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Passing light repositions glass, doesn't it?

  1. Apr 11, 2014 #1
    I'm starting with the Einstein box. It's the one that's floating in space and has photons moving from one side to the other. The idea being that every time a photon is sent from one side to the other, the recoil sets the box in motion until that photon strikes the other side breaking to zero.

    The center of gravity never changes because as the emitting side looses mass, the absorbing side gains and that combined with the shifting position of the box towards opposite to the motion of the photons keeps the center of gravity stationary - as it must.

    So, I now place a pane of glass floating in the middle of this box. As the light passes through the glass, it will travel a bit slower - and so its travel time to the opposite side will be delayed. This, in turn, will allow the box to continue moving a bit longer - so its displacement in the opposing direction will be greater than it was before. So, in order to keep the center of gravity stationary, the glass must become displaced in the direction of travel of the photon.
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  3. Apr 11, 2014 #2


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    I only have a popular level reference for this one.

    If you have light travelling in both directions through the pane of glass, any push (or pull) force of the light on the surface would cancel out, resulting in no net force. So I'm assuming that one considers the forces on a surface of a dielectric due to a unidirectional pulse of light, such as a pulse from a laser beam. This is closely related to your original question, if one takes your original question literally the answer would be that there is no force, but perhaps a compression.

    The popular article in question (which has a link to a peer-reviewed article which is alas, behind a paywall):


    More detail on the "two theories":

  4. Apr 11, 2014 #3
    Thanks for the quick research - I had found nothing.

    From the arguments I gave in the OP, there should be no net force once the light has finished its travel through the glass. I would expect it to apply a forward kick on entry and a equivalent recoil on exit. The net force would be zero, but the effect would leave the glass in a more forward location.
  5. Apr 11, 2014 #4


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    I would agree that if you have a pulse of light, the force will end when the light passes through the glass.

    After some consideration, I suspect you're probably right about the "recoil on exit", but I don't have a reference that talks about that issue. My thoughts so far are that if the light changed the velocity of the dielectric/glass, it would require energy to be taken from the light, but on due consideration the problem appears to be more involved than I thought when I posted my answer.

    I don't understand what role the box plays here at all - it seems to me that the essential elements of the problem are the laser pulse, and a dielectric (in this case glass) sheet. It's possible I'm misunderstanding the problem.

    [add]I've had one more thought. So far we've been ignoring the issue of reflections at the dielectric boundary. This may be part of the problem - in that if we remove all such reflections as "not being of interest", we may arrive at solutions where the interaction between the light beam and the dielectric isn't being correctly modelled to describe the phenomenon of interest (transfer of momentum between the beam and the dielectric).
    Last edited: Apr 11, 2014
  6. Apr 12, 2014 #5


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    Since the conditions are symmetric, the piece of glass does not change the COM.

    The photons passing through the glass (or any transparent media) are repeatedly scattered - forward coherent scattering. This is a totally elastic process, so there is no energy loss to the photons ... otherwise you would notice spectral shifts when light passes through air or glass!

    The calculation of the impulse delivered requires a detailed analysis of the Huyghens wavelets which make up the phase front. But even without going into the details I fail to see why there would be any recoil when the light exits the glass.
  7. Apr 12, 2014 #6

    Ben Niehoff

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    Recoil on exit depends on the shape of the glass. For your thought experiment, you just have a flat plane of glass, so I think you have nothing to worry about.

    However, if you have a lens that focuses rays of light, then incoming parallel rays will be focused on exit. The light cannot change frequency, so the magnitude of its momentum is the same. However, the direction of the momentum changes, on exit from the glass, due to refraction. Which means the net momentum in the original direction of travel changes, and the glass must make an equal, opposite change.

    I heard about this in a biophysics talk a few weeks ago. The principle is used in laser traps to actually lift up individual bacteria and place them down where you want them to go.
  8. Apr 12, 2014 #7
    IF light is scattered and moves to various directions inside the glass as claimed in post #5, then when light leaves the glass and moves into one direction, then if post #6 is correct, there is a force between the light and the glass when light leaves glass.

    Let's say light goes through an optical fibre. Then we shake the middle part of the fibre. This causes a delay of passing of the light through the fibre. And it causes some forces on the fibre caused by the light, those forces should be easy to calculate.
  9. Apr 12, 2014 #8
    If you don't understand the box, you're missing my argument completely. I'm borrowing it from Einstein. He used it to demonstrate how the photon needed to push against the box in order for the system as a whole (box and contents) to maintain it center of gravity.
    Let me draw it out for you:
  10. Apr 12, 2014 #9
    Although I described it as a forward kick on entrance and a recoil on exit, its probably better to describe it in a bit more detail. In order to maintain the center of gravity, the shifting process must occur continuously as the light passes through the glass.

