Generally, text books cover the recoil of a target after absorption of a photon. What happens when a target, it might be an atom, recoil after emission of a photon? The scientific literature shows a mathematical treatment, however not in case the recoil velocity is relativistic. Therefore, I am approaching this problem: in the first place, I follow the classic treatment for then, in the second place, approaching the relativistic part when we can assume the the recoil velocity approaches the speed of light.
I define $${Q_0}=h{ν_0}$$ the incoming photon before absorption $${Q_1}=h{ν_1}$$ the emitted photon $${m_0}$$ the rest mass of the target $${m_1}$$ the mass of the recoiling target $${m_2}$$ the rest mass of the recoiling target $${v_1}$$ the recoil velocity after emission
Mom. conserv.: $$\frac{Q_1}{c}={m_1}{v_1}(Eq.1)$$ Energy conserv.(emission): $${m_o}c^2={m_1}c^2+{Q_1}(Eq.2)$$ Rest energy: $${m_o}c^2-{m_2}c^2={Q_0}(Eq.3)$$ Gordon Eq.: $${E_1}^2=(c{p_1})^2+{E_2}^2(Eq.4)$$ $$({m_1}c^2)^2={Q_1}^2+({m_2}c^2)^2(Eq.5)$$ From Eqs. 2 and 3 we can extract the first therm in LHS and second term in RHS so that $$({m_o}c^2-{Q_1})^2={Q_1}^2+({m_o}c^2)^2-2{m_o}c^2{Q_0}+{Q_0}^2(Eq.6)$$ from which $${Q_1}={Q_0}(1-\frac{Q_0}{2{m_o}c^2})=h{v_1}<h{v_0}(Eq.7)$$ Energy conserv.(absorption): $${m_o}c^2+{Q_1}={m_1}c^2(Eq.8)$$ from which $${m_1}={m_o}+\frac{Q_1}{c^2}(Eq.9)$$ Recoil velocity after emission: $$β=\frac{v_1}{c}=\frac{Q_1}{{m_1}c}\frac{1}{c}(Eq.9)$$ $${v_1}=\frac{Q_1}{{m_1}c}(Eq.10)$$ Due to Eq.9: $${v_1}=\frac{{Q_1}c}{{m_0}c^2+{Q_1}}(Eq.11)$$
Now, let´s assume that, depensing on the energy of incoming photon and of the rest mass of thetarget, we calculate that the recoil velocity of the target in emission is almost the speed of light $${v_1}≈c(Eq.12)$$
From here, the relativistic part starts by the gamma factor $$γ=\frac{1}{\sqrt{1-\frac{{v_1}^2}{c^2}}} (Eq.13)$$
I proceed as follows
I introduce the new rest mass (relativistic) as the gamma factor multiplied the rest mass of the target $${m_3}=γ{m_0}(Eq.14)$$ Mom. conserv.: $$\frac{Q_1}{c}=γ{m_1}{v_3}(Eq.15)$$ Please, notice, I am calling the previous recoil velocity v1 now with v3 as in the stationary ref. frame we should measure another recoil velocity. Energy conserv.(emission): $$γ{m_o}c^2={m_1}c^2+{Q_1}(Eq.16)$$ Rest energy: $$γ{m_o}c^2-{m_2}c^2={Q_0}(Eq.17)$$ Gordon Eq.: $${E_1}^2=(c{p_1})^2+{E_2}^2(Eq.18)$$ $$({m_1}c^2)^2={Q_1}^2+({m_2}c^2)^2(Eq.19)$$ From Eqs. 16 and 17 we can extract the first therm in LHS and second term in RHS so that $$(γ{m_o}c^2-{Q_1})^2={Q_1}^2+(γ{m_o}c^2)^2-2γ{m_o}c^2{Q_0}+{Q_0}^2(Eq.20)$$ from which $${Q_1}={Q_0}(1-\frac{Q_0}{2γ{m_o}c^2})=h{v_1}<h{v_0}(Eq.21)$$ The difference between Eq.7 and Eq.21 is the gamma factor at the denominator which it might make sense in case the calculation is correct. Recoil velocity after emission: $$β=\frac{v_3}{c}=\frac{Q_1}{γ{m_1}c}\frac{1}{c}(Eq.22)$$ $${v_3}=\frac{Q_1}{γ{m_1}c}(Eq.23)$$ Due to Eq.15: $${v_3}=\frac{{Q_1}c}{γ({m_0}c^2+{Q_1})}(Eq.24)$$
I don't know if it is correct, but I get this result. Do you get the same result?