I'm investigating the thermo-migration failure mechanism in nanoscale ICs interconnects. Typically, a nano wire under thermal stress suffers from material/mass migration or void nucleation if it experiences a large amount of tensile hydrostatic stress (e.g. $\sigma_{critical}=40MPa$ for copper).

It's known that a temperature change of $\Delta T$ with respect to a base or reference level produces a thermal strain:

$$\epsilon_T=\alpha \Delta T$$

where $\alpha$ is the coefficient of thermal dilatation (i.e. thermal expansion coefficient). Thus stress can be obtained by $\sigma=E\epsilon$.

In my system, which is a nanoscale copper wire (e.g. $T=64nm$, $W=32nm$ and $L=2um$), temperature distribution is not linear as it is shown in the following figures:

Wire Temperature Distribution

In other words, while the $\Delta T_{max}$ is around $5K$, the temperature gradient in some local regions (e.g. corners) can be up to $10K/\mu m$. The following figure shows the temperature gradient of one particular distribution, in which the gradients are very large at the end of the wire where it connects to other wires.

enter image description here

It has been experimentally reported that in a presence of a large local gradient in temperature (even with a small $\Delta T_{max}$), failure might happen (i.e. mass transports and void nucleates as it's shown in the following figure).

enter image description here

It can be seen that this phenomenon cannot be explained directly by $\epsilon_T=\alpha \Delta T$. I'm wondering how this large local gradient of temperature can be related to strain/stress properly. Because, as I mentioned above, the thermo-migration mortality of a wire is judged based on the critical hydro-static stress. Simply speaking, I'm hoping to find a relationship like:

$$\epsilon_T = \beta \nabla T$$

PS Can I simply conclude that: $\nabla \sigma = \alpha E \nabla T$?


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