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Aug
23
answered Let $F$ be a vector field in $\mathbb{R}^3$. If $F$ is divergence free, we may deform the surface. Why?
Aug
22
comment Stuck at the derivation of divergence in Cartesian coordinates.
I expand the series in terms of all three coordinates; the book only does so on the coordinates that vary on each surface. The approaches are equivalent; expand the $E(x+\Delta x, y, z)$ term as a series in $x$ and the derivative appears, without needing to use the limit definition. - The signs of the terms depend on the orientation of the integration surface. Surfaces I and II must be oppositely oriented.
Aug
22
comment Stuck at the derivation of divergence in Cartesian coordinates.
Yes, but you should take care not to use the same notation for partial derivatives as for vector components.
Aug
22
comment Stuck at the derivation of divergence in Cartesian coordinates.
You should Taylor expand first and then integrate.
Aug
22
revised Stuck at the derivation of divergence in Cartesian coordinates.
fixed missing component index
Aug
22
comment Stuck at the derivation of divergence in Cartesian coordinates.
Yes, though I will have to correct it all to use $E_x$, as the surface integrals should result in scalars.
Aug
22
answered Stuck at the derivation of divergence in Cartesian coordinates.
Aug
5
answered Concise description of why rotation quaternions use half the angle
Aug
2
comment curl-free, conservative vector fields in complex analysis
While you may see that it is fruitless to try to relate fields on $\mathbb R^2$ to functions of $z$ (and not $\bar z$), the connection between complex analysis and vector calculus is still quite worthwhile to consider. For example, see that a divergence- and curl-free vector field can be described everywhere given its values on a closed surface, and compare this to the notion that a holomorphic function is determined by its values on a closed curve. This is not a mere coincidence; they are both consequences of generalized Stokes' theorem.
Jul
24
comment Coordinate Systems Transformation(Rectangular to Cylindrical)
What is $a$? Is it some arbitrary number? Or are you trying to write a subscript?
Jul
21
comment Components of a vector product as an antisymmetrical rank 2 tensor
Apply the method I have described to $F \times r$ instead of just $F$.
Jul
20
comment Components of a vector product as an antisymmetrical rank 2 tensor
If Landau and Lifshitz wrote exactly what you wrote, then what they said is unclear or misleading. With any vector $F$ there is a rank-2 tensor $f$ defined as I defined it (with no reference to $r$). The cross product $F \times r$ can be written in terms of the $f_{ij}$ as $\sum_j f_{ij} x^j$. This is without a doubt what L&L mean, even though what they say seems to imply otherwise.
Jul
19
comment Components of a vector product as an antisymmetrical rank 2 tensor
The tensor $A(v,w)$ is fully described by the total set of the components $A_{ij}$ because you can write $A(v,w) = \sum_{i,j} v^i w^j A_{ij}$ for any $v,w$. I'm trying to get across that you do not need to know $A(v,w)$ for every possible $v$ and $w$. Using the basis vectors is enough because of linearity.
Jul
19
comment Components of a vector product as an antisymmetrical rank 2 tensor
Superscript vs. subscripts: components written with superscripts go with the usual tangent basis vectors, which may be denoted $e_i$ for instance. These are not necessarily unit (but they are in Cartesian coordinates), hence I did not use hats with them. The tangent basis vectors are the vectors tangent (parallel) to coordinate lines. - Components with subscripts go with the cotangent basis vectors, which are normal ("cotangent") to surfaces of constant coordinate. In Cartesian coordinates, the tangent and cotangent basis vectors are identical, and subscripts are the same as superscripts.
Jul
17
answered Components of a vector product as an antisymmetrical rank 2 tensor
Jul
17
comment Components of a vector product as an antisymmetrical rank 2 tensor
So folks have a starting point for their answers (and so those answers don't go over your head or explain stuff you already know): do you know what a tensor is? An antisymmetric tensor? A tensor of rank two?
Jul
11
comment Why Is My Distance Travelled Overshooting?
I don't see how that scheme would work; it still uses the new velocity to update the position.
Jul
11
comment Why Is My Distance Travelled Overshooting?
Because you updated the velocity first and then used the new value to update the position, which is not what the scheme I wrote does.
Jul
11
comment Why Is My Distance Travelled Overshooting?
Don't know if there's anything else I can do from here; the scheme is sound as far as I know, so any further issues with computing the correct position/velocity are out of the scope of this answer--e.g. having to do with implementation, perhaps. - Maybe you could describe the exact situation that you're trying to model and give numerical data for what you have versus what the scheme predicts?
Jul
11
answered Stokes' theorem generalized the FTC part 2. Is there a known generalization for part 1?