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## Hot answers tagged notation

22

This is not really an answer to your question, essentially because there isn't (currently) a question in your post, but it is too long for a comment. Your statement that A co-ordinate transformation is linear map from a vector to itself with a change of basis. is muddled and ultimately incorrect. Take some vector space $V$ and two bases $\beta$ and ...

13

It is, in fact, a double integral! The first notation used $$\varPhi_E = \oint_S \vec{E} \cdot \mathrm{d}\vec{A} = \oint_S \vec{E} \cdot \hat{n} \ \mathrm{d}A$$ is simply a more compact notation. It's much easier to write $\mathrm{d} \vec{A}$ instead of, say, $r \ \mathrm{d}r \ \mathrm{d}\theta$ all the time. Furthermore, it's more general, as $\mathrm{d} ... 5 There are two more points that can be made here. Sorry if I repeat someone. In a way you are right that if you have a vector space and its dual there is no intrinsic way to say which space is the original and which is the dual. This is because there is a canonical isomorphism between a vector space and the dual of its dual. In other words if$V$is a vector ... 5 For doing matrix multiplication, yes, we usually put the dummy indicies together, but this is (assuming one is not working over$\mathbb{H}$) convention really. Don't forget that this is shorthand Einstein convention. When you write $$\mathbf{A} \cdot \mathbf{B} = A_{ij} B_{jk} = \sum_{j=1}^n A_{ij} B_{jk}$$ you are writing a shorthand version in ... 3 It is definitely an ambiguous notation, but one that is quite conventional. You should interpret it as:$(\partial_a X_\mu)(\partial^a X^\mu)$. For instance, often the Klein-Gordon Lagrangian is written as: $$\mathcal{L} = \frac{1}{2} \partial_\mu \phi \partial^\mu \phi + \cdots$$ which should be interpreted as: $$\mathcal{L} = \frac{1}{2} (\partial_\mu ... 3 The general formula is indeed a double integral, so the most technically correct way to write it is$$\Phi_E = \iint_S \vec{E}\cdot\mathrm{d}^2\vec{A}$$But when formulas start to involve four, five, or more integrals, it gets tedious to write them all out all the time, so there's a notational convention in which a multiple integration can be designated by ... 2 I) No, it is important to distinguish between covariant and contravariant tensors. OP's link mentions differential geometry. If one has only studied those objects in the context of pseudo-Riemannian manifolds (M;g), which comes equipped with an (invertible) metric (0,2) tensor g, then the existence of the musical isomorphism may perhaps unnecessarily ... 2 You are right that \langle A\rangle_{\psi} is a number and not an operator. However, people often write just a number when they actually mean the identity operator times that number. So in the right hand side, \langle A\rangle_{\psi} should actually be \langle A\rangle_{\psi} \mathbf{1}_H, where \mathbf{1}_H is the identity operator on your hilbert ... 2 The Kronecker delta is used in the first term, not the second. In the first term, replace \sum\limits_i \omega_i^2 with \sum\limits_{i,j}\omega_i\omega_j\delta_{ij}. For the second term, rearranging the summations is a pretty common thing to do. The identity that you are interested in is essentially the distributive property. As a simple example,$$ ... 2 The notion of co- and contravariance depends on context: If you wanted to be as clear as possible, you should actually mention with respect to what the components transform co- or contravariantly. In case of the algebraic dual of finite-dimensional vector spaces, the implied context is a change of basis of the vector space. Then, we can look at how the ... 1 All the other answers that say the "single" integral is simply a shorthand notation are right, but it is well to remember that one can indeed construe the integral as a single integral as a Lebesgue integral (if you do nothing else, look up Lebesgue's very cute little half paragraph summary (on the Wiki page) of his idea in a letter to Paul Montel). If ... 1 We expect a vector to change in a certain way when we change the scale we use to measure distance. Consider the vector $$\vec{x}=(1, 0, 0)\,\mathrm{m}$$ If we change scale and now measure in centimeters this vector becomes $$\vec{x}=(100, 0 ,0)\,\mathrm{cm}$$ Now consider a vector representing a force: $$\vec{F}=(1,0,0)\,\mathrm{J/m}$$ where I've chosen ... 1 There is an official convention for a positive muon being a nucleus, that a positive muon with one electron is muonium (Mu) and a positive muon with two electrons is muonide (Mu-). See Names for Muonic and Hydrogen Atoms and Their Ions. For a negative muon replacing an electron in helium, I see both$He\mu$and$^{4.1}H\$ in the same paper: Kinetic Isotope ...

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