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# Archive for October, 2013

## Huygens diffraction

Posted by peeterjoot on October 23, 2013

[Click here for a PDF of this post with nicer formatting]

We were presented with a diffraction result, that the intensity can be expressed as the Fourier transform of the aperture. Let’s review a derivation of that based on the Huygens principle. Consider the aperture of fig. 2.1.

Fig 1.1: Diffraction aperture

The Huygens principle expresses the amplitude of a wave $U(\mathbf{r})$ in terms of it’s amplitude $U_\circ$ at $\mathbf{r} = 0$ as

\begin{aligned}U(\mathbf{r}) \propto \frac{U_\circ e^{i k \left\lvert {\mathbf{r}} \right\rvert}}{\left\lvert {\mathbf{r}} \right\rvert}.\end{aligned} \hspace{\stretch{1}}(2.1)

For multiple point diffraction, the diffracted wave is a superposition of such contributions from all points in the aperture. For example, two exponentials are summed when considering a two slit diffraction apparatus. For a more general aperture as above, we’d form

\begin{aligned}U(\mathbf{R}') \propto U_\circ \int_{\mathrm{A}} d^2\mathbf{r} \frac{e^{i k \left\lvert {\mathbf{r} - \mathbf{R}} \right\rvert}}{\left\lvert {\mathbf{r} - \mathbf{R}} \right\rvert} \frac{e^{i k \left\lvert {\mathbf{R}' - \mathbf{r}} \right\rvert}}{\left\lvert {\mathbf{R}' - \mathbf{r}} \right\rvert}.\end{aligned} \hspace{\stretch{1}}(2.2)

Note that this the Huygens result is an approximation in many ways. Fresnel later fixed up the proportionality factor and considered the angular dependence of the between the source and diffracted rays (the obliquity factor). That corrected result is known as the Huygens-Fresnel principle. Kirchhoff later considered solutions to the scalar wave equations for the electromagnetic field components $\mathbf{E}$ and $\mathbf{B}$, ignoring the Maxwell coupling of these fields. See section 8.3.1 [1], section A.2 section 10.4 of [3], or section 5.2 of [2] for details. See section 9.8 [4] for a vector diffraction treatment.

Let’s proceed with using eq. 2.2 to obtain our result from class. For simplicity, first position the origin in the aperture itself as in fig. 2.2.

Fig 1.2: Diffraction aperture with origin in aperture

Now we are set to perform a first order expansion of the vector lengths $\left\lvert {\mathbf{r} - \mathbf{R}} \right\rvert$ and $\left\lvert {\mathbf{r} - \mathbf{R}'} \right\rvert$. It’s sufficient to consider just one of these, expanding to first order

\begin{aligned}\left\lvert {\mathbf{r} - \mathbf{R}} \right\rvert=\sqrt{(\mathbf{r} - \mathbf{R})^2} &= \sqrt{\mathbf{R}^2 + \mathbf{r}^2 - 2 \mathbf{r} \cdot \mathbf{R}} \\ &= R\sqrt{1 + \frac{\mathbf{r}^2}{\mathbf{R}^2} - 2 \mathbf{r} \cdot \frac{\mathbf{R}}{\mathbf{R}^2}} \\ &= \approx R \left( 1 + \frac{1}{{2}} \frac{\mathbf{r}^2}{\mathbf{R}^2} - \mathbf{r} \cdot \frac{\mathbf{R}}{\mathbf{R}^2} \right) \\ &= R + \frac{1}{{2}} \frac{\mathbf{r}^2}{R} - \mathbf{r} \cdot \frac{\mathbf{R}}{R}.\end{aligned} \hspace{\stretch{1}}(2.3)

Assume that both $\mathbf{R}$ and $\mathbf{R}'$ to be far enough from the aperture that we can approximate the $\left\lvert {\mathbf{R} - \mathbf{r}} \right\rvert$ and $\left\lvert {\mathbf{R}' - \mathbf{r}} \right\rvert$ terms downstairs as $R = \left\lvert {\mathbf{R}} \right\rvert$ and $R' = \left\lvert {\mathbf{R}'} \right\rvert$ respectively. Additionally, ignore the second order term above, significant for Fresnel diffraction where the diffraction pattern close to the aperture is examined, but not in the far field. This gives

\begin{aligned}U(\mathbf{R}') &\sim \frac{U_\circ }{R R'}\int_{\mathrm{A}} d^2\mathbf{r} e^{i k \left( R - \mathbf{r} \cdot \hat{\mathbf{R}} \right)} e^{i k \left( R' - \mathbf{r} \cdot \hat{\mathbf{R}}' \right)} \\ &= \frac{U_\circ }{R R'} e^{i k (R + R')}\int_{\mathrm{A}} d^2\mathbf{r} e^{-i k \mathbf{r} \cdot \hat{\mathbf{R}}} e^{-i k \mathbf{r} \cdot \hat{\mathbf{R}}'}.\end{aligned} \hspace{\stretch{1}}(2.4)

Finally write

\begin{aligned}\begin{aligned}\mathbf{k}_s &= - k\hat{\mathbf{R}} \\ \mathbf{k} &= k\hat{\mathbf{R}}',\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.5)

for

\begin{aligned}U(\mathbf{R}') \sim\frac{U_\circ }{R R'}e^{i k (R + R')}\int_{\mathrm{A}} d^2\mathbf{r} e^{-i \mathbf{r} \cdot (\mathbf{k} - \mathbf{k}_s)}.\end{aligned} \hspace{\stretch{1}}(2.6)

Finally, write

\begin{aligned}\mathbf{K} = \mathbf{k} - \mathbf{k}_s,\end{aligned} \hspace{\stretch{1}}(2.7)

and introduce an aperture function $\rho(\mathbf{r})$ (we called this the scattering density) that is unity for any regions of the aperture that light is allowed through, and zero when light is blocked, and some value in $[0,1]$ for translucent regions of the aperture. We can now expand the integral over the surface containing the aperture

\begin{aligned}\boxed{U(\mathbf{R}') \sim\frac{U_\circ }{R R'} e^{i k (R + R')}\int\rho(\mathbf{r}) e^{-i \mathbf{r} \cdot \mathbf{K}}.}\end{aligned} \hspace{\stretch{1}}(2.8)

# References

[1] M. Born and E. Wolf. Principles of optics: electromagnetic theory of propagation, interference and diffraction of light. Cambridge university press, 1980.