    If we work with a green photon passing through glass:
    Glass, SiO2 weight is 60.08amu
    SiO2 specific gravity is 2.648 g/cc
    So, using Avogadro's number, SiO2 molecules are about 0.335nm apart from each other.
    Green photon (500nm) is 2.660 x 10^-10 amu
    Green photon/SiO2 index of refraction is 1.549
    So every time a green photon travel past a SiO2 molecule (0.335nm), it is held back by a that distance time its index of refraction minus one (0.549).
    So we need to compensate by 0.1840nm.
    Our SiO2 molecules are more massive that then photon by 60.08/(2.660 x 10^-10).

    So our SiO2 molecules need to move about 8x10^-22 meters (total) each time a photon crosses an SiO2 molecule. Basically, there is a miniscule mechanical wave created by the photon which displaces a trail of molecules as it passes through the glass. It doesn't leave any energy behind - just displaced molecules.
  11. Apr 13, 2014 #10


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    If its easy, I'll be glad to let you do it!
  12. Apr 13, 2014 #11
    I forgot that light travels in a dielectric in an optical fibre:redface:

    But let's say collimated light beam goes through these 2 lenses:

    Code (Text):

              ------------->            )(               ()            
    I guess we agree that the first lens experiences a force to the right. And second lens experiences a force to the left.

    To be more specific:

    When the beam is turned on, then first the lens on the left starts experiencing a force, then after a while the lens on the right starts experiencing a force.

    When the beam is turned off, then first the lens on the left stops experiencing a force, then after a while the lens on the right stops experiencing a force.

    The momentum stored in the system of the two lenses is: F * t - F' * t'
    where t and t' are the times that lenses L1 and L2 have been experiencing forces F and F'.
  13. Apr 13, 2014 #12
    The first lens will cause a spreading of the light beam, reducing it's average forward momentum, so it will experience a force to the right. Depending on the power of the second lens, it might reduce the spread, leave it the same, or increase it. The result would be a net force to the left, no net force, or a force to the right, respectively.
  14. Apr 13, 2014 #13


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    Thanks for the drawing.

    My research on the literature seems to indicate it's in quite a mess for a good solid answer.

    My current (not well supported) opinion on what should happen is this. (Downgrade this to a non-science-advisor answer :( please ).

    "All the photons on the left" is a quantum description, classically we would say that "there is a narrow laser pulse bouncing around the box.

    When the pulse enters the glass slab, some of it reflects. This will lead to long term problems with your scenario, because the narrow pulse will start to broaden. Eventually, I'd expect that the box would be filled with a uniform distribution of photons, a "photon gas".

    This reflection process will also transfer momentum to the slab. It's not clear if we can ignore this problem and mantain consistency (by assuming an idealized non-reflection coating, for instance) but I think maybe we can.

    Assuming the abraham expression is correct, the electromagnetic part of the momentum drops when it enters the dielectric. This means there should be a net mechanical motion to compensate.

    Logically, the electromagnetic energy has to drop too - some of it should go to shaking the dipoles in the dielectric, some of it should go to cause the afforementioned mechanical motion. I'm currently having some difficulties with confirming this, though. If it's wrong the rest of the argument is suspect.

    Realistically there will be some loss in the process. If we idealize and ignore this again, though, the mechanical energy of motion is returned (mostly) to the EM wave when it exits the glass, and the glass stops moving.

    In quantum terms, what I think is happening is this.

    Atoms are absorbing the photons, and re-emitting them. Momentum is conserved, so when the atom absorbs the light, it tries to move. In a lattice, you'd create a quantized lattice vibration, a phonon. In glass - it's less clear to me what happens. In any case I think there should be a net transmission of EM momentum to mechanical motion that occurs when the photon is absorbed and when the re-emission occurs, the mechanical momentum is transfered back to EM momentum.

    So the bottom line - in my opinion, the glass does move while the pulse travels through it, then stops when the pulse exits the glass. It moves back in he other direction when the pulse reflects from the other side of the box, if you have a box.
  15. Apr 14, 2014 #14


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    Light passes through a transparent medium via "Forward Coherent Scattering"; there is no absorption/emission process.

