Why does a mirror work?

Discussion in 'Physics & Math' started by Omnignost, Apr 23, 2004.

  1. oxymoron Registered Senior Member

    Correct. Metals are not molecular. And there definately is a difference between reflection off a metal and reflection off a dielectric; now that you mention it.

    Reflection off a metal is referred to as Reflectance and it is a property that is unique to a surface that is conducting - like what you said.

    But before I delve into this some housekeeping is in order.

    k is the direction vector (also called the propagation vector). It points in the direction of a plane wave travelling through free space. We will need this vector.

    u is the unit vector normal to the interface pointing in the direction from the incident to the transmitting medium.

    E the electric field vector and B the magnetic field vector.

    Notice that u is normal to the interface regardless of the direction of the electric field.

    Okay, now lets get into it. Lets assume for starters a basic example: An electric field E perpendicular to the plane-of-incidence and that B is parallel to it. So from electromagnetism (remember E = vB) so..

    k x E = vB

    k . E = 0

    Now I wont go into the maths but what we try to do here is work out the formula for the reflection coefficient will depend on...

    1. the angle of incidence and transmission
    2. the refractive indicies of the two media

    This makes sense. But the formula is a little complicated. Ultimately you end up with E_r / E_t (subscripts are for reflection and transmission). This the the quotient of the electric field vector of the reflecting wave and the electric field vector of the transmitted wave. Now you have to express each vector as a sum of its angular components (because we know the angle changes through a medium).

    Reflection Coefficient = [n_i cosθ_i - n_t cosθ_t]/[n_i cosθ_i + n_t cosθ_t]

    You may recognise these expressions for E as the Fresnel Equations. Perhaps you might look this up!

    However, in this thread we are not really interested in the mathematics but rather the physical implications. In particular we are interested in the reflection (and transmission; they come hand in hand). Now we get into the meaty bits.

    Consider a beam of light incident on a surface. This illuminates a spot of area A. The power per unit area crossing the surface (in a vacuum of course) whose normal is parallel to the Poynting vector (S) is given by...

    S = c²ε_0 E x B

    This is from second year electromagnetism (first year advanced physics perhaps). From this relationship we work out the radiant flux density (a very important concept used throughout advanced electromagnetism at uni).

    I = <S> = 1/2 c&epsilon;_0 E&sup2;

    This is the average energy per unit time crossing the area on the surface normal to S. We let I_i , I_r, and I_t be the incident, reflected and transmitted flux densities respectively.

    Can you see that the cross section of a circular wavefront impinging on a surface will trace out a circular illumination? The cross-sectional area of the incident beam will be Acos&theta;_i. And for the relflected and transmitted cross-sections they will be Acos&theta;_r and Acos&theta;_t.

    Similarly the incident power is I_i Acos&theta;_i and so on for reflection and transmission. This is the energy per unit time flowing in the beams, and it is therefore the power arriving in our cross-sections.

    Now we are in a position to define reflectance (R) to be the ratio of the reflected power (flux) to the incident power.

    R = flux_r / flux_i = I_r Acos&theta;_r / I_i Acos&theta;_i

    Now we know what Relfectance is and how it is related to the Fresnel equations of electric fields. Reflectance can be applied to any media. But now we want to apply it to metals.


    You know that the main characteristic of a metal is its conducting property; that is, the presence of a great number of free electric charges. For metals these charges are electrons and their motion is electric current. The current per unit area is the conductivity of the medium &rho;.

    Housekeeping done. Now imagine an electric field plane wave impinging on a metal surface. Lets calculate the reflectance of metal...

    R = I_r / I_i. Take n_i = 1 (air) and n_t = n (a complex index). Therefore

    R = [(n-1)/(n+1)][(n-1)/(n+1)]*

    Since n = n_r - in_i

    R = [(n_r - 1)&sup2; + n_i&sup2;]/[(n_r + 1)&sup2; + n_i&sup2;]

    Notice how the conductivity of the metal goes to zero. Whereas in the case of a dielectric (with real indices). But why are the indices complex for metals?

