A proton, mass m, accelerated in a proton synchrotron to kinetic energy K strikes a second (target) proton at rest in the laboratory. If we assume that the collision is entirely inelastic in that the rest energy of the two protons, plus all the kinetic energy, consistent with the law of conservation of momentum, is available, then this energy can be conveniently found by going to the centre of mass (cm) frame.

Let v be the velocity of the incident proton in the laboratory frame. In the centre of mass frame the incident proton and the target proton will be moving toward each other with the same speed . In the laboratory frame the target proton is at rest and in the cm frame its speed is , the speed of the cm frame with respect to the laboratory frame will be . The speed of the incident proton in the cm frame will be. We use the relativistic velocity addition theorem to find the .

Roots of this quadratic equation are

As

Total energy of the incident proton and the target proton in the cm frame will be the available energy for particle production. It is

Substituting and carrying out algebraic simplifications, we find

Using , expression for can be expressed in the form

(b)

From the above result we calculate the available energy when 100-GeV protons are used. It will be

(c)

Proton energy required to make 100 GeV available can be found using the relation derived above. It is given by the expression

This gives