Lecture 3d -- Lossy Dielectrics

This lecture discusses electromagnetic waves in lossy dielectrics and the meaning of the wave various wave parameters when they are made to be complex.

[@inboccaallupo14]

I am now convinced that these are the best videos to learn electromagnetics. I have only one doubt: When we have very high conductivity this means if the wave passes through it will get very fast attenuated by the conductive losses, but since there is a reflection from this lossy medium the wave actually doesn't experience the losses. Firstly, why does it apply more for lower frequencies? Also, in general, we get reflection when there is an impedance change in the medium, so does it mean that we assume we enter from a medium with a particular impedance to a highly conductive material implying a different impedance ( assuming we have almost all reflected the difference should be big). So is conductivity proportional to the characteristic impedance of the medium ( if this is possible to answer with these complicated equations)? Lastly, I can not express how grateful I am for these lectures - enormous knowledge with a very intuitive and straightforward approach.

[@empossible1577]

It is great to hear you are getting a lot out of the videos! Thank you!! I think the best way to understand the frequency response of conductivity is through the Drude model. This starts by modeling charges at the atomic scale like a mass on a spring. This is called the Lorentz oscillator model. In conductors, charges are free so the electrostatic restoring force is set to zero and the Lorentz model reduces to the Drude model. As frequency goes to zero, the dielectric properties approach infinity. To learn about this, along with lots of visualizations, checkout Topic 2 in “21st Century Electromagnetics.” Here is a link to that course: empossible.net/academics/21cem/ Impedance is a complex number because it relates the amplitude and phase of the electric and magnetic fields. Without loss (i.e. conductivity), impedance would be purely real. So conductivity absolutely affects impedance. You are watching the correct video about this. See slide 15. You will see conductivity affects both magnitude and phase of impedance. While on this subject, let me point you to the official course website where you can download the notes, get links to the videos and other learning resources. The notes are ahead of the videos in terms of revisions, corrections, and improvements. You are watching a video in Topic 6. Here is the course website: empossible.net/academics/emp3302/

[@inboccaallupo14]

@@empossible1577 Thank you for the response, I will check the resources and try to gain enough knowledge and intuition before I proceed with the online courses. Again I am truly grateful for your effort and dedication to create these videos and also for sharing your astonishing knowledge and experience!

[@versatile_gentlemen]

It seems there are very few people in this world who are excellent teachers as well as researchers. And you are one of those. Thank you for sharing these lectures.

[@empossible1577]

Thank you!!!

[@karielf8947]

Excellent course, thanks a lot. very clear and well summurized

[@empossible1577]

Thank you!

[@retefotonica]

You are amazing! I understood everything I've been confused about

[@empossible1577]

Awesome!

[@isee.]

Beautiful! Thank you so much! Professor Rumpf!

[@ManojKumar-cj7oj]

Your channel + your notes are a blessing for every graduate student. This shows your hardwork and creativity . You Will get a special place in heaven 😁❤️❤️❤️❤️❤️❤️

[@empossible1577]

Thank you!

[@salmagamal5676]

this is extremely useful! thank u

[@khaledalhorani7965]

I've reached to the point that I like your video even before watching it :D

[@empossible1577]

Thank you!!

[@trejkaz]

I'm curious about how Snell's Law is interpreted when the refractive index is complex. I've heard of three separate interpretations so far: 1. You ignore the imaginary part and use just the real part when calculating the angle of transmission, then use the imaginary part only for calculating decay. 2. You do all the maths keeping the complex values, get a complex angle out, and then somehow derive the real angle from that. 3. You use the imaginary part (and the angle of incidence) to manipulate the _effective_ refractive index in the material, and then use that to compute everything. And presumably you still get to choose whether to do #1 or #2 after doing this. So maybe this makes 4 possibilities; or maybe one of these possibilities turns out the same as the maths you'd get doing #2.

