According to my textbook, the power radiated of a small hole in a cavity (an ideal blackbody) is given by
$$R = \frac{1}{4}cU$$ where $U$ is the total energy density per unit volume, $R$ is the power, and $c$ is the speed of light. First question: By power do they mean intensity? The units of $R$ are $\frac{J}{m^2s}$, which is power per unit area, or intensity.
Next my book says that the spectral distribution of the power (intensity?) emitted from the hole is proportional to the spectral distribution of the energy density in the cavity.
$$R(\lambda) = \frac{1}{4}cu(\lambda)$$
The only thing they changed was that they made $R$ a function of $\lambda$ and $U$ into $u(\lambda)$. Question 2: So $U$ is the total energy per unit volume (the energy density) and it does not depend on wavelength while $u(\lambda)$ is the energy density, but describes the energy density depending on what type of wavelength is trapped in the cavity? Just wanting to make sure because my next question is the important one.
My books says that $u(\lambda)d\lambda$ is the energy density from $\lambda$ to $\lambda + d\lambda$. So,
$$\int_0^\infty u(\lambda)d\lambda = U$$
Regarding the statement about $u(\lambda)d\lambda$, I can kind of make sense of it if I think about calc I. In calc I, $\int f(x)dx$ meant to find the function of $f(x)$ between $x$ and $x + dx$ and then multiply that value by $dx$ to get the area of a rectangle. Question 3: So in general, does writing $g(t)dt$ mean the function $g(t)$ between $t$ and $t + dt$. Question 4: Why are we finding the area under $u(\lambda)$. Unit wise, if we think of the calc I rectangle thing, $u(\lambda)d\lambda$ has units of $J/m^3 \times m = J/m^2$, which is a different type of energy density.
