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Differential operations with scalars

The major notations of scalars, vectors, and tensors and their operations presented in the text are summarized in Tables A1 through A5. Table A1 gives the basic definitions of vector and second-order tensor. Table A2 describes the basic algebraic operations with vector and second-order tensor. Tables A3 through A5 present the differential operations with scalar, vector, and tensor in Cartesian, cylindrical, and spherical coordinates, respectively. It is noted that in these tables, the product of quantities with the same subscripts, e.g., a b, represents the Einstein summation and < jj refers to the Kronecker delta. The boldface symbols represent vectors and tensors. [Pg.540]

Differential operations with scalars and vectors. The gradient or grad of a scalar field is... [Pg.166]

A function of the position vector x is called a field. We can have a scalar field, a vector field or a tensor field. Derivatives with respect to position vectors are performed using the vector differential operator V, know as the del operator. It is written as 0/6xi in the Cartesian tensor notation. The operator can be treated as a vector but it cannot stand alone. It must operate on a scalar, vector or a tensor. [Pg.151]

In an alternative formulation of the Redfield theory, one expresses the density operator by expansion in a suitable operator basis set and formulates the equation of motion directly in terms of the expectation values of the operators (18,20,50). Consider a system of two nuclear spins with the spin quantum number of 1/2,1, and N, interacting with each other through the scalar J-coupling and dipolar interaction. In an isotropic liquid, the former interaction gives rise to J-split doublets, while the dipolar interaction acts as a relaxation mechanism. For the discussion of such a system, the appropriate sixteen-dimensional basis set can for example consist of the unit operator, E, the operators corresponding to the Cartesian components of the two spins, Ix, ly, Iz, Nx, Ny, Nz and the products of the components of I and the components of N (49). These sixteen operators span the Liouville space for our two-spin system. If we concentrate on the longitudinal relaxation (the relaxation connected to the distribution of populations), the Redfield theory predicts the relaxation to follow a set of three coupled differential equations ... [Pg.54]

Solve the differential equation (D.ll), with the aid of the normal ordering procedure according to which it is possible to pass from operators equations that are functions of the noncommutative Bosons to scalar equations. This is possible with the help of the N 1 operator, which us allow to write the following transformations [54] ... [Pg.403]

Differentiation of a tensor with respect to a scalar does not change its rank. The spatial differentiation of a tensor raises its rank by unity, and identical to multiplication by the vector V, called del or Hamiltonian operator or the nabla... [Pg.687]

It follows from Eq. (A 124) that the vector derivative operator changes the grade of the object it operates on by 1. For example, the vector derivative of the scalar X(xa) is a vector (because a X = 0 for any scalar X, so aX = a A X), and the vector derivative of the vector f(xa) is a scalar plus a bivector. The differentiation with respect to the vector variable xa greatly resembles the differentiation with respect to some scalar variable xa. For example, the vector differentiation is distributive,... [Pg.340]

To convert the derivatives of scalars with respect to x,y,z into derivatives with respect to r,9,z, the chain rule of partial differentiation is used. The derivative operators are thus related as ... [Pg.1161]

The mathematical operator consisting of the dot product, or scalar product, of the differential vector V with another vector is called the divergence operator. For instance, the charge concentration p in a volume is linked to the electric displacement D through one of Maxwell s relationships ... [Pg.104]

In contradistinction with the previous discussion about reducing scalar state variables, the relationship with the next localized variable cannot be obtained by a differential vector operator, but by an ordinary derivation, becanse here the two variables U/ and U/ are both scalars. Notwithstanding, this is possible between the two last levels by a gradient operator ... [Pg.126]

The vertical path at this node is the inductive force given by Equation H2.3 whereas the oblique path is the friction force Fq expressed by Equation H2.1. By relating with Equation H2.2 this latter to the velocity and by combining the conductive and inductive operators into a time constant iRL (which can be a scalar, especially in the case of linearity of the property operators), the resulting differential equation is obtained ... [Pg.522]

The operator, exp (k/), is symmetric in the entropic scalar product. This enables the formulation of symmetry relations between observables and initial data, which can be validated without differentiation of empirical curves and are, in that sense, more robust and closer to direct measurements than the classical Onsager relations. In chemical kinetics, there is an elegant form of symmetry between A produced from B and B produced from A their ratio is equal to the equilibrium coefficient of the reaction A B and does not change in time. The symmetry relations between observables and initial data have a rich variety of realizations, which makes direct experimental verification possible. This symmetry also provides the possibility of extracting additional experimental information about the detailed reaction mechanism through dual experiments. The symmetry relations are applicable to all systems with microreversibility. [Pg.194]

The Laplace operator, V, is defined in Appendix 1. This partial differential equation characterizes wave and relaxation phenomena. Again, we assume the jc-axis to be aligned with the Poynting vector, so that Ex = 0. To simplify matters further, we rotate the coordinate system around the jc-axis until the y-axis coincides with the direction of the electric field strength, so that =0 also. Only the y-component of E remains and Eq. (1.3.6) becomes a scalar equation for the unknown... [Pg.6]


See other pages where Differential operations with scalars is mentioned: [Pg.1502]    [Pg.102]    [Pg.1502]    [Pg.532]    [Pg.16]    [Pg.176]    [Pg.293]    [Pg.18]    [Pg.42]    [Pg.493]    [Pg.436]    [Pg.571]   
See also in sourсe #XX -- [ Pg.166 ]




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