[@empossible1577]

Great question. My group actually just submitted a paper yesterday about scattering at a complex interface. We consider complex permittivity, complex permeability, loss, gain, negative index, positive index, negative impedance and positive impedance. Snell's law, law of reflection, Fresnel equations, and power conservation is all handled. Multiple ways can be made to work for Snell's law, but I think the simplest is to just use regular Snell's law with complex angles. Getting the angle of the ray from this is not as easy as just taking the real or imaginary part. I recently created a video on complex angles that explains and gives equations about how to determine the actual angle. Here is a link: kzbin.info/www/bejne/l36VmGSAZq9ritE

[@engineeres9283]

u r z best sir

[@Astartos22]

Thanks so much for this video, it was very helpful. Sorry if I’m misunderstanding but does this mean that for a lossless material, there will never be dispersion? Are loss and dispersion intrinsically connected?

[@empossible1577]

All materials (not vacuum) will fundamentally have loss and therefore dispersion. There are materials with low enough loss that the loss can be ignored for most simulations. I think in these situations, the dispersion would be low enough to also be ignored. Loss and dispersion are connected. However, more than loss contributes to the dispersion. If you want to get the deeper story behind all this, work through the videos in Topic 2 here: empossible.net/academics/21cem/

[@Cybernetic1]

Lots and lots of thanks from 🇮🇳 India

[@empossible1577]

Greetings India!! Great to have you here!

[@user-si1zn3ir7x]

Amazing! I have a question sir. in 3:27 you have complex permittivity. But in some text book they write epsilon=1+sigma/(epsilon0*omega). are these same equation?

[@empossible1577]

Great question. The correct form of the equation you gave is... eps_r = 1 + sigma/(j*omega*eps0) Notice the j term which makes eps_r a complex number. This equation, however, is a common approximation for the relative permittivity of good conductors. It is not a general equation. The general form of this equation is eps_r = er + sigma/(j*omega*eps0) This equation lets you control both the real and imaginary part of the complex permittivity. Now instead of relative permittivity eps_r, the above equation can be written as permittivity eps. eps = eps0*er + sigma/(j*omega) This is the equation given at 3:27. It is only equivalent to the equation you gave if you divide by eps0 to get relative permittivity and also assume the real part is 1. You also missed a j term. Hope this helps!!

[@user-si1zn3ir7x]

​@@empossible1577 I have thought about this for a while but there still remains a question. You said in 1:10 that sig and eps are both real. However, from our toy model (Lorentz,Drude) we know that both sig and eps are complex number. And we know that Im[eps] and Re[sig] have to do with loss. But in your point of view then, sig have to contain imaginary part of eps because eps cannot be imaginary. And eps contains real part of sig likewise. Is my understanding correct?

[@empossible1577]

@@user-si1zn3ir7x This is confusing and there are many mathematically correct ways to look at this. In my mind, two make the most sense. In one model of materials there is only a complex permittivity. In the second model there is only a real-valued permittivity and a real-valued conductivity. This is what I teach in this lecture, but you can find other models. To me it is confusing, for example, to have a complex permittivity and then also a conductivity because there are redundant ways to incorporate loss.

[@meghjitmajumder3468]

Fantastic video once again... But out of the context, I'd like to say that yesterday I noticed that there is one particular area in electrostatics which I don't understand really well and that is Method of Images. So, it'll be very kind of you, if you think of making a video on it.

[@empossible1577]

I could perhaps create a slide or two on that. When a charge is near a conducting surface, the field looks as if there is an opposite charge on the exact opposite side of the mirror. That really is it.

[@alexandermuller8858]

Thank you, Lots of very good stuff in here. Still i have a question regarding the complex permittvity. You mentioned that one should not mix complex permittivity and real conductivity. Here is my question: the imaginary part of the complex permittivity represents the losses. But these are only losses due to sigma. What about the losses due to changing of polarization in dielectric material in case of dynamic E-Fields? They also have to be included into the imaginary part or not? For fast time varying E-Fields there should be phase difference between Polarization field and E field or not? Also, if my omega is 0, sigma/omega becomes infinite? So for DC i have infinite losses? Im a bit confused.

