Two questions:

  1. I read online that all objects emit the majority of their thermal radiation in the radio, and that this emission increases with temperature. Does this mean that all glowing bodies emit the majority of their radiation in the radio, regardless of their temperature?

  2. Wien’s law is used to find the peak wavelength of a black body. Does this mean that there is zero emission at any wavelength shorter than this peak wavelength?

Thank you in advance for your help.

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    $\begingroup$ Only very cold objects (e.g., $T < 10$ K) would emit most of their radiation in the radio. $\endgroup$ Aug 16, 2019 at 15:12
  • $\begingroup$ +1 for asking a clear question and following up with several comments. This is how Stack Exchange should be used. In the future if you find another passage that you'd like to understand e.g. "I read online that...", it would be better if you included a short quote. $\endgroup$
    – uhoh
    Aug 17, 2019 at 22:53

1 Answer 1

  1. It would be good to specify what "majority of radiation" means. If it means "radiative energy", then this premise is surely wrong. To find the energy emitted per wavelength, take the Planck function $B(T, \lambda)$ at a certain temperature $T$ and multiply with the wavelength $\lambda$. You will find that the short-wavelength radiation always dominates for a black-body.
  2. Wien's law gives you the peak of the Planck function. Radiation at all other wavelengths exists.

Note that objects predominantly loose most of their energy, if photons and matter are in equilibrium. If this equilibrium is disturbed, or particle accelerations play a role (like in the presence of magnetic fields), a whole zoo of non-thermal processes can also lead to important radiative energy loss.
Non-thermal processes in space can be radio emission from gyrating particles, line radiation, Masers, Bremsstrahlung, Compton radiation... Those processes then do not show a Planck spectrum, but other characteristic spectra.

Edit: I've added a plot of the Planck function at two different temperatures. In a linear plot (on the left) it might look as if there's a cut-off, but in a logarithmic plot (on the right), one sees what mathematical analysis already betrays: The function is defined everywhere. It just becomes really small on the high-energy end, as a commentator pointed out correctly.

enter image description here

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    $\begingroup$ @JohnSmith What, no? Why would it be? The Planck function is a continuous function, defined on the whole space of real numbers, there is no cutoff. Just plot it to see this. $\endgroup$ Aug 16, 2019 at 15:03
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    $\begingroup$ Wait, so a black body at ANY temperature will have nonzero emission at ALL wavelengths? $\endgroup$
    – John Smith
    Aug 16, 2019 at 15:30
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    $\begingroup$ @JohnSmith At high frequencies quantization (shot noise) becomes an issue. The intensity may be so low that it amounts to less than one expected photon in the duration of your experiment, so you would expect to see no emmission at those wavelengths. $\endgroup$ Aug 16, 2019 at 15:35
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    $\begingroup$ @JohnSmith the Planck function is widely available (e.g. Wikipedia). It isn't exactly zero at any frequency $\endgroup$
    – ProfRob
    Aug 16, 2019 at 18:45
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    $\begingroup$ One point wasn't explained adequately. The OP thought "peak wavelength" referred to the highest wavelength. Instead, it refers to the wavelength at which the highest amount of energy is radiated. $\endgroup$
    – Bit Chaser
    Aug 16, 2019 at 22:51

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