[2] G.R. Fowles. Introduction to modern optics. Dover Pubns, 1989.

[3] E. Hecht. Optics. 1998.

[4] JD Jackson. Classical Electrodynamics Wiley. John Wiley and Sons, 2nd edition, 1975.

Posted in Math and Physics Learning. | Tagged: , | Leave a Comment »

## What is the justification for such a harsh and expensive bylaw?

Posted by peeterjoot on October 20, 2013

Letter to my Markham council representative, after obtaining an “overnight street parking” ticket.  I don’t expect that my city “representation” will be able to change anything, especially since this insanity likely represents a nice revenue stream.

I live in the Cornell area, which has parking areas cut into segments of the boulevard for street parking that allow for parking without obstructing any flow of traffic.

For some reason there is a by-law that prohibits overnight (2:30 am – 6:00 am) parking in these areas.  These cut in segments are considered Markham roads, and included in the generic Markham no-overnight parking laws.

In winter months the no overnight parking on roads makes sense.  It could obstruct road clearing, and a blanket policy against such parking likely simplifies things.  Since the Markham plows clear out these cut out parking segments too, I don’t object to such a policy for the Cornell cut in parking segments in winter too.  However, what possible sense does this make in other seasons?

Yesterday it was convenient to park in the front of the house, especially since I had intended to go out again afterwards.  Plans changed and I ended up forgetting to move my car to the backyard driveway.  This was a very expensive memory lapse, and cost me 50! I am fully aware of the web form that allows for occasional use of this parking area (for visitors, driveway maintenance, …) but the penalty for simply forgetting to move the car when plans to go out change, is extremely harsh. Especially when we are parking in an area that does not obstruct flow of traffic and the season isn’t one for which snow plowing is relevant. If nothing else, the penalty for this is exorbitant. I can also imagine that this could potentially be abused. Somebody with multiple vehicles could use such an on street parking area as a personal extension to their driveway. This could introduce fairness issues, preventing other taxpaying neighbours from using a space that should also be available to them. This extreme scenario could easily be resolved without a harsh and default blanket penalty, since there is a complaint mechanism available for parking issues. I can’t help but feel that Markham uses this parking by-law as a revenue source, with no real non-monetary justification. This seems especially true in the Cornell area where there is no obstruction to traffic by these pseudo on-road parking areas. Do we not pay enough in property taxes, that we can’t occasionally park in front of our own houses without oppressive punishment for a failure to move the vehicle to our personal off street parking areas (driveways and garages)? Posted in Incoherent ramblings | Tagged: , , , , | 1 Comment » ## Exponential solutions to second order linear system Posted by peeterjoot on October 20, 2013 [Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)] # Motivation We’re discussing specific forms to systems of coupled linear differential equations, such as a loop of “spring” connected masses (i.e. atoms interacting with harmonic oscillator potentials) as sketched in fig. 1.1. Fig 1.1: Three springs loop Instead of assuming a solution, let’s see how far we can get attacking this problem systematically. # Matrix methods Suppose that we have a set of $N$ masses constrained to a circle interacting with harmonic potentials. The Lagrangian for such a system (using modulo $N$ indexing) is \begin{aligned}\mathcal{L} = \frac{1}{{2}} \sum_{k = 0}^{2} m_k \dot{u}_k^2 - \frac{1}{{2}} \sum_{k = 0}^2 \kappa_k \left( u_{k+1} - u_k \right)^2.\end{aligned} \hspace{\stretch{1}}(1.1) The force equations follow directly from the Euler-Lagrange equations \begin{aligned}0 = \frac{d{{}}}{dt} \frac{\partial {\mathcal{L}}}{\partial {\dot{u}_{n, \alpha}}}- \frac{\partial {\mathcal{L}}}{\partial {u_{n, \alpha}}}.\end{aligned} \hspace{\stretch{1}}(1.2) For the simple three particle system depicted above, this is \begin{aligned}\mathcal{L} = \frac{1}{{2}} m_0 \dot{u}_0^2 +\frac{1}{{2}} m_1 \dot{u}_1^2 +\frac{1}{{2}} m_2 \dot{u}_2^2 - \frac{1}{{2}} \kappa_0 \left( u_1 - u_0 \right)^2- \frac{1}{{2}} \kappa_1 \left( u_2 - u_1 \right)^2- \frac{1}{{2}} \kappa_2 \left( u_0 - u_2 \right)^2,\end{aligned} \hspace{\stretch{1}}(1.3) with equations of motion \begin{aligned}\begin{aligned}0 &= m_0 \dot{d}{u}_0 + \kappa_0 \left( u_0 - u_1 \right) + \kappa_2 \left( u_0 - u_2 \right) \\ 0 &= m_1 \dot{d}{u}_1 + \kappa_1 \left( u_1 - u_2 \right) + \kappa_0 \left( u_1 - u_0 \right) \\ 0 &= m_2 \dot{d}{u}_2 + \kappa_2 \left( u_2 - u_0 \right) + \kappa_1 \left( u_2 - u_1 \right),\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.4) Let’s partially non-dimensionalize this. First introduce average mass $\bar{{m}}$ and spring constants $\bar{{\kappa}}$, and rearrange slightly \begin{aligned}\begin{aligned}\frac{\bar{{m}}}{\bar{{k}}} \dot{d}{u}_0 &= -\frac{\kappa_0 \bar{{m}}}{\bar{{k}} m_0} \left( u_0 - u_1 \right) - \frac{\kappa_2 \bar{{m}}}{\bar{{k}} m_0} \left( u_0 - u_2 \right) \\ \frac{\bar{{m}}}{\bar{{k}}} \dot{d}{u}_1 &= -\frac{\kappa_1 \bar{{m}}}{\bar{{k}} m_1} \left( u_1 - u_2 \right) - \frac{\kappa_0 \bar{{m}}}{\bar{{k}} m_1} \left( u_1 - u_0 \right) \\ \frac{\bar{{m}}}{\bar{{k}}} \dot{d}{u}_2 &= -\frac{\kappa_2 \bar{{m}}}{\bar{{k}} m_2} \left( u_2 - u_0 \right) - \frac{\kappa_1 \bar{{m}}}{\bar{{k}} m_2} \left( u_2 - u_1 \right).