The momentum of a photon is $\ p=E/c.$

When a photon reflects off a mirror, it is elastic scattering. Elastic scattering should keep the energy of the photon.

But radiation pressure states, that part of the momentum of the photon will be transferred to the mirror, this is how the photon exerts pressure on the mirror.

Now if $\ p=E/c\ $ and the momentum of the photon changes (part of it gets transferred to the mirror), and the momentum of the photon depends on the wavelength $\ p = h/\lambda$.


Elastic scattering is a form of particle scattering in scattering theory, nuclear physics and particle physics. In this process, the kinetic energy of a particle is conserved in the center-of-mass frame, but its direction of propagation is modified (by interaction with other particles and/or potentials). Furthermore, while the particle's kinetic energy in the center-of-mass frame is constant, its energy in the lab frame is not. Generally, elastic scattering describes a process where the total kinetic energy of the system is conserved.


In Rayleigh scattering a photon penetrates into a medium composed of particles whose sizes are much smaller than the wavelength of the incident photon. In this scattering process, the energy (and therefore the wavelength) of the incident photon is conserved and only its direction is changed. In this case, the scattering intensity is proportional to the fourth power of the reciprocal wavelength of the incident photon.

Now this is a contradiction. How can the energy of the photon be kept, and at the same time how can the photon exert pressure on the mirror, thus losing momentum, and change its wavelength?

$p=E/c$, so the momentum and energy of the photon cannot change without the other. If the photon's energy is kept during elastic scattering (mirror reflection), and the photon still exerts radiation pressure on the mirror, then the photon's momentum has to change (part of it needs to get transferred to the mirror), so the energy needs to change too.


  1. Does the wavelength of the photon change during elastic scattering (mirror reflection)?
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    $\begingroup$ Note that the direction of the momentum changes! $\endgroup$ – mmeent Sep 18 '19 at 13:11
  • $\begingroup$ @mmeent - this may be the OP's only real question - realizing the momentum can change sign and leave energy the same ... I added as a note to my answer $\endgroup$ – Paul Young Sep 18 '19 at 15:11

If a mirror reflection affected the energy of the photons to a large extent the colors would change, and it would not be a "true mirror"

The fact that the colors do not change for a "true" mirror , means that the interaction of the photons is elastic, i.e no energy is lost in our frame, the lab, reference. Elastic scattering keeps the energy of the photon the same in the center of mass system "Photon + mirror" Because the mass of the mirror is so very large, the lab frame is also the center of mass for the "photon+mirror" scattering, the tiny $ΔE$ due to the momentum given to the mirror is not discernible or computable to give a different center of mass frame to the lab frame.

  • $\begingroup$ thank you so much! $\endgroup$ – Árpád Szendrei Nov 30 '19 at 16:49

The answer is "almost no" - the wavelength of the photon is virtually unchanged (in the initial rest frame of the mirror, the "lab frame"). Because the mirror is much more "massive" than the photon, it serves as a "momentum sink" and picks up almost no energy.

The best way to develop an intuition for this is to consider a collision between two balls: one lighter (with mass $m$) and initially moving (at velocity $v_1$) and one more massive (with mass $M$) and initially at rest. After scattering, the lighter ball leaves the scene at velocity $v_3$ and the more massive ball leaves the scene at velocity $v_4$.

Set $v_2 = 0$ in the following worked-out example (see page 3):


We obtain $v_3 = \frac{(M-m)v_1}{M+m}$ and $v_4 = \frac{2mv_1}{M+m}$

In the limit that $M >> m$ the fraction of the initial kinetic energy picked up by the massive object goes to zero, but it acquires twice as much momentum (and in the opposite direction) as the lighter object initially had. Thusly can momentum be transferred but (almost) no energy.

NOTE - In the "center of mass frame" the wavelength will be completely identical, but I believe it is the "lab frame" that supplies the intuition you seek. In the center of mass frame the momentum just changes sign and your equation should really be $|p| = E/c$, which admits a sign change of $p$ while conserving $E$. This is why there is a $2$ in the equation for $v_4$ - flipping the sign of momentum imparts double the initial momentum to the mirror.

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    $\begingroup$ A note of caution: the intuition you can get from considering the light ball and the heavy ball is helpful, but don't trust it too far. Treating an elastic collision between a massless particle and a massive particle in special relativity involves conservation of the relativistic versions of momentum & energy, rather than the Newtonian versions used in the worked example. $\endgroup$ – Michael Seifert Sep 18 '19 at 15:51
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    $\begingroup$ @michael-seifert - I cannot agree with you more, you are 100 pct correct. Intuition is a strange thing, but we all rely on it ... so .... newtonian world to the mass/energy of a photon?? Maybe helpful, maybe not ... $\endgroup$ – Paul Young Sep 18 '19 at 18:17
  • $\begingroup$ its.caltech.edu/~teinav/Lectures/Ph%201a/… is dead link? $\endgroup$ – Shing Nov 27 '19 at 11:58
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    $\begingroup$ @Shing - fixed (by linking to web.archive.org) $\endgroup$ – Paul Young Nov 30 '19 at 16:30
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    $\begingroup$ the web tells me Tal graduated and is now at The Hutch in Seattle $\endgroup$ – Paul Young Dec 1 '19 at 19:05

The reflected photons are entirely identical to the incident photons except for the change in direction. The energy lost due to the pressure exerted on the mirror can be accounted for by comparing the number $N_i$ of incident photons to the number $N_r$ of reflected photons. I.e., $N_i > N_r$.


In the consideration of the photon behaving like a particle and reflecting from the surface of the mirror, the momentum is conserved and the wavelength does not change.

From conservation of momentum we know that the total momentum of the photon and the mirror is the same before and after the collision (considering the photon and mirror to be a closed system). In the simplest of considerations of refelection the quantum of the momentum of the photon is the same, but the direction is different. Resolving the momentum into two parts, one parallel to the surface of the mirror and one normal to the surface of the mirror then the momentum in the direction normal to the surface is reversed. To conserve momentum the mirror itself must have a balancing momentum equal and opposite to the change in momentum of the photom in the direction normal to the surface of the mirror. (You can see this effect in a rotating 'solar windmill' toy left in sunlight on a window sill - the reflective 'sails' reflect light and turn).

I stated above 'in the simplest of considerations' because the process of interaction may not be straight forward. There are different ways in which the photon and the mirror interact. For example Compton scattering may occur and that will change the wavelength of the light. (See https://en.wikipedia.org/wiki/Compton_scattering for more details on Compton Scattering).

So in a simple 'particle' model of reflection momemtum is conserved even though the direction of the photon changes.


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