Simplified approximation to a single Taylor screen atmosphere

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The full thesis can be downloaded with high quality figures using the links on the Contents page. To begin with I will look at some basic imaging properties of the simple single-lens telescope shown in Figure A.1. Point $p$ in the figure is within the image plane of the telescope. The contribution to the optical flux at this point can be determined by first selecting a plane $X$ which is perpendicular to the line of sight from $p$ (joining the point $p$ and the centre of the lens in Figure A.1). The integral of the complex wavefunction across the plane $X$ determines the contribution to the wavefunction amplitude at point $p$. The photon flux is proportional to the square of this amplitude.

Figure A.1: Plane waves are focussed by a single-lens telescope onto a point $p$ in the image plane. Plane $X$ is perpendicular to the direction of propagation of the light focussed onto point $p$. The optical flux at point $p$ can be determined by integrating the optical wavefunction across plane $X$, and squaring the amplitude of the result. The phase of the wavefunction $\phi$ at plane $X$ is represented schematically in the figure by the distance to the nearest wavefront peak.

Figure A.2 shows the same situation as Figure A.1, but the incoming wavefronts have been perturbed by the atmosphere. The phase fluctuations will affect the integral of the wavefunction over plane $X$ altering the measured intensity at point $p$.

Figure A.2: The planar wavefronts of Figure A.1 are replaced by atmospherically perturbed wavefronts. The magnitude of the integral across plane $X$ will be reduced due to the phase perturbations in the wavefunction, reducing the optical flux measured at point $p$.

For the case of a single wind-blown Taylor screen, the phase perturbations introduced by the atmosphere will be translated laterally across the telescope aperture by the wind with no change in the structure of these perturbations. The modulus of the integral of the wavefunction over the plane $X$ is not directly affected by the lateral motion of the phase fluctuations, but the motion of the screen introduces new phase perturbations at the upwind side of the aperture and removes phase perturbations at the downwind side. This can be seen clearly if the outline of the telescope aperture is projected along the line of sight from $p$ onto the Taylor screen, as shown by the solid circle in Figure A.3. After time has elapsed and the wind-blown Taylor screen has moved a distance $\rho$, the outline of telescope aperture will be projected onto the dotted circle. Area $B$ is common to both timepoints and will contribute equally to the amplitude of the wavefunction at point $p$ in the image plane, but the contribution from area $A$ will be lost, and a new contribution from area $C$ will be included at the later timepoint. In reality, the atmospheric phase in areas $C$ and $A$ will remain correlated to that in area $B$ over a region extending approximately $r_{0}$ from the boundary of area $B$, but for the particular case of large diameter apertures this only has a small effect on our calculations and will be neglected in this simple approximation.

Figure A.3: For observations at the zenith, the solid circle represents the original projection of the telescope aperture onto the Taylor screen, while the dotted circle represents the same projection after the Taylor screen has been blown a distance $\rho$ by the wind. $d$ is the diameter of the telescope primary. For observations away from the zenith, projection effects will slightly elongate the circles.

We can write the contributions $\psi_{A}$, $\psi_{B}$ and $\psi_{C}$ to the wavefunction at point $p$ from areas $A$, $B$ and $C$ respectively as:

$$\psi_{A} = \chi_{A} e^{i\phi_{A}}$$ (A.1)

$$\psi_{B} = \chi_{B} e^{i\phi_{B}}$$ (A.2)

$$\psi_{C} = \chi_{C} e^{i\phi_{C}}$$ (A.3)

where $\chi_{A}$, $\chi_{B}$ and $\chi_{C}$ describe the amplitudes and $\phi_{A}$, $\phi_{B}$ and $\phi_{C}$ the phases of these contributions. If the linear dimensions of areas $A$, $B$ and $C$ in Figure A.3 are much larger than $r_{0}$, then the phases $\phi_{A}$, $\phi_{B}$ and $\phi_{C}$ will not be correlated with each other, as the structure function of Equation 1.3 indicates that the typical phase variation between points separated by distances much greater $r_{0}$ will be many cycles in magnitude.

If the linear dimensions of areas $A$, $B$ and $C$ are much larger than $r_{0}$ then the ensemble average amplitudes $\left < \chi_{A} \right >$, $\left < \chi_{B} \right >$ and $\left < \chi_{C} \right >$ will be proportional to the square root of the areas of $A$, $B$ and $C$ respectively. This can most clearly be seen if we imagine utilising a telescope whose aperture has the same size and shape as one of these three regions. The atmospheric seeing will generate an image with a FWHM of approximately $\lambda/r_{0}$ regardless of the aperture size and shape (as long as the aperture is much larger than $r_{0}$), with an average intensity proportional to the area of the aperture.

