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One-forms eat vectors and spit out scalars. That's what they do. In the special case of the one-forms $dU$ for scalar functions $U$, given a vector $v_q$, the definition of $d$ gives us that $dU_q(v_q)$ is equal to $v_q(U)$. Which is to say, the directional derivative of $U$ at $q$ along $v_q$. This is true at any specific point, and it's true for vector / ...
$I(a) = \int_0^\infty \frac {e^{-ax}\sin x}{x} \ dx$ We need to find $I(3-i)$ Differentiation under the integral sign. $I'(a) = \int_0^\infty -e^{-ax}\sin x \ dx\\ I'(a) = \frac {e^{-ax}(a\sin x + \cos x)}{a^2+ 1}|_0^\infty\\ I'(a) = -\frac {1}{a^2+ 1}\\ I(\infty) - I(3-i) = \int_{3-i}^\infty -\frac {1}{a^2+ 1}\ da =\arctan (3-i) - \frac{\pi}{2}\\ I(3-i) ... 3 We assume the extremes in pressure are given by$P_1$and$P_2=rP_1$where$r$is the pressure compression ratio so $$P_0=\frac{P_2+P_1}2=\frac{r+1}2P_1$$ and then $$P_{max}-P_0=P_2-P_0=rP_1-\frac{r+1}2P_1=\frac{r-1}2P_1$$ So on the ellipse $$P=\left(\frac{r+1}2+\frac{r-1}2\sin\theta\right)P_1$$ Similarly the extremes in volume are$V_1$and$V_2=sV_1$where ... 3 A lot of this comes down to the linearity of the time-dependent Schrödinger equation. The linearity means that it's usually straight-forward to prove well-posedness (in some Hilbert space), and uniqueness and existence of solutions. You already have existence of the original equation, $$i\hbar \dfrac{\partial}{\partial t} \Psi = \Big[-\dfrac{\hbar^2}{2m}\... 2 I posted this question last year on brilliant.org by modifying it just a little bit, however it is just the same problem as from the book you referred but I just added more calculations. I appreciate that you tried this problem. Here, is my way of solution including the solution of the numerical method: Consider right half of the section as we need to ... 2 The general formula for the family of catenary curves between fixed points (\pm \ell/2, 0) is$$y=a \cosh \frac{x}{a} - a \cosh \frac{\ell}{2a} $$where higher values of the parameter a (which has units of length) correspond to tauter curves. By some standard results on catenaries, if we denote the gravitational constant by g, the mass per unit length ... 2 Allthough i see no mentionings of F(v) in the text cited by you, the physical meaning of F(v) is the power consumed by the system, when forced into a movement according to q with \dot{q} = v. This is easily derived from the formula for the work$$ W = \int\limits_s F_q(dq) = \int\limits_0^T F_{q(t)}(\dot{q}) dt $$performed on the system driven to a ... 2 This may not relate to quantum physics, but sometimes the square root of a probability isolates the variable of interest when it's multiplied with itself to produce side effects. Take dominant and recessive alleles, for example. Let the dominant allele frequency is p and the recessive frequency is 1-p, or q. If the dominant phenotype (observed effect)... 2 Isomorphic Lie algebras have equivalent representations. The problem here seems to be that people are sloppy in keeping track of what is a representation of either the complexified Lie algebras or the original real Lie algebras. Here, the representations of the complex Lie algebras \mathfrak{so}(3,1)_{\Bbb C} and \mathfrak{so}(4)_{\Bbb C} are equivalent ... 2 Then answer is yes. Here is a proof: For each j=1,\ldots,n put f_j:=\partial_jg_1 \chi_{\Omega_1}+\partial_j g_2 \chi_{\Omega_2}. Clearly, f_j\in L^2(\Omega'). If we can show that the weak (distributional) derivative of f coincides with f_j, we obtain f\in H^1(\Omega'). To this end, we simply verify for any \phi\in C^\infty_0(\Omega') that: \... 2 You want the lifted action to be by exact symplectomorphisms (actually it turns out even better, the lift gives a Hamiltonian action of G on T^*M, but that is a bonus). Suppose you demand that the lifted action commutes with projection and is linear on the fibers. Then on each fiber it is given by a linear map A_q: T^*_q M\to T^*_{\rho_q} M. If we ... 2 Strictly speaking, it’s not a set of functions. It’s a set of equivalence classes of functions where the equivalence is square integrable functions that differ only by sets of measure zero. See md2perpe’s answer. Since you tagged mathematical physics, a reason this comes up so much in quantum mechanics is because L^p(\mathbb{R}^d) is a Hilbert space and ... 2 Scientists tend to be more concerned with inspecting maps from a group to vector space automorphisms rather than properties of groups themselves. Here are some concrete examples for you: For simple-enough molecules, representations of dihedral groups are a good place to start. They can describe discrete mirror/rotational transformations that leave regular ... 2 Following the dimensionaly correct equation on wikipedia, you have$$\frac{d^2 u}{d\theta^2}+u=\frac{r_s c^2}{2h^2}+\frac{3r_s}2 u^2$$with$$r_s=\frac{2G\mu}{c^2}$$Now u is inverse distance, so in order to make it dimensionless you need to multiply it with a constant with units of distance. The obvious choice would be r_s. Then for x=ur_s, you ... 1 This fitting problem can be equivalently rewritten as fitting function of form:$$ f(x) = K \sin(\omega x) + L \cos(\omega x) + C $$And your original A is just A =\sqrt{K^2+L^2} This reduces it to just ordinary least squares problem. We get least squares estimators for K,L from the equation$$\begin{bmatrix} K \\ L \\ ... 1 Yes, $$\langle\ x\rvert \hat M\lvert\ x'\rangle=M(x)\langle\ x\lvert\ x'\rangle=M(x)\delta(x-x'),$$ provided you understand what you claim you do, namely $$\langle\ x\rvert \hat p\lvert\ x'\rangle=-i\hbar \partial_x \langle\ x\lvert\ x'\rangle=-i\hbar \partial_x\delta(x-x'),$$ so then that M(x) represents an operator consisting of functions of x and ... 1 Given that$Y=C_1x+C_2\dfrac{1}{x}\implies xY=C_1 x^2+C_2$Differentiating with respect to$x$,$Y+xY'=2xC_1$. . . . .$(1)$Again differentiating with respect to$x$,$Y'+Y'+xY''=2C_1\implies 2C_1=xY''+2Y'$Putting the value of$2C_1$in equation$(1)$we have ,$Y+xY'=x(xY''+2Y')\implies x^2Y''+xY'-Y=0$This is the required differential equation. ... 1 In Yang-Mills theory, the Jacobi identity, reflecting associativity of covariant derivatives, is but the full antisymmetrization of three covariant derivatives, so, by 7.24, or, in components, we have$\$ D_\mu \tilde G ^{\mu \nu}=\partial_\mu \tilde G ^{\mu \nu} +[A_\mu, \tilde G ^{\mu \nu}] \\ = \epsilon^{\mu\nu\rho\kappa} \Bigl ( \partial_\mu (\...