I'm confused about how you're defining degrees of freedom. For an ideal gas, we had 3 degrees of freedom, but in the metal lattice you show in the derivation of the Dulong-Petit Law, you say there are 6 degrees of freedom? Why the difference? Furthermore, you mention at the end that in future videos there will be elements of QM. You mention that we will now assume particles are indistinguishable, but I thought we were assuming that for a while now? Isn't that what the point of the Gibbs factor (1/N!) is? These videos have been excellent in all other aspects and have hugely helped me navigate this difficult subject.
For the ideal gas, there are indeed 3 DOF. These include: motion in the x direction, motion in the y direction and motion in the z direction. Mathematically, this can be written as a Hamiltonian which has three momentum terms: p_x^2, p_y^2, p_z^2, one for each cartesian coordinate. In the case of an atom in a metal lattice, we model the atom as a harmonic oscillator. This means that in addition to the three momentum motion terms in the Hamiltonian, there are three additional terms which correspond to the elastic potential energy stored when an atom deviates from its equilibrium position in the lattice. Think the formula for the potential energy stored in a spring ½*k*x^2. But because we're in three dimensions, it would be ½*k*(x^2+y^2+z^2). These 3 terms, alongside the 3 momentum terms give rise to 6 degrees of freedom in total in the metal lattice. The key difference is, in an ideal gas there is no potential experienced by the particles, whereas in a lattice the atoms are bound by potentials caused by other atoms. I'll get back to the second part of your question when I rewatch my video!
Note that "degrees of freedom" is used differently in different contexts. Mechanically, the degrees of freedom is the dimensionality of configuration space, not phase space (I think). In this sense a point particle has 3 degrees of freedom. As it happens, the expression for the energy in this case also has three quadratic terms (the square of the velocities in each direction of space, or, the square of the three linear momenta). The energy is accordingly 3kT/2, and the monoatomic ideal gas energy is 3nRT/2. Note that this correspondence between configuration space degrees of freedom and the number of kT/2's is _not_ general. E.g. a one dimensional harmonic oscillator has a single (configuration space) degree of freedom but _two_ quadratic terms in the expression for its energy (position squared and velocity squared). In this case, this number corresponds to the dimensionality of phase space, and the energy is kT.
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Very clear 👍
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I'm confused about how you're defining degrees of freedom. For an ideal gas, we had 3 degrees of freedom, but in the metal lattice you show in the derivation of the Dulong-Petit Law, you say there are 6 degrees of freedom? Why the difference?
Furthermore, you mention at the end that in future videos there will be elements of QM. You mention that we will now assume particles are indistinguishable, but I thought we were assuming that for a while now? Isn't that what the point of the Gibbs factor (1/N!) is?
These videos have been excellent in all other aspects and have hugely helped me navigate this difficult subject.
For the ideal gas, there are indeed 3 DOF. These include: motion in the x direction, motion in the y direction and motion in the z direction. Mathematically, this can be written as a Hamiltonian which has three momentum terms: p_x^2, p_y^2, p_z^2, one for each cartesian coordinate.
In the case of an atom in a metal lattice, we model the atom as a harmonic oscillator. This means that in addition to the three momentum motion terms in the Hamiltonian, there are three additional terms which correspond to the elastic potential energy stored when an atom deviates from its equilibrium position in the lattice. Think the formula for the potential energy stored in a spring ½*k*x^2. But because we're in three dimensions, it would be ½*k*(x^2+y^2+z^2).
These 3 terms, alongside the 3 momentum terms give rise to 6 degrees of freedom in total in the metal lattice. The key difference is, in an ideal gas there is no potential experienced by the particles, whereas in a lattice the atoms are bound by potentials caused by other atoms.
I'll get back to the second part of your question when I rewatch my video!
@@pazzy768Thanks, that makes sense. A very clear and throrough explanation.
@@citra5431 Could you provide a timestamp for the part that you were referring to in the second part of your question? I couldn't find it in the end!
wouldn't there be 6 dof for ideal monoatomic gas? 3 position + 3 momentum?
There are; each of which contributes k_BT/2 of energy per particle
@@pazzy768 oh but then at 29:33 you said there are 3 dof of an ideal gas?
Note that "degrees of freedom" is used differently in different contexts. Mechanically, the degrees of freedom is the dimensionality of configuration space, not phase space (I think). In this sense a point particle has 3 degrees of freedom. As it happens, the expression for the energy in this case also has three quadratic terms (the square of the velocities in each direction of space, or, the square of the three linear momenta). The energy is accordingly 3kT/2, and the monoatomic ideal gas energy is 3nRT/2.
Note that this correspondence between configuration space degrees of freedom and the number of kT/2's is _not_ general. E.g. a one dimensional harmonic oscillator has a single (configuration space) degree of freedom but _two_ quadratic terms in the expression for its energy (position squared and velocity squared). In this case, this number corresponds to the dimensionality of phase space, and the energy is kT.