As the phases $\phi_{A}$, $\phi_{B}$ and $\phi_{C}$ are not correlated with each other, if the light from more than one of the these regions is combined then the ensemble averages of the relevant amplitudes must be added in quadrature to give the total amplitude. It is useful to consider three ensemble average intensities $\left < I_{A} \right >$, $\left < I_{B} \right >$ and $\left < I_{C} \right >$ which describe the contributions from the areas $A$, $B$ and $C$ as follows:

$$\left < I_{A} \right > = \left < \left \vert \psi_{A} \right \vert ^{2} \right >$$ (A.4)

$$\left < I_{B} \right > = \left < \left \vert \psi_{B} \right \vert ^{2} \right >$$ (A.5)

$$\left < I_{C} \right > = \left < \left \vert \psi_{C} \right \vert ^{2} \right >$$ (A.6)

As we have assumed that the wavefunctions $\psi_{A}$, $\psi_{B}$ and $\psi_{C}$ are not correlated (not coherent), the ensemble average intensities $\left < I_{A} \right >$, $\left < I_{B} \right >$ and $\left < I_{C} \right >$ can be summed linearly. As the areas of $A$ and $C$ are equal, the corresponding ensemble average intensities will be equal:

$$\left < I_{A} \right > = \left < I_{C} \right >$$ (A.7)

We are interested in the intensity $I_{AB}$ produced at the first timepoint when light from areas $A$ and $B$ is combined:

$$I_{AB}= \left \vert \psi_{A}+\psi_{B} \right \vert ^{2}$$ (A.8)

and the intensity $I_{BC}$ from areas $B$ and $C$, corresponding to a time when the Taylor screen has moved by a distance $\rho$ in Figure A.3:

$$I_{BC}= \left \vert \psi_{B}+\psi_{C} \right \vert ^{2}$$ (A.9)

The ensemble average of both of these intensities is simply the sum of the relevant ensemble average intensities for the constituent components:

$$\left < I_{AB} \right > = \left < I_{A} \right > + \left < I_{B} \right >$$ (A.10)

$$\left < I_{BC} \right > = \left < I_{B} \right > + \left < I_{C} \right >$$ (A.11)

As the total area of each of the circles in Figure A.3 is independent of $\rho$, the combined intensities $\left < I_{AB} \right >$ and $\left < I_{BC} \right >$ will be independent of the offset $\rho$.

The value of $\left < I_{B} \right >$ is directly dependent on $\rho$. For non-zero values of $\left < I_{B} \right >$, the fluctuations in the instantaneous intensities $I_{AB}$ and $I_{BC}$ will be correlated, as both intensities include a contribution from $\psi_{B}$. Conversely, if the value of $\left < I_{B} \right >$ were zero then the contribution $\psi_{B}$ would be zero, and fluctuations in $I_{AB}$ and $I_{BC}$ would be completely uncorrelated. We are interested in determining the size of the contribution $\left < I_{B} \right >$ for which the correlation between fluctuations in $I_{AB}$ and $I_{BC}$ has dropped by a factor of $1/e$. As the wavefront components $\psi_{A}$, $\psi_{B}$ and $\psi_{C}$ are uncorrelated Rayleigh distributions, this will be true when the ensemble average intensity $\left < I_{B} \right >$ from area $B$ contributes $1/\sqrt{e}$ of the total intensity:

$$\left < I_{B} \right > = \frac{I_{AB}}{\sqrt{e}}$$ (A.12)

The intensity $\left < I_{B} \right >$ will obey Equation A.12 when areas $A$ and $B$ are related as follows:

$$B=\frac{A + B}{\sqrt{e}}$$ (A.13)

If the telescope aperture is described by a function $\chi_{t} \left(\mathbf{r}\right)$ such as the example case in Equation 1.7, then area of overlap $B$ between two offset apertures comes directly from the autocorrelation of this function $R_{\chi} \left ( \rho \right )$:

$$R_{\chi} \left ( \rho \right )\equiv \int_{-\infty}^{\infty}... ...} \left ( r + \rho \right ) \chi_{t}^{*} \left ( r \right ) dr$$ (A.14)

It is useful to note that this remains true regardless of the shape of the telescope aperture described by $\chi_{t} \left(\mathbf{r}\right)$.

For the simple case presented in Figure A.3 of a filled circular aperture of diameter $d$, the area of overlap $B$ can be calculated geometrically. The line joining $q$ and $r$ in Figure A.3 is a chord to both the dotted and filled circles. $B$ is constructed from two symmetric regions either side of this chord, each having an area:

$$\frac{B}{2}=\frac{d^{2}}{4} \arccos \left ( \frac{\rho}{d} \right ) - \frac{\rho}{4}\sqrt{d^{2}-\rho^{2}}$$ (A.15)

The separation at which the area of $B$ is reduced by a factor of $1/\sqrt{e}$ was evaluated numerically as $\rho_{e}=0.31d$ for a telescope of diameter $d$, giving a coherence timescale:

$$\tau_{e}=\frac{0.31d}{\left \vert \mathbf{v} \right \vert}$$ (A.16)

for a constant wind velocity $\mathbf{v}$.