# Page:Popular Science Monthly Volume 82.djvu/304

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THE POPULAR SCIENCE MONTHLY
 ${\displaystyle L'=3.3\times (10)^{10}\ stars\ of\ zero\ magnitude.}$ Hence ${\displaystyle L'={\frac {3.3\times (10)^{10}\times L}{600}}=5.5\times (10)^{7}\times L.}$

The average illumination in intergalactic space is very likely less than one one-hundred-millionth of that of sunlight; but a majority of the stars have less absorbent atmospheres than our sun, and as sunlight at the earth's distance must be increased in the ratio 1: 46,000 to give the light emitted by the surface of the solar sphere, the average radiant energy at stellar surfaces may be assumed as ${\displaystyle (10)^{12}}$ times the average radiant energy in the star-lit ether.

If ${\displaystyle {{\ce {V}}}}$ and ${\displaystyle {{\ce {L}}}}$ are the volume and average illumination of the ether, ${\displaystyle {\ce {V' =}}}$ the total volume of stellar material, and ${\displaystyle {\ce {L' =}}}$ the total light from the combined surfaces of all of the stars, an instantaneous image of the relation between the two bodies—ether and matter—that is to say, a representation of the relation if there were an instantaneous emission of light with an infinite velocity, would give

${\displaystyle VL:V'L'=(10)^{12}\times 1:1\times (10)^{12},}$

or equality. But if the element of time enters, and also the actual velocity of light, the illumination at a given point in the ether will increase with the time. Let the year be the unit of time. After one billion years, supposing that the stellar radiation can have endured as long as this, instead of unity for the ratio ${\displaystyle {\ce {VL / V'L'}}}$ as in the preceding equation, we shall have

${\displaystyle VL=V'L'\times (10)^{12}.}$

Considering the limiting surface of the ether to be, not an imaginary circumscribing sphere, but the sum of the combined stellar surfaces across which the sum total of stellar radiant energy is being constantly transferred from matter to ether, the case stands about like this:

 Volume Radiation (Superficial) Total Radiant Energy (Volumetric) Stars = 1 Stars = (10)12 Stars = (10)12 Ether = (10)12 Ether = (10)12 Ether = (10)24

The large amount of the total radiant energy of the free ether, compared with that of the stars may seem surprising, but it results from the fact that the average illumination of the ether is due to the accumulation of radiant energy from depths of space which are greater as the ether is more transparent. Unless the radiant energy were absorbed, it could not do otherwise than accumulate. The accumulation represents the combined radiation of an immense number of stars whose average distance is to be measured in millions of light-years—how many millions depends upon the time that the stellar radiation remains in the ether before it is all absorbed.

According to what precedes, the average ethereal energy can hardly be less than the radiant energy from the stars within a range of a million light-years, and may amount to many times this figure; and as