Photonics

Photonics

Definition of TE/TM polarizations in FDTD

    • Kazuma Yamashita
      Subscriber

      Hi,

       

      I have a simple question about the definition of TE/TM polarizations in FDTD.

      I think that in general optics, the TE (Transverse Electric) polarization - the electric field is perpendicular to the plane of incidence - corresponds to s-polarization and the TM (Transverse Magnetic) polarization - the magnetic field is perpendicular to the plane of incidence - corresponds to p-polarization.

      However, I found that this relation is opposite in FDTD simulations: when I use a movie monitor in 2D simulations with p-polarization, the output file becomes "file name_monitor name_TE."

      Could anyone please tell me why this is?

       

      Thanks,

      Kazuma

    • Guilin Sun
      Ansys Employee

      In general, TE/TM is not uniquely defined. Actually they can be opposit as you observed.

       

      BTW: I wrote a post in Chinese regarding to this. If you can use google translate please have a look: https://forum.ansys.com/forums/topic/ansys-insight-youguantetmpianzhenyijimoshiguangyuandewenti/#post-176494

    • Kazuma Yamashita
      Subscriber

      Thank you for your reply.

      Now I understand that the reason why 2D FDTD simulations give unexpected definition of polarization comes from the definition of field components in 2D FDTD:

      TE:   Ex, Ey, Hz

      TM:   Hx, Hy, Ez

      But I am wondering why the TE mode can be called TE (transverse electric) in this definition. What is the TE wave (Ex, Ey) transverse to?

       

      Thanks,

      Kazuma

    • Guilin Sun
      Ansys Employee

       

       

      It is simply a definition, called TEz. Meaning transverse to the 3rd dimension z . I used it before 

      Approximate Crank-Nicolson schemes for the 2-D finite-difference time-domain method for TEz waves   IEEE Transactions on Antennas and Propagation ( Volume: 52, Issue: 11, November 2004) 

       

       

    • Kazuma Yamashita
      Subscriber

      Thank you very much for your help. I understand that it is simply a matter of definition in 2D-FDTD.

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