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According to Parr and Weitao (1995), the Dirac exchange-energy formula (1930) comes from

enter image description here

[I am looking only at the last line of the equation and the rest for completeness]

It's an energy, so I'd expect the whole expression $K_D$ to be in untis of energy. Evaluating the dimensionality of the integral, I get $(L^{-3})^{4/3} L^3 = L^{-1}$. From this, I expect $C_x$ to be in $L$ (or even $E L$, where $E$ is a composite for energy?), so length units, which doesn't feel true and I think I also read a few times, it's dimensionless. But if it were dimensionless, $K_D$ would be in $L^{-1}$ which is strange (and false?).

Furthermore, I am curious for the interpretation of a density to the power of 4/3. It sounds very strange. But on the other hand, in thermodynamics for instance, we have exponents of 3/2 all the time, which start to make sense after some stat mech, so I'm sure 4/3 can be squared with some intuition.

TL;DR:

  1. What is the unit of the integral $\int \rho^{4/3} d\mathbf{r}$ and of $C_x$?
  2. What is the interpretation of the exponent 4/3?
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  • $\begingroup$ Please explain what $k_F$ is, have you included $dr$ in your dimensions ? $\endgroup$
    – porphyrin
    Aug 31, 2023 at 7:36
  • $\begingroup$ The derivation starts in Parr&Weitao p 105 but I only look at the last line. The derivation is imo correct and understandable. For completeness: The Fermi wavevector is $k_F = (3π^2\rho(\mathbf{r}))^{1/3}$ and $t=k_F s$. The integral in the second line evaluates to 1/4. And yes, I have included $d\mathbf{r}$, it's the last factor $L^{3}$ in the dimensionality analysis of the integral. $\endgroup$
    – ste
    Aug 31, 2023 at 7:42

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