    1. Forward: the light moves through the media, hence forward
    2. Coherent: the phase front is retained, else we would see a different image before and after
    3. Scattering: absorption/emission has random factors - such as the direction of the emission of the photon! It also requires accessible energy levels - these don't exist if the material is transparent.

    Note that the scattering is elastic - the spectrum does not change from one side to the other; it is a lossless process (though there is always some "absorption" loss, and some reflection loss).

    This scattering is from the electron cloud - the nuclei do not participate; hence there are no phonons.

    You can model this scattering with the Huyghens wavelet mechanism. Thus interference effects play an important role.

    Richard Feynman describes this in some detail in this four lectures: "QED: The strange theory of light and matter".
  16. Apr 14, 2014 #15


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    I don't think there is any difference between saying "absorption and remission" and "scattering", though I may well be missing subtle points, I'm not as familiar with QM - or in this case - QED - as I would like.

    I definitely didn't and do not want to imply that the emission and absorption was atomic absorption and re-emission. In this respect scattering might be a better term. It also might be less clear though.

    A few quotes.

    This quote is why I don't believe there's any significant difference between saying absorption and re-emission and saying scattering.

    As far as phonons go:

    Our PF Faq "Do Photons move slower in a Solid Medium"
    https://www.physicsforums.com/showthread.php?t=511177 [Broken]
    I'm a bit leery of applying the term "phonon" to amorphous solids , though. Wiki defines a phonon as:

    Unfortunately the amorphous structure of glass makes it not periodic :(. If we replaced the amorphous glass with a transparent periodic crystaline solid, then the phonon description would be (AFAIK) spot on.

    I didn't want to get into the process of how coherent interference between photons (which includes them being absorbed by all possible electrons, then re-emitted and coherently combined as part of the all-possible-paths-process) makes them appear to travel in straight lines ala QED. Which is Feynman's description of how it works from QED.

    One reason I don't want to argue the point is I can't currently actually calculate anything with the "all-possible-paths" paradigm that Feynman espouses in QED, though Feynman says his grad students can do it.

    Not being able to actually calculate stuff has unfortunately made the ideas from QED not terribly useful to me in spite of them being intriguing.

    That was one of my inspirations, in my thinking, too. I am also trying to understand the process better from a purely classical view, it's not going as well as I'd like at the moment. In particular, the issue of where the energy would come from is still problematic according to my latest reading. My current guess is that physically in glass there are vibrating dipoles which do have some mechanical energy, but that this energy is neglected in the usual formula for the Poynting vector because that usual calculation shows all power is either transmitted or reflected, leaving none for shaking dipoles and/or moving the dielectric slab.

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  17. Apr 14, 2014 #16


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    This so-called FAQ is incorrect. It should be removed from circulation.
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  18. Apr 14, 2014 #17


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    The Feynman quote is from a non-technical presentation, and is intended to motivate the Feynman diagrams.

    In technical literature absorption and emission have a more definite meaning which comes from spectroscopy; see http://en.wikipedia.org/wiki/Einstein_coefficients

    Note that the photon emitted following an absorption is (a) emitted at a random time (decay rate), and (b) is emitted in a random direction.

    Scattering is an important term; you can read about quantum "Scattering Theory" here:

    The first chapter provides motivation and definition - then follows the rest of the book.

    In the case at hand we have elastic scattering: no energy exchange between the photons and the dielectric material.
  19. Apr 14, 2014 #18


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    Actually you can have phonons in any condensed matter ... it is simply as simple as supporting sound waves, which requires a coherent motion of the material. Quantize that and you have phonons. When the sound loses it's coherence, the phonons are gone ... all you have left is random motion of the atoms - which we call heat.

    But that doesn't make the FAQ correct. Phonons play no significant role in the transmission of light through a dielectric media.

    This is why I always conclude with a reference to the Huyghens wavelet mechanism:

    You can't get anymore classical than that! Integration over "all possible paths" isn't required for a basic understanding, nor for any but the most careful calculations.

    If you want the classical answer from Maxwell's electrodynamics you can try Griffith's "Introduction to Electrodynamics", or any good text on physical optics. The dipoles of the dielectric media are stimulated by the incoming electric field, and their motion (duplicating the incoming frequency) is the generator of dipole radiation for the reflected and the transmitted beams.

    If the process is perfect, there is no left over mechanical vibration, and R+T=1. In reality there is always some loss which is called "absorption".
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