Relate Momentum and Continuity Equation to U uecos Kx wt

Continuity Equation

The time dependent continuity equation including axial dispersion for a fixed bed reactor is given by a partial differential equation (pde) of the parabolic/hyperbolic class.

From: Studies in Surface Science and Catalysis , 2001

A STUDY ON HEAT/MASS TRANSFER FROM RECTANGULAR CYLINDERS USING NAPHTHALENE SUBLIMATION TECHNIQUE

C.H. Chung , ... S.Y. Yoo , in Experimental Heat Transfer, Fluid Mechanics and Thermodynamics 1993, 1993

HEAT/MASS TRANSFER ANALOGY

The governing equations for the flow of a constant property fluid are described by the following dimensionless form.

Continuity Equation

$\frac{\partial }{\partial {\text{x}}_{\text{i}}}({\text{v}}_{\text{i}})=0$

Momentum Equation

$\frac{{\text{Dv}}_{\text{i}}}{\text{Dτ}}=\frac{\partial \text{p}}{\partial {\text{x}}_{\text{i}}}+\frac{1}{{\mathrm{Re}}_0}\frac{\partial }{\partial {\text{x}}_{\text{i}}}\left[\left(1+\frac{\text{ε}}{\text{v}}\right)\left(\frac{\partial {\text{v}}_{\text{i}}}{\partial {\text{x}}_{\text{j}}}+\frac{\partial {\text{v}}_{\text{j}}}{\partial {\text{x}}_{\text{i}}}\right)\right]$

Energy Equation

$\frac{\text{Dt}}{\text{Dτ}}=\frac{1}{{\mathrm{Re}}_0\mathrm{Pr}}\frac{\partial }{\partial {\text{x}}_{\text{i}}}\left[\left(1+\frac{\text{ε}}{\text{v}}\frac{\mathrm{Pr}}{{\mathrm{Pr}}_{\text{t}}}\right)\frac{\partial \text{t}}{\partial {\text{x}}_{\text{i}}}\right]$

Mass Concentration Equation

$\frac{\text{Dw}}{\text{Dτ}}=\frac{1}{{\mathrm{Re}}_0\text{Sc}}\frac{\partial }{\partial {\text{x}}_{\text{i}}}\left[\left(1+\frac{\text{ε}}{\text{v}}\frac{\text{Sc}}{{\text{Sc}}_{\text{t}}}\right)\frac{\partial \text{w}}{\partial {\text{x}}_{\text{i}}}\right]$

The energy and mass concentration equations have the same expressions. Thus, for a given configuration, if the thermal and concentration boundary conditions are the same and if Pr = Sc, the temperature and mass concentration profiles are identical.

The empirical correlations, Nu = C Re^m Prⁿ and Sh = C Re^m Scⁿ, are often used for the forced convection heat transfer and mass transfer studies, respectively. Thus, Nu/Sh = (Pr/Sc)ⁿ is obtained, and this equation is applied to obtain heat transfer coefficient from mass transfer experiment, vise versa.

Read full chapter

URL:

https://www.sciencedirect.com/science/article/pii/B9780444816191500629

Numerical modeling methodologies for friction stir welding process

Rahul Jain , ... Shiv B. Singh , in Computational Methods and Production Engineering, 2017

5.5.3 Governing equation

Continuity, momentum, and energy equations are the conservation equations in Eulerian analysis as specified as follows:

Continuity equation

(5.25) $\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}=0$

Momentum equation

(5.26) $\rho \left(u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}+w\frac{\partial u}{\partial z}\right)=-\frac{\partial p}{\partial x}+\frac{\partial }{\partial x}\left(\mu \frac{\partial u}{\partial x}\right)+\frac{\partial }{\partial y}\left(\mu \frac{\partial u}{\partial y}\right)+\frac{\partial }{\partial z}\left(\mu \frac{\partial u}{\partial z}\right)$

(5.27) $\rho \left(u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y}+w\frac{\partial v}{\partial z}\right)=-\frac{\partial p}{\partial y}+\frac{\partial }{\partial x}\left(\mu \frac{\partial v}{\partial x}\right)+\frac{\partial }{\partial y}\left(\mu \frac{\partial v}{\partial y}\right)+\frac{\partial }{\partial z}\left(\mu \frac{\partial v}{\partial z}\right)$

