Electric Conductivity

Muhammad Qamar Gold nanorod

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When we study the absorption and scattering cross section of sphere. Why we take electrical conductivity of metals like gold, silver, copper ete zero as they have value of electrical conductivity. What is the reason behind this?

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Qamar

3 Replies Last Post 2024年12月14日 GMT-5 07:16

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Posted: 2 weeks ago 2024年12月9日 GMT-5 11:02

Hi Muhammad Qamar,

If the sphere is assumed perfectly conducting like in the RCS benchmark, the corresponding boundary condition does not need to know what material it reflects into (no losses). The sphere is not included in the computational domain. In other models, like the Optical Scattering off a Gold Nanosphere, you give the material properties of the gold and compute losses since the wave penetrates a bit into the material.

I hope it makes sense in your context! Best regards, Eric Favre Renaissance Fusion

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Eric Favre
Renaissance Fusion
Hi Muhammad Qamar, If the sphere is assumed perfectly conducting like in the RCS benchmark, the corresponding boundary condition does not need to know what material it reflects into (no losses). The sphere is not included in the computational domain. In other models, like the Optical Scattering off a Gold Nanosphere, you give the material properties of the gold and compute losses since the wave penetrates a bit into the material. I hope it makes sense in your context! Best regards, Eric Favre Renaissance Fusion

Muhammad Qamar Gold nanorod

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Posted: 2 weeks ago 2024年12月10日 GMT-5 08:08

Hi Eric Favre, Thank you for your reply. I did not understand you answer comp

Hi Muhammad Qamar,

If the sphere is assumed perfectly conducting like in the RCS benchmark, the corresponding boundary condition does not need to know what material it reflects into (no losses). The sphere is not included in the computational domain. In other models, like the Optical Scattering off a Gold Nanosphere, you give the material properties of the gold and compute losses since the wave penetrates a bit into the material.

I hope it makes sense in your context! Best regards, Eric Favre Renaissance Fusion

letly. Can you please elaborate it give reason behind this physics. Best regards, Muhammad Qamar

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Qamar
Hi Eric Favre, Thank you for your reply. I did not understand you answer comp >Hi Muhammad Qamar, > >If the sphere is assumed perfectly conducting like in the RCS benchmark, the corresponding boundary condition does not need to know what material it reflects into (no losses). The sphere is not included in the computational domain. >In other models, like the Optical Scattering off a Gold Nanosphere, you give the material properties of the gold and compute losses since the wave penetrates a bit into the material. > >I hope it makes sense in your context! >Best regards, >Eric Favre >Renaissance Fusion letly. Can you please elaborate it give reason behind this physics. Best regards, Muhammad Qamar

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Posted: 2 weeks ago 2024年12月14日 GMT-5 07:16

The examples I mentioned are available online or using the applications library. Read them carefully! If the incoming wave meets a "perfect" scatterer (meaning fully reflective), then you don't need to know the material of the wall. If the wall absorbs a bit of energy, then of course the amount depends on the wall material properties: you need to take them into account. Analogies would be: * In acoustic propagation of a wave in a tube, finite impedance at the end characterizes if and how much of the wave can leave the domain. Infinite impedance corresponds to a closed tube. Impedance depends on the material at the outlet. * Although it's not a wave, I can think of a simple bouncing ball : elastic collision you don't need to characterize the wall. Inelastic collision you need to put an absorption coefficient that is defined by the mechanical properties of the wall.

In your case Gold absorbs a bit of energy, considering the incoming wavelength. If you need to take into account these losses or not depends on the application of course. If you have no idea, compare both approaches : with Perfect Electric Conductor = without the domain, and with default boundary condition = you give the material properties of the scatterer. Just pay attention to adjust the mesh versus the wavelength. Again, studying the 2 examples given as reference above is not a waste of time.

Good luck!

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Eric Favre
Renaissance Fusion
The examples I mentioned are available online or using the applications library. Read them carefully! If the incoming wave meets a "perfect" scatterer (meaning fully reflective), then you don't need to know the material of the wall. If the wall absorbs a bit of energy, then of course the amount depends on the wall material properties: you need to take them into account. Analogies would be: * In acoustic propagation of a wave in a tube, finite impedance at the end characterizes if and how much of the wave can leave the domain. Infinite impedance corresponds to a closed tube. Impedance depends on the material at the outlet. * Although it's not a wave, I can think of a simple bouncing ball : elastic collision you don't need to characterize the wall. Inelastic collision you need to put an absorption coefficient that is defined by the mechanical properties of the wall. In your case Gold absorbs a bit of energy, considering the incoming wavelength. If you need to take into account these losses or not depends on the application of course. If you have no idea, compare both approaches : with Perfect Electric Conductor = without the domain, and with default boundary condition = you give the material properties of the scatterer. Just pay attention to adjust the mesh versus the wavelength. Again, studying the 2 examples given as reference above is not a waste of time. Good luck!

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