    Envision the metal as a medium full of driven, damped oscillators (electrons affected by an electric field). Some of these oscillators correspond to free electrons and will have zero restoring force whereas others may be bound to the atom. The conduction electrons are however predominant. Now I did an assignment on electron oscillators being driven by an time-varying electric field. What you have is a second order non-homogeneous differential equation and when you solve it you get the solution as

    x(t) = [(q_e / m_e) / (&omega;_0&sup2; - &omega;&sup2

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    ] E cos&omega;t

    The point is the restoring force is 180 degrees out of phase with the displacement. This is unlike a dielectric where they are in-phase.

    So we have our conducting electrons oscillating out of phase with the impinging light and they tend to reradiate little wavelets that tend to cancel the incoming electric disturbances causing them to oscillate in the first place. This lead to a rapidly decaying refracted wave.

    According to quantum mechanics, an electromagnetic wave transfers energy in quantized photons E = hv. Thus E = h-bar &omega;. The spin of a photon is either -h-bar or +h-bar where the signs indicate right or left handed spin. So there are two ways a photon can be 'spun' (hehe). A photon is a spin-1 particle and can either be +1 or -1.

    Whenever a charged particle (electron in a metal) emits or absorbs electromagnetic radiation, it will undergo a change of +- h-bar in its angular momentum (as well as change in energy and momentum of course). So, like you said, there are two orthogonal polarization directions attributed to photon spin.

    Take a pure left-circularly polarized plane wave. This will impart angular momentum to its target as if all the constituent photons in the wave had their spins aligned. Remember that each photon exists in either spin state with equal likelihood.

    LINEARLY POLARIZED LIGHT - No real majority of one type of photon spin.
    LEFT CIRCULARLY POLARIZED LIGHT - Photon spin in one direction
    RIGHT CIRCULARLY POLARIZED LIGHT - Photon spin in the other direction
    ELLIPTICALLY POLARIZED LIGHT - Unequal amounts of photon spin.

    Now you probably know this anyway, but as it is I do not know enough about quantum mechanics to relate it to quantum optics to answer the part about
    but I would say that it probably wouldn't be the Poynting vector but rather the wave propagation direction vector k.

    PS. Good questions! By the way what level of education are you at?
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  3. Omnignost Registered Senior Member

    Thank you for that. I am originally a Ph. D. in organic chemistry but I have worked with Magnetic resonance physics for many years so I am very familiar with spin 1/2 quantum mechanics but not with photons. My physics is a bit spotty since I have read up on what I had to over the years. I have a comment on the last bit of your question. Spin one particles have three spin states +1, 0, -1. Logically the linearily polarized light would correspond to the 0 state. But I don't know if it is so, I'm just guessing. In an other thread, don't remember which one, there was talk about the helicity of light. I have heard about this in some other circumstances too. It seems to be lika some sort of orbital angular momentum but I fail to see how this would be possible for photons. Any ideas?
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  5. oxymoron Registered Senior Member

    A Ph.D in organic chemistry eh? The only part of chemistry I ever liked. Good choice.

    I'm not sure about that either (the linearly polarized light corresponds to 0-spin bit). Maybe someone else on the forum could shed some light on this - but I will look into it anyway.

    I think the talk was about the ellipticity of light. And I can see how this would work. If an electromagnetic wave hit the surface of an object then it would impart its energy AND linear momentum. But if the EM was circularly polarized then wouldn't the electrons be forced into circular motion.

    dE/dt = &omega;&Gamma;

    where &omega; is the angular velocity (circularly polarized light would have angular velocity) of the EM radiation and &Gamma; is the torque.

    dE/dt is the energy transferred per unit time.

    Since &Gamma; = dL/dt where L is angular momentum

    dL/dt = (1/&omega

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    L = E/&omega;

    An electron that absorbs a quantity of energy E from the circularly polarized wave will simultaneously absorb an amount of angular momentum L equal to the above equation.
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  7. Omnignost Registered Senior Member

    Yes, this is well known in atomic physics. If you irradiate rubidium vapor with circularly polarized light at the correct frequency you get spin polarized rubidium vapor that can collide with helium-3 gas. The angular momentum is then transfered to the spin1/2 nucleus of He-3. The spin polarized state lasts for hours so you can put a patient in a MR-scanner, let him breathe a puff of helium-3 and make a beautiful image of the lungs. By the way, the angular momentum can only be transfered in hbar/2 packets of course.
  8. The_Shower_Gnome Registered Member

    Geeez, I'm always amazed at the silly questions that get asked.


