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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

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  • $\begingroup$ What page are you referring to? $\endgroup$ – Aaron Stevens Aug 16 at 22:30
  • $\begingroup$ 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 $\endgroup$ – GGphys Aug 16 at 22:34
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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.

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