Clarifying Planck's law

Question

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?

Edit:

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.

Answer 1

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.

Answer 2

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 $\lambda$.