There are two aspects to current density. First, there is the current due to free charges. Second, there is the current due to movement of bound charges called the displacement current. The relationship you are seeking deals with the second type of current, not the first. Polarization of a medium (not polarization of a wave) describes charges displaced from their equilibrium position. During the transition of the bound charges, there is a tiny momentary current. So, in the process of a material becoming polarized, or losing its polarization, there will be a displacement current. J_disp = d D/ dt Does this make sense?

@@empossible1577 Thank you for the reply. In frequency domain we can write it like J_disp = iwD? And another thing to make sure here D is the polarization. Am I correct?

@@empossible1577 Thank you for the reply. So in frequency domain we can write it like J_disp = jwD? and here we can say D is the polarization? Isn't it?

@@user-nd7rk6dd3p D is the displacement field, or electric flux density. In the frequency domain, w=0. This means for electrostatics you can have a medium that is polarized, but the polarization cannot change or it would be a dynamic problem. If there is no change in polarization, there will be no polarization current.

Thank you!! This video is in a sequence for electrostatics. The magnetic field term appears in a derivative with respect to time. This goes to zero for electrostatics. BTW...here is a a link to the official course website. It has links to the latest versions of the notes, videos, summary sheets, and other learning resources. empossible.net/academics/emp3302/

Very sorry, but I don't understand your question. I "think" you are asking why there seems to be two types of waves later in the simulation. That is because I am visualizing both the electric (blue) and magnetic (red) fields. At first, they overlap so closely you cannot tell them apart. After scattering from the boundaries, they get a different phase shift and they invert from each other. Hopefully this is what you were asking!

@@empossible1577 tq sir

Think of aref as a complex reflection coefficient and atrn as a complex transmission coefficient. They are just for modes in a waveguide instead of plane waves at a flat interface. They are complex quantities because they relate both amplitude and phase of the analyzed modes vs the source mode. In this sense, neither abs() or real() should be used or that would calculate a purely real aref and atrn. I later use abs()^2 because I am calculating fraction of power reflected and transmitted in a guided mode. I am ignoring phase as the abs()^2 operation erases the phase information. The first two equations on Slide 33 are weighing the simulated field against the complex set of guided modes to calculate the complex amplitudes of each of the guided modes that would give the simulated field. So aref are the complex amplitudes of all the modes, such that if all added up would give the simulated field in Eref. Same story for the transmitted modes. You mention “real-time” in your question. Understand that Eref and Etrn are steady-state fields. I am not sure what kind of answer you would get if you are using the instantaneous fields in this calculation. BTW, don’t depend on the numbers in EZR and EZT being purely real. I think they happen to be for this special case, but they can easily become complex numbers. I always treat them as if they are complex so if I ever do anything that will make them truly complex the code will still work. Hope this helps!!

@@empossible1577 Thank you Dr. Rumpf!!!! You are so helpful even after 10 years since uploading the slides! ! I did not realize the aref can be complex amplitudes of all the modes, thought it has to be real... As to the "real-time", I just plot both the real-time field at the record plane, and the real part of the steady state Etrn, they look the same at the same time step (overlay on each other in plot). Not sure why. My source followed slide 24, and it is a real source.

@@yufish6576 Happy to help!! You may also be interested in the paid online FDTD courses. They are next-level in terms of visualization and clarity. They were also designed for self-paced learning instead just being the nodes and videos for my face-to-face class. Here is a video showing the contents of the courses: kzfaq.info/get/bejne/q6iZo9V6s57Llo0.htmlsi=bL2TbIS5P6Rt9cn6 The entire first half of the first course is free. If nothing else, I recommend working through that. Here is a direct link to the courses: empossible.thinkific.com/collections/FDTD-in-MATLAB

The process would be to assemble your data in an array and use a command like imagesc() or pcolor() to image it. I am not sure what "different node RF" is. There is an image processing toolbox available that may make things like object recognition easier. I don't use the toolbox. I am only aware that it exists so I am unsure of the functionality. All that said, you can definitely use MATLAB somehow to find objects in images.

@@empossible1577 can you help me pls ihave many question?

@@user-ss1dp7he8t I can certainly answer quicker questions, but if you need to do any kid of work, I may not be able to help until later September.

@@empossible1577 iwant to know how to use tikhnove or LPP or FBp. With matrix to get image and heat map

This is great to hear! I recently hired a mechanical engineer into my group is is now one of the strongest EM people I have. Always happy to help!

