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I. V. SAVELYEV
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PHYSICS A GENERAL COURSE (In three volumes)
M. B. CABEJIbEB
VOLUME III
KYPC OB~El1 u 2 ' the energy flux from 'J/ V the first cavity into the second one must be greater than the flux in the opposite direction. The walls of the second cavity, as a 31 ~"I result, will absorb more energy than they emit, and their temperature will start growing. The walls of the first cavity, on the 1· ............. ~~12oool( Fig. 1.5 17g0K other hand, will absorb less energy than 16'001( they emit, and they will cool. But two I &'/ I I I I bodies having the same initial temperature cannot acquire different !l 1 2 J A,f"m temperatures as a result of heat exchange with each other-this is Fig. 1.4 forbidden by the second law of thermodynamics. We must therefore acknowledge that our assumption on Ul and u 2 being different is not a definite value of the temperature T of our blackbody. The area en- lawful. The conclusion on the equality of Ul (T) and U 2 (T) covers closed by the curve gives the radiant emittance of the blackbody at each spectral component U (Ul, T). the corresponding temperature. That the equilibrium radiation does not depend on the nature of A glance at Fig. 1.4 shows that the radiant emitta.nce of a blackthe cavity walls can be explained by the following considerations. body grows greatly with the temperature. The maximum of the Blackbody walls would absorb all the energy et>e falling on them and emissivity shifts toward shorter waves with elevation of the temwould emit the same energy flux et>e. Walls with the absorptivity a perature. will absorb the fraction aet>e of the flux et>e falling on them and will reflect a flux equal to (1 - a) et>e. In addition, they will emit the flux aet>e (equal to the absorbed flux). As a result, the walls of the cavity will return the same energy flux et>e = (1 - a) et>e aet>e 1.3. Equilibrium Density of Radiant Energy . to the radiation that blackbody walls would return to it. Let us consider radiation that is in equilibrium with a substance The equilibrium radiant energy density U is related to the radiant For this purpose, let us imagine an evacuated cavity whose walls . emittance of a blackbody R* by a simple expression which we shall are maintained at a constant temperature T. In the equilibrium . now proceed to derive*. state, the radiant energy will be distributed throughout the volume Let us consider an evacuated cavity with blackbody walls. In of the cavity with a definite density u = u (T). The spectral distriequilibrium, a radiant flux of the same density will pass through bution of this energy can be characterized by the function U (Ul, T) every point inside the cavity in any direction. If the radiation were determined by the condition duro = U (Ul, T) dUl, where duro is the to propagate in one given direction (I.e. if only one ray were to pass fraction of the energy density falling within the interval of frethrough a given point), the density of the energy flux at the point quencies dUl. The total energy density U (T) is related to the function being considered would equal the product of the energy density U U (Ul, T) by the formula and the speed of an electromagnetic wave c. But a multitude of rays
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* We have used the symbol R* to stress that we are dealing with the radiant emittance of a blackbody.
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