1 Introduction

The goal of this section is to compute the irradiance at a point on the surface, due to the radiance of all the other points of the surface. We compute this first by assuming that there is no atmosphere, and then by taking the atmospheric absorption into account.

2 Irradiance without atmosphere

Figure 1: Irradiance in a cylinder.

Consider a point $\bp$ on the surface. The irradiance $E(\bp)$ at this point is the integral over all the directions $\bw$ in the hemisphere above $\bp$ of the product of ([1], [2]):

the radiance $L(\bw)$ coming from direction $\bw$,
the cosine $\bw\cdot\bn$ of the angle between $\bw$ and the surface normal $\bn$ at $\bp$,
the infinitesimal solid angle $\diff\bw$.

Using the angles $\varphi$ and $\phi$ defined in Fig. 1 to specify the direction $\bw$, we have $\bw\cdot\bn=\cos\varphi\cos\phi$, and $\diff\bw=\cos\phi\diff\varphi\diff\phi$. The radiance $L(\bw)$ is independent of $\phi$, thanks to the translation invariance hypothesis. Thus, using the angle $\theta=2\varphi$ (see Fig. 1), it can be noted $L(\theta)$. The irradiance $E(\bp)$ is then: \begin{align} E(\bp)&=\int_{-\pi/2}^{\pi/2}L(2\varphi)\left(\int_{-\pi/2}^{\pi/2}\cos^2\phi\,\diff\phi\right)\cos\varphi\diff\varphi\\ &=\frac{\pi}{2}\int_{-\pi/2}^{\pi/2}L(2\varphi)\cos\varphi\diff\varphi\\ &=\frac{\pi}{4}\int_{-\pi}^{\pi}L(\theta)\cos\frac{\theta}{2}\diff\theta \end{align}

We can check the correctness of this result for the case where $L(\theta)$ is constant. In this case the radiance arriving at $\bp$ is independent of $\bw$, and the irradiance should therefore be $\pi L$. This is indeed what we get: $E=L\frac{\pi}{4}\int_{-\pi}^{\pi}\cos\frac{\theta}{2}\diff\theta=\pi L$. Note that in the above equations $\theta$ is measured from the opposite point to $\bp$. If we use instead an absolute coordinate $\theta$, represent $\bp$ by its coordinate $\theta_p$, and define $L$ based on this absolute coordinate, then we get: \begin{equation} E(\theta_p)=\frac{\pi}{4}\int_{2\pi}L(\theta) \vert\cos\frac{\theta-\theta_p+\pi}{2}\vert\diff\theta=\frac{\pi}{4}\int_{2\pi}L(\theta) \vert\sin\frac{\theta-\theta_p}{2}\vert\diff\theta \end{equation}

3 Irradiance with an atmosphere

If there is an atmosphere, the radiance received at $\bp$ is the emitted radiance $L(\theta)$ times the transmittance $\mathfrak{t}(\bp,\bq)$ of the atmosphere between the reception point $\bp$ and the emission point $\bq$ (see Fig. 1). This transmittance is equal to the exponential of the optical depth $\tau(\bp,\bq)$, itself equal to the integral of the absorption and scattering coefficients $k_a(\bx)$ and $k_s(\bx)$ in $m^{-1}$ between $\bp$ and $\bq$: \begin{equation} \mathfrak{t}(\bp,\bq)=\exp(-\tau(\bp,\bq))=\exp\left(-\int_{\bp}^{\bq}k_a(\bx)+k_s(\bx)\,\diff x\right) \end{equation}

Thanks to the translation invariance, the optical depth between $\bp$ and $\bq$ is related to the optical depth between $\bp$ and the projection $\bq^\bot$ of $\bq$ in the plane containing $\bp$ and perpendicular to the axis (see Fig. 1) with $\tau(\bp,\bq)=\tau(\bp,\bq^\bot)/\cos\phi$. The irradiance $E(\bp)$ then becomes: \begin{align} E(\bp)&=\int_{-\pi/2}^{\pi/2}L(2\varphi)\left(\int_{-\pi/2}^{\pi/2}\mathfrak{t}(\bp,\bq^\bot)^{1/\cos\phi}\cos^2\phi\,\diff\phi\right)\cos\varphi\diff\varphi\\ &=\frac{\pi}{4}\int_{-\pi}^{\pi}L(\theta)f_2(\mathfrak{t}(\bp,\bq^\bot))\cos\frac{\theta}{2}\diff\theta \end{align} where $f_2$ is defined by (see Fig. 2): \begin{equation} f_2(x)=\frac{4}{\pi}\int_0^{\pi/2}x^{1/\cos\phi}\cos^2\phi\,\diff\phi \end{equation}

Figure 2: Plot of the function $f_2$.

In the above equations $\theta$ is measured from the opposite point to $\bp$. If we use instead an absolute coordinate $\theta$, represent $\bp$ by its coordinate $\theta_p$, and define $L$ and $T$ based on this absolute coordinate, then we get the result presented in the conclusion.

4 Conclusion

The irradiance $E$ at the surface of an infinite cylinder can be computed from the radiance $L$ emitted at the surface with the following formula: \begin{equation} E(\theta_p)=\frac{\pi}{4}\int_{2\pi}L(\theta)f_2(\mathfrak{t}(r_s,\theta_p,r_s,\theta))\vert\sin\frac{\theta-\theta_p}{2}\vert \diff\theta \end{equation} where $\mathfrak{t}(r_s,\theta_p,r_s,\theta)$ is the transmittance of the atmosphere between the points of cylindrical coordinates $(r_s,\theta_p)$ and $(r_s,\theta)$, in a plane perpendicular to the axis, and $f_2(x)=\frac{4}{\pi}\int_0^{\pi/2}x^{1/\cos\phi}\cos^2\phi\,\diff\phi$.