\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.5) With \begin{aligned}\tau = \sqrt{\frac{\bar{{k}}}{\bar{{m}}}} t = \Omega t\end{aligned} \hspace{\stretch{1}}(1.0.6.6) \begin{aligned}\mathbf{u} = \begin{bmatrix}u_0 \\ u_1 \\ u_2\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(1.0.6.6) \begin{aligned}B = \begin{bmatrix}-\frac{\kappa_0 \bar{{m}}}{\bar{{k}} m_0} - \frac{\kappa_2 \bar{{m}}}{\bar{{k}} m_0} &\frac{\kappa_0 \bar{{m}}}{\bar{{k}} m_0} &\frac{\kappa_2 \bar{{m}}}{\bar{{k}} m_0} \\ \frac{\kappa_0 \bar{{m}}}{\bar{{k}} m_1} &-\frac{\kappa_1 \bar{{m}}}{\bar{{k}} m_1} - \frac{\kappa_0 \bar{{m}}}{\bar{{k}} m_1} &\frac{\kappa_1 \bar{{m}}}{\bar{{k}} m_1} \\ \frac{\kappa_2 \bar{{m}}}{\bar{{k}} m_2} & \frac{\kappa_1 \bar{{m}}}{\bar{{k}} m_2} &-\frac{\kappa_2 \bar{{m}}}{\bar{{k}} m_2} - \frac{\kappa_1 \bar{{m}}}{\bar{{k}} m_2} \end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(1.0.6.6) Our system takes the form \begin{aligned}\frac{d^2 \mathbf{u}}{d\tau^2} = B \mathbf{u}.\end{aligned} \hspace{\stretch{1}}(1.0.6.6) We can at least theoretically solve this in a simple fashion if we first convert it to a first order system. We can do that by augmenting our vector of displacements with their first derivatives \begin{aligned}\mathbf{w} =\begin{bmatrix}\mathbf{u} \\ \frac{d \mathbf{u}}{d\tau} \end{bmatrix},\end{aligned} \hspace{\stretch{1}}(1.0.6.6) So that \begin{aligned}\frac{d \mathbf{w}}{d\tau} =\begin{bmatrix}0 & I \\ B & 0\end{bmatrix} \mathbf{w}= A \mathbf{w}.\end{aligned} \hspace{\stretch{1}}(1.0.9) Now the solution is conceptually trivial \begin{aligned}\mathbf{w} = e^{A \tau} \mathbf{w}_\circ.\end{aligned} \hspace{\stretch{1}}(1.0.9) We are however, faced with the task of exponentiating the matrix $A$. All the powers of this matrix $A$ will be required, but they turn out to be easy to calculate \begin{aligned}{\begin{bmatrix}0 & I \\ B & 0\end{bmatrix} }^2=\begin{bmatrix}0 & I \\ B & 0\end{bmatrix} \begin{bmatrix}0 & I \\ B & 0\end{bmatrix} =\begin{bmatrix}B & 0 \\ 0 & B\end{bmatrix} \end{aligned} \hspace{\stretch{1}}(1.0.11a) \begin{aligned}{\begin{bmatrix}0 & I \\ B & 0\end{bmatrix} }^3=\begin{bmatrix}B & 0 \\ 0 & B\end{bmatrix} \begin{bmatrix}0 & I \\ B & 0\end{bmatrix} =\begin{bmatrix}0 & B \\ B^2 & 0\end{bmatrix} \end{aligned} \hspace{\stretch{1}}(1.0.11b) \begin{aligned}{\begin{bmatrix}0 & I \\ B & 0\end{bmatrix} }^4=\begin{bmatrix}0 & B \\ B^2 & 0\end{bmatrix} \begin{bmatrix}0 & I \\ B & 0\end{bmatrix} =\begin{bmatrix}B^2 & 0 \\ 0 & B^2\end{bmatrix},\end{aligned} \hspace{\stretch{1}}(1.0.11c) allowing us to write out the matrix exponential \begin{aligned}e^{A \tau} = \sum_{k = 0}^\infty \frac{\tau^{2k}}{(2k)!} \begin{bmatrix}B^k & 0 \\ 0 & B^k\end{bmatrix}+\sum_{k = 0}^\infty \frac{\tau^{2k + 1}}{(2k + 1)!} \begin{bmatrix}0 & B^k \\ B^{k+1} & 0\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(1.0.11c) Case I: No zero eigenvalues Provided that $B$ has no zero eigenvalues, we could factor this as \begin{aligned}\begin{bmatrix}0 & B^k \\ B^{k+1} & 0\end{bmatrix}=\begin{bmatrix}0 & B^{-1/2} \\ B^{1/2} & 0\end{bmatrix}\begin{bmatrix}B^{k + 1/2} & 0 \\ 0 & B^{k + 1/2}\end{bmatrix},\end{aligned} \hspace{\stretch{1}}(1.0.11c) This initially leads us to believe the following, but we’ll find out that the three springs interaction matrix $B$ does have a zero eigenvalue, and we’ll have to be more careful. If there were any such interaction matrices that did not have such a zero we could simply write \begin{aligned}e^{A \tau} = \sum_{k = 0}^\infty \frac{\tau^{2k}}{(2k)!} \begin{bmatrix}\sqrt{B}^{2k} & 0 \\ 0 & \sqrt{B}^{2k}\end{bmatrix}+\begin{bmatrix}0 & B^{-1/2} \\ B^{1/2} & 0\end{bmatrix}\sum_{k = 0}^\infty \frac{\tau^{2k + 1}}{(2k + 1)!} \begin{bmatrix}\sqrt{B}^{2 k + 1} & 0 \\ 0 & \sqrt{B}^{2 k+1} \end{bmatrix}=\cosh\begin{bmatrix}\sqrt{B} \tau & 0 \\ 0 & \sqrt{B} \tau\end{bmatrix}+ \begin{bmatrix}0 & 1/\sqrt{B} \tau \\ \sqrt{B} \tau & 0\end{bmatrix}\sinh\begin{bmatrix}\sqrt{B} \tau & 0 \\ 0 & \sqrt{B} \tau\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(1.0.11c) This is \begin{aligned}e^{A \tau}=\begin{bmatrix}\cosh \sqrt{B} \tau & (1/\sqrt{B}) \sinh \sqrt{B} \tau \\ \sqrt{B} \sinh \sqrt{B} \tau & \cosh \sqrt{B} \tau \end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(1.0.11c) The solution, written out is \begin{aligned}\begin{bmatrix}\mathbf{u} \\ \mathbf{u}'\end{bmatrix}=\begin{bmatrix}\cosh \sqrt{B} \tau & (1/\sqrt{B}) \sinh \sqrt{B} \tau \\ \sqrt{B} \sinh \sqrt{B} \tau & \cosh \sqrt{B} \tau \end{bmatrix}\begin{bmatrix}\mathbf{u}_\circ \\ \mathbf{u}_\circ'\end{bmatrix},\end{aligned} \hspace{\stretch{1}}(1.0.11c) so that \begin{aligned}\boxed{\mathbf{u} = \cosh \sqrt{B} \tau \mathbf{u}_\circ + \frac{1}{{\sqrt{B}}} \sinh \sqrt{B} \tau \mathbf{u}_\circ'.