## FANDOM

138 Pages

Consider a luminous object at some point in space, an observer at another point, and in between is a layer of some medium with randomly varying refractive index, like our atmosphere. Let us suppose, further, that the variation in the refractive index are due to Kolmogorov turbulence. Such a turbulence is characterised by some dissipation term $\varepsilon$ with units of energy loss per unit mass, or length squared divided by time cubed

$\left[ \varepsilon \right] = \frac{L^2}{T^3}$

We want to find the velocity structure function, given by

$D_v \left(\vec{r}_1, \vec{r}_2 \right) = <|\vec{v} \left(\vec{r}_1 \right) -\vec{v} \left(\vec{r}_2 \right)|^2>$

Due to translational symmetry $D_v$ can only depend on the magnitude of the difference between two position $r = |\vec{r}_1 - \vec{r}_2|$. From dimensional analysis

$D_v \approx \varepsilon^{2/3} r^{2/3}$. We assume that the turbulence is highly subsonic, so the the velocity perturbations cause temperature and density perturbations, such that the dependence of their structure function $r$ is the same. The refractive index $n$ is some function of the density and temperature, so its structure function also has the same dependence on the distance $D_n \propto r^{2/3}$.

From $D_n$ we proceed to obtain the phase structure function $D_{\phi}$. This relation depend on the dispersion relation. In the case of air, the variation in the refractive index is proportional to the variation in the density. Since we are dealing with small perturbations, $D_{\phi} \propto D_n$. The wave number should appear with the same power as the refractive index, so $D_{\phi} \propto k^2 D_n$. In order to fix the units, we need to multiply by two length scale. We are interested in the difference in phase between two coherent photons that traversed the whole width of the layer, so one length scale should be the width of the layer $\Delta z$. The other length scale must be $r$, so

$D_{\phi} \approx D_n k^2 r \Delta z \propto k^2 r^{5/3}$

The above relation is often written in the form $D_{\phi} \approx \left( r/R_d\right)^{5/3}$ where $R_d \propto k^{-6/5}$ is distance parallel to the screen along which the accumulated random phase adds up to a phase difference of order unity. This scale is also called the Fried length.

Now, suppose that instead of air, we consider waves in a plasma. The dispersion equation for waves in a plasma is

$k^2 = \omega^2 - \omega_p^2$

where $\omega_p^2 = \frac{q^2}{m} n$ is the plasma frequency, $q$ is the elementary charge and $m$ is the electron mass. The difference in the refractive index scales with the difference in density as $d N \propto k^{-2} d n$. The phase structure function in this case is proportional to $D_{\phi} \propto k^{-2} r^{5/3}$, and the Fried length (in this context it is more commonly called the diffraction length) scales as $R_d \propto k^{6/5}$.

We got that in the case of atmospheric scintillation Fried length increases with wavelength, while in the case of interstellar scintillation it decreases with wavelength, and due to some strange coincidence the absolute magnitude of the power law indices is the same.

Another relevant scale is the Fresnel length

$R_f \approx \sqrt{L/k}$

where $L$ is the distance between the observer and the screen. The relation between these two length scales determines the qualitative behaviour of the system. The regime where $R_d \gg R_f$ is called weak scattering, while the regime where $R_f \gg R_d$ is called strong scattering. In analysing these cases, we assume interstellar scintillation, so the dispersion equation is that of plasma.

## Weak Scattering Edit

In this regime the distortion to the image is small. The shape of the Fresnel disc can change due to scintillation, and since the net amount of light does not change, the flux can also change. The relative r.m.s change in the amount of flux is comparable to the r.m.s difference between the centre of the Fresnel disc and its edge

$\frac{\delta f}{f} \propto \sqrt{D_{\phi} \left( r=R_f\right)} \propto k^{-17/10}$

## Strong Scintillation Edit

In this regime, instead of one spot roughly the size of $R_f$, the image is broken up into a multitude of smaller spots of size $R_d$. Some of these spots are lighted, and some are dark, both randomly distributed. Due to the conservation of the net amount of light, the number of spots has to be the square of the number of such spots that fit into the Fresnel zone $R_f^2/R_d^2$. Therefore, these spots occupy a much larger region, with a radius of

$R_r \approx R_f^2/R_d$