TAGGED: LumericalFDTD, rotationalsymmetry


September 23, 2021 at 10:48 pmabackerSubscriber
Hello,
Is there an efficient way to simulate a 3D object possessing rotational symmetry using a 2D FDTD simulation? The tutorial articles show examples of doing small 3D simulations that account for objects that possess inversion symmetry about specific axes, but I am interest in simulating objects that are unchanged by ANY rotation about the y axis.
Thanks,
Adam

September 24, 2021 at 4:28 pmGuilin SunAnsys EmployeeHi, Adam, since most FDTD algorithm is based on Cartesian coordinate system, including Lumerical's FDTD product, it is best to deal with axial symmetry. For rotational symmetry, it will need a special algorithm.
Please note that, even though the the geometry is rotational symmetric, the source polarization is usually NOT. So special care must be taken, as the polarization effect is vital in solving Maxwell's equations. In case both the geometry and the source polarization are rotationally symmetric, you can simply it to be 2D simulation. As far as I know, only some special, user defined sources have such property, which may not work for other situation.

September 24, 2021 at 5:14 pmabackerSubscriberIn order to make the source rotationally symmetric as well, It seems like two separate simulations using TE (z) and TM (x) polarized sources could be carried out, the summed intensity results would then show the result for unpolarized light. Is there a setting that could then be used that transforms the results of the 2D simulation into a 3D rotationally symmetric simulation?

September 24, 2021 at 6:20 pmGuilin SunAnsys EmployeePlease make sure this statement " In order to make the source rotationally symmetric as well, It seems like two separate simulations using TE (z) and TM (x) polarized sources could be carried out, the summed intensity results would then show the result for unpolarized light." is correct.
Unpolarized light has nothing to do with the symmetry! before going more details, please narrow down your thoughts: is the result rotationally symmetric? how do you justify? if you use a linear polarized or circularly  polarized light source , the result will not have rotationally symmetry. What you mentioned is the intensity using TE and TM, but FDTD solves Maxwell's Equations with fields, not intensity. Please make sure the physics is correct before you do any simplification. I believe there is physics behind. Once the physics is resolved, then we can discuss simulation.

September 24, 2021 at 7:06 pmabackerSubscriberIf we kept things simple and assumed the lightsource were radially polarized, would there then be a way to simulate the 3D optic using a 2D simulation?
Thanks

September 24, 2021 at 10:39 pmGuilin SunAnsys EmployeeIt should be ok, as long as the radially polarized light is correctly characterized, and the rotationally symmetry structure interacts with such light the same way in 3D and 3D, you can simply do a 2D simulation. Since every phi has the same result, usually people will only use a cross section, eg, the 2D result. If you want to transform 2D result into 3D, it is doable as long as you have the correct math.

September 25, 2021 at 12:03 amabackerSubscriberDo you have any pointers on how to transform the 2D result into a 3D result with radial symmetry? Is there a way to do this by configuring the 2D FDTD solver a particular way. The 2D result essentially assumes that the structure is invariant out of the simulation plane (along the z axis) which is different from a 3D problem with radial symmetry. The reason I am asking this is because I would like an efficient way to do a 3D simulation with radial symmetry efficiently, without having to do the full 3D FDTD simulation.
Thanks!

September 27, 2021 at 2:49 pmGuilin SunAnsys EmployeeI am not sure what is your simulation setting and device configuration. As you said, the result is radially symmetric. That means it has no field component with phi, only radial fields left. You can use cylindrical to Cartesian coordinate transformation, which can be found in some textbooks. This is post processing so it is up to you to transform the result in 3D. BTW: since you know it is radially symmetric, except for visualization, it is not helpful in 3D.

September 27, 2021 at 4:25 pmabackerSubscriberBasically, my question is how to set up a Lumerical FDTD simulation so that it can simulate a radially symmetric structure as fast and efficiently as possible. If I were to do a complete 3D FDTD simulation, I think it would be a much more computationally intensive simulation than it would need to be. Is there a way to use radial symmetry to do the simulation efficiently?

