I read the article Threshold energy on Wikipedia about colliding particles and their threshold energy.

Consider the case where a particle 1 with lab energy $E_1$ (momentum $p_1$) and mass $m_1$ impinges on a target particle 2 at rest in the lab, i.e. with lab energy and mass $E_2 = m_2$. The threshold energy $E_{1,thr}$ to produce three particles of masses $m_a,m_b,m_c$, i.e.
$1+2 \rightarrow a+b+c$

is then found by assuming that these three particles are at rest in the center of mass frame (symbols with hat indicate quantities in the center of mass frame):

$$E_{cm}=m_ac^{2}+m_bc^{2}+m_cc^{2}=\hat{E}_1+\hat{E}_2=\gamma(E_1-\beta p_1c)+\gamma m_2c^{2}$$

Here $E_{cm}$ is the total energy available in the center of mass frame.

Using $\gamma = \frac{E_1+m_2c^{2}}{E_{cm}}$, $\beta =\frac{p_1c}{E_1+m_2c^2}$, and $\ p_1^2c^2=E_1^2-m_1^2c^4$

one derives that


I don't understand where $\gamma = \frac{E_1+m_2c^{2}}{E_{cm}}$ and $\beta= \frac{p_1c}{E_1+m_2c^2}$ come from. I am also confused by the meaning of the equation
$$\hat{E_1}+\hat{E_2}=\gamma(E_1-\beta p_1c)+\gamma m_2c^{2}.$$

I know it has something to do with conservation of energy but the term $(E_1-\beta p_1c)$ feels like its coming from nowhere

If someone could clarify this that would be great.


1 Answer 1


The trick here is in understanding what $\beta$ and $\gamma$ represent in the first place.

That $\beta$ is the speed of the system's center-of-momentum as measured in the lab frame, and $\gamma$ is the Lorentz factor that goes with it.1

Now $\hat{E}_1$ and $\hat{E}_2$ are the total energies of the incident particles as measured in the CoM frame, and that will let us make sense of the line you are asking about.

Start with the second term: $\gamma m_2c^2$ is the energy of a particle of mass $m_2$ that has Lorentz factor $\gamma$. Well, particle 2 is at rest in the lab frame as so has speed $\beta$ in the CoM frame and this is it's CoM energy (i.e. $\hat{E}_2$).

So the first term $\gamma(E_1 - \beta p c)$ must be $\hat{E}_1$. This is, however, non-trivial as we don't have an expression for the speed of $m_1$ in the CoM frame (except in the easy case where $m_1 = m_2$). But look at the shape of that term2 $$ \gamma \left[ (\text{time-like component}) - \beta (\text{space-like component}) c \right] \;.$$ It's a Lorentz transform of the lab-frame energy of particle 1 into the CoM frame.

Done and dusted.

1 The gamma part is relatively easy to see because it is (total energy)/(rest energy). How do you see the beta part?

2 Recall that $(E, \frac{1}{c}\vec{p})$ is a Lorentz four-vector and transforms just like $(ct, \vec{x})$. (Assuming I got all the factors of $c$ in the right places; I'm used to doing this in natural units!)


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