(5.28) $\rho \left(u\frac{\partial w}{\partial x}+v\frac{\partial w}{\partial y}+w\frac{\partial w}{\partial z}\right)=-\frac{\partial p}{\partial z}+\frac{\partial }{\partial x}\left(\mu \frac{\partial w}{\partial x}\right)+\frac{\partial }{\partial y}\left(\mu \frac{\partial w}{\partial y}\right)+\frac{\partial }{\partial z}\left(\mu \frac{\partial w}{\partial z}\right)$

Energy equation

(5.29) $\rho C_p\left(u\frac{\partial T}{\partial x}+v\frac{\partial T}{\partial y}+w\frac{\partial T}{\partial z}\right)=\frac{\partial }{\partial x}\left(k\frac{\partial T}{\partial x}\right)+\frac{\partial }{\partial y}\left(k\frac{\partial T}{\partial y}\right)+\frac{\partial }{\partial z}\left(k\frac{\partial T}{\partial z}\right)+S_t$

where u, v, and w are the velocity in X, Y, and Z directions, respectively, and S_t is the viscous heat dissipation. It has been reported that viscous heat dissipation is small as compared to the plastic heat dissipation (Nandan et al., 2008) and is given by Eq. (5.30).

(5.30) $S_t=\alpha \varphi$

where α is a constant reflecting the extent of mixing and it is defined as 0.05. ϕ is the viscous dissipation heat as shown in Eq. (5.31).

(5.31) $\varphi =\mu \left[2{\left(\frac{\partial u}{\partial x}\right)}^2+2{\left(\frac{\partial v}{\partial y}\right)}^2+2{\left(\frac{\partial w}{\partial z}\right)}^2+{\left(\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}\right)}^2+{\left(\frac{\partial v}{\partial z}+\frac{\partial w}{\partial y}\right)}^2+{\left(\frac{\partial w}{\partial x}+\frac{\partial u}{\partial z}\right)}^2\right]$

The coupled equations were solved by using pressure-based solver with second-order discretization of momentum and energy equation

Read full chapter

URL:

https://www.sciencedirect.com/science/article/pii/B9780857094810000057

The wave equation and solutions

Leo L. Beranek , Tim J. Mellow , in Acoustics: Sound Fields and Transducers, 2012

2.2.3 The continuity equation

The continuity equation is a mathematical expression stating that the total mass of gas in a deformable "box" must remain constant. Because of this law of conservation of mass, we are able to write a unique relation between the time rate of change of the incremental velocities at the surfaces of the box.

One-dimensional derivation	Three-dimensional derivation
Refer to Fig. 2.2. If the mass of gas within the box remains constant, the change in volume τ depends only on the difference of displacement of the air particles on the opposite sides of the box. Another way of saying this is that, unless the air particles adjacent to any given side of the box move at the same velocity as the box itself, some will cross into or out of the box and the mass inside will change.	If the mass of gas within the box remains constant, the change in incremental volume τ depends only on the divergence of the vector displacement. Another way of saying this is that, unless the air particles adjacent to any given side of the box move at the same velocity as the side of the box itself, some will cross into or out of the box and the mass inside will change; so
In a given interval of time the air particles on the left-hand side of the box will have been displaced ξx . In this same time, the air particles on the right-hand side will have been displaced
${\xi }_x+\frac{\partial {\xi }_x}{\partial x}\Delta x.$
The difference of the two quantities above multiplied by the area ΔyΔz gives the increment in volume τ
(2.11a) $\tau =\frac{\partial {\xi }_x}{\partial x}\Delta x\Delta y\Delta z$ or	(2.11b) $\tau =V_0\text{div}\xi =V_0\nabla ·\xi$
(2.12) $\tau =V_0\frac{\partial {\xi }_x}{\partial x}.$
Differentiating with respect to time yields,	Differentiating with respect to time yields,
(2.13a) $\frac{\partial \tau }{\partial t}=V_0\frac{\partial u}{\partial x},$ where u is the instantaneous particle velocity.	(2.13b) $\frac{\partial \tau }{\partial t}=V_0\nabla ·q,$ where q is the instantaneous particle velocity.

FIG. 2.2. Change in volume of the box with change in position.

From (a) and (b) it is seen that the incremental change in volume of the box is τ  =   (∂ξ_x /∂x) Δx Δy Δz.

