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I read that at a particular temperature, all black bodies emitted all wavelengths of electromagnetic energy but at different intensifies. This would require infinite energy and therefore violates the law of conservation of energy. How is this possible?

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  • $\begingroup$ Yes but if there is an emission for all frequencies, and there are infinite frequencies on a spectrum, there is infinite emission, how does this work? $\endgroup$
    – Person
    Sep 26, 2017 at 8:52
  • $\begingroup$ This is alike the runner and the snail :) $\endgroup$
    – Alchimista
    Sep 26, 2017 at 12:32
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    $\begingroup$ The sum of an infinite number of terms is not necessarily infinite. $\endgroup$
    – garyp
    Sep 26, 2017 at 14:11

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The basic misconception here seems to be a misunderstanding of the meaning of the luminosity function $B_\nu$ (or $B_\lambda$). It is not a power function that gives the power at a frequency (or wavelength), rather, it is a power density function that gives the power (really, power density) per unit of frequency (or wavelength). This can be seen by dimensionally analyzing it, which reveals $B_\nu$ to have units of power per unit area per unit of solid angle per unit of frequency, or simply power density per unit frequency. In particular $h$ has dimension $\mbox{(energy)} \cdot \mbox{(time)}$ and $\nu$ has dimension $\mbox{(time)}^{-1}$ while $c$ has dimension $\mbox{(length)} \cdot \mbox{(time)}^{-1}$ and so combining all these in $\frac{h\nu^3}{c^2}$ yields $\mbox{(energy)} \cdot \mbox{(length)}^{-2}$ which is equivalent to $\mbox{(power)} \cdot \mbox{(length)}^{-2} \cdot \mbox{(frequency)}^{-1}$ since $\mbox{(frequency)} = \mbox{(time)}^{-1}$ and power times time is energy.

So there is not a finite amount of power density at any exact, specific wavelength. To get power density you must integrate, just as to get mass from mass density you must integrate. If you integrate "over a point" you just get 0 - so in a sense there is 0 power at any exact wavelength, just as a single point taken out of a finite mass density would have 0 mass(*).

That is, your mistake is in interpreting/understanding the math correctly. If you instead consider it as being like mass density, then you note that there is mass density at all points in a body, yet there is not infinite mass present. The same mathematical principles that govern that also govern this.

(*) (yes atoms, yes caveat, yes handwave - the point here is this is a mathematical misconception.)

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The intensities are dependent on the frequency indeed. Total luminosity however is finite for any finite temperature.

Classical infinities from the Statistical Mechanics lead to creation of the Planck's law.

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Mathematically, whe can prove the finitness of the power emitted from the blackbody. We start recalling the spectral radiance of a black body, given by the Planck's law $$B(\nu , T) = \frac{2h\nu^3}{c^2}\cdot\Big(e^{\frac{h\nu}{k_B T}}-1\Big)^{-1} $$ that is the power emitted by the black body, per unit area and unit solid angle, relative to a single emitted frequency $ \nu=\frac{c}{\lambda}$; $h$ is the Planck's constant, T the temperature and $k_B$ the Boltzmann constant The total power emitted, then, is found by integrating over the solid angle $\Omega$, the black body surface $S$ , and all the frequencies from 0 to infinity. So $$ P = \int_0^\infty d\nu\int_\Omega d\Omega\int_S dS \frac{2h\nu^3}{c^2}\cdot\Big(e^{\frac{h\nu}{k_B T}}-1\Big)^{-1}$$ This integral can be proved to converge and it is exactly the Stephan-Boltzmann law $$ P = \sigma T^4 $$ Thisshould be enough to convince you that it's only necesary a finite amout of energy. You can also take the limits of $B(\nu, T)$ for $\nu \to \infty$ and for $\nu \to 0$, and see that both tend to $0$. This means that the higher (or lower) the freqencies are, the smaller spectral radiance is. This means that a blck body hardly emis high energy photons, or very low eenrgy photons. This is easily understandable, at least for high frequencies, because to emit very energetic photos you need high energy phenomena,for example nuclear $\gamma$ decay. Moreover, energetic photons may easily pass across the black body without being absorbed or scattered.

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