Past exam of the mathematics course of the University of Cambridge / 2009 / ia / Paper 4 / 10A (Dynamics and Relativity)

For $|\mathbf{v}|<c$, define the Lorentz factor $\gamma =(1-|\mathbf{v}{|}^2/c^2{)}^{-1/2}$. A particle of rest mass $m$ hasThese expressions give the relativistic energy-momentum relation directly:so $\boxed{E^2=c^2\mathbf{p}\cdot \mathbf{p}+m^2{c}^4}$.

The four-momentum iswhere $U^{\mu }$ is the four-velocity, the proper-time derivative of $(ct,\mathbf{x})$. It is a four-vector, so its Minkowski metric norm is invariant under Lorentz transformations. Choose metric signature $(+,-,-,-)$; in the particle's rest frame, $P^{\mu }=(mc,0)$ and $P_{\mu }{P}^{\mu }=m^2{c}^2$. In any frame the same norm is $E^2/c^2-|\mathbf{p}{|}^2$. Equating the two yields the same relativistic energy-momentum relation. With the opposite metric convention both norm expressions change sign, leaving the energy relation unchanged.

The two incoming particles have equal energies and opposite momenta, so their total four-momentum in this frame isIt is already a centre-of-momentum frame. By four-momentum conservation, the final spatial momenta sum to zero and their energies sum to $2E$. Each product has energySumming gives the necessary relativistic collision mass thresholdFor massive products the equality threshold has all of them at rest in this frame. Above it, multiple products can carry balancing momenta and relative kinetic energy. The displayed condition is necessary, as requested; a single final particle must instead have exactly the incoming invariant mass, and additional physical conservation laws could further restrict a real reaction.

Let the incoming projectile momentum be ${\mathbf{p}}_1$. The stationary target has four-momentum $(m_{2}c,0)$, so the total incoming norm isHere the relativistic energy-momentum relation eliminated $|{\mathbf{p}}_1{|}^2$. Write $W$ for the total energy in the centre-of-momentum frame; then $W^2=c^2{P}_{\mathrm{in}}^2$. If $M_f={\sum }_i{\mu }_i$, the relativistic collision mass threshold gives $W\ge M_f{c}^2$. Squaring and rearranging with $m_2>0$ yields the fixed-target production thresholdThe actual incident total energy also obeys $E_1\ge m_1{c}^2$. Thus a negative algebraic reaction bound would not allow a negative-energy projectile.

For a photon, $m_1=0$ and $E_1=h\nu$, where $h$ is the Planck constant. The necessary bound becomesIn the usual massive multi-product channel with $M_f>m_2$, conservation alone permits the threshold configuration in which all products are at rest relative to one another in the centre-of-momentum frame. Boosting that configuration supplies the required laboratory momentum. Hence the photon fixed-target production threshold isFor a multi-product channel with $M_f\le m_2$, there is no positive mass threshold: the lower limit of allowed positive frequencies is zero. The compact expression for that lower limit is $\boxed{{\nu }_{\inf }=\max \{0,M_f^2-m_2^2\}{c}^2/(2m_{2}h)}$; it need not correspond to a zero-frequency incident photon.

For completeness, if the only product is one particle of mass ${\mu }_1$, its invariant mass must match exactly. A positive incident frequency is possible only when ${\mu }_1>m_2$, and is then fixed at $\nu =({\mu }_1^2-m_2^2)c^2/(2m_{2}h)$; frequencies above that value do not produce the same single particle while conserving four-momentum. If a multi-product channel includes massless particles, an all-products-at-rest configuration is impossible for those particles with nonzero energy, so the nominal massive threshold may be only an infimum. These qualifications distinguish the requested necessary inequalities from unrestricted claims of sufficiency.