Why do tetrahedral complexes have approximately 4/9 the field split of octahedral complexes?

I am trying to calculate the relationship between the octahedral field splitting parameter ($\Delta\_\mathrm{o}$) and the square planar field splitting parameter ($\Delta\_\mathrm{sp}$) and thought a good way to start would be to understand why $\Delta\_\mathrm{t}=4/9\Delta\_\mathrm{o}$ for tetrahedral fields but couldn't find any information on it.


How does one arrive at this particular ratio mathematically?




The ratio is derived in [The angular overlap model. How to use it and why](http://pubs.acs.org/doi/abs/10.1021/ed051p633) J. Chem. Educ., vol. 51, page 633-640.


First, relative expressions ([Shaffer Angular Overlap Factors](https://www.iupac.org/publications/pac/24/2/0361/)) are derived for overlap integrals between metal and ligand orbitals as a function of angles. 


For ligand sigma orbitals, the expressions are a functions of 2 angles, specifying the direction of the ligands in the coordinate system of the metal. 


For pi-x and pi-y ligand orbits, there is a third angle for orientation about the ligand-metal axis. 


Then, assuming the ligand-metal distances are the same in both the tetrahedral and octahedral cases, and considering the coordinate geometry of the ligands in the two cases, the relationship:


$$\Delta\_{T} = -4/9 \Delta\_{O}$$


is obtained. 


For another discussion of this ratio in the framework of the angular overlap model, as well as extension to other geometries such as square planar, see [A New Look at Structure and Bonding in Transition Metal Complexes](https://books.google.com/books?id=5D-C9Z1IPrMC&pg=PA123&dq=%22derivation%22%20%224/9%22%20octahedral%20tetrahedral&hl=en&sa=X&ved=0ahUKEwjK44qV3MPPAhVTgiYKHfdWDaEQ6AEIHzAB#v=onepage&q=%22derivation%22%20%224%2F9%22%20octahedral%20tetrahedral&f=false), *Advances in Inorganic Chemistry and Radiochemistry*, Volume 21, pages 113-146.


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