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| bio | website | |
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| location | ||
| age | ||
| visits | member for | 7 months |
| seen | 11 mins ago | |
| stats | profile views | 197 |
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10h |
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Prolate spheroidal coordinates To prove a mapping is a diffeomorphism, you might need the definition of a diffeomorphism. Do you know what that is? |
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1d |
answered | Rotate the axis of rotation of a quaternion by another quaternion |
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2d |
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Why is boundary information so significant? — Stokes's theorem There are Green's functions for other differential operators, but Stokes' theorem relies on using $\nabla F$ (which I replaced by $j$) specifically. Hence, if $\nabla F$ is not specified (or not easily found), it is more difficult to reconstruct $F$. |
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May 17 |
answered | Line Integral of Every Positively Oriented Simple Closed Path - Green's Theorem |
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May 17 |
revised |
Why is boundary information so significant? — Stokes's theorem added a section explaining why only the surface information is necessary and how it can be thought about |
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May 17 |
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Why is boundary information so significant? — Stokes's theorem What do you mean by "more general derivatives"? Do you mean, for instance, fields that obey more complicated differential equations? Or something else? |
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May 17 |
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Why is boundary information so significant? — Stokes's theorem Complex analysis works this way because the holomorphic functions in CA have zero derivative with respect to the vector derivative $\nabla$--they are "sourceless". Meromorphic functions have point sources. In this way, saying that holomorphic functions are differentiable is actually kinda misleading: the analogues of these functions in 3d and beyond have no divergence or curl. |
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May 17 |
answered | Why is boundary information so significant? — Stokes's theorem |
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May 16 |
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What's the Clifford algebra? I've left a note on your talk page. |
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May 16 |
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What's the Clifford algebra? @rschwieb Anything in particular you're looking for input on? Seems like there's been a lot of discussion on what that intro should be. I don't want to step into some points that have already been settled to satisfaction. |
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May 15 |
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Jacobian or No Jacobian - Surface Integrals Yes, you interpret the meaning of "that" correctly. I was referring to #23, which though the dot product is done in Cartesian, the breakdown of $dA$ can be done with derivatives of $\mathbf r$ without writing $\mathbf r$ in terms of Cartesian basis vectors. Of course, if $\mathbf r$ is written in terms of Cartesian basis vectors, you're essentially calculating the Jacobian. |
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May 15 |
answered | Clifford Algebra Multiplication Intuition |
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May 14 |
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Jacobian or No Jacobian - Surface Integrals @LaPrevoyance It's of course perfectly legitimate to use the Jacobian there. Perhaps I should say more that for these kinds of problems, I personally prefer not to. I showed you can calculate $\hat{\mathbf n}$ in terms of spherical basis vectors without writing $(x,y)$ in terms of $(r,\theta, \phi)$ at all--that is the approach the other two solutions use, after all. But both approaches--using the Jacobian, or finding the tangent basis vectors and computing the cross product---are equivalent and essentially do the same thing. One is just a little more abstract than the other. |
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May 14 |
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Wedge product of vector fields I take it $x$ and $y$ are complex coordinates? |
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May 14 |
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What does it mean to be divergence thorem applicable? "Blah" is a placeholder for any condition you might want to verify a vector field obeys. You can do this component by component because vector addition of two smooth fields does not result in an un-smooth field, and each component can be treated as its own field. |
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May 13 |
awarded | Caucus |
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May 13 |
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Solving tensor Identities It's a directional derivative in the direction of $u$. |
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May 12 |
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Jacobian or No Jacobian - Surface Integrals @LaPrevoyance I have added a section answering your supplemental questions. I did make a mistake in describing parameterization with cartesian coordinates, which I have corrected. |
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May 12 |
revised |
Jacobian or No Jacobian - Surface Integrals added a section answering supplemental questions |
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May 10 |
answered | Intuition for Cross Product of Vector with Itself and Vector with Zero Vector |
