What is the origin of translational motion?

[Maurice Morelock]

TL;DR

Take a single atom in a vacuum chamber at T=0 K. The atom is motionless. Add thermal energy to the vacuum chamber. The atom will not gain energy unless it is contact with the container, which will have rot and vib motions, causing motion of the atom. If it is not in contact with the container, no energy will be absorbed. The only force on the atom remaining is gravity. Is gravity the primary force for all movement of mass.

Take a single atom in a vacuum chamber at T=0 K. The atom is motionless. Add thermal energy to the vacuum chamber. Does this cause the atom to move? If so, by what mechanism?

 

[Maurice Morelock]

Molecules have trans, rot, and vib modes of energy. Take away rot and vib modes by considering a single atom. What force causes the atom to move?

 

[PeterDonis]

Summary:: Take a single atom in a vacuum chamber at T=0 K. The atom is motionless.

No, it isn't. This would violate the uncertainty principle.

The atom will not gain energy unless it is contact with the container

"In contact" is meaningless at the quantum level. The atom will interact with the atoms in the container. There is no way to avoid this; the atom can always exchange photons with the atoms in the container.

Is gravity the primary force for all movement of mass.

No.

 

[Maurice Morelock]

My understanding is that at T = 0 K, the degrees of freedom are zero. I am considering a single atom, not a population.

 

[Demystifier]

My understanding is that at T = 0 K, the degrees of freedom are zero. I am considering a single atom, not a population.

What do you mean by "the degrees of freedom are zero"? The notion of temperature does not make much sense for a single atom. More to the point, forces are not causes of motion, they are causes of acceleration; do you know the first Newton law?

 

[Ibix]

Depends how you added the energy. Energy isn't a thing, it's a property of a thing, so you can't just "add heat to the vacuum chamber". How were you thinking of doing it?

 

[mfb]

A single atom does not have a temperature. Temperature is only defined in the thermodynamic limit.

You can have radiation at a given temperature in the vacuum. If you do that then the atom will randomly collide with photons and acquire some kinetic energy that way.

I merged your two threads as they are essentially the same question.

 

[Dale]

I am considering a single atom, not a population.

The concept of temperature doesn’t really apply for single atoms.

Add thermal energy to the vacuum chamber. Does this cause the atom to move? If so, by what mechanism?

Yes, this would cause the atoms to move (assuming a population not a single atom) by the electromagnetic interaction with the black body radiation.

Edit: I see @mfb made both of these points, and faster

 

[vanhees71]

The misconception causing this confusion is pretty deeply rooted in bad popular-science textbooks, mixing the classical point-particle description of particles (let's talk about elementary particles instead of an atom to consider first the simple case) with the quantum description. In addition also a thermodynamic picture is mixed into it.

Let's look at the quantum description of a non-relativistic particle within a container. The most simple model which admits proper description of both position and momentum is to think about the particle being somehow trapped in an harmonic-oscillator potential. That means it's Hamiltonian (the operator representing the energy of the particle) is
$$\hat{H}=\frac{1}{2m} \hat{\vec{p}}^2 + \frac{m \omega^2}{2} \hat{\vec{x}}^2.$$
As one derives pretty soon in the QM 1 lecture the energy eigenstates of the particle are labelled by three values $n_1,n_2,n_3 \in \mathbb{N}_0=\{0,1,\ldots \}$. The energy eigenvalues are
$$E(n_1,n_2,n_3)=\hbar \omega \left (\frac{3}{2} +n_1 +n_2 +n_3 \right).$$
The lowest energy level is for $n_1=n_2=n_3=0$.

The corresponding energy eigenstate is given by the wave function,
$$u_{0,0,0}(\vec{x})=\left (\frac{1}{\pi a^2} \right )^{3/4} \exp\left (-\frac{x^2}{2a^2} \right) \quad \text{with} \quad a=\frac{\hbar}{m \omega}.$$
That means that the particle has not a well-determined position. There's an uncertainty given by the standard deviation of the probability distribution $P(\vec{x})=|u_{000}(\vec{x})|^2$.

