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PHYSICAL explanation of light reflection by metals 
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#1
Jan114, 05:30 PM

P: 41

Electric field of an electromagnetic wave incident on a metal accelerates free electrons, and these accelerating electrons then emit radiation opposite to the electric field of the wave, thus reducing the resultant amplitude in the forward direction. Radiation emitted in the backward direction is reflected light.
Now, this will only happen if the frequency of the EM wave is lower than the plasma frequency ω_{f} of the metal, the metal being transparent to a wave of frequency greater than ω_{f}. The reason given everywhere is that the index of refraction of the metal is imaginary for ω < ω_{f} and real for ω > ω_{f}. In my opinion, that is hardly a physical explanation. I understand what plasma oscillations are and I can derive the equation for ω_{f}. I just don't understand physically why the phenomenon described in my first paragraph stops happening for ω > ω_{f}. I have read that for high frequencies the electrons have trouble following the electric field, which makes sense, but I'd like to understand why the plasma frequency is the critical frequency. Thanks in advance. 


#2
Jan114, 06:31 PM

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Go back one step and focus on the basic physics of the system. The free electrons behave like a sinusoidally forced, linearly damped, simple harmonic oscillator. See for example the following link for a reminder on how it works:
https://www.csupomona.edu/~ajm/classes/phyXXX/dho.pdf Now have a look at the frequency response of the amplitude and phase. You have a constant amplitude response at low frequencies, a huge response at resonance and at high frequencies you get an amplitude that vanishes as [tex]\frac{\omega_{resonance}^2}{\omega^2}.[/tex] So the amplitude of the response vanishes at high frequencies. The resonance frequency is governed by the strength of the restoring force on the one hand and the amount of inertia (mass) on the other hand. Low mass gives you a higher resonance frequency. Also have a look at the phase response. While it is zero for low frequencies, you get a nonzero phase response for frequencies above resonance frequencies. This is what you see for the electrons in a metal, too. 


#3
Jan214, 01:09 AM

P: 41




#4
Jan214, 02:54 AM

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P: 1,641

PHYSICAL explanation of light reflection by metals
Ok, that was a bit brief so let me elaborate. You are right that acceleration is the important thing here as accelerated charges radiate. Acceleration is the second derivative of position. If the driving force is proportional to sin(omega t), so will be the position response. The velocity response will be proportional to cos (omega t) (first derivative) and the acceleration will be proportional to sin(omega t) (second derivative), which gives you the phase you are looking for.



#5
Jan214, 09:07 AM

P: 41

OK I need to be more precise. Let's say that at a given moment the driving force on the electron is upwards. It means that the electric field of the EM wave propagating to the right at the position of the electron is downwards at this moment, since the electron is negative. Like you said, the acceleration of the electron is downwards (180° phase with driving force), but a negative charge that accelerates downwards produces a transverse electric field (radiation) towards the right that is also downwards! We get constructive interference with the electric field of the wave, but we expect destructive interference! Thank you for your help.



#6
Jan214, 10:55 AM

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P: 1,641

Oh, sorry. I completely misinterpreted your question and thought you were not too familiar with the description of the driven harmonic oscillator and wanted to give a simple example so you can figure out the equivalent for your situation yourself. My wording was chosen pretty badly, especially using the word driving force nad my comment on the relative importance of acceleration. Sorry.
What I was intending to say: Start with the case of an oscillator that is not externally driven. Here, the restoring force is the only important one. This one is always out of phase with displacement. Now have a look again at the equations in the link I gave you, especially equation (3). Without driving and damping you will get an acceleration out of phase with displacement. Now consider what happens if you add the driving term. You still keep the term out of phase with displacement, but have another sinusoidally varying term on top of it. Now consider the case of moderate to strong driving and have a look at what happens to the displacement and acceleration terms and the relative phase between them. Sorry for being somewhat cryptic, but it is kind of the spirit of these forums to help people find a solution than just to give them the solution directly. People just tend to understand things better that way. Once you have done that, reconsider all the minus signs and you should arrive at the correct solution. edit: Just to make it (hopefully) a bit more clear. Consider some static electromagnetic field. Do you agree that it will just be screened by the electrons following the field until it is compensated? This is the low frequency response. Now evaluate what happens for slow oscillations and what it means when the acceleration starts to become out of phase with the displacement from the simple static point of view. 


#7
Jan214, 07:59 PM

P: 41

If I apply the driven harmonic oscillator model to an electron bound in an atom of a dielectric (let's say glass) having a resonance frequency higher than the frequency of the incident EM wave, it works alright. I get the acceleration in phase with the electric field of the EM wave, therefore the transverse electric field emitted by the electron will interfere constructively and we have transmission.
This means that the driven harmonic oscillator model must be different in the case of a free electron in a metal of plasma frequency higher than the frequency of the incident EM wave, because we need destructive interference! I can't find where the extra minus sign goes! 


#8
Jan314, 07:14 AM

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P: 1,641

These will give rise to radiation current and displacement current. The importance of the latter is a critical thing in antennas/dipoles and similar stuff, although there is no real flow of charges involved in the displacement current. 


#9
Jan314, 04:43 PM

P: 41

Here is what I could understand, but it does not sound right. If a low frequency pulse of electromagnetic wave is incident on a block of metal, the free electrons will move almost in phase with the electric force and the polarization will cancel a good portion of the electric field of the wave. While the electrons move, they emit a small transverse electric field (radiation due to acceleration) almost in phase with the electric field of the EM wave. So inside the metal, the net electric field is almost zero, but the EM wave will emerge on the other side of the metal with a greater amplitude. So the block of metal does not shield the EM wave for an observer behind... A part of the energy of the field that is "missing" inside the block, since the field is almost zero there, could go in the reflected wave.