Thank you!! Check out our website... empossible.net/academics/

Thank you!!! Checkout our website... empossible.net/academics/

It will still work. In some sense, there are two sets of independently operating material properties. First, there are the actual material properties. Second, there are the PML material properties. The first set you configure according to your semiconductor problem. The second one will always have free space away from the PMLs and therefore will be impedance matched to free space inside of the PMLs.

@@empossible1577 Thank you！ I wish I could have met you 15 years ago. but it is never too late to learn.

The modes are discrete spatially, not by frequency. That is, each mode looks different. They will also propagate at different speeds. Each mode has its own cutoff frequency. Below this frequency that mode is not a guided mode. At any frequency above the cutoff, the mode is propagating. If you pick a frequency "between two modes" then I think you have picked a frequency above one mode's cutoff and below the other's. That will give you just one guided mode. BTW...at frequencies below the cutoff frequency, that mode still exists. It is just that the mode will decay quickly with distance and can usually be ignored. Hope this helps!

You are correct. I misspoke. The wave vectors along the diagonal have a larger magnitude, meaning they resolve the fields better along the diagonal by this factor. For this reason, a truncation scheme makes sense to as to even out the resolving power.

I did my best to mention the math, but not go crazy with it for this tutorial series. If you want more details on coordinate transforms, transformation optics, etc., checkout the series of videos under Topic 6 here: empossible.net/academics/21cem/

They are exactly the same, just that the terms are phasors instead of scalars. The magnetostatic boundary conditions are Lecture 5j here: empossible.net/academics/emp3302/

It does look like I goofed. I think I was so focused on getting the animation to look good enough I missed ensuring it was actually correct! How embarrassing! Thank you for pointing this out!

You are welcome!

Q.1 Exciting guided modes is actually a bit more complicated than it seems. The main problem is that you have to match a free space wave to a guided mode and these do not usually just match. To match them, you need to make the external wave look like the guided mode. A guided more looks a bit like a Gaussian beam so these tend to excite guided modes most effectively. A good model of a guided mode is a wave bouncing back and forth between the interfaces via total-internal-reflection. Only certain angles correspond to guided modes because the other angles would lead to a case where the wave bouncing back and forth would interfere with itself and leak out of the waveguide. The angle of incidence you are asking about corresponds to the allowed angle. Different modes will have different angles. I think a better model is thinking of the guided mode as a single packet of energy. It also has a propagation constant which describes how quickly the guided mode accumulates phase in the longitudinal direction. When you launch a wave at some angle of incidence, you are matching the component of its waveguide vector in the direction of the mode to the phase constant of the guided mode. This whole store gets more complicated when you want to excite higher-order modes, but I will not address that here. The short answer goes back to making the external wave look like the guided mode. In practice, at optical frequencies this would entail precisely aligning a beam using micropositioners to the angle and focusing down to a small spot at the entrance of the waveguide. There are other mechanisms such as grating couplers that can excite guided modes as well. Q.2 If your exciting is not correct, you will see the power lead from the waveguide. At visible frequencies, optical fibers will glow red and dim with distance. It is not usually an exact thing and there will be some tolerance. If you are aligning a beam to a waveguide, take your time and be patient. An alignment could take a while. I know people who spend 30 minutes or more getting an alignment and even then it is not perfect. It is not just about getting the angle correct, but also making the external wave look like the guided mode. If that is not perfect, power will eventually leak out of the waveguide. Q.3 Correct. This is not generally an efficient way to excite a guided mode. You will see people do things to collimate the wave before entering the waveguide in order to prevent what you are talking about. Hope this helps!!

@@empossible1577 Thanks a lot!! Very Detailed and helpful!

When you calculate the FFT, you are calculating the amplitude of all the frequencies making up the original signal (or image). If you then calculate the inverse FFT, you are back to your original signal (or image). In this algorithm, some of the frequencies are set to zero after the first FFT. This leads to filtered signal (or image) after the inverse FFT. I do this to create low frequency noise. If you want to see more that all the Fourier transforms and this noise generating algorithm, checkout Topic 9 here: empossible.net/academics/emp4301_5301/

@@empossible1577 thank you prof. for the explanation

Great to hear!

This is awesome to hear! You have made my day! Welcome to electromagnetics and the crazy world of metamaterials! Have fun with it and don't let the bullies intimidate you!

