Final answer:
To relate the energy density P(v,T) within an enclosed cavity to the power emitted by a blackbody W(v,T), we utilize the Stefan-Boltzmann law of blackbody radiation, which states that the power emitted is proportional to the fourth power of its absolute temperature, and understand that W(v,T) = c/4 × P(v,T).
Step-by-step explanation:
To show that the energy density within an enclosed cavity P(v,T) is related to the power emitted by a blackbody of unit area W(v,T), we need to utilize the concept of blackbody radiation along with Stefan-Boltzmann law. The energy flux, as described by this law, is given by F = σT4, where σ is the Stefan-Boltzmann constant and T is the absolute temperature. The term 'flux' here refers to the flow of power into a unit area, such as the area of a telescope mirror when observing stars.
In a cavity radiator, the intensity I(λ, T) represents the power radiated per unit area per unit wavelength. The energy density P(v,T) within the cavity is related to this intensity, and by considering the blackbody as a perfect emitter and absorber of radiation, it can be inferred that the power emitted W(v,T) is the intensity times the area of the blackbody times the speed of light, divided by 4, due to geometrical considerations of radiation in all directions. Therefore, we derive the relationship W(v,T) = c/4 × P(v,T), with 'c' being the speed of light.