    Science is an illusion. It's ALL magickal little gnomes operating all your "subatomic" and "micro" thingies.

    And we steal your socks from the dryer too (not me, my brother in law, the Dryer Gnome).

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  9. oxymoron Registered Senior Member

    Not the best way to use your first ever post on this forum.
  10. MacM Registered Senior Member

    A related question:

    What is the determining factor that seperates a photon from being reflected vs absorbed and re-admitted, as in a transparancy?

    The question in particular conjures up the case of a hologram of a mirror which is created not of a normal substance but by a transparency which simply combines light waves harmonically (Interference) at a location in free space.

    Or it projects light patterns that replicate a solid, not transparent barrier. That is light seems to act the same with or without any physical substance as a medium with which it reacts.
  11. John Connellan Valued Senior Member

    reflection IS absorption and re-emission (if thats the word ur talking about

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  12. MacM Registered Senior Member

    It would appear so but how about the question regarding holograms where the light seems to acheive the affect without any substance interference but is just an interference patten of light waves?
  13. oxymoron Registered Senior Member

    I never went into holography in my optics course but if you say that a hologram is just an interference pattern of light waves, then so is any object. When we look at something, we are not seeing the object we are seeing the interference pattern that arises from the light waves reflecting off the surface and interfering with each other. If one was to make a perfect hologram then there should be no way to tell if it is a real object or not - unless you tried to interact with it.
  14. Pete It's not rocket surgery Registered Senior Member

    You don't ned to get so complicated.
    A mirror contructed from two planes intersecting at right angles will do just fine!
  15. Max Action Registered Member

    I might have missed something, but just clear something up for me quickly - how did you know that glass's "n" was 1.5? Was it assumed in order to explain the results, or determined experimentally, or other?

    [the above was going to be the whole post, but it started me thinking]

    OK - I hate to be lumped in with the "Shower Gnome," but I've got to get this off my chest. In the spirit of inquiry, not attack ...

    So it seems to me, as a layman in this domain (but no stranger to scientific thought), that when you're honest about it, all these "explanations" of "physical laws" being bandied about are nothing more or less than a bunch of familar metaphors that happen to match the available data. Are these forces and fields and what-not REALLY circlets and waves and shit with three spin states at once?

    I'm not sure what anyone here thinks about it, but I guess I don't think so ... it seems that these terms are just linguistic constructs - or models. Over-simplified and crude metaphors for the much more complex and unknown reality.

    And, frankly, I'm not impressed all that much with the mathematical expression of these models - it seems that all the mathematicians are doing is using an invented-as-needed language of "high math" to accurately model* the metaphors ("circlets," "quantum states," "wave/particles-at-onces," and even "electons") that those creative artists in the physics department have philosophized up - in order to "sensibly" organize the limited data that we're able to collect about what we perceive to be reality. Anyone know what I mean?

    So what I'm saying is - I don't know if anyone has really even SORT of answered the question. To me, Wes's interpretation still seems just as sufficient and just as supported as the ones that invoke a whole bunch of specialized concepts and visual metaphors, and are alleged to be supported by the description of these metaphors in the wildly flexible language of mathematics.

    Anyone find it worth discussing?

    *('accurately enough,' anyway - which makes sense if math, like many (every?) other field, is based upon utility rather than any semblance of "truth.")
  16. Pete It's not rocket surgery Registered Senior Member

    I'm not sure, but I can think of a whole bunch of ways to measure it.

    The most direct would be to measure the time delay for a light beam shining through a chunk of glass, and thus determine the speed of light in the glass.