}\end{aligned} \hspace{\stretch{1}}(1.0.11c) As a check differentiation twice shows that this is in fact the general solution, since we have \begin{aligned}\mathbf{u}' = \sqrt{B} \sinh \sqrt{B} \tau \mathbf{u}_\circ + \cosh \sqrt{B} \tau \mathbf{u}_\circ',\end{aligned} \hspace{\stretch{1}}(1.0.11c) and \begin{aligned}\mathbf{u}'' = B \cosh \sqrt{B} \tau \mathbf{u}_\circ + \sqrt{B} \sinh \sqrt{B} \tau \mathbf{u}_\circ'= B \left( \cosh \sqrt{B} \tau \mathbf{u}_\circ + \frac{1}{{\sqrt{B}}} \sinh \sqrt{B} \tau \mathbf{u}_\circ' \right)= B \mathbf{u}.\end{aligned} \hspace{\stretch{1}}(1.0.11c) Observe that this solution is a general solution to second order constant coefficient linear systems of the form we have in eq. 1.5. However, to make it meaningful we do have the additional computational task of performing an eigensystem decomposition of the matrix $B$. We expect negative eigenvalues that will give us oscillatory solutions (ie: the matrix square roots will have imaginary eigenvalues). ## Example: An example diagonalization to try things out {example:threeSpringLoop:1}{ Let’s do that diagonalization for the simplest of the three springs system as an example, with $\kappa_j = \bar{{k}}$ and $m_j = \bar{{m}}$, so that we have \begin{aligned}B = \begin{bmatrix}-2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2 \end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(1.20) A orthonormal eigensystem for $B$ is \begin{aligned}\left\{\mathbf{e}_{-3, 1}, \mathbf{e}_{-3, 2}, \mathbf{e}_{0, 1} \right\}=\left\{\frac{1}{{\sqrt{6}}}\begin{bmatrix} -1 \\ -1 \\ 2 \end{bmatrix},\frac{1}{{\sqrt{2}}}\begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix},\frac{1}{{\sqrt{3}}}\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}\right\}.\end{aligned} \hspace{\stretch{1}}(1.21) With \begin{aligned}\begin{aligned}U &= \frac{1}{{\sqrt{6}}}\begin{bmatrix} -\sqrt{3} & -1 & \sqrt{2} \\ 0 & 2 & \sqrt{2} \\ \sqrt{3} & -1 & \sqrt{2} \end{bmatrix} \\ D &= \begin{bmatrix}-3 & 0 & 0 \\ 0 & -3 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.22) We have \begin{aligned}B = U D U^\text{T},\end{aligned} \hspace{\stretch{1}}(1.23) We also find that $B$ and its root are intimately related in a surprising way \begin{aligned}\sqrt{B} = \sqrt{3} iU \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}U^\text{T}=\frac{1}{{\sqrt{3} i}} B.\end{aligned} \hspace{\stretch{1}}(1.24) We also see, unfortunately that $B$ has a zero eigenvalue, so we can’t compute $1/\sqrt{B}$. We’ll have to back and up and start again differently. } Case II: allowing for zero eigenvalues Now that we realize we have to deal with zero eigenvalues, a different approach suggests itself. Instead of reducing our system using a Hamiltonian transformation to a first order system, let’s utilize that diagonalization directly. Our system is \begin{aligned}\mathbf{u}'' = B \mathbf{u} = U D U^{-1} \mathbf{u},\end{aligned} \hspace{\stretch{1}}(1.25) where $D = [ \lambda_i \delta_{ij} ]$ and \begin{aligned}\left( U^{-1} \mathbf{u} \right)'' = D \left( U^{-1} \mathbf{u} \right).\end{aligned} \hspace{\stretch{1}}(1.26) Let \begin{aligned}\mathbf{z} = U^{-1} \mathbf{u},\end{aligned} \hspace{\stretch{1}}(1.27) so that our system is just \begin{aligned}\mathbf{z}'' = D \mathbf{z},\end{aligned} \hspace{\stretch{1}}(1.28) or \begin{aligned}z_i'' = \lambda_i z_i.\end{aligned} \hspace{\stretch{1}}(1.29) This is $N$ equations, each decoupled and solvable by inspection. Suppose we group the eigenvalues into sets $\{ \lambda_n < 0, \lambda_p > 0, \lambda_z = 0 \}$. Our solution is then \begin{aligned}\mathbf{z} = \sum_{ \lambda_n < 0, \lambda_p > 0, \lambda_z = 0}\left( a_n \cos \sqrt{-\lambda_n} \tau + b_n \sin \sqrt{-\lambda_n} \tau \right)\mathbf{e}_n+\left( a_p \cosh \sqrt{\lambda_p} \tau + b_p \sinh \sqrt{\lambda_p} \tau \right)\mathbf{e}_p+\left( a_z + b_z \tau \right) \mathbf{e}_z.\end{aligned} \hspace{\stretch{1}}(1.30) Transforming back to lattice coordinates using $\mathbf{u} = U \mathbf{z}$, we have \begin{aligned}\mathbf{u} = \sum_{ \lambda_n < 0, \lambda_p > 0, \lambda_z = 0}\left( a_n \cos \sqrt{-\lambda_n} \tau + b_n \sin \sqrt{-\lambda_n} \tau \right)U \mathbf{e}_n+\left( a_p \cosh \sqrt{\lambda_p} \tau + b_p \sinh \sqrt{\lambda_p} \tau \right)U \mathbf{e}_p+\left( a_z + b_z \tau \right) U \mathbf{e}_z.