Example 2.1. In the steady state, that is,

$\partial u/\partial t=j\omega \tilde{u}=\sqrt{2}{u}_{rms},$

determine mathematically how the sound pressure in a plane progressive sound wave (one-dimensional case) could he determined from measurement of particle velocity alone.

Solution. From Eq. (2.4a) we find in the steady state that

$-\frac{\partial p_{rms}}{\partial x}=j\omega {\rho }_0{u}_{rms}.$

Written in differential form,

$-\Delta p_{rms}=j\omega {\rho }_0{u}_{rms}\Delta x.$

If the particle velocity is 1 cm/s, ω is 1000   rad/s, and Δx is 0.5   cm, then

$\begin{array}{c}\Delta p_{rms}=-j0.005\times 1000\times 1.18\times 0.01\\ =-j0.059\text{Pa}.\end{array}$

We shall have an opportunity in Chapter 5 of this text to see a practical application of these equations to the measurement of particle velocity by a velocity microphone.

Read full chapter

URL:

https://www.sciencedirect.com/science/article/pii/B9780123914217000026

Fully Coupled Solver for Incompressible Navier-Stokes Equations using a Domain Decomposition Method

Jerome Breil , ... Tadayasu Takahashi , in Parallel Computational Fluid Dynamics 2002, 2003

2.3 Pressure Equation

Instead of the continuity equation we use an equation for pressure by modifying the well-known "Poisson equation for pressure" obtained from (1);

(9) $\mathrm{\Delta }\mathit{p}+\mathit{\rho }\left(\mathbf{\nabla }\cdot \mathit{C}\left[\mathbf{u}\right]-\mathbf{f}\right)=0\text{.}$

The equation (9) can be balanced by multiplying by μ and adding the divergent equation

(10) $-\mathit{\nu }\mathbf{\Delta }\mathit{p}+\frac{\mathrm{\gamma }}{\mathit{\rho }}\mathbf{\nabla }\cdot \mathbf{u}=\mathit{\nu \rho }\mathbf{\nabla }\cdot \left(\mathit{C}\left[\mathbf{u}\right]-\mathbf{f}\right)\text{,}$

where γ is a parameter. This form of pressure equation has been proposed in [2]. The formulation with equation for pressure is equivalent to the original system only if the continuity equation is satisfied also on the boundary, namely the following additional boundary condition has to be used

(11) $\mathbf{\nabla }\cdot {\left.\mathbf{u}\right|}_{\partial \mathit{\Omega }}=0\text{.}$

Now the problem is equivalent to the original equations.

Read full chapter

URL:

https://www.sciencedirect.com/science/article/pii/B9780444506801500322

Special Volume: Computational Methods for the Atmosphere and the Oceans

Bennert Machenhauer , ... Peter Hjort Lauritzen , in Handbook of Numerical Analysis, 2009

3.2.1 Explicit HIRLAM-DCISL

The explicit continuity equation for moist air is solved for each model layer as described in Section 3.1.3. (see Eq. (3.21)). Hybrid trajectories determine the irregular upstream departure area δ _k Aⁿ , and an "upstream integration" determines the horizontal mean of Δ_k $\overline{p}$ ⁿover the departure area δ _k Aⁿ (3.29). Here Δ_k $\overline{p}$ ⁿis defined as

(3.62) ${\Delta }_k{\overline{p}}^n={\overline{p}}_{k+1/2}^n-{\overline{p}}_{k-1/2}^n.$

The departure cells are the same for all tracers, including water vapor, and Lagrange interpolations between the hybrid trajectory departure points determine the departure points for temperature Tand the velocity components uand v. In HIRLAM-DCISL, two alternative upstream integration methods are available, the method of Nair and Machenhauer [2002] and that of Nair, Scroggs and Semazzi [2002]. The mean top pressures of the arrival cells ${\overline{\widehat{p}}}_{k-1/2}^{n+1}$ are determined hydrostatically from Eq. (3.22), i.e., from the Lagrangian pressure thicknesses δ_k ${\overline{\widehat{p}}}^{n+1}$ in Eq. (3.33). Together with Eq. (3.62), these values determine a mean value of the vertical pressure velocity ω = dp/dtalong the trajectory (Eq. (3.34)). This ω is consistent with the hydrostatic assumption and the horizontal flow, contrary to the inconsistent vertical velocities, based on partly Eulerian solutions to the continuity equation, which are applied in traditional semi-Lagrangian models such as HIRLAM. ω is used in the thermodynamic equation (Eq. (3.2)) in the energy conversion term αω/c_p= R_dT_v/c_pω/p, which is approximated with