You get the same state in position representation by Fourier transforming the position wave function, which is again a Gaussian. This means that also the momentum has not a well-determined value either. The ground state of the harmonic oscillator is a socalled coherent state, and the uncertainty product is minimal here, i.e.,
$$\Delta p_i \Delta x_i =\frac{\hbar}{2} \quad \text{for} i \in \{1,2,3 \}.$$
That implies that if in the ground state the particle is neither at a fixed position nor is it at rest. That's only true for the expectation values since in this state $\langle \vec{x} \rangle=0$ and $\langle \vec{p} \rangle=0$, which is what you expect for the classical particle, where the lowest possible total energy is $0$ and this implies that the particle is at rest and located in the origin (the minimum of our harmonic-oscillator potential).

But in quantum mechanics it's impossible to get the particle being exactly at rest at exactly one point. That's because of the position-momentum uncertainty relation, as reflected in the above quantum mechanical analysis. This also explains why the well-determined ground-state energy eigenvalue is not 0 in the quantum case but takes the "zero-point value" $3\hbar \omega/2$.

Now concerning temperature. To define a temperature for our particle we must assume that it can exchange energy with a "heat bath" consisting of very many particles. If this heat bath is in thermal equilibrium this defines a temperature and after a sufficiently long time the particle is in thermal equilibrium with this heat bath and thus its state is described by the temperature given by the heat bath. This is described by the quantum version of the Maxwell-Boltzmann distribution, i.e., the Statistical Operator
$$\hat{\rho}=\frac{1}{Z} \exp \left (-\frac{\hat{H}}{k T} \right)$$
with the partition sum
$$Z=\mathrm{Tr} \exp \left (-\frac{\hat{H}}{k T} \right).$$
You can evaluate this using the energy eigenvalues. Thanks to the normalization factor the zero-point energy cancels and in the limit $T \rightarrow \infty$ the only non-vanishing contribution in the decomposition of this statistical operator in terms of the energy eigenvectors is the contribution of the ground state, i.e.,
$$\rho_0=|u_{000} \rangle \langle u_{000}|,$$
i.e., the particle is in the ground state at temperature $T=0$. So the properties of a particle coupled to a heat bath kept at zero temperature are described by the particle in the ground state with the corresponding conclusions explained above, i.e., the particle doesn't "stand still" at 0 temperature, but it gets into the ground state, where both position and momentum are fluctuating. That's also called zero-point fluctuations and is a specific quantum phenomenon which cannot be in any way be described in terms of classical physics.

 

[PeterDonis]

My understanding is that at T = 0 K, the degrees of freedom are zero.

Your understanding is wrong. The degrees of freedom that a particular quantum system has have nothing to do with its temperature; they are the same whether its temperature is 0 K or 1 million K. The only difference is which degrees of freedom contain how much energy, on average.

 

[vanhees71]

I think the misunderstanding comes from the idea of "active degrees of freedom" in thermodynamics. Indeed at not too high temperatures the usual model for an ideal gas of monatomic, two-atomic, and $n$-atomic molecules simply counts the degrees of freedom as if these molecules were rigid rotators. So you have $3$ "active" degrees of freedom entering the energy quadratically namely the kinetic energy $\vec{p}^2/(2m)$, for two-atomic ($n$-atomic) molecules you have 2 (3) additional rotational degrees of freedom. This gives, using the famous equipartition theorem
$$U=\frac{f}{2} N k T \quad \text{with} \quad f=3,5,6$$
for 1-, 2-, and $n$-atomic molecules.

At higher temperatures, however this doesn't work anymore, because also vibrational modes become "active". The reason that they are "frozen" at lower temperatures simply is that their energy is quantized and the corresponding excitation energies are considerably larger than the typical "thermal energy scale", $k T$.