# Tag Info

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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 ...

27

The thing is that $\mathrm{dm}$ is a single symbol, not a combination of two symbols. Yes, it can be understood in terms of a prefix and a base indicator, but it is still a single symbol. An analogy to the concatenation of variable is inappropriate. Reference to an authoritative statement: The grouping formed by a prefix symbol attached to a unit ...

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It is an ångström, a unit of length commonly used in chemistry to measure things like atomic radii and bond lengths. Although not an official SI unit, it has a simple relationship to the metric units of length: $$1\:\mathrm{ångström} = 1\:\mathrm{Å} = 10^{−10}\:\mathrm{m} = 0.1\:\mathrm{nm} = 100\:\mathrm{pm}.$$

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Typically: $\rm d$ denotes the total derivative (sometimes called the exact differential):$$\frac{{\rm d}}{{\rm d}t}f(x,t)=\frac{\partial f}{\partial t}+\frac{\partial f}{\partial x}\frac{{\rm d}x}{{\rm d}t}$$This is also sometimes denoted via $$\frac{Df}{Dt},\,D_tf$$ $\partial$ represents the partial derivative (derivative of $f(x,y)$ with respect to $x$ ...

15

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} ... 13 If I saw the word "amp" written as such in a paper in my field (astrophysics) it would strike me as a bit informal. I would expect to see the full "ampere" written. That said, it is rare to actually write out the full name of a unit; usually it follows a number and is given its standard abbreviation. When abbreviated to e.g. "$5\ \mathrm{A}$", I would ... 11 Technically, apparently, your teacher is correct. BIPM and NIST In the official brochure from the Bureau international des poids et mesures (BIPM, the keepers of SI units) in §5.1 Unit symbols we find: It is not permissible to use abbreviations for unit symbols or unit names, such as sec (for either s or second), sq. mm (for either mm2 or ... 11 It's c for constant or celeritas, which means speed in Latin. Everyone uses it because it's convention. You could use$\xi$or$\zeta$or$\gamma$or any other symbol you wanted, but then you'd have to explain what it meant, and people would have to go through the trouble to remember this every time they read your papers. Better to go with convention and ... 10 The wedge product has its roots in exterior algebra. Exterior algebra lets you talk about objects like planes or volumes as algebraic elements of their own, separate from ordinary vectors, but still obeying the same notions of being "vectors" in their own vector spaces. The wedge product of two vectors is a bivector, and many concepts you may have been ... 10 Quite often everything inside bra or ket is just a label. In this particular case the meaning of$|λ,m_l⟩$is "a state with the square of the angular momentum being equal to$λ$(in atomic units, where$\hbar=1$) and with the projection of the angular momentum in some direction ($z$-axis conventionally) being equal to$m_l$". That is,$|λ,m_l⟩$state is ... 10 Capital$\mathrm{C}$, in upright font, is the symbol for the coulomb. Lowercase$c$, italicized, is the speed of light in vacuum. Thanks to Einstein's equation, we can switch between mass and energy ($\mathrm{MeV}$is a unit of energy) by using factors of$c^2$, and sometimes it's more convenient to know the energy equivalent of a particle's mass rather than ... 9 They're not used because it's ugly to read such texts with parentheses and it's time-consuming to write it down. A decimeter is indeed a "product" of "deci" and a meter, so the origin is analogous to the product of two real numbers$ab$. But once we define the new derived unit${\rm dm}$, we treat it as a single object, so it really means what you would ... 8 A physicist would write your first equation$x^a = x^\mu e_\mu^a$. The notation$x^a$is invariant in your terminology. The$a$is an abstract index. It is ostensibly not supposed to be thought of as ranging over a set of numerical values, but is just a marker that indicates that$x$is a vector (i.e., rank 1,0 tensor.) Similarly for each$\mu$,$e^a_\mu$is ... 8 Canonically, the wedge product is distinct from the cross product and should not be confused, however in three dimensions they are inextricably linked. The wedge product (or outer product) comes from exterior algebra, first due to Grassmann who generalized vector products to arbitrary dimensions. This would later be extended by Clifford into the Clifford ... 