# Which Formula for the Internal Elastic Energy is Correct?

I am confused by the internal energy of a crystal. Lets say I have a isotropically stressed crystal. The total differential of the internal energy per unit volume V is defined as

$$du = \sigma d\epsilon$$

$\epsilon$ being the strain and $\sigma$ the stress. Because the internal energy is a homogeneous function, this implies

$$u=\sigma \epsilon$$

But if I integrate $du = \sigma d\epsilon$ from $0$ to some strain $\epsilon$ using the constitutive relation $\sigma = C \epsilon$, $C$ being some constant, I arrive at

$$u = \frac{1}{2} \sigma \epsilon$$

Now I am very confused. Which equation is the correct one, $u=\frac{1}{2}\sigma \epsilon$ or $u=\sigma \epsilon$. Or which equation applies to which situation? Probably I didnt get some really important point here :)

Thanks for the help.

UPDATE

Ok, I think I now have understood my problem and solved it. In case anyone else ever wonders about this I will post my answer her. But first I want to reframe the question and clarify some assumptions. I will then post my answer beneath.

So, let there be a system given with some time dependent volume $V$ that we want to compress or expand isothermically and reversibly. That is $dT = 0$ and $dS=0$. Furthermore, for simplicity of the argument, I will assume a homogeneous crystal and homogeneous deformation, so the work of the deformation is $-pdV$. That said, we have for the change of internal energy

$$dU = -p dV \tag{1} \label{eq1}$$ After division by the reference volume $V_0$ with the definition $J=V/V_0$ and $u=U/V_0$, this reads $$du=-pdJ' \tag{2} \label{eq2}$$ We will assume a constitutive relation for the pressure as being $$p(J) = - K ln(J) \tag{3} \label{eq3}$$ Where $K$ is the bulk modulus. Using the small strain approximation, that is $J\approx 1$, this reads $$p(J) = - K (J-1) \tag{4} \label{eq4}$$

We now integrate eq. \eqref{eq2} from $J'=1$ to $J'=J$ and get $$\Delta u = K/2(J-1)^2 \tag{5} \label{eq5}$$ On the other hand, $U$ is a homogeneous function of order $1$, that is $U(\lambda V) = \lambda U(V)$ and therefor, according to Eulers Theorem, we got $$U=-pV \tag{6} \label{eq6}$$ This holds independently of the function $p(J)$ and after devision by $V_0$ and usage of \eqref{eq4} we end up with $$\Delta u = K(J-1)^2 \tag{7} \label{eq7}$$ Which, obviously, is not the same as \eqref{eq5}. Why is that so? Which formula is correct?

• First of all, if $\sigma$ is a tensile stress (given by $\sigma=E\epsilon$), then the equation for internal energy change should read $du=+\sigma d\epsilon+Tds$ – Chet Miller Jul 1 '18 at 11:57
• sorry,i updated the question accordingly – Stefanowitschko Jul 1 '18 at 12:02

First of all, the equation $6$ in the question is not the whole truth. Because $U$ is also a homogeneous function of $S$ and $N$, $N$ being the total number of particles in the system, the function $U$ can be written as $$U = TS-pV+N\mu \tag{8} \label{eq8}$$ $\mu$ is the chemical potential of the system. Keeping in mind, that $dS=0$ because of reversibility and $dT=0$ and $dN = 0$ we calculate for the change of internal energy between the states $0$ and $1$, $1$ being the finale state of volume $V$ and $0$ the initial state of voume $V_0$ $$\Delta u = -p_1 J + p_0 + N \Delta \mu = -p(J) J + p(1) + N \Delta \mu = -p(J)J + N \Delta \mu \tag{9} \label{eq9}$$ So we have to calculate $\Delta \mu$ to get the full picture. This can be done by considering the Gibbs-Duhem relation. Since $du = TdS-pdJ+\mu d\rho$, $\rho := N/V_0$, the Gibbs-Duhem relation reads $$sdT-Jdp+\rho d \mu = 0 \tag{10} \label{eq10}$$ Since $dT=0$ and using equation $4$ from the question, this then leads to $$d \mu = \frac{J}{\rho} dp = \frac{J}{\rho} p_J dJ' = -K \frac{J}{\rho} dJ' \tag{11} \label{eq11}$$ After integration from $J'=1$ to $J'=J$ this gives $$\Delta \mu = - \frac{K}{2 \rho} (J^2-1) \tag{12} \label{eq12}$$ Inserting \eqref{eq12} and the equation $4$ from the question into \eqref{eq9} we end up with $$\Delta u = K/2(J-1)^2 \tag{13} \label{eq13}$$ So, in the end the equations are, of course, consistent.