Boundary conditions and time domain electromagnetic waves

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Boundary conditions & time domain electromagnetic waves: does classical model fit?

Consider two propagating media: a lossy dielectric medium and a lossless dielectric medium. Thus, the interface that separates them has two tangential components of electric field, one for each medium. One of them, the component of the lossy medium, decays with time, say, ET1 = C * exp (-a * t) * sin (b * t). The other one, by being in a lossless medium, does not decay with time: let's say ET2 = D * sin (d * t), for example.

Classical electromagnetism states that the tangential electric field must be continuous at the interface of separation the two media, ie: C * exp (-a * t) * sin (b * t) = D * sin (d * t) for all t> 0. However, these two equations will never equal for all t due to the factor exp (-a * t), present in Et1 and absent in ET2.

Conclusion: there will never be transmission of electromagnetic energy between two media.

Obviously I am wrong. But where?
 
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  • #2
Andy Resnick
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I wonder if the problem is that the loss term should go as (for a wave propagating in 'z') exp (-az) instead of exp(-at). Then there's no problem.

Having a loss term like exp(-at) is a way to introduce resonance in a damped oscillator, but I don't recall seeing that in the context of classical E&M.
 
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Ok, I will explain the problem a little deeper.

In frequency domain, it is usual to analyse a lossless medium by solving maxwell wave equation assuming no sources of electric or magnectic field. This is fine; for a lossless medium.

The problem arrive when dealing with lossy media. Consider solving wave equation for a circular cavity which has lossy inner dielectric (finite conductivity). Of course, for a medium with sources, there is no damped vibration in time, since it is possible to design the source in a way it compensates the energy losed due to Joule Effect on inner dielectric. However, let's consider the analysis without sources.

In the case of inner lossy medium without sources of electromagnectic field, the wave amplitude MUST decay in time, since the system is losing electric / magnectic energy (again, because of Joule effect). In this case, if we try to solve wave equation in frequency domain, we will find that there are no solutions with real number ressonant frequencies for the circular cavity. It's obvious! Since frequency domain assumes undumped waves (exp{jwt}), it will never solve our problem! In fact, the ressonant frequencies we find by solving the problem in frequency domain are complex numbers; then exp{jwt} = exp{j*(wr + j*wi)*t} = exp{(-wi + j*wr)*t}, that is, a dumped wave!!

Therefore, the certain way to solve this problem is using time domain Maxwell wave equations. In time domain, the factor exp{-at} appears. Furthermore, if we have two media, one of them being lossy (finite conductivity) and the other one being lossless, there is no way to apply tangential electric (or tangential magnectic) bondary conditions (1st post of this topic). Why?
 
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  • #4
Andy Resnick
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I'm a little confused by your post. As best I know, the wave equation for a lossy medium is something like:

[tex]\nabla \times \nabla \times E = -\mu \frac{\partial J}{\partial t} - \mu\epsilon \frac{\partial^{2} E}{\partial t ^{2}}[/tex]

And then substituting Ohm's law J = [itex]\sigma[/itex] E :

[tex]\nabla \times \nabla \times E = -\mu \sigma \frac{\partial E}{\partial t} - \mu\epsilon \frac{\partial^{2} E}{\partial t ^{2}}[/tex]

I'd have to look up the solution in a book- but I know there is an analytical solution. Matching this solution to the source-free solution:

[tex]\nabla \times \nabla \times E = - \mu\epsilon \frac{\partial^{2} E}{\partial t ^{2}}[/tex]

at a boundary should be straightforward, have you tried?

Am I misunderstanding your post?
 
  • #5
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actually, equation is:

[tex]\nabla^{2} E = \mu \sigma \frac{\partial E}{\partial t} + \mu\epsilon \frac{\partial^{2} E}{\partial t ^{2}}[/tex]

Suppose there is only [itex] E_{x} [/itex], so:

[itex] E = \hat{a}_{x} E_{x} [/itex]

and equation can be rewriten as

[tex]\nabla^{2} E_x = \mu \sigma \frac{\partial E_x}{\partial t} + \mu\epsilon \frac{\partial^{2} E_x}{\partial t ^{2}}[/tex]

This equation is separable in cartesian, cylindrical, spherical and some other coordinate systems. If we choose [tex] E_x = f(x)g(y)h(z)T(t) [/tex] and rewrite this equantion, we will find that:

[tex] \mu \sigma \dot{T} + \mu \epsilon \ddot{T} = -\tau^{2} T \quad \tau \in \mathbb{R} [/tex]

The solution is a damped wave on time. The same procedure can be applied for the lossless medium, giving a sinusoidal wave for T(t). Now suppose the boundary is at z = 0. Therefore, the tangential component of the electric field is [tex] E_x [/tex] and boundary conditions state that, for both media, they should equal. However, T1 will never be equal to T2 for all T, because T1 is dumped and T2 is not.
 