\end{aligned} \hspace{\stretch{1}}(1.31) We see that the zero eigenvalues integration terms have no contribution to the lattice coordinates, since $U \mathbf{e}_z = \lambda_z \mathbf{e}_z = 0$, for all $\lambda_z = 0$. If $U = [ \mathbf{e}_i ]$ are a set of not necessarily orthonormal eigenvectors for $B$, then the vectors $\mathbf{f}_i$, where $\mathbf{e}_i \cdot \mathbf{f}_j = \delta_{ij}$ are the reciprocal frame vectors. These can be extracted from $U^{-1} = [ \mathbf{f}_i ]^\text{T}$ (i.e., the rows of $U^{-1}$). Taking dot products between $\mathbf{f}_i$ with $\mathbf{u}(0) = \mathbf{u}_\circ$ and $\mathbf{u}'(0) = \mathbf{u}'_\circ$, provides us with the unknown coefficients $a_n, b_n$ \begin{aligned}\mathbf{u}(\tau)= \sum_{ \lambda_n < 0, \lambda_p > 0 }\left( (\mathbf{u}_\circ \cdot \mathbf{f}_n) \cos \sqrt{-\lambda_n} \tau + \frac{\mathbf{u}_\circ' \cdot \mathbf{f}_n}{\sqrt{-\lambda_n} } \sin \sqrt{-\lambda_n} \tau \right)\mathbf{e}_n+\left( (\mathbf{u}_\circ \cdot \mathbf{f}_p) \cosh \sqrt{\lambda_p} \tau + \frac{\mathbf{u}_\circ' \cdot \mathbf{f}_p}{\sqrt{-\lambda_p} } \sinh \sqrt{\lambda_p} \tau \right)\mathbf{e}_p.\end{aligned} \hspace{\stretch{1}}(1.32) Supposing that we constrain ourself to looking at just the oscillatory solutions (i.e. the lattice does not shake itself to pieces), then we have \begin{aligned}\boxed{\mathbf{u}(\tau)= \sum_{ \lambda_n < 0 }\left( \sum_j \mathbf{e}_{n,j} \mathbf{f}_{n,j}^\text{T} \right)\left( \mathbf{u}_\circ \cos \sqrt{-\lambda_n} \tau + \frac{\mathbf{u}_\circ' }{\sqrt{-\lambda_n} } \sin \sqrt{-\lambda_n} \tau \right).}\end{aligned} \hspace{\stretch{1}}(1.33) Eigenvectors for eigenvalues that were degenerate have been explicitly enumerated here, something previously implied. Observe that the dot products of the form $(\mathbf{a} \cdot \mathbf{f}_i) \mathbf{e}_i$ have been put into projector operator form to group terms more nicely. The solution can be thought of as a weighted projector operator working as a time evolution operator from the initial state. ## Example: Our example interaction revisited {example:threeSpringLoop:2}{ Recall that we had an orthonormal basis for the $\lambda = -3$ eigensubspace for the interaction example of eq. 1.20 again, so $\mathbf{e}_{-3,i} = \mathbf{f}_{-3, i}$. We can sum $\mathbf{e}_{-3,1} \mathbf{e}_{-3, 1}^\text{T} + \mathbf{e}_{-3,2} \mathbf{e}_{-3, 2}^\text{T}$ to find \begin{aligned}\mathbf{u}(\tau)= \frac{1}{{3}}\begin{bmatrix}2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2\end{bmatrix}\left( \mathbf{u}_\circ \cos \sqrt{3} \tau + \frac{ \mathbf{u}_\circ' }{\sqrt{3} } \sin \sqrt{3} \tau \right).\end{aligned} \hspace{\stretch{1}}(1.34) The leading matrix is an orthonormal projector of the initial conditions onto the eigen subspace for $\lambda_n = -3$. Observe that this is proportional to $B$ itself, scaled by the square of the non-zero eigenvalue of $\sqrt{B}$. From this we can confirm by inspection that this is a solution to $\mathbf{u}'' = B \mathbf{u}$, as desired. } # Fourier transform methods Let’s now try another item from our usual toolbox on these sorts of second order systems, the Fourier transform. For a one variable function of time let’s write the transform pair as \begin{aligned}x(t) = \int_{-\infty}^{\infty} \tilde{x}(\omega) e^{-i \omega t} d\omega\end{aligned} \hspace{\stretch{1}}(1.0.35.35) \begin{aligned}\tilde{x}(\omega) = \frac{1}{{2\pi}} \int_{-\infty}^{\infty} x(t) e^{i \omega t} dt\end{aligned} \hspace{\stretch{1}}(1.0.35.35) One mass harmonic oscillator The simplest second order system is that of the harmonic oscillator \begin{aligned}0 = \dot{d}{x}(t) + \omega_\circ^2 x.\end{aligned} \hspace{\stretch{1}}(1.0.35.35) Application of the transform gives \begin{aligned}0 = \left( \frac{d^2}{dt^2} + \omega_\circ^2 \right)\int_{-\infty}^{\infty} \tilde{x}(\omega) e^{-i \omega t} d\omega= \int_{-\infty}^{\infty} \left( -\omega^2 + \omega_\circ^2 \right)\tilde{x}(\omega) e^{-i \omega t} d\omega.\end{aligned} \hspace{\stretch{1}}(1.0.35.35) We clearly have a constraint that is a function of frequency, but one that has to hold for all time. Let’s transform this constraint to the frequency domain to consider that constraint independent of time. \begin{aligned}0 =\frac{1}{{2 \pi}}\int_{-\infty}^{\infty} dt e^{ i \omega t}\int_{-\infty}^{\infty} \left( -{\omega'}^2 + \omega_\circ^2 \right)\tilde{x}(\omega') e^{-i \omega' t} d\omega'=\int_{-\infty}^{\infty} d\omega'\left( -{\omega'}^2 + \omega_\circ^2 \right)\tilde{x}(\omega') \frac{1}{{2 \pi}}\int_{-\infty}^{\infty} dt e^{ i (\omega -\omega') t}=\int_{-\infty}^{\infty} d\omega'\left( -{\omega'}^2 + \omega_\circ^2 \right)\tilde{x}(\omega') \delta( \omega - \omega' )=\left( -{\omega}^2 + \omega_\circ^2 \right)\tilde{x}(\omega).\end{aligned} \hspace{\stretch{1}}(1.0.35.35) How do we make sense of this? Since $\omega$ is an integration variable, we can’t just mandate that it equals the constant driving frequency $\pm \omega_\circ$. It’s clear that we require a constraint on the transform $\tilde{x}(\omega)$ as well. As a trial solution, imagine that \begin{aligned}\tilde{x}(\omega) = \left\{\begin{array}{l l}\tilde{x}_\circ & \quad \mbox{iflatex \left\lvert {\omega – \pm \omega_\circ} \right\rvert < \omega_{\text{cutoff}}} \\ 0 & \quad \mbox{otherwise}\end{array}\right.\end{aligned} \hspace{\stretch{1}}(1.0.35.35)\$

