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B3NR4Y
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I wasn't sure where to post this, and I hope this is the right place. I've been reading ahead of my lectures, and I've gotten a book that introduces tensors. It very quickly introduces Einstein Summation Convention, which I think I understand, [itex] \sum_{i=1}^{3} x_{i} y_{i} = x_{i} y_{i} = x \cdot y [/itex], or, for instance the gradient of a scalar function, φ, [itex] \nabla \varphi = \sum_{i=1}^{3} \frac{1}{h_{i}} \partial_{ i } \varphi = \frac{1}{h_{i}} \partial_{ i } [/itex] . I feel like I understand this well enough to use it, but when it introduces the curl it uses a symbol, εijk that is 1 when ijk is an even permutation of (1,2,3) and -1 when ijk is an odd permutation of (1,2,3), and zero if two or more indices are equal. I thought I understood this well enough, for instance ε2,1,3 = -1, but I guess I don't or I'm too confused. Directly after that it says εijk = δilδjm-δjlδlm, and I'm confused from here on. Where do m and l come from? It mentions these are just "dummy indices" but I'm unsure what even that means. Am I summing the repeated indices or something? I'm sorry if I sound dumb, but I'm teaching ahead and don't really have a professor to ask these questions. It uses this epsilon to define the cross product symbolically, and here is where another confusion hits.
[itex] \vec{e_{ i }} \times \vec{e_{ j }} = \epsilon_{ijk} \vec{e_{k}} [/itex], I get this, the permutation of ijk is even, and it's positive so it makes sense. Then it goes on to define a left-handed basis as follows [itex] \vec{e_{ i }} \times \vec{e_{ j }} = -\epsilon_{ijk} \vec{e_{k}} [/itex], why not just replace the ijk with jik or some other odd permutation and drop the negative sign? Or is the epsilon there for some other reason and the negative is applied to the k basis vector?
And the last thing is when it uses this to define the curl, it does as follows [itex] (\nabla \times \vec{V})_{ i } = \epsilon_{ijk} \partial_{j} \vec{V_{k}} [/itex], but I know from previous works, the curl has three terms and the sign alternates. I assumed the alternating sign would come from the epsilon function, and more confusion hits me. Am I summing multiple indices? so
[tex] \sum_{ i = 1}^{3} \sum_{ j = 1}^{3} \sum_{ k = 1}^{3} \epsilon_{ijk} \partial_{j} \vec{V}_k =\epsilon_{ijk} \partial_{j} \vec{V}_k = [\partial_{2}(\vec{V_{3}})-\partial_{3}(\vec{V_{2}})]\vec{e_{1}} + [\partial_{3}(\vec{V_{1}})-\partial_{1}(\vec{V_{3}})]\vec{e_{2}} + [\partial_{1}(\vec{V_{2}})-\partial_{2}(\vec{V_{1}})]\vec{e_{3}}[/tex]
Or am I way off? Sorry for the lengthy post
[itex] \vec{e_{ i }} \times \vec{e_{ j }} = \epsilon_{ijk} \vec{e_{k}} [/itex], I get this, the permutation of ijk is even, and it's positive so it makes sense. Then it goes on to define a left-handed basis as follows [itex] \vec{e_{ i }} \times \vec{e_{ j }} = -\epsilon_{ijk} \vec{e_{k}} [/itex], why not just replace the ijk with jik or some other odd permutation and drop the negative sign? Or is the epsilon there for some other reason and the negative is applied to the k basis vector?
And the last thing is when it uses this to define the curl, it does as follows [itex] (\nabla \times \vec{V})_{ i } = \epsilon_{ijk} \partial_{j} \vec{V_{k}} [/itex], but I know from previous works, the curl has three terms and the sign alternates. I assumed the alternating sign would come from the epsilon function, and more confusion hits me. Am I summing multiple indices? so
[tex] \sum_{ i = 1}^{3} \sum_{ j = 1}^{3} \sum_{ k = 1}^{3} \epsilon_{ijk} \partial_{j} \vec{V}_k =\epsilon_{ijk} \partial_{j} \vec{V}_k = [\partial_{2}(\vec{V_{3}})-\partial_{3}(\vec{V_{2}})]\vec{e_{1}} + [\partial_{3}(\vec{V_{1}})-\partial_{1}(\vec{V_{3}})]\vec{e_{2}} + [\partial_{1}(\vec{V_{2}})-\partial_{2}(\vec{V_{1}})]\vec{e_{3}}[/tex]
Or am I way off? Sorry for the lengthy post
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