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  • #6
Andy Resnick
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actually, equation is:

[tex]\nabla^{2} E = \mu \sigma \frac{\partial E}{\partial t} + \mu\epsilon \frac{\partial^{2} E}{\partial t ^{2}}[/tex]
Hang on- [itex]\nabla \times \nabla \times E = \nabla(\nabla \bullet E) - \nabla^{2} E[/itex]. This can be reduced to [itex]\nabla \times \nabla \times E = \nabla\rho - \nabla^{2} E[/itex].

Since we are allowing mobile charges, I can see that the charge density gradient will not be zero, and could allow a discontinuity in the slope of E.
 
  • #7
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Hang on- [itex]\nabla \times \nabla \times E = \nabla(\nabla \bullet E) - \nabla^{2} E[/itex]. This can be reduced to [itex]\nabla \times \nabla \times E = \nabla\rho - \nabla^{2} E[/itex].

Since we are allowing mobile charges, I can see that the charge density gradient will not be zero, and could allow a discontinuity in the slope of E.
You are missing the part that I am assuming a source free media, hence no mobile charges allowed.
 
  • #8
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actually, equation is:

[tex]\nabla^{2} E = \mu \sigma \frac{\partial E}{\partial t} + \mu\epsilon \frac{\partial^{2} E}{\partial t ^{2}}[/tex]

Suppose there is only [itex] E_{x} [/itex], so:

[itex] E = \hat{a}_{x} E_{x} [/itex]

and equation can be rewriten as

[tex]\nabla^{2} E_x = \mu \sigma \frac{\partial E_x}{\partial t} + \mu\epsilon \frac{\partial^{2} E_x}{\partial t ^{2}}[/tex]

The solution can still be written as
[tex] E(r,t)=E_0 e^{i(kr-\omega t)} [/tex]
with [tex] k^2=\mu\epsilon\omega^2 + i \mu\sigma\omega [/tex]

Whoose Im part is
[tex] Im{k}=\omega\sqrt{\epsilon\mu}(\sqrt{1+\sigma^2/\epsilon^2\omega^2}-1) /2 [/tex]
responsible for the damping inside the metal.
 
  • #9
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The solution can still be written as
[tex] E(r,t)=E_0 e^{i(kr-\omega t)} [/tex]
with [tex] k^2=\mu\epsilon\omega^2 + i \mu\sigma\omega [/tex]

Whoose Im part is
[tex] Im{k}=\omega\sqrt{\epsilon\mu}(\sqrt{1+\sigma^2/\epsilon^2\omega^2}-1) /2 [/tex]
responsible for the damping inside the metal.
Sure it can, but it doesn't help answering my question. Diving a little deeper in book "Advanced Engineering Electromgnetics" - Constantine A. Balanis, I ran into the continuity equation, which states:

[tex] \nabla \cdot (J_i + \sigma E) = -\dot{q_{ev}}[/tex]. Therefore, since [itex]\sigma[/itex] is not null, there must be a charge density in the boundary separanting the 2 media, hence, allowing the discontinuity. Andy Resnick was right.
 
  • #10
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I thought the problem was to get the wave's decay correctly through k (and not ω).

What exactly is / was your question ?
 
  • #11
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I was wrong, the problem is yet not solved.

The question is about boundary conditions in the interface of separation of 2 media: one is lossy and the other is lossless.

In the lossless media, the wave is non-decaying. In the lossy media, the wave is decaying in time.
Since I am assuming no magnectic current sources, [itex]E_{1t} = E_{2t}[/itex], that is, the tangential electric field at the boundary MUST be continuous. However, since the medium 1 is lossy, electric field [itex]E_{1t}[/itex] is decaying with time and, therefore, will never equal [itex]E_{2t}[/itex] (not decaying with time).

