Past exam of the symmetry and relativity course of the University of Oxford / 2017 / 4 / e

Let $\epsilon =E_{ph}$ and $\gamma =W/(M_0{c}^2)$. For a head-on photon and particle collision, boost to the particle's initial rest frame. The incident photon energy is ${\epsilon }_{\ast }=\gamma (1+\beta )\epsilon$. On the inverse-Compton branch $\epsilon <\gamma \beta M_0{c}^2$, the maximal outgoing lab energy corresponds to rest-frame backscattering into the particle's forward direction. The exact Compton scattering recoil formula gives ${\epsilon }_{\ast }^{\prime }={\epsilon }_{\ast }/[1+2{\epsilon }_{\ast }/(M_0{c}^2)]$. Boosting the backscattered photon back to the lab givesFor $\gamma \gg 1$ this becomesOnly in the additional Thomson scattering limit $4\gamma \epsilon \ll M_0{c}^2$ may the denominator be neglected, yielding the printed estimate $4[W/(M_0{c}^2){]}^2{E}_{ph}$. Ultrarelativistic motion alone does not justify it. For example $\gamma =1000$ and $\epsilon =M_0{c}^2$ would give $4\times 10^6{M}_0{c}^2$ from that estimate, exceeding the available total energy; the recoil-corrected maximum is about $999.75M_0{c}^2$.

For an arbitrarily hard incident photon, maximizing over direction needs a further qualification: if $\epsilon \ge \gamma \beta M_0{c}^2$, the largest outgoing photon energy is the unchanged $\epsilon$, rather than this amplified backscattering branch. The exact photon energy gain is $\Delta {\epsilon }_{\max }={\epsilon }_{\max }^{\prime }-\epsilon$. Calling the outgoing energy the gain is accurate only when amplification is large. These distinctions prevent the inverse Compton scattering estimate from violating conservation of energy outside its low-rest-frame-photon-energy regime.