This gives us

\begin{aligned}0 = \tilde{x}_\circ\int_{\pm \omega_\circ -\omega_{\text{cutoff}}}^{\pm \omega_\circ +\omega_{\text{cutoff}}}\left( \omega^2 - \omega_\circ^2 \right)e^{-i \omega t} d\omega.\end{aligned} \hspace{\stretch{1}}(1.0.35.35)

Now it is clear that we can satisfy our constraint only if the interval $[\pm \omega_\circ -\omega_{\text{cutoff}}, \pm \omega_\circ + \omega_{\text{cutoff}}]$ is made infinitesimal. Specifically, we require both a $\omega^2 = \omega_\circ^2$ constraint and that the transform $\tilde{x}(\omega)$ have a delta function nature. That is

\begin{aligned}\tilde{x}(\omega) = A \delta(\omega - \omega_\circ)+ B \delta(\omega + \omega_\circ).\end{aligned} \hspace{\stretch{1}}(1.0.35.35)

Substitution back into our transform gives

\begin{aligned}x(t) = A e^{-i \omega_\circ t}+ B e^{i \omega_\circ t}.\end{aligned} \hspace{\stretch{1}}(1.0.35.35)

We can verify quickly that this satisfies our harmonic equation $\dot{d}{x} = -\omega_\circ x$.

Two mass harmonic oscillator

Having applied the transform technique to the very simplest second order system, we can now consider the next more complex system, that of two harmonically interacting masses (i.e. two masses on a frictionless spring).

F1

Our system is described by

\begin{aligned}\mathcal{L} = \frac{1}{{2}} m_1 \dot{u}_1^2+ \frac{1}{{2}} m_2 \dot{u}_2^2-\frac{1}{{2}} \kappa \left( u_2 - u_1 \right)^2,\end{aligned} \hspace{\stretch{1}}(1.0.35.35)

and the pair of Euler-Lagrange equations

\begin{aligned}0 = \frac{d}{dt} \frac{\partial {\mathcal{L}}}{\partial {\dot{u}_i}}- \frac{\partial {\mathcal{L}}}{\partial {u_i}}.\end{aligned} \hspace{\stretch{1}}(1.0.35.35)

The equations of motion are

\begin{aligned}\begin{aligned}0 &= m_1 \dot{d}{u}_1 + \kappa \left( u_1 - u_2 \right) \\ 0 &= m_2 \dot{d}{u}_2 + \kappa \left( u_2 - u_1 \right) \end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.35.35)

Let

\begin{aligned}\begin{aligned}u_1(t) &= \int_{-\infty}^\infty \tilde{u}_1(\omega) e^{-i \omega t} d\omega \\ u_2(t) &= \int_{-\infty}^\infty \tilde{u}_2(\omega) e^{-i \omega t} d\omega.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.35.35)

Insertion of these transform pairs into our equations of motion produces a pair of simultaneous integral equations to solve

\begin{aligned}\begin{aligned}0 &= \int_{-\infty}^\infty \left( \left( -m_1 \omega^2 + \kappa \right) \tilde{u}_1(\omega) - \kappa \tilde{u}_2(\omega) \right)e^{-i \omega t} d\omega \\ 0 &= \int_{-\infty}^\infty \left( \left( -m_2 \omega^2 + \kappa \right) \tilde{u}_2(\omega) - \kappa \tilde{u}_1(\omega) \right)e^{-i \omega t} d\omega.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.35.35)

As with the single spring case, we can decouple these equations with an inverse transformation operation $\int e^{i \omega' t}/2\pi$, which gives us (after dropping primes)

\begin{aligned}0 = \begin{bmatrix}\left( -m_1 \omega^2 + \kappa \right) &- \kappa \\ - \kappa &\left( -m_2 \omega^2 + \kappa \right) \end{bmatrix}\begin{bmatrix}\tilde{u}_1(\omega) \\ \tilde{u}_2(\omega)\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(1.0.35.35)

Taking determinants gives us the constraint on the frequency

\begin{aligned}0 = \left( -m_1 \omega^2 + \kappa \right) \left( -m_2 \omega^2 + \kappa \right) - \kappa^2=m_1 m_2 \omega^4-(m_1 + m_2) \omega^2 =\omega^2 \left( m_1 m_2 \omega^2 -\kappa (m_1 + m_2) \right).\end{aligned} \hspace{\stretch{1}}(1.0.35.35)

Introducing a reduced mass

\begin{aligned}\frac{1}{{\mu}} = \frac{1}{m_1}+\frac{1}{m_2},\end{aligned} \hspace{\stretch{1}}(1.0.35.35)

the pair of solutions are

\begin{aligned}\begin{aligned}\omega^2 &= 0 \\ \omega^2 &=\frac{\kappa}{\mu}\equiv \omega_\circ^2.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.35.35)

As with the single mass oscillator, we require the functions $\tilde{u}{\omega}$ to also be expressed as delta functions. The frequency constraint and that delta function requirement together can be expressed, for $j \in \{0, 1\}$ as

\begin{aligned}\tilde{u}_j(\omega) = A_{j+} \delta( \omega - \omega_\circ )+ A_{j0} \delta( \omega )+ A_{j-} \delta( \omega + \omega_\circ ).\end{aligned} \hspace{\stretch{1}}(1.0.35.35)

