I'm working a problem out of d'Inverno's "Introducing Einstein's Relativity", and I'm hitting a funny issue with my algebra. The problem states:

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.

The answers are given in the back of the book as

$$ u=\frac{ch\nu}{h\nu + m_0c^2} \qquad m=\left(m_0^2 + \frac{2h\nu m_0}{c^2}\right)^{1/2}. $$

I've found that I can figure out the velocity by starting with conservation of momentum and energy

$$ \frac{h\nu}{c} = \gamma m_0u \qquad h\nu + m_0c^2 = \gamma m_0c^2, $$

(where of course $\gamma=\left[1-\frac{u^2}{c^2}\right]^{-1/2}$) then eliminating the relativistic mass $\gamma m_0$ and solving for $u$. Strangely though, in my first couple of attempts I started with conservation of only energy or momentum, and obtained answers close to, but not quite the same as, the above:

$$ u=\frac{c(2m_0c^2h\nu+h^2\nu^2)^{1/2}}{h\nu + m_0c^2} \qquad (\textrm{from conservation of energy}) $$

$$ u=\frac{ch\nu}{(h^2\nu^2+m_0^2c^4)^{1/2}} \qquad (\textrm{from conservation of momentum}) $$

Is there simply some mistake in my algebra that I haven't managed to suss out, or is the error in assuming that I can proceed from only one conservation law?

As for the mass, I attempted to simply substitute the velocity $u$ into the relativistic mass $\gamma m_0$, but quickly got a monstrosity of $m_0$'s, $h$'s, $\nu$'s, and $c$'s that bore no resemblance to the answer. Is this the correct way to approach this part of the problem, or should I start from some other relation?


Sofia answered the second half of my question regarding the mass of the atom. However, I'm still curious about the first part. That is, why do I obtain different results for the velocity with different equations? One approach may be more difficult or roundabout than another, but if they're all based on the same physical principles, I feel that I should get the same result regardless of which relation I begin with.

  • $\begingroup$ I am trying to solve your problem, but there is a certain approximation done in the solution, and that I don't understand why they did it. You will see in my answer. $\endgroup$
    – Sofia
    Jan 19, 2015 at 23:30

1 Answer 1


The solution in the book doesn't consider the recoil velocity $u$ as relativistic. So, no need of $\gamma$. But they consider simply that the mass of the atom increases, instead of $m_0$ to $m$. So, this is what they do, I mean, their equations of conservation:

(1) $h \nu /c = mu, \ \ \ $ linear momentum conservation,

(2) $(m_0c^2 + h\nu)^2 = m^2c^4 + m^2u^2c^2, \ \ \ $ energy conservation and Klein-Gordon equation.

Now, from eq. (1) I obtain

$u = \frac {h \nu}{mc}$,

and I introduce in eq. (2).

$(m_0c^2 + h\nu)^2 = m^2c^4 + h^2 \nu ^2$.

From the last equation we can extract $m$,

$m_0^2 + \frac {2h\nu m_0}{c^2} = m^2$.

The quantity on the LHS can be eventually completed to a square

$m_0^2 + \frac {2h\nu m_0}{c^2} + \frac {(h \nu)^2}{c^4} = m^2$.

  • $\begingroup$ Hmmm, I'm not so certain. I'm getting something that, again, looks like the answer, but isn't quite there, even if I were to try and approximate. And on that note, in my experience introductory undergraduate level texts usually don't require approximation unless explicitly stated as part of the problem. $\endgroup$
    – Itserpol
    Jan 20, 2015 at 1:42
  • $\begingroup$ Also, in equation (2), it seems that the right side includes both the kinetic energy twice. That is, the first term is the total energy, and so the second should be unnecessary. Unless, you had meant to type m0? I understand that you feel the problem seems to approximate away the relativistic recoil velocity, but in that case wouldn't the mass also approximate down to the original mass before the collision? $\endgroup$
    – Itserpol
    Jan 20, 2015 at 1:46
  • $\begingroup$ @Itsrpol , I will try to help. As a matter of fact, I obtained the desired result. Wait a bit. $\endgroup$
    – Sofia
    Jan 20, 2015 at 1:46
  • $\begingroup$ @Itserpol, yes I see. These people apply for the energy after the absorption, the Klein-Gordon equation, $E^2 = p^2c^2 + m^2c^4$. Let's try if the things go well this way, and after that we will think why Klein-Gordon. O.K.? $\endgroup$
    – Sofia
    Jan 20, 2015 at 1:58
  • $\begingroup$ @Itserpol : look now ! The mass gets O.K. $\endgroup$
    – Sofia
    Jan 20, 2015 at 2:10

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