# Newton's third law in relation to a mirror reflecting light

I am not too versed in the physics realms other than a couple courses from college, but my question mainly pertains to newton's third law of motion.

If for every action there is an equal and opposite reaction, why does the act of a mirror reflecting (or even partially deflecting) photons moving at the speed of light not cause an infinite reaction force in the opposite direction of the reflection?

I assume it has to do with photons being considered mass-less particles and newton's laws failing at quantum levels but I guess I hoped there was a better explanation than this.

• Why do you think there would be an "infinite reaction force?" Jan 19, 2017 at 21:49
• Well to get an object to the speed of light would take infinite energy from my understanding, so stopping and reflecting that object in the opposite direction would technically have an infinite reactionary force wouldn't it? Maybe this is where my logic is flawed Jan 19, 2017 at 21:52
• For example if you tossed a baseball at a wall in a vacuum (or space) after the collison the wall and baseball would be both moving away from each other in opposite directions correct? So why does this not apply to photons colliding with the mirror's surface? Jan 19, 2017 at 21:56
• Photons are massless so they do not have an infinite amount of either energy or momentum. Thus, no need for an infinite reaction force.
– user126422
Jan 19, 2017 at 22:02
• Try calculating the force : en.wikipedia.org/wiki/Radiation_pressure Jan 19, 2017 at 22:07

The energy of a photon is not infinite, but is given by $$E=hf=\dfrac{hc}{\lambda},$$ where $h$ is Planck's constant ($6.626\times 10^{-34}$ joule$\cdot$seconds), $f$ is the frequency of the photon, $c$ is the speed of light, and $\lambda$ is the wavelength of light.

The momentum of a photon is related to the energy by $E=pc$, where $p$ is the momentum. Combining these relationships, we see that the photon momentum can be calculated by $$p=\dfrac{h}{\lambda}.$$

If a photon is reflected from a mirror, we can estimate that the momentum of the photon changes to $p$ in the opposite direction. Ignoring outside itneractions, conservation of momentum tells us (the mirror is initially at rest) $$p + 0 = -p + p_{\mathrm{mirror}}.$$ This tells us the momentum of the mirror after the reflection should be (if the mirror is not attached to anything else, like a table with friction)$$p_{\mathrm{mirror}}=2\dfrac{h}{\lambda}.$$

Let's assume some numbers: a high energy visible photon has $\lambda = 4\times10^{-7}$ meters, and assume a very small, low-mass mirror with $m=1$ nanogram = $1\times 10^{-12}$ kilograms. This would result in a recoil speed for the mirror of about 3 femtometers per second.

I believe it's safe to assume that the resulting energy would easily be absorbed by frictional forces from objects to which the table is attached.

As you said, photons are massless. So getting a photon to travel at the speed of light (which, of course, it always is travelling at) does not take "infinite energy". This idea only applies to objects with mass.

However, photons do have momentum, which is defined by the equation:

$Momentum$= $Energy$ / $c$

with c being the speed of light. Of course, the energy of a typical photon is relatively very small and the speed of light is very large. Therefore, compared to usual objects in motion, a photon will have an extremely small momentum, which is why it has no large scale effect on something such as a mirror.

• So is harnessing this small amount of energy dispersed basically how solar sails work, or is that something differently entirely? Jan 19, 2017 at 22:04
• Yep, solar sails work on this same idea that photons have some momentum. I believe the sails are built so that they reflect light very efficiently so that they can get the most momentum as possible, (think Newton's third Law). You can read about solar sails here if you'd like: en.wikipedia.org/wiki/Solar_sail Jan 19, 2017 at 22:08

i think that photons suffer elastic collisions as they completely reflect back with same velocit that is c therefore : 1=(v2-u2)/(u1-v1)= coefficient of restitution which must be equal to 1 for elastic collissions.(u1,v1,u2,v2 are initial and final velocities of photon and mirror respectively

therefore         u1+u2=v1+v2____________A


and according to

                  p+0=-p+pm  (pm=momentum of mirror)___________B

NOW compare equation A and B We get:
u2=0,v1=u1-u and v2=vm                                                          (vm=velocity of mirror) and u is the infinitesimal velocity change of photons reflecting back from mirror .


therefore therefore * **u1=u1-u+vm **
vm=u,which is at the order of fermimeters

  and hence very small and get absorbed by the mirror due to some frictional forces

• i think this is the correct reason may be Jan 19, 2017 at 23:06