Timeline for Maxwell in multiple dimensions: What happens to curl?
Current License: CC BY-SA 3.0
9 events
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| Aug 18 at 5:10 | comment | added | David Z | @James The tensor/symbol only exists if the number of indices is the same as the number of dimensions. So you could never have $\epsilon^{134}$; in three dimensions, the indices can only be 1, 2, and 3. | |
| Jul 28 at 9:50 | comment | added | James | thank you very much. I think I am stumped by the notation $\epsilon^{\alpha \beta \gamma}$ when the indexes are not consecutive, for e.g. what is the value of $\epsilon^{134}$? If it's easier perhaps you could just write out explicitly the full component sum $(\nabla \times B)^\alpha=\epsilon^{\alpha \beta \gamma} \frac{\partial B_\beta}{\partial x_\gamma}$ for $\alpha=1$, please? | |
| Jul 27 at 23:45 | comment | added | David Z | @James I think that's covered by the latter part of my answer, but if you want to know more, it might be good material for a question on Mathematics. | |
| Jul 25 at 3:32 | comment | added | James | Nice answer! (+1) Based on $\nabla \times \vec E = \begin{bmatrix} \frac{dE_z}{dy} - \frac{dE_y}{dz} \\ \frac{dE_x}{dz} - \frac{dE_z}{dx} \\ \frac{dE_y}{dx} - \frac{dE_x}{dy} \end{bmatrix}$ do you think there is a natural way we can define the curl of a 4-dimensional $\nabla \times \begin{bmatrix} E_w \\ E_x \\ E_y \\ E_z \end{bmatrix}$? | |
| Apr 13, 2017 at 12:39 | history | edited | CommunityBot |
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| Mar 2, 2012 at 18:33 | comment | added | David Z | You know the identity $\vec\nabla\times(\vec{v}\times\vec{F})=(\vec\nabla\cdot\vec{F} + \vec{F}\cdot\vec\nabla)\vec{v} - (\vec\nabla\cdot\vec{v} + \vec{v}\cdot\vec\nabla)\vec{F}$? It's basically something like that. $\vec{B}$ is already cross-product-like, so when you take the curl of it, you get something that is more like a divergence. | |
| Mar 2, 2012 at 12:57 | comment | added | Manishearth | Awesome answer!. I didn't understand what you meant by "The curl is not a curl at all". Is it because we have a much simpler expression than the normal curl expression? And in the same expression, is $\mu$ being Einstein-summed (It looks like $\alpha$ isn't)? I doubt we'd get our 3D curl unless it was being Einstein summed. I usually get confused in the summation notation between tensor indices and summation. (My knowledge of tensors is half-baked) | |
| Mar 2, 2012 at 12:49 | vote | accept | Manishearth | ||
| Mar 2, 2012 at 9:22 | history | answered | David Z | CC BY-SA 3.0 |