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Don't understand dispersive medium

by facenian
Tags: dispersive
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facenian
#1
Jul18-11, 10:53 AM
P: 208
Hello, I have a problem understanding wave propagation in dispersive medium because the prescription to solve the wave equation for electromagnetics waves is this :
a) suppose [itex]\mu,\epsilon[/itex] are function of frequency only
b) solve the wave equation [itex]\Delta\textbf{E}-\mu\epsilon\frac{\partial^2\textbf{E}}{\partial t^2}=0[/itex], for each Fourier component
c) use superposition to get (sum each solution) to obtain the general solution,ie,
[tex]\textbf{E}(\textbf{r},t)=\int_{-\infty}^{\infty}\textbf{E}(\textbf{r},\omega)e^{-i\omega t}dw[/tex]
How can this superposition be valid since there is no linearity?, or in any case, what is the wave equation this prescribed expression is solution to?
It must be simple and clear since this is the ussual way to do it, but I'm not seeing it.
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xts
#2
Jul18-11, 11:02 AM
P: 882
Quote Quote by facenian View Post
How can this superposition be valid since there is no linearity?
Wrong! There is linearity! Maxwell's equations (also in dispersive medium) are linear in regard to E/charge.
[itex]\mu[/itex] and [itex]\epsilon[/itex] result from charges (electrons) and magnetic momenta of atoms/grains of the medium, responding to the fields calculated "in vacuum". All those responses (unless you go to very strong fields) are linear. So it is always true that sum of solutions is also a valid solution.
But (for dispersive media) they may vary with frequency at which we force them to oscilate.
You may build your final solution as a sum of any partial solutions chosen at any scheme as you like. As it is easy to compute solution for single frequency, Fourier decomposition is a convenient approach to build general solution.
chrisbaird
#3
Jul18-11, 11:52 AM
P: 617
The term "linear" is overloaded, so we have to be careful. Maxwell's equations are linear. Electromagnetic fields obey linear superposition, meaning that they sum together. But, there are materials that have a significant non-linear electromagnetic response to applied fields. These materials are rare. However, the wave equation in its text-book form requires linear materials in order to be derived. So if you see the wave equation, it is understood that linear materials is already implied.

facenian
#4
Jul18-11, 12:31 PM
P: 208
Don't understand dispersive medium

Quote Quote by chrisbaird View Post
However, the wave equation in its text-book form requires linear materials in order to be derived. So if you see the wave equation, it is understood that linear materials is already implied.
That's why I don't understand, can you tell me what is the equation for which the expression
[tex]\textbf{E}(\textbf{r},t)=\int_{-\infty}^{\infty}\textbf{E}(\textbf{r},\omega)e^{-i\omega t}dw[/tex]
is solution to?
The expression under the integral symbol is solution for the equation [itex]\Delta\textbf{E}-\mu\epsilon\frac{\partial^2\textbf{E}}{\partial t^2}=0[/itex]
where [itex]\mu,\epsilon[/itex] are function of frequency.
xts
#5
Jul18-11, 01:22 PM
P: 882
There is no simple representation of such equation, as in dispersive medium you can't write down Maxwell's equations abstracting from frequency. That is why you had to split your problem into possible-to-formulate-easy-to-solve Fourier composition of monochromatic waves.
facenian
#6
Jul18-11, 02:00 PM
P: 208
Quote Quote by xts View Post
There is no simple representation of such equation, as in dispersive medium you can't write down Maxwell's equations abstracting from frequency. That is why you had to split your problem into possible-to-formulate-easy-to-solve Fourier composition of monochromatic waves.
We only know the equation and solution for each Fourier component.
How do we justify the the superposition of such solucions is a solution when we don't even know what equation is to be satisfied?
xts
#7
Jul18-11, 03:06 PM
P: 882
We can't write down the formula, but we know the formula is linear regarding electric field strength. The unwritten formula is equivalent to Maxwell's equations in vacuum, but in presence of additional charges and currents, resulting from atoms responding to the field. Maxwell's equations in vacuum are linear. We assume now that atoms in our medium respond linearily to attached field. Thus their response is also linear, and the final unwritable equation must also be linear.

