# infinitesimal strain and atomic concentration

In short: I need to drive:

$$\text d\varepsilon= \frac{\text d\sigma}{K} =-\frac{\text dc}{c}$$

where $\varepsilon$ and $\sigma$ are the volumetric strain and stress respectively, $K$ is the bulk modulus and $c$ is the atomic concentration.

My effort:

Long ago a galaxy far far away I showed that above equation is true. The left-hand side equality is obvious. In overall, it says the relative density change $\frac{\text dc}{c}$ correspond to the strain increment $\text d\varepsilon$. It is also widely used in material science of metals. I came across this again, tried to obtain it but I could not :( I'm getting old. At least, I know the following things ($V$ is the volume):

• Incremental strain (Ludwik): $\delta \varepsilon=\frac{\delta V}{V}$
• True strain: $\varepsilon=\frac{\Delta V}{V}$
• Logarithmic strain: $\varepsilon=\ln(\frac{V}{V_0})$
• $\text dy=\frac{\text dy}{\text dx}\text dx$

Yet, I could not manage it to use my aforementioned knowledge to obtain that expression again. A quick reverse engineering says $\varepsilon=\ln(\frac{1}{c})$. I'm stuck. Can you please help me to drive the beloved equation again?

• Never mind, I just got it. Holy power of writing! – Ali Abbasinasab Aug 19 '17 at 11:16
• Am I right? Considering an incremental strain (Ludwik): $$\delta \varepsilon =\frac{\delta V}{V}$$ And plugging $V=\frac{m}{c}$, where $m$ is the mass, yields: $$\delta \varepsilon =\frac{\delta (\frac{m}{c})}{\frac{m}{c}}=\frac{\delta (\frac{1}{c})}{\frac{1}{c}}$$ Thus, $$\text d \varepsilon =-\frac{\text dc}{c}$$ Any comment, fix or improvement is appreciated. – Ali Abbasinasab Aug 19 '17 at 11:25