This is wierd! And I just don't know where I am wrong.
 
  • #12
Andy Resnick
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You are missing the part that I am assuming a source free media, hence no mobile charges allowed.
Then why are you keeping the current J?
 
  • #13
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Then why are you keeping the current J?
I'm keeping a conduction current, not a source current. There's a difference. Anyways, you are wrong: The tangential electric field is discontinuous if, and only if, there is an impressed magnectic current, which is not the case. It has nothing to do with electric charge density. Actually, boundary conditions state that: [itex]n \times (E_1 - E_2) = M_i[/itex]. Therefore, nothing to do with free electric charges.

I still think the classical model can do with this. Actually, a discontinuity in conductivity is not physcally realizable. What happens in reality is that, when we put 2 materials together, the boundary contains a mix of molecules of both media that interact together to create a smooth (continuous and differentiable) transition of medium conductivity in space.
 
  • #14
Andy Resnick
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I'm keeping a conduction current, not a source current. There's a difference. Anyways, you are wrong: The tangential electric field is discontinuous if, and only if, there is an impressed magnectic current, which is not the case. It has nothing to do with electric charge density. Actually, boundary conditions state that: [itex]n \times (E_1 - E_2) = M_i[/itex]. Therefore, nothing to do with free electric charges.

I still think the classical model can do with this. Actually, a discontinuity in conductivity is not physcally realizable. What happens in reality is that, when we put 2 materials together, the boundary contains a mix of molecules of both media that interact together to create a smooth (continuous and differentiable) transition of medium conductivity in space.
You really lost me here.... in any case, my goal was to help you resolve your conceptual difficulty, so I'm glad that happened.
 
  • #15
Born2bwire
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And why wouldn't the field in the lossless medium not be decaying with time? If the incident field is coming from a lossy medium that is decaying in time, then obviously the transmitted field would also be decaying in time since the incident field is gradually losing its amplitude.
 
  • #16
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And why wouldn't the field in the lossless medium not be decaying with time? If the incident field is coming from a lossy medium that is decaying in time, then obviously the transmitted field would also be decaying in time since the incident field is gradually losing its amplitude.
You have a point. However, since I'm assuming no sources, it has no meaning in saying "the incident field is coming from a lossy medium". Moreover, a decaying wave cannot solve Maxwell equation [itex] \nabla^2 E_x = \mu \epsilon \partial^2 {E}_x / \partial t^2[/itex] - the equation for the lossless media.
 
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You really lost me here.... in any case, my goal was to help you resolve your conceptual difficulty, so I'm glad that happened.
Your help was extraordinary. Thankx!
 
  • #18
Andy Resnick
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I was able to get my copy of Jackson this morning- in my edition (2nd), section 7.7 is devoted to waves in lossy media. To summarize, the dielectric constant is separated into real and imaginary components [itex]\epsilon[/itex] and [itex]i \sigma[/itex]. The wavenumber k is also then complex, becoming (for a poor conductor)

[tex]k = \sqrt{\mu\epsilon}\frac{\omega}{c}+ i \frac{2\pi}{c}\sqrt{\frac{\mu}{\epsilon}}\sigma[/tex]

Which I'll condense to [itex]k = \kappa + i \alpha[/itex].

The electric field is

[tex]E = E_{0} exp[-\alpha n\bullet x] exp[i(\kappa n\bullet x) -\omega t)] [/tex]

So the boundary condition between lossless (real dielectric constant) and lossy (complex dielectric constant) is still continuous for all times t.

Jackson references Stratton's "Electromagnetic Theory" for a detailed discussion of reflection and refraction and interfac4es between lossy and lossless media.
 
  • #19
Born2bwire
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You have a point. However, since I'm assuming no sources, it has no meaning in saying "the incident field is coming from a lossy medium". Moreover, a decaying wave cannot solve Maxwell equation [itex] \nabla^2 E_x = \mu \epsilon \partial^2 {E}_x / \partial t^2[/itex] - the equation for the lossless media.
It has to be coming from somewhere. Either it is coming from the lossy to the lossless, from the lossless to the lossy or both. It doesn't matter that you have removed your sources, there is an inherent causality that you cannot ignore.