Now you'll ask why we assumed that atoms in our medium respond linearily to attached field...
That is the first order approximation, valid for most materials and light densities present in everyday life. This assumption about linearity had already been done implicitely by your teacher, who told you that [itex]\mu,\epsilon[/itex] are function of frequency (but not of E). Such model ([itex]\mu,\epsilon[/itex] depends only on frequency) results from modelling the medium as an ensamble of harmonic oscillators of various own frequencies, responding to attached field. Harmonic oscillators are linear.
chrisbaird
#8
Jul18-11, 04:01 PM
P: 617
Quote Quote by facenian View Post
That's why I don't understand, can you tell me what is the equation for which the expression
[tex]\textbf{E}(\textbf{r},t)=\int_{-\infty}^{\infty}\textbf{E}(\textbf{r},\omega)e^{-i\omega t}dw[/tex]
is solution to?
The expression under the integral symbol is solution for the equation [itex]\Delta\textbf{E}-\mu\epsilon\frac{\partial^2\textbf{E}}{\partial t^2}=0[/itex]
where [itex]\mu,\epsilon[/itex] are function of frequency.
This is the solution to the wave equation as a function of time. The integrand is the solution to the wave equation as a function of frequency. With a frequency-dependent (but still linear) material, the wave equation is a bear to solve directly. So instead we Fourier transform the wave equation into frequency space, solve that, then Fourier transform the solution back into the time domain.
facenian
#9
Jul18-11, 06:24 PM
P: 208
Quote Quote by xts View Post
This assumption about linearity had already been done implicitely by your teacher, who told you that [itex]\mu,\epsilon[/itex] are function of frequency (but not of E). Such model ([itex]\mu,\epsilon[/itex] depends only on frequency) results from modelling the medium as an ensamble of harmonic oscillators of various own frequencies, responding to attached field. Harmonic oscillators are linear.
My problem is that I think the equation is not linerar since it depends on the frequency of the field,i.e., it depends on the field itself and not simply on the time. If the coeficients in the equation depended merely on the space and time coordenates then the equation could be linear. I know I must be wrong however it does not help not knowing the equation to which the Fourier superposition is solution to.
matonski
#10
Jul18-11, 11:47 PM
P: 166
In Fourier space, you are referring to the equation
[tex]\left[ \nabla^2 +\mu(\omega)\epsilon(\omega) \omega^2 \right] \mathbf E(\mathbf r, \omega)=0[/tex]
To find out what that equation is in terms of time, you would have to take its inverse Fourier transform. Then, the functions of [itex]\omega[/itex] will become complicated functions involving time derivatives.
facenian
#11
Jul19-11, 05:08 AM
P: 208
Quote Quote by matonski View Post
In Fourier space, you are referring to the equation
[tex]\left[ \nabla^2 +\mu(\omega)\epsilon(\omega) \omega^2 \right] \mathbf E(\mathbf r, \omega)=0[/tex]
To find out what that equation is in terms of time, you would have to take its inverse Fourier transform. Then, the functions of [itex]\omega[/itex] will become complicated functions involving time derivatives.
Ok, suppose you take the inverse Fourier transform then you will get the differential equation the field satisfies. What does this equation have to do with fundamental laws of electromagnetics? It seems to me that supperposition principle have been used incorrectly, shouldn't the solution satisfy Maxwell Equations or some equation (like the wave eq.) derive from them?
xts
#12
Jul19-11, 05:25 AM
P: 882
You are confused because you mix two meanings of 'nonlinearity'

Take a mechanical example: harmonic oscillator (e.g. guitar string) driven by external force (e.g. sound from nearby loudspeaker). It is strongly nonlinear regarding frequency (response chages as the frequency of driving force changes), but (unless you go to very high amplitudes) it is linear regarding amplitude: if you apply twice bigger force (regardless of its freq. composition) the string responds with vibrations of twice bigger amplitude.

Actually that is our case, as a classical way to introduce [itex]\epsilon[/itex] of real media is to treat them as a set of harmonic oscillators: massive electric charges at the springs (see e.g. great lecture in Feynman's Lectures on Physics)
facenian
#13
Jul19-11, 06:02 AM
P: 208
Quote Quote by xts View Post
You are confused because you mix two meanings of 'nonlinearity'
Yes, I'm confused with linearity because [itex]\epsilon,\mu[/itex] does not depend on time but depend on frecuency. I guess one could argue that frecuency is a characteristic of the field and so they depend indirectly on the fields and thereby they are no longer linear. But never mind now I think I got it. My principal problem was how to jutify the method and now, thank to all of you, I realized it's just a mathematical problem: first Fourier transform the wave equation then solve it in the frequency domain and finally Fourier transform back to the time domain.
Thank you folks!
chrisbaird
#14
Jul19-11, 09:33 AM
P: 617
The term "linearity" in electrodynamics typically refers to the fact that there is a linear material response between the induced and total field:

D = ε E and B = μ H

as opposed to a nonlinear response:

D = ε1 E + ε2 |E|2 x+ ... and B = μ1 H + μ2 |H|2 x + ...

The linearity of the material is what allows us to create the standard wave equation in the first place.

The term "dispersive" is what we use to describe a material that has a permittivity that is significantly non-constant as a function of frequency:

D = ε(ω) E and B = μ(ω) H

Dispersive materials can still be linear, and thus propagate waves according to the standard wave equation. The frequency-dependent response means that different frequency components of a wave packets travel at different speeds in the material. This causes a wave packet to disperse, or spread out.

Physically speaking, the wave equation as a function of time, and the wave equation as a function of frequency are the same fundamental equation, directly deduced from Maxwell's equations. They are just different mathematical representations.


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