In space, a photon with momentum $P_1$ is reflected off a mirror, accelerating it slightly. Now there is a reflected photon with momentum $P_2$ in the other direction. So, the mirror must have momentum $P_1 + P_2$ in order to conserve the momentum of the entire system, because it has to cancel out the momentum of the reflected photon in addition to maintaining the momentum of the initial photon.

But, in this example, the total magnitudesof the momentum vectors has increased, which just feels bonkers to me. How could an object hit another object and impart a momentum onto it that is greater than the one it originally carried? This seems to create energy out of nowhere.

For example, if this was true, you could set up a bunch of these objects in a chain reaction and the tiniest push could send something further down the chain flying. What am I not understanding here?


1 Answer 1


When calculating (elastic) collisions, both energy and momentum conservation is taken into account. The photon initially has the momentum $p_{\gamma i} = \hbar k$ and the energy $E_{\gamma i} = \hbar \omega$, with $k$ being the wave number and $\omega$ the frequency, and the mirror has the momentum $p_{mi}=0$ and the Energy $E_{mi} = mv^2/2 = 0$. With $i$ denoting initial and $f$ denoting final quantities, energy and momentum conservation are $$ p_{\gamma i} + p_{mi} = p_{\gamma f} + p_{mf} \quad \Leftrightarrow \quad \hbar k_i = -\hbar k_f + m v_f \quad \Leftrightarrow \quad \frac \hbar c \omega_i = - \frac \hbar c \omega_f + m v_f~, \tag{1} $$ $$ E_{\gamma i} + E_{mi} = E_{\gamma f} + E_{mf} \quad \Leftrightarrow \quad \hbar \omega_i = \hbar \omega_f + \frac 12 m v_f^2 \quad \Rightarrow \quad \omega_f = \omega_i - \frac{m}{2\hbar} v_f^2~. $$ The negative sign in $p_{\gamma f} = -\hbar k_f$ is needed, because the photon propagates in the opposite direction, so the wave vector points in the opposite direction, too.

Combining the above equations, there follows $$ \frac \hbar c \omega_i = - \frac \hbar c \omega_i + \frac{m}{2c} v_f^2 + m v_f \quad \Leftrightarrow \quad v_f^2 + 2 c v_f - \frac{4\hbar}{m} \omega_i = 0 \quad \Leftrightarrow \quad (v_f + c)^2 - c^2 - \frac{4\hbar}{m} \omega_i = 0~. $$ $$ \Rightarrow v_f = \sqrt{c^2 + \frac{4\hbar}{m}\omega_i} - c~. \tag{2} $$ And using (1) yields $$ \omega_f = -\omega_i + \frac{cm}{\hbar} v_f~. $$ This means, $|\omega_f| < |\omega_i|$, so the photon looses the energy which the mirror gets and energy conservation is not violated.


The last paragraph of the original post is not wrong, but I got a little confused there and, since $c/\hbar$ is very large, it does not really help in pointing out, why $|\omega_f| < |\omega_i|$. So, let's try again:

It holds $$ \frac{cm}{\hbar} v_f \overset{\text{(2)}}= \frac{cm}{\hbar} \left( \sqrt{c^2 + 4 \frac{\hbar \omega_i}{m}} - c \right) = \frac{c^2 m}{\hbar} \left( \sqrt{1 + 4 \frac{\hbar \omega_i}{c^2 m}} - 1 \right)~. $$ The Taylor-expansion of $\sqrt{1+4\epsilon}$ at $\epsilon = 0$ is $$ \sqrt{1 + 4 \epsilon} = \sqrt 1 + \left[ \frac{4}{2 \sqrt{1 + 4 \epsilon}} \right]_{\epsilon =0} \epsilon + \frac 12 \left[ - \frac{16}{4 \sqrt{1 + 4 \epsilon}^3} \right]_{\epsilon =0} \epsilon^2 + \mathcal O(\epsilon^3) = 1 + 2 \epsilon - 2 \epsilon^2 + \mathcal O(\epsilon^3)~. $$ Setting $\epsilon = \hbar \omega_i / (c^2 m)$ (which is very small for every reasonable $\omega_i$ and $m$), we get $$ \omega_f = - \omega_i + \frac{cm}{\hbar} v_f = - \omega_i + \frac{c^2 m}{\hbar} \left( \sqrt{1 + 4 \epsilon} - 1 \right) \\= - \omega_i + \frac{c^2m}{\hbar} \left( 1 + 2 \frac{\hbar \omega_i}{c^2m} - 2 \left(\frac{\hbar \omega_i}{c^2 m} \right)^2 + \mathcal O(\epsilon^3) -1 \right) \\= -\omega_i + 2\omega_i - \underbrace{2 \frac{\hbar\omega_i^2}{c^2 m}}_{\text{very small}} + \mathcal O\left( \left( \frac{\hbar \omega_i}{c^2 m} \right)^3 \right)~. $$ $$ \Rightarrow \omega_f = \omega_i - \underbrace{2 \frac{\hbar\omega_i^2}{c^2 m}}_{\text{very small}}+ \mathcal O\left( \left( \frac{\hbar \omega_i}{c^2 m} \right)^3 \right)~. $$

  • 1
    $\begingroup$ I think you need to elaborate that final paragraph because $c/\hbar$ is very large, ~$2.84×10^{42}\,\mathrm{kg^{-1}m^{-1}}$ $\endgroup$
    – PM 2Ring
    Aug 11, 2021 at 18:45
  • $\begingroup$ @PM2Ring, you are right, thanks, seems like I was to quick here and in my head turned $c/\hbar$ into $\hbar/c$. I should have noticed, because $\omega_f$ must not be negative, otherwise $E_{\gamma f}$ would be, too. Plugging in a few example numbers yields that $\omega_f < \omega_i$, but they are almost the same, which agrees with my intuition. I will have to think about how to show this in general and then edit the post. $\endgroup$
    – nu.
    Aug 13, 2021 at 7:31
  • $\begingroup$ That's much better. :) Another option is to use $$\frac{\omega_1}{\omega_0} = \frac{2c-v}{2c+v}$$ $\endgroup$
    – PM 2Ring
    Aug 13, 2021 at 15:37

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