(3.63) $\Delta t{\left[{\left(\frac{R_d{T}_{\upsilon }}{c_p}\frac{\omega }{p}\right)}_k\right]}^{n+1}=\frac{R_d}{c_p}{\left[T_{\upsilon }^n+{\tilde{T}}_{\upsilon }^{n+1}\right]}_k\left[\frac{{\overline{\widehat{p}}}_k^{n+1}-{\overline{\left({\overline{p}}_k^n\right)}}_{*}^{\delta }}{{\overline{\widehat{p}}}_k^{n+1}+{\overline{\left({\overline{p}}_k^n\right)}}_{*}^{\delta }}\right].$

The hydrostatic mean surface pressure (Eq. (3.23)) is the weight of all NLEV model layers above the surface:

(3.64) ${\overline{p}}_s^{n+1}=\sum _{l=1}^{\text{N}\text{L}\text{E}\text{V}}{\delta }_k{\overline{\widehat{p}}}^{n+1},$

determining the top pressure of Eulerian cells (Eq. (3.24))

(3.65) ${\overline{p}}_{k-1/2}^{n+1}=A_{k-1/2}+B_{k-1/2}{\overline{p}}_s^{n+1}.$

The explicit continuity equations for passive tracers (Eq. (3.40)) and water vapor (Eq. (3.43)) are

(3.66) ${\left({\overline{\stackrel{⌢}{q}}}_i^{\delta }\right)}_k^{n+1}{\delta }_k{\overline{\widehat{p}}}^{n+1}\Delta A={\overline{{\left({\overline{\stackrel{⌢}{q}}}_i^{\Delta }\right)}_k^n{\Delta }_k{\overline{p}}^n}}^{\delta }\delta A_k^n$

and

(3.67) $\begin{array}{lll}{\left({\overline{\stackrel{⌢}{q}}}_{\upsilon }^{\delta }\right)}_k^{n+1}{\delta }_k{\overline{\widehat{p}}}^{n+1}\Delta A& =& {\overline{{\left({\overline{\stackrel{⌢}{q}}}_{\upsilon }^{\Delta }\right)}_k^n{\Delta }_k{\overline{p}}^n}}^{\delta }\delta A_k^n\\ & +& \Delta t{\overline{{\left({\overline{\stackrel{⌢}{P}}}_{q_{\upsilon }}^{\Delta }+{\overline{\stackrel{⌢}{K}}}_{q_{\upsilon }}^{\Delta }\right)}_k^{n+1/2}{\delta }_k{\overline{p}}^{n+1/2}}}^{{\delta }_{n+1/2}}\delta A_k^{n+1/2},\end{array}$

respectively, determine updated specific concentrations, ${\left({\overline{\stackrel{⌢}{q}}}_i^{\delta }\right)}_k^{n+1}$ and ${\left({\overline{\stackrel{⌢}{q}}}_{\upsilon }^{\delta }\right)}_k^{n+1}$ , in Lagrangian arrival cells (δV= δ $\stackrel{\wedge }{p}$ ΔV) from ${\left({\overline{\stackrel{⌢}{q}}}_i^{\Delta }\right)}_k^n$ and ${\left({\overline{\stackrel{⌢}{q}}}_{\upsilon }^{\Delta }\right)}_k^n$ plus Eq. (3.62). Finally, the updated specific concentrations, ${\left({\overline{\stackrel{⌢}{q}}}_i^{\Delta }\right)}_k^{n+1}$ and ${\left({\overline{\stackrel{⌢}{q}}}_i^{\delta }\right)}_k^{n+1}$ , in the Eulerian cells (ΔV= Δp ΔA) are determined from ${\left({\overline{\stackrel{⌢}{p}}}_i^{\delta }\right)}_k^{n+1}$ and ${\left({\overline{\stackrel{⌢}{p}}}_{\upsilon }^{\delta }\right)}_k^{n+1}$ by 1D vertical remappings.

