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I'm solving a problem 4.11 in <Introducing Einstein's Relativity, Ray D'Inverno>. The problem states as follows:

An atom of rest mass $m_0$ is at rest in a laboratory and absorbs a photon of frequency $\nu$. Find the velocity and mass of the recoiling particle.

I tried with momentum & energy conservation, but I immediately ran into contradiction.

\begin{align} \tag{1}\text{momentum conservation : }\frac{h\nu}{c}&=\gamma m_0 v\\ \tag{2}\text{energy conservation : }h\nu+m_0 c^2&=\gamma m_0 c^2 \end{align}

The velocity is easy: dividing (1) by (2) and multiplying $c^2$, we have $$\tag{3}v=\frac{h\nu c}{h\nu+m_0 c^2}$$ The mass can be obtained in two ways: first method is just dividing (2) by $c^2$: $$\tag{4}m=\gamma m_0=\frac{h\nu+m_0 c^2}{c^2}$$ Second method is substituting (3) into $m=\gamma m_0$: $$\tag{5}m=\gamma m_0=\frac{m_0}{\sqrt{1-v^2/c^2}}=\frac{m_0(h\nu+m_0 c^2)}{\sqrt{2h\nu m_0 c^2+m_0^2 c^4}},$$ which is evidently different from (4). I suspect that this is because the recoiling particle is off-shell, so that the energy conservation does not hold. After all, the answer in the textbook was neither of them: $$\tag{6}v=\frac{h\nu c}{h\nu+m_0 c^2}$$ $$\tag{7}m=\sqrt{m_0^2+\frac{2h\nu m_0}{c^2}}$$ My question is following:

  1. If my suspection were correct, then the velocity (6) must be also different from (3), since (2) is no longer true! Am I misunderstanding the problem? or what's wrong?
  2. The process described in the problem seems to convert the entire energy of the photon to the kinetic energy of the atom. Is is equivalent to the statement that the absorbed energy raises the energy levels of bound electrons (or nucleus)? How is it related to the kinetic energy of the atom?
Qmechanic
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Mr. Anomaly
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1 Answers1

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Your Equation (3) is the exact same expression as Equation (6) and both are correct. However, the new rest mass must be computed another way, and Equations (4) and (5) are wrong: $$ \begin{align} \tag 8 m&=\frac1{c^2}\sqrt{E^2-p^2c^2}\\ \tag 9 &=\frac1{c^2}\sqrt{\left(m_0c^2+h\nu\right)^2-\left(\frac{h\nu}c\right)^2c^2}\\ \tag{10}&=\frac1{c^2}\sqrt{\left(m_0c^2+h\nu\right)^2-(h\nu)^2}\\ \tag{7 obtained}&=\sqrt{m_0^2+2m_0h\nu/c^2} \end {align} $$ This is really the invariant total energy in the new rest frame of the atom, representing that the atom is now slightly more energetic and thus heavier. There must be a new energy level for the atom at precisely this energy.

naturallyInconsistent
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