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Bulk Modulus and its derivative in a fcc lattice

  1. Feb 17, 2015 #1
    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. jcsd
  3. Feb 18, 2015 #2
    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.
     
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