With a transformation back to time domain, we have functions of the form

\begin{aligned}\begin{aligned}u_1(t) &= A_{j+} e^{ -i \omega_\circ t }+ A_{j0} + A_{j-} e^{ i \omega_\circ t } \\ u_2(t) &= B_{j+} e^{ -i \omega_\circ t }+ B_{j0} + B_{j-} e^{ i \omega_\circ t }.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.35.35)

Back insertion of these into the equations of motion, we have

\begin{aligned}\begin{aligned}0 &=-m_1 \omega_\circ^2\left( A_{j+} e^{ -i \omega_\circ t } + A_{j-} e^{ i \omega_\circ t } \right)+ \kappa\left( \left( A_{j+} - B_{j+} \right) e^{ -i \omega_\circ t } + \left( A_{j-} - B_{j-} \right) e^{ i \omega_\circ t } + A_{j0} - B_{j0} \right) \\ 0 &=-m_2 \omega_\circ^2\left( B_{j+} e^{ -i \omega_\circ t } + B_{j-} e^{ i \omega_\circ t } \right)+ \kappa\left( \left( B_{j+} - A_{j+} \right) e^{ -i \omega_\circ t } + \left( B_{j-} - A_{j-} \right) e^{ i \omega_\circ t } + B_{j0} - A_{j0} \right)\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.35.35)

Equality requires identity for all powers of $e^{i \omega_\circ t}$, or

\begin{aligned}\begin{aligned}0 &= B_{j0} - A_{j0} \\ 0 &= -m_1 \omega_\circ^2 A_{j+} + \kappa \left( A_{j+} - B_{j+} \right) \\ 0 &= -m_1 \omega_\circ^2 A_{j-} + \kappa \left( A_{j-} - B_{j-} \right) \\ 0 &= -m_2 \omega_\circ^2 B_{j+} + \kappa \left( B_{j+} - A_{j+} \right) \\ 0 &= -m_2 \omega_\circ^2 B_{j-} + \kappa \left( B_{j-} - A_{j-} \right),\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.35.35)

or $B_{j0} = A_{j0}$ and

\begin{aligned}0 =\begin{bmatrix}\kappa -m_1 \omega_\circ^2 & 0 & - \kappa & 0 \\ 0 & \kappa -m_1 \omega_\circ^2 & 0 & - \kappa \\ -\kappa & 0 & \kappa -m_2 \omega_\circ^2 & 0 \\ 0 & -\kappa & 0 & \kappa -m_2 \omega_\circ^2 \end{bmatrix}\begin{bmatrix}A_{j+} \\ A_{j-} \\ B_{j+} \\ B_{j-} \end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(1.0.35.35)

Observe that

\begin{aligned}\kappa - m_1 \omega_\circ^2=\kappa - m_1 \kappa \left( \frac{1}{{m_1}} + \frac{1}{{m_2}} \right)=\kappa \left( 1 - 1 - \frac{m_1}{m_2} \right)= -\kappa \frac{m_1}{m_2},\end{aligned} \hspace{\stretch{1}}(1.0.35.35)

(with a similar alternate result). We can rewrite eq. 1.0.35.35 as

\begin{aligned}0 =-\kappa\begin{bmatrix}m_1/m_2 & 0 & 1 & 0 \\ 0 & m_1/m_2 & 0 & 1 \\ 1 & 0 & m_2/m_1 & 0 \\ 0 & 1 & 0 & m_2/m_1 \end{bmatrix}\begin{bmatrix}A_{j+} \\ A_{j-} \\ B_{j+} \\ B_{j-} \end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(1.0.35.35)

It’s clear that there’s two pairs of linear dependencies here, so the determinant is zero as expected. We can read off the remaining relations. Our undetermined coefficients are given by

\begin{aligned}\begin{aligned}B_{j0} &= A_{j0} \\ m_1 A_{j+} &= -m_2 B_{j+} \\ m_1 A_{j-} &= -m_2 B_{j-} \end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.35.35)

\begin{aligned}\boxed{\begin{aligned}u_1(t) &= a+ m_2 b e^{ -i \omega_\circ t }+ m_2 c e^{ i \omega_\circ t } \\ u_2(t) &= a- m_1 b e^{ -i \omega_\circ t }- m_1 c e^{ i \omega_\circ t }.\end{aligned}}\end{aligned} \hspace{\stretch{1}}(1.0.35.35)

Observe that the constant term is not really of interest, since it represents a constant displacement of both atoms (just a change of coordinates).

Check:

\begin{aligned}u_1(t) - u_2(t) = + (m_1 + m_2)b e^{ -i \omega_\circ t }+ (m_1 + m_2)c e^{ i \omega_\circ t },\end{aligned} \hspace{\stretch{1}}(1.0.35.35)

\begin{aligned}m_1 \dot{d}{u}_1(t) + \kappa (u_1(t) - u_2(t) )=-m_1 m_2 \omega_\circ^2 \left( b e^{ -i \omega_\circ t } + c e^{ i \omega_\circ t } \right)+(m_1 + m_2) \kappa \left( b e^{ -i \omega_\circ t } + c e^{ i \omega_\circ t } \right)=\left( -m_1 m_2 \omega_\circ^2 + (m_1 + m_2) \kappa \right)\left( b e^{ -i \omega_\circ t } + c e^{ i \omega_\circ t } \right)=\left( -m_1 m_2 \kappa \frac{m_1 + m_2}{m_1 m_2} + (m_1 + m_2) \kappa \right)\left( b e^{ -i \omega_\circ t } + c e^{ i \omega_\circ t } \right)= 0.\end{aligned} \hspace{\stretch{1}}(1.0.35.35)

# Reflection

We’ve seen that we can solve any of these constant coefficient systems exactly using matrix methods, however, these will not be practical for large systems unless we have methods to solve for all the non-zero eigenvalues and their corresponding eigenvectors. With the Fourier transform methods we find that our solutions in the frequency domain is of the form

\begin{aligned}\tilde{u}_j(\omega) = \sum a_{kj} \delta( \omega - \omega_{kj} ),\end{aligned} \hspace{\stretch{1}}(1.63)

or in the time domain

\begin{aligned}u(t) = \sum a_{kj} e^{ - i \omega_{kj} t }.\end{aligned} \hspace{\stretch{1}}(1.64)

We assumed exactly this form of solution in class. The trial solution that we used in class factored out a phase shift from $a_{kj}$ of the form of $e^{ i q x_n }$, but that doesn’t change the underling form of that assumed solution. We have however found a good justification for the trial solution we utilized.