#10
Jan414, 12:12 AM

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#11
Jan414, 10:43 AM

P: 41

First I want to thank you for your patience.
By low frequency I just meant a frequency lower than the plasma frequency. Thus for a typical metal the low frequency could be visible light. The visible light reflected by the block of metal is the radiation emitted backwards by the electrons oscillating with the field of the EM wave. But those electrons emit the same radiation in the forward direction. Now when the incident wave arrives on the other side of the block, it is no longer cancelled by the polarization, so it is added to the radiation emitted in the forward direction by the electrons. Since this radiation is almost in phase with the EM wave, the amplitude has to increase, no? Let me rephrase my question. Let's suppose we have a plane EM wave of visible light travelling to the right in vacuum towards an observer. At time t=0 we put a sheet of metal perpendicular to the direction of the wave between the observer and the source. What would the observer detect at time t = d/c, where d is the distance between the sheet and the observer? I thought the answer was nothing (if the sheet is thick enough). Because of the principle of superposition, that would mean that an EM wave out of phase with the original wave had to be emitted by the sheet, but this is not what the driven oscillator model gives me! Is it possible that a simple sheet of metal cannot shield an EM wave? If instead of an observer behind a sheet of metal we have an observer inside a metal box, then I understand that the EM wave would be canceled by the polarization. 


#12
Jan414, 11:16 AM

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#13
Jan414, 11:23 AM

P: 41

http://www.feynmanlectures.caltech.edu/I_28.html 


#14
Jan414, 11:59 AM

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http://en.wikipedia.org/wiki/Displacement_current 


#15
Jan514, 02:34 PM

P: 41




#16
Jan614, 04:41 AM

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P: 1,641

Not necessarily. You need accelerating charges, but not necesserily electrons.
Sorry, I am not good at explaining. So here is another try on how to look at it. Consider the usual em wave equation: [tex]\frac{\epsilon_r}{c^2}\frac{\partial^2 E(r,t)}{\partial t^2}=\nabla^2 E(r,t).[/tex] You can verify that this has the standard propagating wave solution [tex]E(r,t)=E(\omega) e^{i (kr\omega t)} [/tex] if [tex]\omega^2 \epsilon_r=c^2 k^2[/tex]. Now [tex]\epsilon_r=1+\chi=\frac{N p(\omega)}{\epsilon_0 E(\omega)}[/tex], where p is the polarization/dipole moment. As this is directly opposed to the driving field (as discussed earlier), that is just: [tex]p(\omega)=\frac{e^2}{m \omega^2} E(\omega)[/tex]. That gives: [tex]\epsilon_r=1\frac{N e^2}{\epsilon_0 m \omega^2}[/tex] or simply: [tex]\epsilon_r=1\frac{\omega_p^2}{\omega^2}.[/tex] This is the dispersion relation for plasmons. As you see, you get a problem at frequencies below the plasma frequency. The epsilon will be negative, k will be imaginary and the wave cannot propagate. Instead you get a field that decays exponentially. Still there are valid solutions of Maxwell's equations for low frequencies. As the field is exponentially decaying, you need to consider the surface rather than the bulk and treat the system as a dielectricmetal interface. From there on you can use the Fresnel equations, but note that it is quite messy as the quantities of interest are now complex. 


#17
Jan814, 01:12 PM

P: 41

Your explanation is clear and makes sense, but I am looking for an explanation in terms of free electrons interacting with fields and producing their own. I wanted to avoid mathematical definitions like susceptibility, dielectric constant, or even index of refraction...
I see a little better now. In summary: First, what does a free electron in a metal do when a EM wave of frequency ω passes it? If we assume Drude model, if ω is small, collisions between electron and ions will happen frequently enough during a cycle, and electron velocity will be in phase with the electric force (thus 180° out of phase with the electric field of the EM wave). This is what happens for a static electric field, which is a limiting case (Ohm's law J = σE then prevails). If ω is high, collisions will be rare (average of none per cycle) and we can ignore damping in the equation of motion for the electron, and we get as a solution that the position of the electron will be 180° out of phase with the electric force (thus in phase with the electric field). The electron velocity will then be 90° out of phase with the electric field of the EM wave. Also, we find that the velocity decreases with frequency. Second, what radiation is emitted by an infinite plane of free electrons oscillating in phase with the same amplitude on an axis perpendicular to the plane? We start with the equation giving the transverse electric field emitted by an electron, which is proportional to its acceleration, and we integrate. The integral is hard but the result is interesting: the transverse electric field at a given moment is parallel to the velocity of the electrons at that moment and its amplitude is proportional to the velocity. So if ω is low the electrons velocity is 180° out of phase with the electric field of the wave, so the radiated electric field is also out of phase and destructive interference occurs in the forward direction (radiation emitted backwards is reflected light). The plane of electrons has thus shielded an observer behind from the source. But if ω is high, the radiated electric field and the electric field of the EM wave will be 90° out of phase, but the radiated electric field will be small since velocity of electrons decreases with frequency. Thus the amplitude will not vary much and the observer behind the plane will not be shielded from the source. However, the interference will result in a phase shift, which will be interpreted by the observer as a slowing down of the wave (this is also what happens in dielectrics). What I want to know is: what determines if ω is low or high? I expect plasma frequency to be part of the answer, but I cannot see how. 


#18
Jan814, 02:01 PM

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P: 1,641

Frequencies lower than this resonance frequency are low and those higher are high. 


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