Isothermal compressibility and spring constant solids

[PhizKid]

Homework Statement

Consider a solid of compressibility $\kappa$. Assume that the atoms in this solid are arranged on a regular cubic lattice, the distance between their nearest neighbors being $a$. Assume further that a restoring force $-k_0 \Delta a$ acts on a given atom when it is displaced by a distance $\Delta a$ from its nearest neighbor. Use simple reasoning to find an approximate relation between the spring constant $k_0$ and the compressibility $\kappa$ of this solid.

Homework Equations

$\kappa = \frac{1}{V}\frac{\partial V}{\partial p}|_{T}$.

The Attempt at a Solution

Consider a cubic lattice of $N+1$ atoms on each edge. Take $N \gg 1$ so that $N+1 \approx N$. The area of any given face of the lattice is $A = N^2 a^2$. Imagine now that the face is pushed inward by the amount $\Delta a$ where $\frac{\Delta a}{a} \ll 1$. Then there is an increase in the total force exerted on this face due to the restoring force $F = -k_0 \Delta a$ on each of the $(N+1)^2$ atoms in this face; the increase in total force on this face will be given in magnitude by $\Delta F = (N+1)^2 \Delta a k_0 \approx N^2 \Delta a k_0$.

Then $\kappa = \frac{1}{V}\frac{\partial V}{\partial p}|_T = \frac{1}{N a}\frac{\partial a}{\partial p}|_T \approx \frac{1}{a}\frac{\Delta a}{\Delta p} = \frac{1}{a}\frac{N^2 a^2 \Delta a}{N^2 \Delta a k_0} = \frac{a}{N k_0}$.

According to the book the correct answer is $\kappa = \frac{a}{k_0}$ which is not what I got. However I am not sure of how to fix my solution. I could I suppose take a parallelipiped with sides of length $Na$ whose faces are to be pushed inwards, and sides of length $a$ which are parallel to the force applied to push the aforementioned faces inwards. However this looks to be a very contrived escape in fixing my solution on top of the assumptions I already made that make my solution above look quite non-rigorous. For example I took the solid to be a cubic lattice as opposed to any crystal lattice and I also took $N \gg 1$ to make $(N+1)^2 \approx N^2$, neither of which are assumptions made in the problem statement. Could anyone help me in fixing my solution. Also, could any comment on whether or not I could make the calculation more rigorous? Any hints on how to do so? Thanks!

 

[TSny]

Then $\kappa = \frac{1}{V}\frac{\partial V}{\partial p}|_T = \frac{1}{N a}\frac{\partial a}{\partial p}|_T$

Can you explain how you get the second equality here?

 

[PhizKid]

Can you explain how you get the second equality here?

$\kappa = \frac{1}{V}\frac{\partial V}{\partial p} = \frac{1}{A L}\frac{A \Delta a}{\Delta p} = \frac{a N^2 \Delta a}{N (N^2 k_0 \Delta a)} = \frac{a}{k_0 N}$

 

[TSny]

$\kappa = \frac{1}{V}\frac{\partial V}{\partial p} = \frac{1}{A L}\frac{A \Delta a}{\Delta p} = \frac{a N^2 \Delta a}{N (N^2 k_0 \Delta a)} = \frac{a}{k_0 N}$

V = AL. When V changes, it appears you are assuming that only L changes and not A. Shouldn't A also change?

That is, how do you get $\partial V = A \Delta a$?

 

[PhizKid]

V = AL. When V changes, it appears you are assuming that only L changes and not A. Shouldn't A also change?

That is, how do you get $\partial V = A \Delta a$?

In the calculation I was simply pushing one face of the cube inwards by some small amount so the area of the face is the same since the push is normal to the face itself.

 

[TSny]

I believe the usual definition of compressibility $\kappa$ involves the change in volume when the pressure is increased over the entire surface. So, for a cube, you would consider the change in volume when the pressure is increased on all six faces.