September 27, 2021 at 5:15 pmGuilin SunAnsys EmployeeSince you know it is radially symmetric and want to do 2D simulation, please just start with a 2D simulation (only in xy plane) and see if the result is expected, where the cylinder become a rectangle with diameter and height. You can also use 3D simulation with the true cylinder, but only set the other axis with periodic BCs, and one mesh cell only. eg, if the cylinder is along z axis and its cross section is in xy plane, you can set y axis in periodic BC. Now it is not the geometry nor the simulation determine the 3D, it is your post processing. please try one of the two methods and check the result. Those two methods will show different cylinder orientation, but the post processing result should be the same, except that due to the curve in y there is a small error in representing it.

September 27, 2021 at 5:23 pmabackerSubscriberI think that just doing a 2D simulation without any additional configuration is incorrect: The 2D simulation would assume that the structure is unchanged along the z axis (out of the simulation plane) which is different from a structure that is radially symmetric about the yaxis. Is there a way to properly configure the 2D simulation to account for radial symmetry?
I saw the help article on how to do a smaller 3D simulation with periodic BCs as you have described, but this will only reduce the overall simulation time by a factor of ~4x because you still have to do a 3D simulation (over a smaller volume).
What I would really like to know is how to configure the 2D simulation for radial symmetry, instead of a structure which is invariant along the z axis.

September 27, 2021 at 5:34 pmGuilin SunAnsys EmployeeI suggest that you carefully consider what I suggested: it is the post processing that matters. take the cylinder along z as an example: any cut parallel to and through the cylinder axis has the same result, isn't it? in this cut, can you do a 2D simulation? if yes, just take the cut geometry size and do a 2D simulation, you WILL get what you want as long as your assumption is correct.
Please think it over.
If you do not agree with the above two suggestions, then we may not have other solutions for you.

September 27, 2021 at 5:45 pmabackerSubscriberThis is incorrect. The 2D simulation you have shown assumes that the structure is infinitely long in z. It will not give the same result as a cylinder or a square that has finite dimensions in z, even for the central xycrosssection at z=0. Is there a Lumerical FDTD help line where I can schedule a quick phoneconversation to discuss my issue? Thanks!

September 28, 2021 at 6:50 pmGuilin SunAnsys Employeeok. It is up to you doing what you expect to do. What I know is the invariant is on phi. We do not have any example or any documents on this topic.
if you have premium support please send us email.

March 28, 2022 at 4:46 pmstefan.appelSubscriberHi, I just stumbled upon your topic. Indeed, running a 2d (rzcut) simulation will not give you the same results as a full (time, memory and CPUconsuming) 3d simulation for a rotational symmetric body.
The problem is that light propagation in radial direction is different from light propagation in any rectangular dimension. This is because in radial geometry, the differential operators (like nabla, div and rot) look different than in rectangular coordinates. In essence, this is why the periodic zbounds of the 2d rectangular simulation will not give the correct results
There is lots of software other than Lumerical which incorporates according 2D radial solvers, some even open source freeware like Meep. I wish there was such a feature in Lumerical, too. In such other software, you will have to specify the rotational symmetry m of the fields (how many oscillations there are in one revolution), and then you can just simulate in 2d rzcuts. As you anticipated, different field components may have different rotational symmetries which will require multiple simulations, but this is still much faster than any 3D FDTD.
There may even be some software which can handle periodic/discrete rotational symmetry (structures symmetric under finite rotation angles, like honeycomb photonic crystals which are 6fold rotational symmetric), but as far as I know no freeware among them.
You could try to work around this: Essentially, you have to convert the differences in the differential operators (radial vs rectangular) into differences/scales of distances, permittivities and/or susceptibilities. This is similar to how PMLs work (they scale the infinite space beyond into finite space by altering material properties).
However, for radial geometries, this will create nonisotropic permittivities and susceptibilities (even if your inital structure will only contain isotropic refractive index media), which additionally do gradually scale with radius. And this combination unfortunately is also not supported by Lumerical, so you do not need to bother finding out the complex way how to actually scale everything.
One other solution may be possible for structures at large radii (compared to effective wavelength) with small radius span (compared to the radius of interest). There, you can use periodic (bloch) boundaries to simulate rotational (discrete/periodic) symmetric structures. Basically, there, the curvature is small enough there such that its effect is hopefully negligible.

March 28, 2022 at 5:07 pmabackerSubscriberThanks for the pointers! Will certainly look into the potential workarounds you have suggested.

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