6 What Goldstein means by$\nabla_iV_i$is $$\nabla_iV_i=\left(\frac{\partial}{\partial x_{1,i}}\hat{x}_{1,i}+\frac{\partial}{\partial x_{2,i}}\hat{x}_{2,i}+\frac{\partial}{\partial x_{3,i}}\hat{x}_{3,i}\right)V_i$$ which is indeed a vector. Here,$\mathbf r_i=(x_{1,i},\,x_{2,i},\,x_{3,i})$is the position of the$i$th particle (with respect to the origin), ... 6 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 ... 6 The Lane-Emden is really a non-dimensional form of the Poisson's equation with spherical symmetry: $$\nabla^2f(r)\equiv\frac1{r^2}\frac{d}{dr}\left(r^2\frac{df}{dr}\right)$$ This should be clear to all that, since we have two factors of$d/dr$on the right hand side of the above, it must be a 2nd order differential equation. This 2nd-order derivative ... 5 The square brackets mean antisymmetrization. That is: $$X_{[a_1a_2\dots a_n]} = \frac{1}{n!}\sum_{P\in S(n)} \text{Sign}(P) X_{a_{P(1)}a_{P(2)}\dots a_{P(n}}$$ where$S(n)$is the set of permutations of$n$elements, and$\text{Sign}(P)$is the sign of the permutation$P$, that is,$\text{Sign}(P)=-1$if you need an odd number of element exchanges, and ... 5 The way I imagine it is that the left side of the direct product is exclusively reserved for hilbert space 1 and the right side is for Hilbert space 2. So that the total hilbert space you are working in is written as: $$H=H_1⊗H_2$$ And so when you have a wavefunction in H you write: $$|\psi\rangle = |r_1\rangle \otimes |r_2 \rangle$$ And then: ... 5 It is just a more compact notation. It is implied by the integration element$dA$that you are integrating over the surface. 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$\nabla_\sigma$is the covariant derivative.$\nabla^\sigma$means$g^{\sigma\rho}\nabla_\rho$. It's pretty much the same as raising any other index. The covariant derivative when acting on any tensor adds a down index, and you can raise it as with any other index. Since the covariant derivative of the metric is 0, you can work with either$\nabla_\sigma$or ... 5 The (anti)symmetrization simply acts on all the enclosed indices (at the same "height" which are really enclosed between the brackets), regardless of their belonging to the same tensor or different tensors. For example, $$\delta^{[\alpha}{}_{[\gamma} R^{\beta]}{}_{\delta]} = \frac 12 \left(\delta^{[\alpha}{}_{\gamma} R^{\beta]}{}_{\delta} - ... 5 Is this based off some new, obscure research in string theory? No. This is fringe science. There is no mathematical connection to string theory. It is remarkably bad behavior for this man to set up himself up as a public intellectual -- giving TED talks, making a website juxtaposing himself with Einstein, writing a popular science book -- when he has no ... 5 This is typical to see in situations where the potential is a function of the coordinates of more than one particle:$$ V=V(\mathbf r_1,\ldots,\mathbf r_N)=V(x_1,y_1,z_1,\ldots,x_N,y_N,z_N). $$The force produced by such a potential on the ith particle is the gradient of this function with respect to that particle's coordinates, while holding all the other ... 5 This is the same notation that you'll find in Weinberg's books.$$(\psi, \chi)$$is the inner product of the two states \psi and \chi, and corresponds to$$\langle \psi \mid \chi \rangle$$. So, the above corresponds literally to$$ \frac{1}{\sqrt{\langle \psi_k \mid \psi_k \rangle}} \left| \psi_k \right>$$This new object is just the normalized ... 5 This is notation for the imaginary part of a complex number. It is a fraktur letter I, and its counterpart for the real part is a fraktur letter R. Thus, if z=x+iy and x,y are real, one writes$$ \mathfrak{R}\,z=x\ \ \text{ and }\ \ \mathfrak{I}\,z=y. $$A good chart of the fraktur alphabet is in this Yale resource, which includes handwriting guidance, ... 5 Each set of coordinates comes with its own set of coordinate differentials and its own coordinate expression for the metric tensor: Given coordinates (t,x,y,z), the coordinate differentials are dt,dx,dy,dz and$$ (\eta_{\mu\nu})=\begin{pmatrix} c^2&0&0&0\\ 0&-1&0&0\\ 0&0&-1&0\\ 0&0&0&-1 \end{pmatrix}$$... 4 There are a few conventions out there. The short answer is: use whatever will make sense to you. I personally don't like single subscripts: It's not as easy to match 3rd law pairs. If you do want to use a single subscript, I'd go with the object that is exerting the force, otherwise every label will be the same, and what's helpful about that since you've ... 4$1\text{ dm}^3$can be seen as a shorthand way to write$1\text{ dm}\cdot 1\text{ dm}\cdot1\text{ dm}$. The motivation behind this is that the quantity$1\text{ dm}^3$represents the volume of a cube with each side of length$1\text{ dm}$, and the way to find the volume is to multiply the three sides:$LWH=(1\text{ dm})\cdot(1\text{ dm})\cdot(1\text{ ...

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