But why would the wave be decaying INSIDE the lossless medium? When the wave is travelling inside the lossy medium, it decays. When it is in the lossless medium, it does not decay. You ask then how does the wave in the lossless medium demonstrate the effects of the decay? That is done by the fact that any wave that impinges onto the lossless medium from the lossy medium was decaying as it travelled through the lossy medium. Thus the transmitted field is weaker than any transmissions from transmissions at an earlier time (assuming that the incident waves have sourced at a common point in time and thus the later time indicates a longer path of transmission through the lossy medium). The transmission coefficient is not going to give rise to any loss (power is conserved), the loss is inherent due to the wave travelling through the lossy material in the first place.

Perhaps it would be better if you corrected your expressions for your waves. We do not express a wave as decaying as exp(-at). We express a wave as decaying over its spatial phase variance exp(ikr) if we assume time-harmonic waves with time dependence exp(-i \omega t). The material properties, \epsilon and/or \mu, contain the lossy terms in the form of imaginary parts and give rise to a spatially dependent loss term (this was by the squeezed state's post above). So, let us separate the real and imaginary parts of the wave vector so that,

[tex] k = \alpha+i\beta [/tex]
[tex] \mathbf{E}_1 (\mathbf{r}) = \mathbf{E}_1 e^{i\alpha_1\hat{k}_1\cdot\mathbf{r}} e^{-\beta_1\hat{k}_1\cdot\mathbf{r}} [/tex]
[tex] \mathbf{E}_2 (\mathbf{r}) = \mathbf{E}_2 e^{i\alpha_2\hat{k}_2\cdot\mathbf{r}} [/tex]

We can see, assuming our boundary lies along the x-y plane where z=0 and that the direction of propagation is along the x-z plane (k_y = 0), that

[tex] E_{x1} e^{i\alpha_1 k_{x1} x} e^{-\beta_1 k_{x1} x} = E_{x2} e^{i\alpha_2 k_{x2} x}} [/tex]
[tex] E_{y1} = E_{y2} [/tex]

are the constraints upon the tangential electric field from our boundary conditions.

EDIT: It appears that Andy has beat me to the punch as I was typing this. I have also taken the inspired choice towards added confusion by choosing alpha as my real part of the wave number while Andy has used it for his imaginary part.
 
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  • #20
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Perhaps it would be better if you corrected your expressions for your waves. We do not express a wave as decaying as exp(-at). We express a wave as decaying over its spatial phase variance exp(ikr) if we assume time-harmonic waves with time dependence exp(-i \omega t). QUOTE]

I'm not assuming time harmonic waves, because I need to "see" the transients. That's why I'm working on time domain.
 
  • #21
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I was able to get my copy of Jackson this morning- in my edition (2nd), section 7.7 is devoted to waves in lossy media. To summarize, the dielectric constant is separated into real and imaginary components [itex]\epsilon[/itex] and [itex]i \sigma[/itex]. The wavenumber k is also then complex, becoming (for a poor conductor)

[tex]k = \sqrt{\mu\epsilon}\frac{\omega}{c}+ i \frac{2\pi}{c}\sqrt{\frac{\mu}{\epsilon}}\sigma[/tex]

Which I'll condense to [itex]k = \kappa + i \alpha[/itex].

The electric field is

[tex]E = E_{0} exp[-\alpha n\bullet x] exp[i(\kappa n\bullet x) -\omega t)] [/tex]

So the boundary condition between lossless (real dielectric constant) and lossy (complex dielectric constant) is still continuous for all times t.

Jackson references Stratton's "Electromagnetic Theory" for a detailed discussion of reflection and refraction and interfac4es between lossy and lossless media.
That's perfect! for a time harmonic wave. Take a closer look, and you'll find out that your waves are still time harmonic (exp{jwt}). My question is about time domain because I am working with transients. Books assume every wave in the world can be represented by a fouries series expanded using [itex]e^{jnwt}[/itex] bases, which is not true for time decaying waves. Actually, only functions with compact support can be expanded in this way. Transients like [itex]e^{-at} \quad t>0[/itex] have no compact support, and, therefore, cannot be expressed by a fouries series of exp{jwt}. To do so, one has to expand using Legendre polynomials as bases (take a look at Fourier-Legendre series) or solve on time domain. In time domain, boundary equations just don't fit and I still don't have an answer why...
 
  • #22
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Ok, i think I am not explaining myself clearly. I tried to simplify the problem, but I'll post it entirely here, because I need help. It's for my msc program.

