) is done by:

$$p^\mu=m v^\mu= (\gamma m c, \gamma m \vec{v}) = (\gamma m c, \gamma \vec{p})\text{.}$$

Inserting Einsteins energy mass relation \(E = \gamma m c^2\) we get:

$$\begin{align}

\mathbf{P} = p^\mu & = (E/c, \gamma m \vec{v}) \\

& = (E, \mathbf{p})\text{,}

\end{align}$$

with inserting the relativistic momentum \(\mathbf{p} = \gamma m \vec{v}\) and normalizing to the speed of light with \(c = 1\).

## Momentum Conservation

Incoming momentum equals outgoing momentum, valid for a three-momentum as well as a four-momentum:

$$\mathbf{P}_\text{in} = \mathbf{P}_\text{out}$$

The incoming four momentum are those of the incoming particles, for my case the sum of the antiproton beam and the proton target, \(\mathbf{P}_\bar{p} + \mathbf{P}_p\).

For my case, two particles are outgoing of the process, the \(\Lambda\) and the \(\Sigma\). Since we want to produce them at rest, they shouldn’t have momentum but all energy is their rest mass. All this leads to the following relation, which is squared for calculating with it more cleverly in the steps after that:

$$(\mathbf{P}_\bar{p} + \mathbf{P}_p)^2 = (m_\Lambda + m_\Sigma)^2$$

Solving the left side:

$$(\mathbf{P}_\bar{p} + \mathbf{P}_p)^2 = \mathbf{P}_\bar{p}^2 + \mathbf{P}_p^2 + 2 \mathbf{P}_\bar{p} \mathbf{P}_p$$

Using the Minkowski metric to calculate the square of a four vector, the energy-mass-relation can be deduced: \(\mathbf{P}^2 = p_\mu p^\mu = E^2 – \vec{p}^2 c^2 = m^2 c^4 \underset{c=1}{=} m^2\). With this, the previous equation simplifies to:

$$\mathbf{P}_\bar{p}^2 + \mathbf{P}_p^2 + 2 \mathbf{P}_\bar{p} \mathbf{P}_p = \\

m_\bar{p}^2 + m_p^2 + 2 \mathbf{P}_\bar{p} \mathbf{P}_p = \\

2 m_p^2 + 2 (E_\bar{p} E_{p} – \vec{p}_\bar{p} \vec{p}_p)\text{,}$$

using in the last step that the antiproton and the proton have the same mass \(m_p\) and the metric to multiply the two four vectors \(\mathbf{P}_\bar{p} \mathbf{P}_p\).

As we’re at a fixed-target experiment, the momentum of the proton target, \(\vec{p}_p\), can be neglected and set to zero. All its energy is in the rest mass of the particle, leading to \(E_p = m_p\).

Putting everything together:

$$(m_\Lambda + m_\Sigma)^2 = s = 2 m_p^2 + 2 E_\bar{p} m_p = 2 m_p (m_p + E_\bar{p})\text{,}$$

with \(s\) being a Mandelstam variable and the square of the center of mass energy \(\sqrt{s}\). Solving this for \(E_\bar{p}\), the energy of the antiproton, yields:

$$E_\bar{p} = (m_\Lambda + m_\Sigma)^2 / (2 m_p) – m_p \text{.}$$

Using the energy-momentum-relation (e.g. from squaring the four-momentum, or, EDIT, see below),

$$E^2 = (\vec{p} c)^2 + (m_0 c^2)^2\text{,}$$

and setting \(c=1\), the three-momentum of the beam is

$$\vec{p}_\text{beam}= \sqrt{E_\bar{p}^2 – m_p^2}\text{.}$$

With \(E_\bar{p}\) from above, we finally get the desired beam momentum to produce our two particle, \(\Lambda\) and \(\Sigma\) at threshold / rest:

$$\begin{align}

\vec{p}_\text{beam} &= \sqrt{\left((m_\Lambda + m_\Sigma)^2 / (2 m_p) – m_p\right)^2 – m_p^2} \\

& = \sqrt{(m_\Lambda + m_\Sigma)^4 / (2 m_p)^2 – (m_\Lambda + m_\Sigma)^2}\\

& = (m_\Lambda + m_\Sigma) \sqrt{\left(\frac{m_\Lambda + m_\Sigma}{2 m_p}\right)^2 -1}\text{.}

\end{align}$$

Tadaaa!

## Λ Σ Antiproton Numbers

Plugging in some numbers for my case:

$$\begin{align}

m_\Lambda & = 1.115\,GeV/c^2\\

m_\Sigma & = 1.192\,GeV/c^2\\

m_p & = 0.938\,GeV/c^2\\

\Rightarrow p_\text{beam} & = 1.651\,GeV/c^2

\end{align}

$$

## Edit: Appendix – Derivation of the Energy-Momentum Relation

We know that a particle’s energy is connected to it’s mass by Einsteins famous relation \(E = m c^2\). The \(m\), though, is the particle’s *relativistic* mass. It’s connected to its rest mass by multiplying a gamma factor as seen in the following formula:

$$E = m c ^2 = \gamma m_0 c^2\text{,}$$

with \(\gamma = 1 / \sqrt{1-\beta^2}\) and \(\beta = \vec{v}/c\), the velocity normalized to the speed of light.

Let’s square and rearrange a bit:

$$

\begin{align}

E^2 = m^2 c^4 = \gamma^2 m_0^2 c^4 &= \frac{m_0^2 c^4}{1 – \vec{v}^2/c^2}\\

\Leftrightarrow m^2 c^4 (1-\vec{v}^2/c^2) & = m_0^2 c^4\\

\Leftrightarrow m^2 c^4 – m^2 \vec{v}^2 c^2 & = m_0^2 c^4\\

\Leftrightarrow m^2 c^4 & = m_0^2 c^4 + m^2 \vec{v}^2 c^2

\end{align}$$

On the left side we have the \(E\) we started off with. On the right side, there’s \(m \vec{v}\), the relativistic momentum — connected to the classical momentum by \(\vec{p_r} = m \vec{v} = \gamma m_0 \vec{v} = \gamma \vec{p}_c\).

So, putting everything together:

$$

E^2 = (m_0 c^2)^2 + (\vec{p} c)^2\text{.}

$$