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 Aug 26 comment Looking for a simple proof of symmetry of linear susceptibility tensor I think it is. millersville.edu/~jdooley/macro/derive/elpol/alphasym/… Aug 21 comment Looking for a simple proof of symmetry of linear susceptibility tensor @ValterMoretti I agree with the math, but disagree with the interpretation. This whole $P\propto E$ relation is a linear response theory. It is derived the other way around: The energy, to leading order in $\vec E$, is given by $U\approx U_0 +\chi_{ij}E_iE_j$. The lack of a linear term is because $U$ is minimal when $\vec E=0$. Therefore, WLOG $\chi$ can be chosen to be symmetric. $P_i$ is defined as $\partial U/\partial E_i$. Aug 21 answered Looking for a simple proof of symmetry of linear susceptibility tensor Jul 15 awarded Scholar Jul 15 accepted Interpretation of dispersion relation Jul 14 comment Why plane stress condition is taken for thin plates As I wrote in my answer, this is not a proof, nor a consistent derivation. Plane-stress equations are an APPROXIMATION, which is not exactly valid. Specifically for the profile you suggested, it is not enough in order to determine whether it is a possible solution. You need the other stress components, and if they satisfy the force-balance equation $$\sum_j \frac{\partial \sigma_{ij}}{\partial j}=0$$ then you're good. Jul 14 answered Why plane stress condition is taken for thin plates Jul 14 asked Interpretation of dispersion relation Dec 15 awarded Yearling May 17 comment How can I determine whether the mass of an object is evenly distributed? This is true, but the OP only wants to know if it's evenly distributed or not, which is an easier prblem than "find the distribution". For example, you could measure the moment of inertia around an axis, and compare that to what you'd get if it were homgenously distributed. If there's a discrepancy - you can catch it. I wonder whether one can find a counterexample of a body that has the full tensorial moment of inertia of a homogeneously distributed one, but is actually not. May 2 comment Will a hole cut into a metal disk expand or shrink when the disc is heated? @jamesdlin I agree that it's heuristic and as such one could argue differently. The real solution is sketched by David Z. If you want more justification, you can say that the thermo-elastic equations with $T=const$ and stress-free BC will result in a stress free configuration (easily verified). Therefore, having the disc in place or cut out has no effect on the surrounding - the "interaction" is the stress, and therefore stress free boundary conditions are equivalent to not having a disc at all. Again, I stress that this can be solved analytically and then there's no ambiguity. Mar 18 revised From momentum to solid angle edited tags Feb 7 comment When driving uphill why can't I reach a velocity that I would have been able to maintain if I started with it? @Gugg In my experience, the phenomena is pretty common. Also, there are surely many other factors involved (the computer regulating fuel injection, exauhst, wind/friction dissipation...) so this simplified model does not capture the difference between "hard" and "impossible". Feb 7 comment When driving uphill why can't I reach a velocity that I would have been able to maintain if I started with it? @dmckee I agree, but I think the car is not yet at this regime. Feb 7 comment When driving uphill why can't I reach a velocity that I would have been able to maintain if I started with it? I'd say that the newtonian-gravity is not relevant. Feb 7 answered When driving uphill why can't I reach a velocity that I would have been able to maintain if I started with it? Feb 6 comment How does the correlation length of weather emerge? I agree, but that's a totally different effect. It would be awefully wrong to say that the typical length-scale of the waves in lakes stems from the lake size, scaling-wise. Feb 6 comment How does the correlation length of weather emerge? The wave length in lakes is determined by the lake size? This sounds weird to me. The wavelength is typically a few orders of magnitude smaller. Feb 6 asked How does the correlation length of weather emerge? Jan 27 comment Scattering from a box potential of width $L$ doesn't reproduce a step potential in the limit $L \rightarrow \infty$ @Joe Well, unless your function grows exponentially for large $x$, you have no freedom in choosing the sign of $k$.