7 Appendix

In the following the notations $X=\mbox{e}^{b\cdot(x-x_0)}$ and $Y=\mbox{e}^{p\cdot(x-x_0)}$ will be used, as well as the functions $U=\frac{X}{X+\frac{1}{X}}$ and $V=\frac{Y}{Y+\frac{1}{Y}}$. In these notations the intensity I(x) in Eq. (2) can be written as:

 

$$I(x)=c+4\cdot a\cdot U\cdot V.$$ (3)

It is necessary, now, to specify that this model cannot fit the entire section of the real solar image by each CCD line as this function stretch to $c+4\cdot a$ or to c when x stretch to $\pm\infty$, following the sign of b.

To have a correct representation of a CCD line section on the image, we can use the better representation:

 

$$I(x)=c+16\cdot a\cdot U_0\cdot V_0\cdot U_1\cdot V_1$$ (4)

where the function $4\cdot U_1\cdot V_1$ contains the same parameters as U₀ and V₀ but with opposite signs for b and p and a different value for the x₀, as $x^{\prime}_0$. The difference $\mid x^\prime_0-x_0\mid$ is nothing else that the solar radius when the section is obtained along a line which contains the solar center. This model (Eq. 4) gives a very satisfactory representation of the solar image sections but is unnecessary in our case because the images obtained by CCD observations are always near the limb of the Sun (the total field of the CCD covers about $6\hbox{$^\prime$}$). When the CCD is entirely cover by the Sun image one can see only about 20% of the total solar image. In this case the function $4\cdot U_1\cdot V_1$ (or $4\cdot U\cdot V$ following the observed part of the limb) is practically constant and equal to 1, and the differences between the models 3 and 4 around the inflection point is, with a large approximation, equal to 0. Consequently, we will use the model given by Eqs. (2) or (3) in the following.

The necessary formula to find the limb position, as defined, are the successive derivatives of the Eq. (3). They are:

$$\frac{{\rm d}I}{{\rm d}x}=4a\left(\frac{{\rm d}U}{{\rm d}x}\cdot V+U\cdot\frac{{\rm d}V}{{\rm d}x}\right)$$

or, as $\frac{{\rm d}X}{{\rm d}x}=b\cdot X$, and the same for $\frac{{\rm d}Y}{{\rm d}x}$,with p, becomes:

$$\frac{{\rm d}I}{{\rm d}x}=8a\left({\frac{b}{\left({X+\frac{1... ...rac{1}{Y}}\right)^2}\cdot \frac{X}{X+\frac{1}{X}}}\right) \cdot$$

The second derivative can be written as:

 

$$\frac{{\rm d}^2I}{{\rm d}x^2}=4a\left({V\cdot\frac{{\rm d}^2... ...{\rm d}V}{{\rm d}x}+U\cdot\frac{{\rm d}^2V}{{\rm d}x^2}}\right)$$ (5)

or, after calculation:

$$\begin{eqnarray} \frac{{\rm d}^2I}{{\rm d}x^2} & = & -16a\cdot\left({b^2\cdot\fr... ...Y+\frac{1}{Y}}\right)^3}\cdot\frac{X}{X+\frac{1}{X}}}\right) \cdot\end{eqnarray}$$ (6)

Some quantities as the maximum (or minimum) derivative value or the derivative width need to be calculated. It is quite evident that solving the equation $\frac{{\rm d}^2I}{{\rm d}x^2}=0$ is not easy. One way to obtain approximate values of useful quantities is to expand the series which can represent the model I(x).

The calculation, to the fifth order, gives:

If x₀, b, p and the other parameters are obtained, it is possible to derive information about the images. By derivation of the preceding formula, which is easy now, until the third order, the derivative extremum may be found by solving iteratively the equation $\frac{{\rm d}^2I}{{\rm d}x^2}=0$. We find that the derivative peak abscissa is:

$$x_{\rm s}=x_0+\frac{p}{b^2}-\frac{2}{3}\cdot\frac{p^3}{b^4} \cdot$$

In the same way, using the third derivative, the first derivative width can be calculated. This parameter may be defined in two way:

1.

The width is defined by the distance between the two points in which the intensity is half of the one of the extremum. We will note it as $w_{\rm h}$;

2.

The width may be also defined as the distance between the two inflection points of the first derivative. This definition needs to calculate the zeros of the third derivative. We will note it as $w_{\rm p}$.

We obtain: $w_{\rm h}=\frac{1}{b}\cdot\sqrt{\frac{7}{2}}\approx\frac{1.871}{b}$ and $w_{\rm p}=\frac{2}{b}$, exactly.

If we use the perfect model, which represent a perfect solar image, the parameter p is equal to zero (Eq. 1). The same quantities may be computed and, we have naturally $x_{\rm s}=x_0$, so:

$$w_{\rm h}=\frac{1}{b}\ln\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\... ...\frac{1}{2b}\ln\left(\frac{2+\sqrt{3}}{2-\sqrt{3}}\right) \cdot$$

which gives $w_{\rm h}=\frac{1.763}{b}$ and $w_{\rm p}=\frac{1.317}{b}$.