Great point. In principle your approach is the most accurate. I think your approach is needed for high contrast. For low contrast, I tend not worry about this and just ensure proper convergence. When you get a 2D, I teach a method that I call the "2x grid technique." It is a great way to sort all of this out. I am not sure if you are aware of the online FDTD courses that we offer. The entire first half of the 1D FDTD course is completely free. If nothing else, I recommend working through that. You will learn a lot. I teach the 2x grid technique in the 2D FDTD course. Here is a link about the course: kzfaq.info/get/bejne/q6iZo9V6s57Llo0.htmlsi=DUgqX0aQDue82JtS Here is a direct link to the courses: empossible.thinkific.com/collections/FDTD-in-MATLAB Hope this helps!

@@empossible1577 Thank you professor！ I bought your book but does not have time to read it yet. I am a semiconductor laser engineer. FDTD is very useful in our industry at this moment. I never got chance to learn it myself. Your course is very helpful！I recommend it to my friends already. The companies never give us time to learn，just project, timeline and budget...I will try to learn all your courses！ Thank you

You can analyze a rectangular waveguide using just a 2D grid. It is just the cross section that is needed to calculate the modes. I think the best resource for this is Chapter 6 in my recent book: empossible.net/fdfdbook/ Hope this helps!!

Checkout the videos in Topic 2a here. This is where I cover these types of skills. empossible.net/academics/emp4301_5301/ You may also be interested in Chapter 1 of my recent book where I also cover these types of skills. empossible.net/fdfdbook/

I am away from my computer on vacation, but hopefully I can still add some helpful words. There are multiple ways to make this all work. I think I am using this approach because the handedness is more intuitive to me. Power flows in the direction of ExH. If the two components are Ey and Hx, there would need to be a negative sign to keep power flowing in the +z direction. Use whatever approach you like. As long as your backward source is several orders of magnitude smaller than the forward wave, everything should work fine for you. Hope this helps!

There is also energy velocity! LOL Phase velocity routinely exceeds the speed of light. In fact, I recently made a video about this for guided modes in a rectangular metal waveguide. kzfaq.info/get/bejne/oNClbMWbyJaaZ6M.htmlsi=CeQOqdf74hUg6Jqi I am aware that group velocity can also exceed the speed of light, but I am unaware of energy or information moving faster than the speed of light.

There are two ways to store electric energy: (1) in the field itself, and (2) in matter at displaced charge. A capacitor would still store energy even if it had only vacuum between the plates. Added a dielectric increases capacitance because it introduces the second way to store electric energy. The ability to store electric energy in the field or in matter is all quantified by the permittivity. The relative permittivity is not zero in vacuum, it is one to account for the ability to store energy in the field itself. Anything above a value of 1 is combining the ability to store energy in the field and in matter. Hope this helps!!!

The main skill you are missing is building arbitrary geometries into your grid. I teach these skills in my Computational Methods course. See Topic 2 here: empossible.net/academics/emp4301_5301/ With these skills, I think you will see how it can be done. If you think you are still weak on the numerical aspects of the finite-difference method, work through Topics 6 and 7 in this same course. Then go back through this capacitor video and I think you will understand everything. Hope this helps!!

Usually the substrate is made to be the transmission region. The RCWA presented here cannot have loss. This is something we are working on. If you substrate has significant loss, you may need to make your last layer the substrate in order to account for the loss. You would then set the transmission region to air or something that matches the substrate material. Hope this helps!

@@empossible1577 okey! this is really helpful! thanks

Thank you! Happy to help!

Happy to hear the materials are helping you! I am not sure if you are only going by this channel or not. I recommend using the course website as your main portal to the information. You can see the information organized, download the notes, get links to the latest version of everything, and get links to other learning resources and help. empossible.net/academics/emp5337/

Thank you!! I am looking forward to seeing what others do with this.

You could design it so it is the other way around. It really is the same thing. It is generally thought of as H being ahead of E. This delay comes with how the finite-differences are handled. It is best to stagger E and H in both time and space. The space staggering leads to the Yee grid. The time staggering leads to things like what you are commenting on. Hope this helps!! BTW...I also talk through this in the paid FDTD courses. The entire first half of the first course is free. I highly recommend checking it out. I think you will learn a lot and the material is newer, more visual, and next level compared this video. empossible.thinkific.com/collections/FDTD-in-MATLAB

@@empossible1577 Thanks a lot. I am going to check it.

That is awesome to hear! Thanks for sharing!

Thank you!! I do not, but next time I teach Computational Methods this is on my to-do list as a great example for finite-difference time-domain. BTW...I am not sure if you are aware of the online FDTD courses on Thinkific. They present MUCH better information and do it much more clearly with greatly improved visuals. While they are paid, the entire first half of the 1D FDTD course is completely free. I recommend checking that out. Here is a link: empossible.thinkific.com/collections/FDTD-in-MATLAB