# Why can we equate these two integrals related to blackbody radiation?

I was reading this Wikipedia article which describes how Planck’s Law of blackbody radiation is derived. Letting $$B(v,T)$$ represent the energy emitted at frequency $$v$$ and temperature $$T$$, the article states that we may “[equate] the integral of spectral radiance in per unit wavelength to that of per unit frequency” like this:

$$\int_{\lambda_1}^{\lambda_2} B(\lambda,T)d\lambda=\int_{v(\lambda_1)}^{v(\lambda_2)}B(v,T)dv$$

From here, we can easily solve for $$B$$ by converting the above equality into a differential equation.

My question is: why is this equation true in the first place? Why is it a valid assumption that “the integral of spectral radiance in per unit wavelength [equals] that of per unit frequency?” What property of radiation implies that this must be true?

• Minor comment: That should be a Greek letter $\nu$, not a Roman letter $v$. Commented Jun 25, 2020 at 1:40

These are the definitions of the two spectral radiances. $$B(\lambda, T)$$ is defined by stating that the intensity emitted between wavelengths $$\lambda_1$$ and $$\lambda_2$$ is
$$I = \int_{\lambda_1}^{\lambda_2} B(\lambda, T)\ d\lambda,$$
and similarly for $$B(\nu, T)$$. Or, in physicist language, the intensity in a small interval $$d\lambda$$ is $$B(\lambda, T)\ d\lambda$$. Therefore, we have two different formulas for the intensity emitted between two wavelengths or between their corresponding frequencies, and so the formulas have to be the same.