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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?
Jul
11
comment Stokes' theorem generalized the FTC part 2. Is there a known generalization for part 1?
I second @JyrkiLahtonen - it is probably best to state explicitly what you think Stokes' generalizes and what you would like to see generalized in turn.
Jul
9
comment Why Is My Distance Travelled Overshooting?
Not "exponential", but definitely not linear either. (In point of fact, it's quadratic.)
Jul
9
comment Why Is My Distance Travelled Overshooting?
You would want $x(t+1) = x(t) + v(t) + a/2$ and $v(t+1) = v(t) + a$. You can substitute $a = v(t+1)-v(t)$ into the $x$ equation and get the scheme from @JyrkiLahtonen as a result. - Take note while $v(t+1) = v(t) + a$ for constant $a$, it is not true that $x(t+1) = x(t) + v(t)$ because the velocity is not constant over the interval.
Jul
9
answered Why Is My Distance Travelled Overshooting?
Jul
9
answered Calculate the divergence of the polar coordinate vector field $\partial_\phi$
Jul
9
comment Calculate the divergence of the polar coordinate vector field $\partial_\phi$
Ok, so the issue is the divergence. What is the definition of divergence that you should use? Are you given an expression that should compute the divergence in polar coordinates?
Jul
9
comment Calculate the divergence of the polar coordinate vector field $\partial_\phi$
You've said you have a problem with the notation. What is the actual question you're being asked?
Jul
9
comment Calculate the divergence of the polar coordinate vector field $\partial_\phi$
@kaos In what basis is that coordinate tuple written in? I ask because $\partial_\phi$ is not in general a unit vector--this is a major difference between how a basis is chosen in higher mathematics versus elementary vector calculus.
Jul
9
answered why is representing rotations through quaternions more compact and quicker than using matrices??