[@empossible1577]

I don’t like the model of mixing conductivity and complex permittivity, although some people do it. To me it is redundant and confusing, so I never use that convention. Sometimes it is done to separate DC conductivity from the more dispersive conductivity at higher frequencies (polarization as you state it). It is a mistake to think the imaginary part of permittivity represents loss. When it is zero there is no loss. However, when it is nonzero, both the real and imaginary parts of permittivity, along with permeability, contribute to loss. Loss due to both DC conductivity and polarization can be lumped into the permittivity. If you want to learn more about this, study the Lorentz model of dielectrics and Drude model for metals. I have some older videos and notes on this subject in Topic 2 here: empossible.net/academics/21cem/ When it comes to the polarization, there is a term called susceptibility and it is a measure how easily a material becomes polarized due to an applied electric field. The polarization P is related to the electric field E and susceptibility X through P = e0*X*E, where e0 is the free space permittivity. If there is loss, X is a complex number and P and E can be out of phase as you asked. As for your last question, you are mistakenly interpreting the imaginary part of permittivity as loss. While permittivity and permeability are the fundamental electromagnetic parameters, they are not very insightful about the actual properties of the medium or how they will affect the propagation of waves. Instead, we have parameters that consolidate all of the information into more intuitive parameters that quantify things like loss. For example, the attenuation coefficient consolidates all of the loss information from permittivity, permeability and conductivity into a single term. If you look at the equation to calculate attenuation, you will it essentially has omega/sqrt(omega). So in the limit as omega approaches zero, the attenuation actually goes to zero. However, the attenuation coefficient describes waves. Is it really a wave at zero frequency? By the way, this part of the notes has been revised a bit and I have not yet updated the videos. Here is a link to the course website with the latest version of everything. empossible.net/emp3302/

[@alexandermuller8858]

@@empossible1577 Thank your for your reply! I will check out both drude and lorentz model. Regarding the complex premittivity: actually this is exactly what i thought about. The complex permittvity includes both, losses due to conduction and polarisation. D=e0*er×E and er is already complex because from P=e0*X*E the X is also complex. If i put this equation into Maxwell ampers law i have: rot H= sigma*E+e0*er*jw*E=jw*E*e0*(er-sigma/(omega*e0)) now if i plug in the already complex er=er'-j*er'' due to complex X i should get rot H=e0*jw*E*(er'-j(er"+sigma/(omega*e0))). So that way i have both polarisation and conduction. Im not Sure if i can write it like this.

[@empossible1577]

@@alexandermuller8858 There should be a curl operation somewhere in that last equation, correct? Otherwise, it seems correct in terms of the complex permittivity.

[@alexandermuller8858]

​@@empossible1577 Yes, thank you again. I was just not sure about that permittivity part. Watched some videos on youtube and i missed that er'' in the equation (only saw that part with sigma/(omega)). In the script from my professor it is also a bit confusing, because he mentiones that both effects can be lumbed into the imaginary part of er. The equation contains only the sigma/omega part though. As for the total losses, thank you again for that information. Imaginary part=0 means no losses, but if it is not equal to 0, both real and imaginary part "contribute" to losses. One can think that the real part has to contribute, because it influences the E-Field in the material. All in all as you said, the parameter alpha is much more intuitive to think about this situation, instead of permittivity, pearmebility..

[@empossible1577]

@@alexandermuller8858 When I teach this stuff, I explain it exactly like this. Permittivity, permeability and conductivity are the fundamental parameters to Maxwell's equations, but they are not very intuitive in terms of how they affect waves. Instead, we have parameters like refractive index, impedance, attenuation coefficient, phase constant, etc.. These are not the fundamental parameters, but they consolidate information from permittivity, permeability, and conductivity into terms that explain intuitively how they affect waves. Glad to hear this model of thinking helps!

[@user-ub1qi3ye5k]

Dear sir, do you have a video about losses in capacitor and inductor?

[@empossible1577]

Unfortunately I do not. Very sorry!!

[@ManojKumar-cj7oj]

In good conductos E field leads H fields by 45°

[@empossible1577]

I touch on that in the following lecture.

[@egghead55425]

Since the speed of light is 1/sqrt(mu*ep), doesn't a complex dielectric imply the lossy part of the wave moves into a new dimension? The output is in units of meters/sec, not heat. Is ehat we experience as heat loss the manifestation of the energy moving into a new dimension?

