8

When light reflects from a surface, at least the direction of its momentum changes. Since the total momentum must be conserved, there has to be something going on within the atoms of the surface.

So my question is that does reflected light increase the internal energy of the surface even if it is a really really tiny amount?

P.S. I am not talking about the fraction of light that gets absorbed by the surface. I know the energy of that fraction contributes to the internal energy of the surface. My concern is only about the photons being reflected.

physicsguy19
  • 1,492
  • 10
  • 28

3 Answers3

7

Yes, this is the principle behind Doppler radar. The frequency/energy increases if the surface is moving towards the source. The frequency/energy decreases if the reflecting surface is moving away from the source.

The only time the frequency/energy will be unchanged is if the surface initially has the opposite momentum of the light.

Dale
  • 99,825
  • So the frequency of reflected light is less than the incoming light evem if the surface is stationary? – physicsguy19 Oct 21 '18 at 00:54
  • 1
    If the surface is stationary then there will be a slight decrease. If the surface has the opposite momentum of the photon then the energy/frequency will be unchanged. I will add this to the answer – Dale Oct 21 '18 at 01:00
  • 1
    To clarify, upon reflection from a stationary object, the object recoils, giving it kinetic energy. The photon’s energy then needs to decrease (redshift) in order to account for the kinetic energy. The amount by which the frequency of the photon shifts is the Doppler factor. Thus both energy and momentum are conserved. – Gilbert Oct 21 '18 at 04:34
  • Whenever light hits the surface, the reflected light is redshifted. Of course less redshifted, if the obstacle moves in the same direction as the light. No redshift only in the (theoretical) case of a velocity of the obstacle equal to c. – HolgerFiedler Oct 21 '18 at 07:17
  • 1
    @HolgerFiedler that is incorrect. There is no redshift in the center of momentum frame. If the surface is moving towards the source faster than in the center of momentum frame then there will be blueshift. – Dale Oct 21 '18 at 13:08
  • The only time the frequency/energy will be unchanged is if the surface initially has the opposite momentum of the light. This is wrong. In the case of elastic scattering, that is reflection, and with a perfect mirror, the photons keep their energy level. – Árpád Szendrei Oct 21 '18 at 23:34
  • 1
    @ÁrpádSzendrei That is incorrect. Do the math: if the photon keeps its energy in once reference frame, its energy changes in another, since the Lorentz boost acts differently on the photon depending on the direction it is traveling. The photon only keeps its energy during reflection as measured in the center of momentum frame. – Chris Oct 21 '18 at 23:48
  • @ Chris, so you are saying, that when I look in the mirror, the image's photons' energy level is changed, so it is not the same as the incident photons' energy level that popped off of me originally? So how do I see my exact mirror image? – Árpád Szendrei Oct 21 '18 at 23:52
  • 1
    In the reference frame where the mirror is initially at rest that is correct, the energy level is not the same. To understand why you don’t notice it you should calculate the expected frequency difference. – Dale Oct 22 '18 at 01:12
0

When a photon interacts with an atom, three things can happen:

  1. elastic scattering, the photon keeps its energy and phase, and changes angle, this is the case of a mirror, reflection

  2. inelastic scattering, the photon gives part of its energy to the atom and changes angle, this happens when infrared light transfers kinetic energy to the vibrational motion of molecules (heats up)

  3. absorption, the photon gives all its energy to the atom, and the absorbing electron moves to a higher energy level as per QM

In your case, reflection, is elastic scattering, and this is the only way to keep the energy level of photons, and to build a mirror image.

Of course this is assuming a stationary reflecting surface (relative to the observer), mirror.

As the other answers say, when the reflecting surface is traveling towards or away from the observer, the energy level of the photons can change.

It is very important to talk about specular reflection, like a mirror, where the relative angle of the photons is kept too. And differentiate it form diffuse relfection, where the relative angle of the photons is not kept.

You are asking whether the momentum of the photons can be transferred to the surface. Yes, photons can exert pressure on the surface of the mirror.

Please see here:

Can something without mass exert a force?

Árpád Szendrei
  • 28,452
  • You misunderstand how elastic scattering works. In elastic scattering the photon’s energy is not necessarily conserved, it is only the total kinetic energy that is conserved. In the case of a stationary mirror, conservation of momentum requires that it accelerate and gain KE. Therefore the photon’s energy is decreased. Elastic scattering does not in any way imply that the photon’s energy must be unchanged – Dale Oct 22 '18 at 00:27
0

The answer is yes, but the effect is extremely small. In an elastic collision of a photon with a massive object which is initially at rest, the relative loss of energy of the photon will be $$ \frac{\Delta E}{E} = \frac{E}{E + mc^2/2}. $$ Assuming $mc^2\gg E$ this reduces to $$ \frac{\Delta E}{E} = \frac{2E}{mc^2}. $$ For a single electron ($m=511\,\textrm{keV}/c^2$) and a typical photon in the visible range ($E=2\,\textrm{eV}$) this would amount to roughly $1/100,000$. This is an upper bound of the effect. Outside of experiments in atom traps visible light is not reflected off single electrons but off solids. In this case the recoil momentum is absorbed by the much heavier bulk of the solid. Typically, the immediate interaction of a light wave with the surface of a solid involves the atoms in a volume whose dimensions are given by the wavelength. For visible light ($\nu=O(500\,\textrm{nm}$) you easily end up with hundreds of thousands of atoms (a typical spacing is $10\,\textrm{nm}$ collectively absorbing the recoil, each thousands of times heavier than the single electron from above. Taking hydrogen atoms with the numbers as above, you thus end up with an additional suppression factor of $\approx 125,000,000$ rendering the impact of the recoil momentum totally negligible.

It is worth pointing out that this collective absorption of recoil by a crystal is what led to the experimental technique of Mößbauer spectroscopy. Unlike in your question, here the phtons aren't reflected off the crystal but instead emitted by excited nuclei inside the crystal. The logic is the same, though.

You asked about light, but the picture changes going to higher photon energies and thus shorter wavelengths. Once the wavelength of the photon becomes smaller than the size of atoms, reflection of photons by single electrons becomes an important effect. For photon energies where the electron mass is negligible, the entirety of the photon's energy can be absorbed by the electron. This is the opposite situation to the one I described above for visible light! These kinds of processes are a type of what is known as Compton scattering.

tobi_s
  • 1,299