The discretized explicit momentum and thermodynamic equations are straightforward grid-point semi-Lagrangian and finite difference approximations to Eqs. (3.1) and (3.2), respectively (see KÄllén [1996] and Undén et al. [2002]), except that in the thermodynamic equation the consistent energy conversion term is approximated consistently with (3.63). Regarding the addition of the physics in Eq. (3.67): since DMI-HIRLAM adds the physics at the arrival level (no averaging along the trajectory), that procedure was also adopted in HIRLAM-DCISL. Of course, it should ideally be done as indicated in Eq. (3.67).

Read full chapter

URL:

https://www.sciencedirect.com/science/article/pii/S1570865908002019

Segregation and Component Distribution

Aleks G. Ostrogorsky , Martin E. Glicksman , in Handbook of Crystal Growth: Bulk Crystal Growth (Second Edition), 2015

25.4.1.2 Two- and Three-Dimensional Flow Patterns

The steady state continuity equations for incompressible fluid flow are

(25.11) $\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0\left(\text{rectangular}\text{coordinates}\right)$

(25.12) $\frac{\partial u}{\partial r}+\frac{u}{r}+\frac{\partial w}{\partial z}=0\left(\text{cylindrical}\text{coordinates}\right)$

These equations state that one-dimensional flows are possible if, and only if, components u and v are either constant or zero. One-dimensional flows are observed only in pipes or ducts, where u  =   const. and v  =   0. Convection occurring in a melt can never result in a one-dimensional flow.

Typical flow patterns occurring near a growing crystal growth interface are depicted in Figure 25.3. Crystal growth configurations for (1) CZ, (2) vertical Bridgman (VB), and (3) horizontal Bridgman (HB) are shown. Note especially that perpendicular flows carry rejected impurities toward the interface, whereas lateral flows transport solute along the interface and eventually back into the bulk melt.

FIGURE 25.3. Schematic of flow patterns near the growth interface.

(a), (b) axisymmetric flows, and (c) nonaxisymmetric flows.

Read full chapter

URL:

https://www.sciencedirect.com/science/article/pii/B9780444633033000250

The Electromagnetic Field Equations

Jerry B. Marion , Mark A. Heald , in Classical Electromagnetic Radiation (Second Edition), 1980

4.4 Maxwell's Modification of Ampères Law

We have previously found that under steady-state conditions Ampères law may be expressed as [see Eq. (1.54)]

(4.14) $curlH=\frac{4\pi }{c}J$

Let us now examine the validity of this equation in the event that the fields are allowed to vary with time. If we take the divergence of both sides of Eq. (4.14), then since the divergence of the curl of any vector vanishes identically, we have

$\text{div}J=0$

Now, the continuity equation [ Eqs. (4.4)] states that in general div J equals − ∂ρ/∂t and will therefore vanish only in the special case that the charge density is static. Consequently, we must conclude that Ampères law as stated in Eq. (4.14) is valid only for steady-state conditions and is insufficient for the case of time-dependent fields. It was Maxwell who sought to modify Ampères law so that it would apply under time-varying conditions as well. His solution to the problem was to make the substitution ^*

$J\to J+\frac{1}{4\pi }\frac{\partial D}{\partial t}$

so that the modified form of Ampères law becomes

(4.15) $curlH=\frac{4\pi }{c}J+\frac{1}{c}\frac{\partial D}{\partial t}$

If we now take the divergence of both sides of this equation, we obtain

$0=\frac{4\pi }{c}\text{div}J+\frac{1}{c}\text{div}\left(\frac{\partial D}{\partial t}\right)$

Interchanging the space and time derivatives of D, we find

$\text{div}J+\frac{1}{4\pi }\frac{\partial }{\partial t}\text{div}D=0$

Using Gauss' law [Eqs. (1.20)] to substitute 4πρ for div D, we have finally

$\text{div}J+\frac{\partial \rho }{\partial t}=0$

and the continuity equation is recovered intact. That is, Maxwell's modification of Ampères law is compatible with conservation of charge, whereas Eq. (4.14) is not.

The term which Maxwell added to Ampères law, viz., (1/4π)∂D/∂t is called the displacement current, and corresponds, for example, to the "current" which must flow in the space (even a vacuum) between a pair of capacitor plates when the charged plates are connected by an external circuit. There is a displacement current even though no charge moves across the space. In order to illustrate this, consider the circuit in Fig. 4-1 which consists of a source of alternating current and a capacitor. The circuit is looped by the line Γ which bounds the surface S. If a current I flows in the circuit, Ampères law states that

FIG. 4-1.

