22.

Since the energy radiation is propagated in the medium with a finite velocity $q$, there must be in a finite space a finite amount of energy. We shall therefore speak of the "space density of radiation," meaning thereby the ratio of the total quantity of energy of radiation contained in a volume-element to the magnitude of the latter.

Let us now calculate the space density of radiation $u$ at any arbitrary point of the medium. When we consider an infinitely small element of volume $v$ at the point in question, having any shape whatsoever, we must allow for all rays passing through the volume-element $v$. For this purpose we shall construct about any point $O$ of $v$ as center a sphere 1 of radius $r$, $r$ being large compared with the linear dimensions of $v$ but still so small that no appreciable absorption or scattering of the radiation takes place in the distance $r$ ([fig:1]Fig. 1). Every ray which reaches $v$ must then come from some point on the surface of the sphere. If, then, we at first consider only all the rays that come from the points of an infinitely small element of area $d\sigma$ on the surface of the sphere, and reach $v$, and then sum up for all elements of the spherical surface, we shall have accounted for all rays and not taken any one more than once.

Let us then calculate first the amount of energy which is contributed to the energy contained in $v$ by the radiation sent from such an element $d\sigma$ to $v$. We choose $d\sigma$ so that its linear dimensions are small compared with those of $v$ and consider the cone of rays which, starting at a point of $d\sigma$, meets the volume $v$. This cone consists of an infinite number of conical elements with the common vertex at $P$, a point of $d\sigma$, each cutting out of the volume $v$ a certain element of length, say $s$. The solid angle of such a conical element is ${\displaystyle \frac{f}{r^2}}$ where $f$ denotes the area of cross-section normal to the axis of the cone at a distance $r$ from the vertex. The time required for the radiation to pass through the distance $s$ is: $\tau =\frac{s}{q}.$

From expression [eqn:(6)] (6) we may find the energy radiated through a certain element of area. In the present case $d\Omega ={\displaystyle \frac{f}{r^2}}$ and $\theta =0$; hence the energy is: $\tau d\sigma \frac{f}{r^2}K=\frac{fs}{r^{2}q}·Kd\sigma .\mathrm{\backslash Label}[eqn](19)\text{upshape (19)}$ This energy enters the conical element in $v$ and spreads out into the volume $fs$. Summing up over all conical elements that start from $d\sigma$ and enter $v$ we have $\frac{Kd\sigma }{r^{2}q}\sum fs=\frac{Kd\sigma }{r^{2}q}v.$ This represents the entire energy of radiation contained in the volume $v$, so far as it is caused by radiation through the element $d\sigma$.

In order to obtain the total energy of radiation contained in $v$ we must integrate over all elements $d\sigma$ contained in the surface of the sphere. Denoting by $d\Omega$ the solid angle ${\displaystyle \frac{d\sigma }{r^2}}$ of a cone which has its center in $O$ and intersects in $d\sigma$ the surface of the sphere, we get for the whole energy: $\frac{v}{q}\int Kd\Omega .$ The volume density of radiation required is found from this by dividing by $v$. It is $u=\frac{1}{q}\int Kd\Omega .\mathrm{\backslash Label}[eqn](20)\text{upshape (20)}$

Since in this expression $r$ has disappeared, we can think of $K$ as the intensity of radiation at the point $O$ itself. In integrating, it is to be noted that $K$ in general depends on the direction $(\theta ,\varphi )$. For radiation that is uniform in all directions $K$ is a constant and on integration we get: $u=\frac{4\pi K}{q}.\mathrm{\backslash Label}[eqn](21)\text{upshape (21)}$