The Lorentz-invariant flux term for the process \(a+b \rightarrow 1+2\) in the centre-of-mass frame was shown to be \(F=4 p_{j}^{*} \sqrt{5}\), where \(p_{j}^{*}\) is the momentum of the intial-state particles. Show that the corresponding expression in the frame where \(b\) is at rest is $$ F=4 m_{b} p_{e} $$

The center-of-mass frame is an important concept in physics, especially in particle physics. In this frame, the total momentum of a system is zero. This makes it simpler to analyze collisions and reactions.
Imagine two particles moving towards each other. In the center-of-mass frame, the system is 'balanced out'. This means each particle's momentum is exactly opposite to the other's.
In our Lorentz-invariant flux term problem, calculating the value in the center-of-mass frame helps us normalize the results. Given the term is expressed as \( F=4 p_{j}^{*} \sqrt{s} \), we consider momenta (\( p_{j}^{*} \)) in center-of-mass frame to better understand energy and momentum distribution during particle interaction.

Rest mass energy is the energy a particle has when it is not moving. This concept is derived from Einstein's famous equation, \( E=mc^2 \). Here, \( E \) is the rest mass energy, \( m \) is the rest mass, and \( c \) is the speed of light.
For our problem, knowing the rest mass of particle \( b \) when it is at rest helps convert the Lorentz-invariant flux term. In simpler terms, we transition from the center-of-mass frame to the rest frame using the rest mass of particle \( b \).
This provides a reference to measure energy and makes it easier to compute the flux using \( F = 4 m_{b} p_{a} \).

Momentum calculation is crucial for understanding particle interactions. In physics, momentum \( p \) is the product of mass \( m \) and velocity \( v \): \( p = mv \.\)
In our problem, we need to account for the momentum of particles in different frames. Momentums in the center-of-mass frame can convert into momentums in the rest frame (where particle \( b \) is at rest).
The task simplifies by recognizing that in the rest frame, particle \( b \)'s momentum is zero. We then only need to consider the momentum of particle \( a \). The final expression of Lorentz-invariant flux term shifts from center-of-mass momentum \( p_{j}^{*} \) to simplified rest frame momentum \( p_{a} \).

Particle physics deals with fundamental particles and forces governing their interactions. In this field, terms like Lorentz invariance, rest mass, and momentum are commonly used.
Understanding particle interactions, like the process \( a+b \rightarrow 1+2 \), involves calculations such as the flux term. The Lorentz-invariant flux term helps describe the likelihood of these interactions occurring.
The exercise and solution provided delve into this concept by breaking down the process into different frames and utilizing rest mass energy and momentum calculations.
By mastering these core principles, students can better understand the dynamic world of particle physics and applications in real-world scenarios.