${\oint }_{\Gamma }H\cdot dl=\frac{4\pi }{c}{\int }_{S}J\cdot nda=\frac{4\pi }{c}I$

Clearly, this result must be independent of the particular manner in which we construct the surface S. But consider the construction shown in Fig. 4-2. Now, the conduction current I does not flow through the surface and we are forced to conclude that

FIG. 4-2.

${\oint }_{\Gamma }H\cdot dl=0$

The two situations can be made to yield the same result if we include the displacement current, since then

(4.15a) ${\oint }_{\Gamma }H\cdot dl=\frac{4\pi }{c}{\int }_{S}J\cdot nda+\frac{1}{c}\frac{d}{dt}{\int }_{S}D\cdot nda=\frac{4\pi }{c}I$

For a perfectly conducting wire and a vacuum between the plates of the capacitor, the integral of J · n contributes only in the event that the surface S cuts the circuit and the integral of D · n contributes only in the event that the surface passes between the capacitor plates; the value of the integral in either instance is 4πI/c (see Problem 4-6).

It must be emphasized that the ultimate justification for Maxwell's assumption is in the experimental verification. Indeed, the effects of the displacement current are difficult to observe directly except at very high frequencies. In some microwave phenomena, however, the displacement currents are sufficiently large so that the magnetic effects can actually be measured. Indirect verification is afforded by the prediction of many effects which are confirmed by experiment; we shall treat some of these in later chapters. We may therefore consider that Maxwell's form of Ampères law has been subjected to experimental tests and has been found to be generally valid.

Read full chapter

URL:

https://www.sciencedirect.com/science/article/pii/B9780124722576500082

BASIC PRINCIPLES OF ELECTROMAGNETIC FIELDS

M.V.K. Chari , S.J. Salon , in Numerical Methods in Electromagnetism, 2000

1.11.4 Electric Vector Potential, T–Ω Formulation

Just as the continuity equation for the magnetic flux, ∇ · $\overrightarrow{B}$ = 0, allowed us to represent $\overrightarrow{B}$ by a vector potential such that ∇ × $\overrightarrow{A}$ = $\overrightarrow{B}$ , the continuity of current ∇ · $\overrightarrow{J}$ = 0 allows us to define a current vector potential often referred to as the electric vector potential, $\overrightarrow{T}$

(1.185) $\nabla \times \overrightarrow{T}=\overrightarrow{J}$

We note that ∇ × $\overrightarrow{H}$ = $\overrightarrow{J}$ as well, so $\overrightarrow{H}$ and $\overrightarrow{T}$ differ by the gradient of a scalar and have the same units. So

(1.186) $\overrightarrow{H}=\overrightarrow{T}=\nabla \Omega$

Using $\overrightarrow{E}$ = ρ $\overrightarrow{J}$ , from Faraday's law

(1.187) $\nabla \times \overrightarrow{E}=-\frac{\partial \overrightarrow{B}}{\partial t}=\nabla \times \rho \nabla \times \overrightarrow{T}$

Here ρ is the resistivity, the reciprocal of the conductivity. Substituting for the magnetic induction

(1.188) $\overrightarrow{B}=\mu \overrightarrow{H}=\mu \left(\overrightarrow{T}-\nabla \Omega \right)$

we obtain

(1.189) $\nabla \times \rho \nabla \times \overrightarrow{T}+\mu \frac{\partial \overrightarrow{T}}{\partial t}-\mu \nabla \frac{\partial \Omega }{\partial t}=0$

In current-free regions we can find the magnetic field from the scalar potential

(1.190) $\overrightarrow{H}=-\nabla \Omega$

where Ω can be found from Laplace's equation

(1.191) $-\nabla \cdot \mu \nabla \Omega =0$

The solution can be made unique by setting a gauge condition, for example the Coulomb gauge ∇ · $\overrightarrow{T}$ = 0. With this choice equation (1.189) becomes [7]

(1.192) $\nabla \times \rho \nabla \times \overrightarrow{T}-\nabla \rho \nabla \cdot \overrightarrow{T}+\mu \frac{\partial \overrightarrow{T}}{\partial t}-\mu \nabla \frac{\partial \Omega }{\partial t}=0$

An equation relating the vector and scalar potential may be found by using ∇ · $\overrightarrow{B}$ = 0. From (1.188) we have

(1.193) $\nabla \cdot \mu \left(\overrightarrow{T}-\nabla \Omega \right)=0$

Read full chapter

URL:

https://www.sciencedirect.com/science/article/pii/B978012615760450002X

Relativistic Electrodynamics

Jerry B. Marion , Mark A. Heald , in Classical Electromagnetic Radiation (Second Edition), 1980

13.5 Four-Vectors in Electrodynamics

Having established some of the basic formalism of relativity theory, we now turn our attention exclusively to electromagnetic matters.

