Relativistic case recoil of the target by emission of a photon

Question

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?

Answer 1

You should simply NOT introduce relativistic mass. The only well-defined quantity is rest mass, and working in terms of it will be much easier than you think. You should learn the way I am organising this computation, using 4-vectors, because it will keep everything simple.

Let us consider the most simplest case, the centre-of-momentum frame's view of the collision between an atom and a photon. The atom has energy $E=+\sqrt{m^2c^4+p^2c^2}$ if it has momentum $p$, i.e. satisfying $E^2-p^2=m^2$ always, and the photon, using your notation, has energy $Q=h\nu$ and momentum $\frac Qc$, and I have chosen for all of these quantities to be in the centre-of-momentum frame. In 4-vector form, this collision looks like this: $$\tag1 \begin{pmatrix}+\sqrt{m^2c^4+Q^2}\\+Q/c \end {pmatrix}+ \begin{pmatrix}+Q\\-Q/c \end {pmatrix}\Rightarrow \begin{pmatrix}+\sqrt{m^2c^4+Q^2}\\-Q/c \end {pmatrix}+ \begin{pmatrix}+Q\\+Q/c \end {pmatrix} $$ where initially the atom is moving rightwards with equal and opposite momentum to the photon, and later they just exchanged their momenta. You can literally read off the 4-vectors and know that the left quantity is always the atom, and the right quantity is the photon, they always obey the Einstein energy-momentum-mass relation $E^2-p^2=m^2$

Now we move to the laboratory frame, where the initially the atom was at rest. In this case, we obtain $$\tag2 \begin{pmatrix}+mc^2\\0 \end {pmatrix}+ \begin{pmatrix}+Q_0\\-Q_0/c \end {pmatrix}\Rightarrow \begin{pmatrix}+\sqrt{m^2c^4+(Q_0+Q_1)^2}\\-(Q_0+Q_1)/c \end {pmatrix}+ \begin{pmatrix}+Q_1\\+Q_1/c \end {pmatrix} $$ You can check that, in this way of writing things, I have guaranteed that

1. linear momentum is strictly conserved.
2. Each particle still strictly obeys $E^2-p^2=m^2$ appropriate for itself.

We just have to ensure that energy is conserved. i.e. just the top components $$ \begin{align} \tag3mc^2+Q_0&=+\sqrt{m^2c^4+(Q_0+Q_1)^2}+Q_1\\ \tag4(mc^2+Q_0-Q_1)^2&=m^2c^4+(Q_0+Q_1)^2\\ \tag5m^2c^4+2mc^2(Q_0-Q_1)+(Q_0-Q_1)^2&=m^2c^4+(Q_0+Q_1)^2\\ \tag62mc^2(Q_0-Q_1)&=(Q_0+Q_1)^2-(Q_0-Q_1)^2\\ \tag7&=2Q_0(2Q_1)=4Q_0Q_1\\ \tag8\therefore\qquad Q_1&=\frac{mc^2}{mc^2+2Q_0}Q_0\quad<Q_0 \end {align} $$ Now, my Equation (8) is very different from your Equations (7) and (21), and it should be clear that it is yours that is wrong. The 2 is in the numerator, not in the denominator.

To get the recoil velocity, we just need to use $\frac vc=\frac{pc}E$, and to do that, it helps if we first have $Q_0+Q_1$, which, using the above, simplifies to $\frac{2mc^2+2Q_0}{mc^2+2Q_0}Q_0$ and so $$\tag9v=c\frac{\frac{2mc^2+2Q_0}{mc^2+2Q_0}Q_0}{\sqrt{m^2c^4+\left(\frac{2mc^2+2Q_0}{mc^2+2Q_0}Q_0\right)^2}}$$ This is actually a function of $\frac{Q_0}{mc^2}$; expand it this way to get the usual limit. If you want the atom to recoil relativistically, it is obvious then that $Q_0\gg mc^2$ so that you should expand this in powers of $\frac{mc^2}{Q_0}$ and see what happens.

The entire above analysis assumes that the recoiling atom is still in its initial state. As the photon increases in energy, it is more likely for the photon to excite the atom to a higher state. Then you cannot assume that the rest mass is the same afterwards. That is also an interesting thing to consider and compute, in roughly the same manner.

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After many days of back and forth, it is now clear what the OP wants is something completely else. We now consider the problems as covered by AP French. For the situation of a photon being completely absorbed by an atom, we have $$\tag{10} \begin{pmatrix}+Q_0\\+Q_0/c \end {pmatrix}+ \begin{pmatrix}+mc^2\\0 \end {pmatrix}\Rightarrow \begin{pmatrix}+mc^2+Q_0\\+Q_0/c \end {pmatrix} $$ That is, the exact full SR velocity of the recoil of the atom is $v=c\frac{Q_0}{mc^2+Q_0}$, and this formula is in the book (and seems to be in OP's question too). However, it is important and interesting to consider the invariant rest energy of the resulting atom, which is $\sqrt{(mc^2+Q_0)^2-Q_0^2}=\sqrt{m^2c^4+2mc^2Q_0}=mc^2+E_\text{excitation}$; For a rough estimate, consider that the rest energy of the Hydrogen atom is $938.27208816\times10^6\,e$V whereas the maximum energy that a Hydrogen atom can absorb and yet still stay an atom, the binding energy, is the famous $13.6\,e$V, and you can immediately tell that $Q_0\ll mc^2$ for the above formula to be applicable. i.e. the recoil velocity of the atom is necessarily non-relativistic.

Similarly, we can consider the emission of a photon. Now the excited atom is stationary and transitions to the ground state. The smart thing to do is to take the invariant rest energy level from earlier and deduce what the new photon energy is. Namely, $$\tag{11} \begin{pmatrix}+\sqrt{m^2c^4+2mc^2Q_0}\\0 \end {pmatrix}\Rightarrow \begin{pmatrix}+\sqrt{m^2c^4+Q_1^2}\\-Q_1/c \end {pmatrix}+ \begin{pmatrix}+Q_1\\+Q_1/c \end {pmatrix} $$ $$ \begin{align} \tag{12}m^2c^4+2mc^2Q_0-2Q_1\sqrt{m^2c^4+2mc^2Q_0}+Q_1^2&=m^2c^4+Q_1^2\\ \tag{13}Q_1&=\frac{Q_0}{\sqrt{1+\frac{2Q_0}{mc^2}}} \end {align} $$ Here, the recoil velocity is $v=c\frac{Q_1}{\sqrt{m^2c^4+Q_1^2}}=c\frac{Q_0}{mc^2+Q_0}$, which is already an interesting result. Needless to say, there is yet again constraints on how energetic the photon can be, and so the recoiling atom must be in the non-relativistic regime.

Anyway, between my Equation (8) and my Equation (13), this problem is very completely solved. There is not much more to say.