# Bulk Modulus and its derivative in a fcc lattice

1. Feb 17, 2015

### Hermes Chirino

The bulk modulus B = - V (∂P/∂V). At constant temperature the pressure is given by P= -∂U/∂V, where U is the total energy. We can write B in terms of the energy per particle u = U/N and volume per particle
v = V/N :

B = v (∂/∂v) (∂u/∂v) Eq (1)

The volume per particle v in a fcc lattice is v = a^3/4, where the side a of the conventional cubic cell is related to the nearest-neighbor separation r by a = √2r. We may therfore write:

v = r^3/√2 ; thus: ∂/∂v = (√2 / 3r^3)(∂/∂r) Eq (2)

And rewrite the bulk modulus as:

B = (√2/9) r (∂/∂r) 1/r^2 (∂/∂r) u Eq (3)

Questions:
How the three equations were derived (and I am very familiar with differential calculus and still not get it)

Thanks, any help will be appreciate !

2. Feb 18, 2015

### nasu

Do you know the "chain rule" for the derivatives?
I suppose you understand eq 1, it's just substituting the quantities per atoms in the definition.