In ordinary three-dimensional space the gradient operator is

(13.47) $\text{g}\text{r}\text{a}\text{d}\equiv e_j\frac{\partial }{\partial x_j}$

We may also define a four-dimensional gradient operator according to ^*

(13.47a) $\begin{array}{l}\text{g}\text{r}\text{a}\text{d}\equiv e_j\frac{\partial }{\partial x_j}-\frac{i}{c}{e}_4\frac{\partial }{\partial t}\\ =e_{\mu }\frac{\partial }{\partial x_{\mu }}\end{array}$

By forming the scalar product of Grad with itself, we obtain the four-dimensional version of the Laplacian operator, the so-called d' Alembertian operator, denoted by □²:

(13.48) $\begin{array}{ll}{\blacksquare }^2& =\frac{{\partial }^2}{\partial x_{\mu }\partial x_{\mu }}\\ & =\frac{{\partial }^2}{\partial x_j\partial x_j}-\frac{1}{c^2}\frac{{\partial }^2}{\partial t^2}\\ & ={\nabla }^2-\frac{1}{c^2}\frac{{\partial }^2}{\partial t^2}\end{array}$

Therefore, the wave equation

(13.49) ${\nabla }^2\Psi -\frac{1}{c^2}\frac{{\partial }^2\Psi }{\partial t^2}=0$

may be expressed as

(13.49a) ${\blacksquare }^2\Psi =0$

The quantity □² is a Lorentz-invariant operator (see Problem 13-2).

The mathematical statement of the experimental fact that charge is conserved is contained in the continuity equation [ Eqs. (4.4)]:

(13.50) $\text{div}\text{J}+\frac{\partial \rho }{\partial t}=0$

In relativity theory it is clear that current density and charge density cannot be distinct and completely separable entities since a charge distribution that is static in one reference frame will appear to be a current distribution in a moving reference frame. Therefore, we group together the current density J and the charge density ρ according to

(13.51) $J=\left(\text{J},ic\rho \right)$

Then the scalar product of the four-dimensional gradient operator and J is

$\begin{array}{l}\frac{\partial {\text{J}}_{\mu }}{\partial x_{\mu }}=\frac{\partial {\text{J}}_{\text{J}}}{\partial x_{\text{J}}}+\frac{\partial \left(ic\rho \right)}{\partial \left(ict\right)}\\ =\text{div}\text{J}+\frac{\partial \rho }{\partial t}=0\end{array}$

Therefore, the continuity equation may be expressed in four-dimensional form as

(13.52) $\text{Div}J=0$

where Div is the four-dimensional divergence operator. ^*

In writing Eq. (13.51) there is the implicit assumption that J is a four-vector as defined in Section 13.4. We shall now show that this is actually the case. In the reference system K, in which the charge is all at rest, an element of charge dq is given by the product of the charge density ρ₀ and an element of volume dV:

$\begin{array}{cc}dq={\rho }_{0}dV,& \text{in}K\end{array}$

If charge is to be conserved, then the charge dq, when viewed from a moving system K′ will remain unchanged; that is,

$dq={\rho }_{0}dV=\rho d{V}^{\prime }=d{q}^{\prime }$

where ρ₀ and ρ are the charge densities in K and in K′ respectively, and where

$\begin{array}{cc}dV=d{x}_{1}d{x}_{2}d{x}_3,& \text{in}K\\ d{V}^{\prime }=d{x^{\prime }}_{1}d{x^{\prime }}_{2}d{x^{\prime }}_3& \text{in}{K}^{\prime }\end{array}$

