I've edited this. The 2 answers i've received so far were confused as to my meaning.

I'm in an endless argument with a guy who insists that Planck's law predicts that a black body where $T>0$ radiates at all wavelengths.

I insist that a black body can only radiate the wavelengths that it receives from it's surroundings because it's in thermal equilibrium.

So who's right?


I think that you guys are trying to answer this question without knowing the definition of a black body. A black body is called "black" because it's indistinguishable from it's surroundings. It's in thermal equilibrium with it's surrounding environment. I don't see how any of you guys can say that real-world objects behave like black bodies since A) black bodies are an idealized state, and B) black bodies can not exist in nature.

Given a black body where T=300K (15C). The other guy claims that the BB will radiate at all wavelengths, while I say that radiation will peak at ~10um. He says that any BB, regardless of temperature will radiate the full spectrum including visible light and x-rays. That is simply ludicrous. What he's saying is that we can get more out of a system than we put into it.

Given all that, if you have the interest to answer then would you please stick to the question and not bring real world objects like the Earth and Sun into it. Neither are black bodies. Please stick to black bodies.

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    $\begingroup$ If blackbodies radiate at all wavelengths, then, since most things are approximate blackbodies, the "surroundings" contain radiation of all wavelengths! It's not clear to me why you think the condition of thermal equilibrium would restrict the possible wavelengths. $\endgroup$ – ACuriousMind Nov 10 at 14:48
  • $\begingroup$ The definition of a black body is that it is thermal equilibrium with its surroundings. That's why it's called "black" because it's indistinguishable from it's surroundings. $\endgroup$ – Oortcloud Nov 11 at 11:52
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    $\begingroup$ The short answer: The other guy is right and you are wrong. Black body radiation is a continuous spectra (probability distribution, if you like) from $\lambda = 0$ to $\lambda= +\infty$. $\endgroup$ – J. Manuel Nov 11 at 14:47
  • $\begingroup$ I don't know much as I haven't been started thermodynamics but once while reading about black bodies and their continuous spectrum I found an analogy from University Physics which states:... $\endgroup$ – The Last Airbender Nov 11 at 16:18
  • $\begingroup$ ...A tuning fork emits sound waves of a single definite frequency (a pure tone) when struck. But if you tightly pack a suitcase full of tuning forks and then shake the suitcase, the proximity of the tuning forks to each other affects the sound that they produce. What you hear is mostly noise, which is sound with a continuous distribution of all frequencies. In the same manner. isolated atoms in a gas emit light of certain distinct Frequencies when excited, but if the same atoms are crowded togethe or liquid they produce a continuous spectrum of light. $\endgroup$ – The Last Airbender Nov 11 at 16:20

A blackbody will radiate at all wavelengths. The radiation is determined by the body's temperature, and is not related to incoming radiation.

For example, the Sun radiates a blackbody spectrum with a peak around the wavelengths of visible light. But it hardly absorbs any radiation at all (probably mostly from the CMB, which has a peak in the microwave region).

Another example: The Earth absorbs sunlight, but since it is colder than the Sun, it radiates a blackbody spectrum with a peak in the infrared. The difference in outgoing and incoming radiation is exactly what makes the greenhouse effect work.

Edit: Since you edited your question, I will explain more. The Earth and the Sun are in fact blackbodies, to a good approximation; see here. The essential feature of blackbody radiation is that some physical system (the blackbody) is in equilibrium with a photon gas. It is assumed that the small amount of radiation that escapes into the surroundings does not significantly affect the thermal equilibrium. By the Planck law, this photon gas will indeed contain some photons of all wavelengths, including visible light, x-rays and even gamma rays.

However, you are also right that the spectrum has a peak at a temperature-dependent wavelength ($\lambda_\max = \frac{2.9\, \mathrm{mK\,m}}{T}$). There will be comparatively few photons with wavelengths far away from this peak.

A blackbody does not need to be "indistinguishable from its surroundings". It is called "black" because it absorbs all light that hits it; it does not reflect anything. So in a very narrow way you are right that it "can only radiate the wavelengths that it receives from its surroundings": In thermal equilibrium, all wavelengths are present, and the blackbody also absorbs and emits all wavelengths; therefore it only emits those wavelengths it absorbs.

  • $\begingroup$ Thanks, but that's not quite what I was asking. Please see my edit. $\endgroup$ – Oortcloud Nov 11 at 12:18
  • $\begingroup$ @Oortcloud I have added some more discussion in response to your edits. $\endgroup$ – Elias Riedel Gårding Nov 11 at 15:50

A blackbody will radiate at any wavelength, as you can infer from the Planck's law $$ B_\lambda(\lambda, T) =\frac{2hc^2}{\lambda^5}\frac{1}{ e^{\frac{hc}{\lambda k_\mathrm B T}} - 1} $$ which is different from zero for any $\lambda$.

  • $\begingroup$ Thanks, but that's not quite what I was asking. Please see my edit. $\endgroup$ – Oortcloud Nov 11 at 12:18
  • $\begingroup$ Thanks for the comment. What do you think is the problem with this answer? Do you agree that this formula gives the spectrum of the radiation emitted by a blackbody, or not? Because if you do, then this gives a clear answer to your question I think.. $\endgroup$ – pp.ch.te Nov 12 at 17:51

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