[@empossible1577]

No, but that would be cool! The velocity of light is actually a more complicated subject than you may think. For example, which velocity are you asking about, phase velocity, group velocity, or energy velocity? From your equation, that is phase velocity, but it is not exactly correct as you have written it. The more rigorous and intuitive way to calculate this is to first calculate the complex refractive index n = sqrt(ur*er). Second, the phase velocity is v = c0/Re(n). The real operation Re() gets rid of the imaginary part that characterizes loss. BTW, phase velocity can exceed the speed of light in vacuum. The speed of a wave in a rectangular metal waveguide is the classic example of this.

[@egghead55425]

@@empossible1577 thank you for your detailed reply! Metamaterials can have effective properties near zero making nearly infinite phase velocity. Did I understand your video correctly that the group velocity can be made to refract based on the effective properties?

[@egghead55425]

I thought group velocity was the same as the energy velocity. Have you made a video covering differences?

[@empossible1577]

@@egghead55425 Yes. You can predict refraction with effective properties. Remember, phase and power can refract differently! Crazy, huh? The planes of equal phase will refract one way and the beam itself can refract in a different direction. Negative refraction in a photonic crystal is a classic example of that. This is not due to negative refractive index.

[@charanpreetiitd]

Sir, great lecture, thank you, but I'm very confused about the correct mathematical formula for calculating intrinsic impedance of a lossy medium, which has both permeability and permittivity as complex numbers. I have been using Zin = Zo * sqrt(μr'-μr''/εr'-εr'') * tanh [ j 2πf.d. sqrt((μr'-μr'')(εr'-εr''))/c]. I then use it to calculate Reflection loss as R = 20.log (Zin-Zo)/(Zin+Zo). However, my answers never match that of the authors in several published articles, which quote the same formula. This is a lengthy explanation but I wanted to be explicit to be able to get an answer I have been searching for so long. Would really appreciate your response, possibly with a quick problem example. Thanks in advance.

[@empossible1577]

We are actually working on a paper about this. It is mostly written, but not finished yet. The problem with a lossy interface is that there is energy in multiple directions and the ordinary way of analyzing things ignores half the problem. Thus, the answers are incorrect. There is too much to describe in this message. For now, you can search for things like "Fresnel Equations" and "Lossy Media."

[@charanpreetiitd]

@@empossible1577 Thank you. Do share any published articles that would be of relevance.

[@empossible1577]

@@charanpreetiitd Will do! You can see a little bit of it in Lecture 7b at the official course website: empossible.net/academics/emp3302/

[@TheMorningbirdFoundation]

Where is the energy going in a lossy material?

[@empossible1577]

Great question! The answer is heat. Electric fields put forces on free electrons in conductors making them move and produce an electrical current. The loss mechanisms involve some of this push being on the atoms and causing mechanical vibrations. The mechanical vibrations are heat. Generally the heat is very low, but it can be measured. At high power it can even become a problem.

[@TheMorningbirdFoundation]

Thank you for your reply. I am working on a project that calls for a material where the imaginary part of the dielectric is greater than the real part. It turns out that an antifreeze (Ethylene glycol) is such a material. Is it conceivable that there is a reciprocal relationship in that the heat energy is transformed to electromagnetic in such a material and thus why it is able to dissipate heat efficiently? I am using this material because of the impact on the energy density.

[@empossible1577]

@@TheMorningbirdFoundation I don't know your answer. I know that materials which are electrically conductive tend to also be thermally conductive, although there are some exceptions like diamond. If the imaginary part is large, that is something relatively conductive. It does not surprise it is also thermally conductive, but I cannot explain the relationship. When you figure it out, let me know!

[@motobacktoconstitution4138]

Thank you shearing sir. I couldnt time to watch all. Please make shorter video or please lead me to short e r video please I wanted understand what you are saying but too long I have to do my job. So couldn't here all sorry.

[@empossible1577]

It is an interesting idea to make a short summary video for each lecture. I will think about that. There are one-page summaries provided on the official course website as well as the electronic notes that you can skim through faster. Here is the website: empossible.net/academics/emp3302/