I have a structure made by coaxial cylinders of radius a < b. The cylinders are PEC material (perfect electrically conductor). Moreover, the cylinders are semi infinite, but from z = 0 to z = -L (L = length > 0) there's a lossy dielectric between them. For z < -L to z = -infinity, there is a perfect dielectric. At the plane at z=0, there is a closing shield made of PEC material.

I want to find equations for TM waves inside both dielectrics. Lets start out using time harmonic waves.

The boundary conditions, writen in cylindrical coordinates, are:

1) at z=0, [itex]E_{\phi1} = E_{\rho1} = 0[/itex] for medium 1
2) at [itex]\rho = a, \rho=b \quad E_z=E_{\phi}=0[/itex] for both media
3) at z=-L, [itex]E_{phi1} = E_{\phi2}\quad E_{\rho1}=E_{\rho2} [/itex] (continuous tangential electric field)
4) at z=-L [itex] H_{\phi2} - H_{\phi1} = J_{superficial} [/itex]
5) at z=-L [itex] \epsilon_2 E_{z2} - \epsilon_1 E_{z1} = q_{superficial}[/itex] (q is the electric charge density at z=-L)

Are we ok with the boundary conditions?
 
  • #23
Andy Resnick
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That's perfect! for a time harmonic wave. Take a closer look, and you'll find out that your waves are still time harmonic (exp{jwt}). My question is about time domain because I am working with transients. Books assume every wave in the world can be represented by a fouries series expanded using [itex]e^{jnwt}[/itex] bases, which is not true for time decaying waves. Actually, only functions with compact support can be expanded in this way. Transients like [itex]e^{-at} \quad t>0[/itex] have no compact support, and, therefore, cannot be expressed by a fouries series of exp{jwt}. To do so, one has to expand using Legendre polynomials as bases (take a look at Fourier-Legendre series) or solve on time domain. In time domain, boundary equations just don't fit and I still don't have an answer why...
Ok- now I am wondering if you are confusing boundary conditions and initial conditions.

Let's go back to the (1-D) original expression:

[tex]k^{2} \frac{\partial^{2} E}{\partial x ^{2}} = \mu \sigma \frac{\partial E}{\partial t} + \mu\epsilon \frac{\partial^{2} E}{\partial t ^{2}} [/tex]

This is the same expression for a free (transverse) vibration of a stretched damped string, and it is solved the same way. That is, it's a damped oscillator, with a damped response that goes as exp (-at).

On the boundary, however, the electric field must be continuous- so the lossless medium acts as a driving force for the damped oscillator (assuming the field in the lossless medium is a constant E exp(iwt)). Then the lossy medium acts as a driven damped oscillator.

This gets interesting when resonant effects are considered (for example, the atoms in a laser cavity can be modeled as a damped driven oscillator).

If the lossy medium is excited with an impulse- a short flash of light- the material response will be the same as a plucked string. Again, if you want to consider the transfer of energy into and out of the lossy medium, the electric field at the boundary will always be continuous, and the lossy medium will leak some power out into free space.

Does that help?
 
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  • #24
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actually, confused me a lot!

Andy, my apologies, but a damped wave cannot be a solution for wave equation in the lossless media. Its math now, not physics. How can a dumped wave be a solution for [itex] \partial^2 E / \partial x^2 = \mu \epsilon \partial E / \partial t [/itex]
 
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  • #25
Andy Resnick
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actually, confused me a lot!

Andy, my apologies, but a damped wave cannot be a solution for wave equation in the lossless media. Its math now, not physics. How can a dumped wave be a solution for [itex] \partial^2 E / \partial x^2 = \mu \epsilon \dot{E} [/itex]
Eh?

[tex]\frac{\partial^{2} E}{\partial x ^{2}} = \mu \sigma \frac{\partial E}{\partial t} + \mu\epsilon \frac{\partial^{2} E}{\partial t ^{2}} [/tex]

is for a wave in a *lossy* medium- that's why there's the extra [itex]\mu \sigma \frac{\partial E}{\partial t}[/itex] term. In the lossless medium, [itex]\frac{\partial^{2} E}{\partial x ^{2}} = \mu\epsilon \frac{\partial^{2} E}{\partial t ^{2}} [/itex].

The damped wave is only in the lossy medium.
 

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