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I am reading the lecture notes of a friend. He has written about the cold (3-20 K) molecular $H_2$ cosmic gas, that "practically doesn't radiate, due to symmetry". Can someone confirm that $H_2$ has this feature of radiating much less electromagnetic radiation than other molecules? And how can this be explained with the symmetry of the molecule? Excuse me, for making you interpret vague statements. Nevertheless, I am very curious

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    $\begingroup$ Possibly the lack of rotational transitions? Homonuclear diatomics do not undergo purely rotational transitions since their symmetry means they have no dipole moment. $\endgroup$ Jun 10, 2021 at 18:20
  • $\begingroup$ @JohnRennie Thanks, this is helpful! Can you explain a bit more why dipole moments are required for rotational transitions? $\endgroup$
    – NeStack
    Jun 11, 2021 at 8:59

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Yes, it is correct that cold $H_2$ radiates much less electromagnetic radiation than other molecules (for example, carbon monoxide).

Here's one way to think about this intuitively. On macroscopic scales, electromagnetic radiation is produced when charged objects accelerate. The atoms in a hydrogen molecule consist of charged particles (protons and electrons). These particles are certainly capable of accelerating. However, since momentum is conserved and the two hydrogen atoms are identical, if one of the atoms accelerates, the other atom must accelerate by exactly the same amount in the opposite direction. Therefore, the radiation from the two atoms cancels out. (This assumes that the molecule does not interact with other molecules, which is very reasonable in the interstellar medium.)

At molecular scale, the relative acceleration of the two atoms in a diatomic molecule becomes quantized as rotational and vibrational transitions. But the same intuition applies: if the molecule rotates or vibrates, conservation of momentum implies that there is no net radiation.

Of course, the protons and electrons in the two atoms can still accelerate relative to each other. Since protons and electrons have very different mass-to-charge ratios, this does produce radiation. However, the proton-electron interaction is much stronger than the atom-atom interaction. Using quantum mechanics, that means that the available energy levels (the electronic levels) are much more widely separated than the rotational and vibrational levels. At the very low temperatures of the interstellar medium, there simply isn't enough energy available to jump between different electronic levels at any significant rate. As such, interstellar hydrogen can't produce much radiation through this process either.

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  • $\begingroup$ Thanks for the long explanation! Does your first paragraph mean that the 2 atoms in H2 radiate, but the radiation interferes destructively? Can you also comment on John Rennie's comment - that rotational transitions don't occur in symmetric molecules and why? $\endgroup$
    – NeStack
    Jun 11, 2021 at 8:38
  • $\begingroup$ @NeStack Yes, I think that "the atoms radiate, but the radiation interferes destructively" is a reasonable way to think about what is happening. However, I would caution against taking the classical description overly literally here – it can easily give you the wrong intuition. $\endgroup$
    – Thorondor
    Jun 16, 2021 at 0:47
  • $\begingroup$ @NeStack I agree with John that homonuclear diatomics do not undergo purely rotational transitions. To expand a little, if the atoms were different, the molecule would have an electric dipole moment (because one atom would attract the electrons more strongly). Therefore, rotation would result in a net acceleration of charge and thus EM radiation. $\endgroup$
    – Thorondor
    Jun 16, 2021 at 0:48

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