# Reflection of particles from a cylindrical wall

Let us consider that a particle from the starting position $$r_0 = (x_0, y_0)$$, in time $$\Delta t$$, moving with velocity $$v = (v_x, v_y)$$, crosses the boundary of the circle of radius $$R$$ in one time step $$\Delta t$$. If there is no boundary the particle will reach to point $$r_e$$.

First step in calculating the reflection is to find the point where the particle and the wall touch. If we define this point as $$r_T$$ ($$|r_T| = R$$), and the time of impact $$T = \Delta t \cdot \tau$$ we have the following equation:

\begin{equation*} (x_0 + v_x T)^2 + (y_0 + v_y T)^2 = R^2. \end{equation*}

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from IPython.display import Image
Image(filename='./cylindrical_reflection.png')

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If we solve the quadratic equation for T we get:

\begin{split}\begin{align*} & (v_x^2 + v_y^2)T^2 + 2(x_0v_x + y_0v_y)T + x_0^2 + y_0^2 - R^2 = 0, \\ \\ & T = \frac{-b + \sqrt{b^2 - 4ac} }{2a} \, \Rightarrow \, \tau = T/\Delta t. \end{align*}\end{split}

The particle moves from $$r_0$$ to the point of impact $$r_T$$, and reflects reaching $$r_r$$ in time $$\Delta t$$. Since the particle reaches the boundary in time less than $$\Delta t$$ we have $$0 < \tau < 1$$. If the boundary didn’t existed, the particle would reach $$r_e$$.

In order to get the final position under reflection we first rotate the coordinate system so that $$r_T = (x_T, y_T)$$ in the transformed system is $$r'_T = (x'_T, y'_T) = (R, 0)$$. The rotation is performed for angle $$-\theta$$ and the rotation cosine and sine can be expressed as:

\begin{split}\begin{align*} cos(\theta) &= \frac{x_T}{R} = C, \\ \\ sin(\theta) &= \frac{y_T}{R} = S. \\ \end{align*}\end{split}

Where the rotation matrices for a negative and positive angle rotations are:

\begin{split}\begin{align*} \begin{bmatrix} x' \\ y' \end{bmatrix} &= \begin{bmatrix} C & S \\ -S & C \end{bmatrix}\cdot\begin{bmatrix} x \\ y \end{bmatrix} \\ \\ \begin{bmatrix} x \\ y \end{bmatrix} &= \begin{bmatrix} C & -S \\ S & C \end{bmatrix}\cdot\begin{bmatrix} x' \\ y' \end{bmatrix} \end{align*}\end{split}

## Calculation of velocity and position after reflection

Calculation of $$r_r$$ is performed by simple push from the point of reflection ($$r_T$$), with velocity $$v_r$$ (velocity after reflection), during time $$\Delta t (1 - \tau)$$.

We calculate the velocity after reflection in the rotated coordinate system ($$v'_r$$), since it is easily calculated as a simple inversion of the $$v'_x$$ component.

\begin{split}\begin{align*} v'_r = \begin{bmatrix} v'_x \\ v'_y \end{bmatrix}_{\text{after reflection}} = \begin{bmatrix} - v'_x \\ v'_y \end{bmatrix}_{\text{before reflection}} \end{align*}\end{split}

Which gives the following velocity vector in the original coordinate system:

\begin{split}\begin{align*} v_r = \begin{bmatrix} v_x \\ v_y \end{bmatrix}_{\text{after reflection}} = \begin{bmatrix} - C v'_x - S v'_y \\ - S v'_x + C v'_y \end{bmatrix}_{\text{before reflection}} \end{align*}\end{split}

Now we can directly calculate $$r_r$$ in the original coordinate system.

\begin{align*} r_r = \left(v \cdot \tau + v_r \cdot (1 - \tau)\right)\Delta t \end{align*}

## Check for multiple reflection

At this point we have new location $$r_r$$ which can be outside the cylindrical boundaries. In this case we repeat the same steps as above. The difference is that we have the initial position at $$r_T$$, with veloctiy vector $$v_r$$, and timestep set to $$\Delta t_{new} = (1 - \tau)\Delta t$$. The algorithm is repeated as many times as needed - until the $$r_r$$ stays inside the geometry.