Now, if K′ moves along the x ₃-axis of K with a velocity v = (0, 0, ν), then dx′₁ = dx′₁ and dx′₂ = dx ₂; but, as may be seen from Eq. (13.25), $dx{\prime }_m=d{x}_3\sqrt{1-{\beta }^2}$ . (This is the so-called FitzGerald-Lorentz contraction of length in the direction of motion. ^* ) Therefore,

$\begin{array}{l}{\rho }_{0}dV=\rho d{V}^{\prime }=\rho d{x^{\prime }}_{1}d{x^{\prime }}_{2}d{x^{\prime }}_3\\ \rho d{x}_{1}d{x}_{2}d{x}_3\sqrt{1-{\beta }^2}\\ =\rho dV\sqrt{1-{\beta }^2}\end{array}$

Thus, the charge density ρ in a moving system is related to the proper charge density in the same way that mass and proper mass are related. The conservation law therefore applies to total charge, but not to charge density. Since the ordinary current density is given by J = ρu, the quantity J may be expressed as

$\begin{array}{ll}J& =\left(\text{J},ic\rho \right)\\ & =\left(\rho \text{u},ic\rho \right)\\ & ={\rho }_0\left(\frac{\text{u}}{\sqrt{1-{\beta }^2}},\frac{ic}{\sqrt{1-{\beta }^2}}\right)\end{array}$

Since ρ₀ is a scalar invariant and $\mathcal{U}$ is a four-vector, J must possess the transformation properties of U and must therefore be a four-vector.

We have previously found it convenient to represent the magnetic field vector B as the curl of the vector potential A. Since A is not completely determined by the specification of its curl alone, we are at liberty to choose the divergence of A; that is, we choose a gauge for the potential. A particularly useful choice is the Lorentz gauge [Eqs. (4.26)] in which

(13.55) $\text{div}A+\frac{1}{c}\frac{\partial \Phi }{dt}=0$

If we define

(13.56) $A\left(A,i\Phi \right)$

the Lorentz condition is expressed as

(13.57) $\text{Div}A=0$

In free space the potentials A and Φ satisfy inhomogeneous wave equations [see Eqs. (4.34) and (4.35)]:

(13.58a) ${\nabla }^2\text{A}-\frac{1}{c^2}\frac{{\partial }^2\text{A}}{\partial t^2}=-\frac{4\text{π}}{c}\text{J}$

(13.58b) ${\nabla }^2\Phi -\frac{1}{c^2}\frac{{\partial }^2\Phi }{\partial t^2}=-4\text{π}\rho$

By using the four-vector potential A and the four-vector current density J, these two equations may be expressed simply as

(13.59) ${\square }^2+A=-\frac{4\text{π}}{c}J$

The space portion of this equation is just Eqs. (13.58a) and the fourth component is Eqs. (13.58b). From Eq. (13.59) it is clear that A is indeed a four-vector since J is a four-vector and the operator □² is Lorentz invariant.

Read full chapter

URL:

https://www.sciencedirect.com/science/article/pii/B9780124722576500173

Conservation Equations for Compressible Turbulent Flows

TUNCER CEBECI , in Analysis of Turbulent Flows, 2004

Problems

2.1

Show that the continuity equation (2.2.1) can be regarded as a "transport" equation for the density, in the same sense that Eq. (2.2.2) is a transport equation for the momentum per unit volume, Qu _i .

2.2

Take the x _i derivative of Eq. (2.2.2), i.e. "take the divergence" of the set of equations for the three components of momentum, and show that if Q is constant the result is nominally a transport equation for the divergence $\partial u_i/\partial x_i$ . Further show that if $\partial u_i/\partial x_i$ is set to zero (which Eq. (2.2.1) shows is the correct value in constant-density flow) the transport equation reduces to a Poisson equation for the pressure.

2.3

Show that the rate of viscous dissipation given in Eq. (2.8.11) is the mean product of the fluctuating viscous stress and the fluctuating rate of strain, $(\partial {u^{\prime }}_i/\partial x_k+\partial {u^{\prime }}_k/\partial x_i)/2.$ . Show that this is the mean rate at which the turbulance does work against viscous stresses, and that the first law of thermodynamics confirms that this is truly the rate at which turbulent kinetic energy is converted into thermal internal energy.

Read full chapter

URL:

https://www.sciencedirect.com/science/article/pii/B9780080443508500026

clarklier1965.blogspot.com

Source: https://www.sciencedirect.com/topics/mathematics/continuity-equation

Share this post