    This could be done reasonably easily by splitting a laser beam and recombining it. If one of the split beams is sent through a sheet of glass, you'll get interference when the beams recombine. From the interference, you can easily calculate the time difference between the beams.
  17. Quantum Quack Life's a tease... Valued Senior Member

    Max, Whilst I understand what you are saying I can only ask you whether you can think of any other way to describe, predict and come to understand what is being observed. Is there another language or method of communication that does this job?
  18. MacM Registered Senior Member

    I agree. But the question becomes why objects cause various responses to light. i.e. transparent or opaque, etc and the refelected light patterns giving an image of the object vs creating the image of an object by creating light interference.

    In other words it seems objects are superflous it is all a function of light.
  19. Dunnoyet Registered Senior Member

    Wow. I am amazed. My head feels stretched. Yes--do continue.
  20. oxymoron Registered Senior Member

    What makes surfaces look the way they are? (rhetorical question).

    To be honest I have no idea. However, think about what happens when you touch something. We are touching a surface that is made up mostly of space, no? The distance between the atoms is massive compared to the atoms themselves. So why don't we feel these spaces? The thing that determines what a surface feels like is not actually real! It is all down to what our brain decides we should feel. The same thing applies with light. Why do metallic objects appear grey and shiny, why not blue and fuzzy? The objects themselves are superfluous I believe. It is the nature of light and the way it reacts to certain surfaces, our brain interprets these slight differences/interferences and differentiating different kinds of surfaces. Inasmuch as when light reflects off a shiny mirror, there must be something in the way that light reflects off that certain kind of atomic arrangement that triggers something in our brain which makes us realise we are looking at a shiny mirror.

    I believe too that reflections must come down to some function of light - partly due to the atomic interactions/arrangements on the surface. But I really have no idea - I'm just throwing up my thoughts for now.
  21. bradguth Banned Banned

    Photons dot NOT move, they conduct via atoms?

    I can't offer any fancy forumla, or even for that matter simple formula, as to expain this conjecture but, if you'd care to think of the photon/packet as in NOT MOVING about the country, but of instead as a soup of a great many photons as merely resting, as in trillions upon trillions/m3 (minus whatever portion of said space atoms take up), whereas such, all that's needed is the packet instructions as to whatever needs to be conducted/transfered through the photon soup of the day.

    Obviously of the more atoms within the soup (such as diamond) the slower and potentially greater loss of ennergy transpires, whereas a perfect mirror (a somewhat auto-alignment of spinning atoms) is capable of redirecting the packet instructions as to transfering the necessary FIFO message, thereby transfering the packet energy off in an another direction from the point of interacting with the atoms of that mirror.

    Since I'm not even close to being "all knowing", or even that I'm all that great at terminology or of asking important questions, as it seems I'm just continually learning all sorts of nifty stuff on the fly, such as this tid bit from "eshal", as his rendition of the relationship of the photon and atom seems quite realistic, though so does a good number of very artificial attributes as situated upon Venus.

    From: eshal (un ty@yahoo.com.au)
    (the speed of light - since that is what is spinning to form the particle)

    Spin frequency of electron = 1.236e20/sec

    "If we "load down" a photon (give it energy to increase its frequency) with the rest mass energy of an electron we get a photon with an energy of (since E=mc^2): 9.109e-31 kg x c^2 = 8.198e-14 joules. Dividing by Plank's constant we find the resonant frequency of this photon is: 8.198e14 j / 6.626e-34 j/sec. = 1.236e20/sec = photon resonant frequency."

    "The reason you haven't heard that much about this is because it completely contradicts Relativity and academia would rather believe Relativity than reality"

    Obviously there's something of worth to being kindly said about the reality of what "eshal" has to share of "photonic energy to matter", and perhaps there's an even better reality of the ever expanding coulomb zone or outer most spinning shells of influence that'll yield upon the likes of what the NEC/Wang FTL outcome has suggested, as into a "c+" packet adventure that'll eventually make for a LD call to Sirius in no time at all.
  22. dagr8n8 Registered Senior Member

    Oxymoron, your jsut so dam smart i cant even bleave it!!!!!
  23. oxymoron Registered Senior Member

    why thank you

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