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As far as I know every atom vibrates / oscillates / I don't know what it's called in English with a frequency, specific for its element. I don't mean the velocity of the atom, which defines temperature. Although, honestly the two might be related in a way I don't know.

For example, I know that the proof that time is a real physical dimension and not just something people made up with clocks is that you take an atomic clock at sea level and one on a high mountain and one will run behind.

I know atomic clocks don't measure time by the atoms' vibration but by a more complicated mechanism but I just thought that the other mechanism might be affected by the difference in temperature (which I am sure they account for, if needed) but still, I thought to myself if you just measure the rate of oscillation / vibration then that shouldn't be affected by temperature and then realised I am not sure about that.

tl,dr:

  1. Is the vibration / oscillation of an atom with a frequency, strictly specific to that element, different from the atom's velocity?
  2. Is said vibration / oscillation not affected by temperature? Difficult question to formulate, as it's more than atoms define temperature than the other way around.
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    $\begingroup$ It has hard to follow the reasoning especially in relation to time. However the vibrations within a molecules or of atoms/molecules in a lattice depend on T. $\endgroup$
    – Alchimista
    Sep 16 at 10:42
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    $\begingroup$ @Alchimista I believe the vibrations refer to "the thing which atomic clocks measure" i.e. spectral lines. The question is "Do spectral lines depend on temperature?" with a side helping of "Do atoms naturally vibrate at their spectral line frequencies?" $\endgroup$
    – user253751
    Sep 16 at 11:24
  • $\begingroup$ @user253751 indeed your comment made me think to understand more of OP question, now. Above I referred to vibrations as commonly intended in physical chemistry, real motions. Now I think that the answer to OP is NOT at least at first approximation, as for the intrinsic lines in atoms are at high energy scale. A specialist would provide a real answer. Of course, atomic clocks need T control for a series of reasons but not for anything intrinsic to the inner atomic structure. $\endgroup$
    – Alchimista
    Sep 16 at 11:54
  • $\begingroup$ @Alchimista, yeah, sorry, it probably is really confusing because I learnt about the "vibrations" not in English, so I'm not sure if I'm using the proper term, as well as I learnt about them in high school and in uni we never spoke about them, so either I was taught some concept too elementary in high school and cannot relate it to anything due to in uni not meeting the same terms or there is something else amiss. Sorry but I really can't articulate it any better. $\endgroup$
    – mummy
    Sep 22 at 16:07
  • $\begingroup$ @user253751, yeah, sorry, it probably is really confusing because I learnt about the "vibrations" not in English, so I'm not sure if I'm using the proper term, as well as I learnt about them in high school and in uni we never spoke about them, so either I was taught some concept too elementary in high school and cannot relate it to anything due to in uni not meeting the same terms or there is something else amiss. Sorry but I really can't articulate it any better. $\endgroup$
    – mummy
    Sep 22 at 16:07
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Atoms do not have vibrational degrees of freedom, but rather the eletronic excitations. On the other hand, molecules (which consist of multiple atoms) do have vibrational and rotational degrees of freedom. As the question seems to be motivated by statistical physics, I will further make the remarks only about the latter.

The question (implicitly) refers to the equipartition theorem, which tells us that energy per degree of freedom in a gas is approximately $$\frac{k_B T}{2}$$ It is clear, according to the OP, how this is applied to the translational degrees of freedom, i.e., the velocities of the molecular center-of-mass. It is probably also clear how this applies to the molecular rotations, since these can be characterized by velocities. When it comes to oscillationsof atoms in respect to each other, each degree of feredom can be modeled as a harmonic oscillator with energy: $$ E=\frac{mv^2}{2}+\frac{m\omega^2x^2}{2}, $$ where $x$ and $v$ refer to the relative displacement of the two parts of teh molecule. It is the energy of such socillation that is given by $k_BT$. As the amplitude of the oscillations is related to their energy, one could equally express the equipartition in terms of the oscillator velocity at its minimum (at $x=0$) where we have: $$ E= \frac{mv^2}{2}=\frac{k_B T}{2}. $$

Remarks:

  • Rotational and vibrational degrees of freedom is the readon why the heat capacity of molecular gases is higher than that of monoatomic gases, see here and here.
  • When treated quantum mechanically, the vibrational degrees of feredom of molecules usually have high excitation energy and are not involved at low temperatures. Thus, depending on temperature, the number of degrees of freedom that need to be taken into account changes, and the heat capacity of a molecular gas changes with temperature as well!
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    $\begingroup$ I upvoted but I think you took the Q as I was doing initially. See comments under the question. $\endgroup$
    – Alchimista
    Sep 16 at 12:03
  • $\begingroup$ "Atoms do not have vibrational degrees of freedom" have a look at this chem.libretexts.org/Bookshelves/… $\endgroup$
    – anna v
    Sep 16 at 12:29
  • $\begingroup$ @annav the link is about the vibrational degrees of freedom of molecules - it is atoms in a molecule that move, not isolated atoms. $\endgroup$ Sep 16 at 12:41
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    $\begingroup$ @annav I think it is misleading: 1) even classically, a molecule would have vibrational degrees of freedom, i.e., one does not need to invoke the uncertainty principle; 2) a molecule can be treated as a group of atoms bound by springs, and its normal modes can be found - some of these will be (mostly) displacement of single atoms, while others may correspond to the oscillations of whole blocks of atoms. Strictly speaking, none of these is a motion of an atom alone. $\endgroup$ Sep 16 at 13:14
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    $\begingroup$ yeah, sorry, it probably is really confusing because I learnt about the "vibrations" not in English, so I'm not sure if I'm using the proper term, as well as I learnt about them in high school and in uni we never spoke about them, so either I was taught some concept too elementary in high school and cannot relate it to anything due to in uni not meeting the same terms or there is something else amiss. Sorry but I really can't articulate it any better. $\endgroup$
    – mummy
    Sep 22 at 16:11

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