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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?

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

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The amount of radiance in an interval of the electromagnetic spectrum is the same regardless of whether that spectral interval is described in terms of a range of wavelengths or a range of frequencies. For example, if you asked how much radiance is in green light, it’s the same whether you define green as being 500-565 nm or 530-600 THz. These are the same spectral interval.

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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.

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