## Discrete Fourier transform

Posted by peeterjoot on October 11, 2013

[Click here for a PDF of this post with nicer formatting]

For decoupling trial solutions of lattice vibrations we have what appears to be a need for the use of a discrete Fourier transform. This is described briefly in [1], but there’s no proof of the inversion formula. Let’s work through that detail.

Let’s start given a periodic signal of the following form

\begin{aligned}a_n = \sum_{k = 0}^{N-1} A_k e^{-2 \pi i k n/N },\end{aligned} \hspace{\stretch{1}}(1.1)

and assume that there’s an inversion formula of the form

\begin{aligned}A_k' \propto \sum_{n = 0}^{N-1} a_n e^{2 \pi i k' n/N }.\end{aligned} \hspace{\stretch{1}}(1.2)

Direct substitution should show if such a transform pair is valid, and determine the proportionality constant. Let’s try this

\begin{aligned}\sum_{n = 0}^{N-1} a_n e^{2 \pi i k' n/N }=\sum_{n = 0}^{N-1} e^{2 \pi i k' n/N }\sum_{k = 0}^{N-1} A_k e^{-2 \pi i k n/N }=\sum_{k = 0}^{N-1} A_k \sum_{n = 0}^{N-1} e^{2 \pi i (k' - k)n/N }.\end{aligned} \hspace{\stretch{1}}(1.3)

Observe that the interior sum is just $N$ when $k' = k$. Let’s sum this for the values $k \ne k'$, writing

\begin{aligned}S_N(k'-k) = \sum_{n = 0}^{N-1} e^{2 \pi i (k' - k)n/N }.\end{aligned} \hspace{\stretch{1}}(1.4)

With $a = \exp( 2 \pi i (k' - k)/N)$ we have

\begin{aligned}S_N(k'-k) = \sum_{n = 0}^{N-1} a^n= \frac{a^N - 1}{a - 1}= \frac{a^{N/2}}{a^{1/2}} \frac{a^{N/2} - a^{-N/2}}{a^{1/2} - a^{-1/2}}= e^{ \pi i (k' - k)(1 - 1/N)} \frac{ \sin \pi (k'-k) }{\sin \pi (k' - k)/N}.\end{aligned} \hspace{\stretch{1}}(1.5)

Observe that $k' - k \in [-N+1, N-1]$, and necessarily takes on just integer values. We have terms of the form $\sin \pi m$, for integer $m$ in the numerator, always zero. In the denominator, the sine argument is in the range

\begin{aligned}[ \pi \left( -1 + \frac{1}{{N}} \right), \pi \left( 1 - \frac{1}{{N}} \right)],\end{aligned} \hspace{\stretch{1}}(1.6)

We can visualize that range as all the points on a sine curve with the integer multiples of $\pi$ omitted, as in fig. 1.1.

Fig 1.1: Sine with integer multiples of pi omitted

Clearly the denominator is always non-zero when $k \ne k'$. This provides us with the desired inverse transformation relationship

\begin{aligned}\sum_{n = 0}^{N-1} a_n e^{2 \pi i k' n/N }= N \sum_{k' = 0}^{N-1} A_k' \delta_{k, k'}= N A_k'.\end{aligned} \hspace{\stretch{1}}(1.7)

Summary

Now that we know the relationship between the discrete set of values $a_n$ and this discrete transformation of those values, let’s write the transform pair in a form that explicitly expresses this relationship.

\begin{aligned}\boxed{\begin{aligned}a_n &= \sum_{k = 0}^{N-1} \tilde{a}_k e^{-2 \pi i k n/N } \\ \tilde{a}_k &= \frac{1}{{N}} \sum_{n = 0}^{N-1} a_n e^{2 \pi i k n/N }.\end{aligned}}\end{aligned} \hspace{\stretch{1}}(1.8)

We have also shown that our discrete sum of exponentials has an delta function operator nature, a fact that will likely be handy in various manipulations.

\begin{aligned}\boxed{\sum_{n = 0}^{N-1} e^{2 \pi i (k' - k)n/N } = N \delta_{k, k'}.}\end{aligned} \hspace{\stretch{1}}(1.9)

# References

[1] S.S. Haykin. Communication systems. Wiley, 1994.

## Letter to Markham ward 5 councillor, re: proposed publicly funded NHL arena

Posted by peeterjoot on October 3, 2013

Here’s my minor attempt at playing the “democracy” game.  I don’t really expect this to have any effect, but figured it wouldn’t hurt to send in my 2c anyways:

Greetings Councillor Cambell,

I’m a resident and Markham taxpayer, living in Ward 5 at <address>

I have I’d like to express that I do not have any interest in funding an arena that is almost surely to end up as a new financial liability for current and future Markham taxpayers.

Given the historical failure of so many public funded arenas at profitability, an extremely convincing case ought to be required for this proposed project.  Additionally, all those that are lobbying for the project ought to be publicly identified, since they stand to benefit substantially from the proposed subsidies and hidden future tax incentives.  Those supporting those lobbyists in city council also ought to be publicly identified so that a future audit for potential benefits received can be performed.

We live in a culture that has a collective case of Alzheimers, so I like to point out that it wasn’t that long ago that Toronto’s Skydome failed spectacularly as an Ontario and city investment, eventually sold again and again at firesale prices until Rogers eventually bought it.  As Ontario tax payers, we are surely still paying for interest charges on the debt incurred for this project, despite the fact that it no longer provides any direct source of income to finance that debt.

I found the following relevant :

This last points out that the Montreal stadium took three decades to pay off!

An arena paid for using Markham city funds and tax benefits, is really an arena that will be paid with taxpayer debt servitude, directly or indirectly.  Allowing our council to fall for this sales pitch will mean that we will almost surely end up, like so many other municipalities, playing out the same old pathetic story of padding the pockets of commercial developers at the expense of the residents.

For non-Markham residents, here’s a couple stories with some of the background.  Also interesting is the survey that can be found on the Markham council website, since it outlines some of the tax shelters that were being provided with the package.

http://www.thestar.com/news/gta/2013/08/07/proposed_markham_arena_draws_chinese_investor_interest.html

http://www.thestar.com/news/gta/2013/08/08/markham_arena_fight_escalates_as_city_council_vote_looms.html