How to propagate $E_x (x,t) = \exp(-t^2/\tau^2-i\omega_0 t) \exp(-x^2/w_0^2)$ in finite difference time domain (FDTD) analysis

Question

Finite difference time domain (FDTD) allows to solve differential equations for time evolution. For example, we can analyze ultra-short pulses in free space by solving the Maxwell's equations.

The pulse above is plane wave with gaussian shape in the time domain that propagates in $z$ direction. The wave $E_x$ does not have have shape in $x$ direction it is simple plane wave, $E_x (x,t) = \exp(-t^2/\tau^2 - i \omega_0 t)$)

How does one add transverse dimensions (just $x$ direction) to the FDTD method to solve for Gaussian pulse in space ($x$ direction) and Gaussian pulse in time that propagates in $z$ direction?

How to propagate $$E_x (x,t) = \exp(-t^2/\tau^2 - i \omega_0 t) \exp(-x^2/w_o^2)~?$$

Answer 1

If you use Lumerical or MEEP to do the FDTD calculation, you can simply add a Gaussian source polarized along $x$-direction but propagating along $z$-direction with the given pulse width and $\tau$ and frequency. Alternatively, you can make both $y$ and $z$ components equal to zero for the source. Is this what you want?

Answer 2

I assume that you are integrating the following equation:

$$\frac{\partial^{2}E}{\partial z^{2}}+\frac{\partial^{2}E}{\partial x^{2}}-\frac{1}{v^{2}(\omega)}\frac{\partial^{2}E}{\partial t^{2}}=0$$

With an "initial" condition: $$E(x,t, z=0) = f(x,t).$$

First thing you have to do is transforming $x$ and $t$ derivatives to finite differences:

$$\frac{\partial^{2}E}{\partial x^{2}}=\frac{E_{k+1}+E_{k-1}-2 E_{k}}{\Delta x^{2}} \equiv X_k(z),$$

$$\frac{\partial^{2}E}{\partial t^{2}}=\frac{E_{m+1}+E_{m-1}-2 E_{m}}{\Delta t^{2}} \equiv T_m(z),$$

Where I introduced shorthand notations $X_k(z)$ and $T_m(z)$ because I'm too lazy to type those double differences on mobile.

Next, you have to discretize your f(x,t) on $xt$ plane: turn it into a grid of $(x_k, t_m)$ points with discretization steps $\Delta x$ and $\Delta t$. Let's assume you have $K$ points along $x$ and $M$ points along $t$. Thus, your second order PDE turns into a system of $K \times M$ coupled second-order ODEs.

$$\left\{ \begin{array}{c} E''+X_1(z)-\frac{1}{v^2} T_1(z) = 0\\ E''+X_2(z)-\frac{1}{v^2} T_1(z) = 0\\ ...\\ E''+X_K(z)-\frac{1}{v^2} T_1(z)=0\\ E''+X_1(z)-\frac{1}{v^2} T_2(z) = 0\\ ...\\ E''+X_K(z)-\frac{1}{v^2} T_M(z) = 0 \end{array}\right.$$

Where $E'' \equiv \frac{d^2 E}{d z^2}$. From here, you can just use standard procedure as in the 1D case.

Answer 3

Your question is unclear. If you add transverse dimensions, your problem is no longer 1-dimensional.

If you mean adding transverse components of your interested quantities, then there is no change needed to the usual FDTD method. For example, if you are analysing the propagation of an electromagnetic plane wave in one-dimension, say the positive $z$ direction, we know that the electric field will be linearly polarized in the $x$ direction and the magnetic field will be linearly polarised in the $y$ direction. Your computational domain will consist of $E_x (z)$ and $B_y(z)$ values, then.

If you are interested in gaussian pulses, you can simply set your source to be a propagating gaussian and run a FDTD simulation as usual.

A good resource for the FDTD method applied to electromagnetism is Computational Electrodynamics, by Allen Taflove and Susan C. Hagness.

Answer 4

Generally to propagate, one needs a term like $\mathbf{k} \cdot \mathbf{x}$ in the exponent (i.e., a finite phase velocity). The last term you showed should not propagate anywhere. It's envelope is a modulated Gaussian (i.e., from the $e^{-x^{2}}$ term) and its amplitude grows in time (i.e., from the $e^{-t^{2}}$ term) while it oscillates as a standing wave (i.e., from the $e^{-i \ \omega \ t}$ term).

Since I do not see any wave numbers ($\mathbf{k}$), I am curious how you are propagating a signal. You could multiply by something simple like $e^{i \ \mathbf{k} \cdot \mathbf{x}}$ to get the finite phase velocity required for the phase fronts to propagate. Then ensure that $\partial \omega / \partial \mathbf{k} \neq 0$ to get the envelope to propagate (i.e., finite group velocity).