# What is the meaning and unit of the integral in the local density approximation (LDA)?

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

[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?
• Please explain what $k_F$ is, have you included $dr$ in your dimensions ? Aug 31, 2023 at 7:36
• 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.
– ste
Aug 31, 2023 at 7:42