# Einstein solid degree of freedom

I was studying from Schroeder's thermal physics book. When it talks about Einstein solids it says that they have 2 degrees of freedom thus $$U=NkT$$

However, I thought when we talk about Einstein solids, we think about harmonic oscillator that oscillates in 3 dimensions. Thus I was expecting Einstein solids to have $$U=3NkT$$ degree of freedom (3 from position space 3 from momentum space).

What is the correct interpretation of 2 degrees of freedom in Einstein solids?

Note: I might be making a big mistake as I have not taken thermo course in my degree yet, and I am studying all by myself.

Note 2: I know that my approach is flawed as $$U=NkT$$ result can be obtained from entropy function of Einstein solid. I'm just trying to use Equipartition theorem

• What page are you referring to? – BioPhysicist Aug 16 '19 at 22:30
• page can different from edition to edition but it is under chapter 3.2 in the sub-chapter that follows "silly analogy" . In my edition it is page 91 – GGphys Aug 16 '19 at 22:34

Each atom is considered to be three oscillators. So each oscillator is actually a 1D oscillator with two degrees of freedom. Therefore, $$N$$ represents the number of 1D oscillators in the Einstein solid (i.e. there are $$N/3$$ atoms in the solid). This is explained in chapter 2 of the book.
Therefore, to get to what you were thinking, just use that the number of atoms $$n$$ relates to the number of oscillators $$N$$ by $$n=N/3$$. Therefore we have $$U=NkT=3nkT$$
Note that this doesn't mean that an Einstein solid has two degrees of freedom like you are proposing the book says. Each 1D oscillator (atom) has $$2$$ ($$6$$) degrees of freedom. Therefore the entire Einstein solid has $$2N$$ ($$6n$$) degrees of freedom.