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seen Mar 23 at 13:04

Mar
20
comment Definition of non-degenerate metric tensor
I have never heard this definition of degenerate matrix. The identity matrix has only one eigenvalue (which is 1) and it corresponds to all the vectors (every vector is an eigenvector of the identity). Is the identity degenerate?
Mar
19
comment Definition of non-degenerate metric tensor
They are related in the same sense that the sentence "two times three is six" is related to the statement $2\times3=6$. It's the same statement, one in English, the other in algebraic notation.
Mar
19
comment Definition of non-degenerate metric tensor
The question is not clear. Is det$\ne0$ not the definition of non-degeneracy? Do you know a different definition and you search for a proof that the two are equivalent?
Mar
19
comment Electric field in a hollow object
Hint: when you write $$ ...=\oint_\Gamma E_i d\Gamma=E_i\oint_\Gamma d\Gamma=... $$ You assume that the electric field is constant over the surface of integration (and perpendicular to it). Is this true in the second example?
Mar
18
comment Insight into Torricelli's Equation ($v^2=u^2+2as$)
Suvat (also, Suvad) is a village in the Lachin Rayon of Azerbaijan.
Mar
16
answered Wave equation - dissipation
Dec
17
awarded  Good Answer
Dec
15
awarded  Yearling
Dec
10
comment Mass particle trajectory on a sphere
@KyleKanos, you've misunderstood the OP question. There's no gravitational field due to the sphere, he's talking about a motion in a constant gravitation constrained to a sphere.
Dec
10
revised Mass particle trajectory on a sphere
addendum
Dec
9
answered Mass particle trajectory on a sphere
Dec
4
answered How can black holes have electric charge and spin?
Sep
21
comment Density of States vs Dispersion
en.wikipedia.org/wiki/…
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