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Chapter 10: Problem 19

Maxwell's equations for electromagnetism in free space (i.e. in the absence of charges, currents and dielectric or magnetic media) can be written (i) \(\nabla \cdot \mathbf{B}=0\) (ii) \(\nabla \cdot \mathbf{E}=0\) (iii) \(\nabla \times \mathbf{E}+\frac{\partial \mathbf{B}}{\partial t}=\mathbf{0}\) (iv) \(\nabla \times \mathbf{B}-\frac{1}{c^{2}} \frac{\partial \mathbf{E}}{\partial t}=\mathbf{0}\). A vector \(\mathbf{A}\) is defined by \(\mathbf{B}=\nabla \times \mathbf{A}\), and a scalar \(\phi\) by \(\mathbf{E}=-\nabla \phi-\partial \mathbf{A} / \partial t\). Show that if the condition (v) \(\nabla \cdot \mathbf{A}+\frac{1}{c^{2}} \frac{\partial \phi}{\partial t}=0\) is imposed (this is known as choosing the Lorenz gauge), then both \(\mathbf{A}\) and \(\phi\) satisfy the wave equations (vi) \(\nabla^{2} \phi-\frac{1}{c^{2}} \frac{\partial^{2} \phi}{\partial t^{2}}=0\), (vii) \(\quad \nabla^{2} \mathbf{A}-\frac{1}{c^{2}} \frac{\partial^{2} \mathbf{A}}{\partial t^{2}}=\mathbf{0}\) The reader is invited to proceed as follows. (a) Verify that the expressions for \(\mathbf{B}\) and \(\mathbf{E}\) in terms of \(\mathbf{A}\) and \(\phi\) are consistent with (i) and (iii). (b) Substitute for \(\mathbf{E}\) in (ii) and use the derivative with respect to time of \((v)\) to eliminate \(\mathbf{A}\) from the resulting expression. Hence obtain (vi). (c) Substitute for \(\mathbf{B}\) and \(\mathbf{E}\) in (iv) in terms of \(\mathbf{A}\) and \(\phi .\) Then use the divergence of (v) to simplify the resulting equation and so obtain (vii).

Short Answer

Expert verified
Given the gauge condition and provided expressions for \(\mathbf{A}\) and \(\phi\), both satisfy their respective wave equations after appropriate substitutions and simplifications.

Step by step solution

01

Verify \(\mathbf{B} = abla \times \mathbf{A}\) with \(abla \cdot \mathbf{B} = 0\)

Calculate the divergence of \(\mathbf{B} = abla \times \mathbf{A}\). Due to the vector calculus identity \(abla \cdot (abla \times \mathbf{A}) = 0\), the expression \(\mathbf{B} =abla \times \mathbf{A}\) satisfies \(abla \cdot \mathbf{B} = 0\).
02

Verify \(\mathbf{E} = -abla \phi - \frac{\partial \mathbf{A}}{\partial t}\) with \(abla \times \mathbf{E} + \frac{\partial \mathbf{B}}{\partial t} = \mathbf{0}\)

Substitute \(\mathbf{E} = -abla \phi - \frac{\partial \mathbf{A}}{\partial t}\) into \(abla \times \mathbf{E} + \frac{\partial \mathbf{B}}{\partial t} = \mathbf{0}\). We get \(abla \times (-abla \phi - \frac{\partial \mathbf{A}}{\partial t}) + \frac{\partial (abla \times \mathbf{A})}{\partial t} = \mathbf{0}\). Using the theorem \(abla \times (abla f) = 0\), we get \(-abla \times \frac{\partial \mathbf{A}}{\partial t} + abla \times \frac{\partial \mathbf{A}}{\partial t} = \mathbf{0}\). Thus, the expression is verified.
03

Substitute \(\mathbf{E}\) in \(abla \cdot \mathbf{E} = 0\)

Substitute \(\mathbf{E} = -abla \phi - \frac{\partial \mathbf{A}}{\partial t}\) into \(abla \cdot \mathbf{E} = 0\). This gives \(- abla^2 \phi - abla \cdot \frac{\partial \mathbf{A}}{\partial t} = 0\).
04

Use derivative of the Lorenz gauge condition to simplify

From condition \(abla \cdot \mathbf{A} + \frac{1}{c^2} \frac{\partial \phi}{\partial t} = 0\), take the time derivative: \(\frac{\partial}{\partial t}(abla \cdot \mathbf{A}) + \frac{1}{c^2} \frac{\partial^2 \phi}{\partial t^2} = 0\). Substitute \(abla \cdot \mathbf{A} = -\frac{1}{c^2} \frac{\partial \phi}{\partial t} \) into the previous step to get \(abla^2 \phi - \frac{1}{c^2} \frac{\partial^2 \phi}{\partial t^2} = 0\). This verifies the wave equation for \(\phi\).
05

Substitute \(\mathbf{B}\) and \(\mathbf{E}\) in \(abla \times \mathbf{B} - \frac{1}{c^2} \frac{\partial \mathbf{E}}{\partial t} = \mathbf{0}\)

Substitute \(\mathbf{B} = abla \times \mathbf{A}\) and \(\mathbf{E} = -abla \phi - \frac{\partial \mathbf{A}}{\partial t}\) into \(abla \times \mathbf{B} - \frac{1}{c^2} \frac{\partial \mathbf{E}}{\partial t} = \mathbf{0}\). We get \(abla \times (abla \times \mathbf{A}) - \frac{1}{c^2} \left( -\frac{\partial(abla \phi)}{\partial t} - \frac{\partial^2 \mathbf{A}}{\partial t^2} \right) = \mathbf{0}\). Using the vector identity \(abla \times (abla \times \mathbf{A}) = abla (abla \cdot \mathbf{A}) - abla^2 \mathbf{A}\), we get: \[ abla (abla \cdot \mathbf{A}) - abla^2 \mathbf{A} + \frac{1}{c^2} \left( abla \frac{\partial \phi}{\partial t} + \frac{\partial^2 \mathbf{A}}{\partial t^2} \right) = 0 \]
06

Use the Lorenz gauge condition again

Simplify using the Lorenz gauge condition \(abla \cdot \mathbf{A} = -\frac{1}{c^2} \frac{\partial \phi}{\partial t}\). We plug this in and obtain: \[ -abla^2 \mathbf{A} + \frac{1}{c^2} \frac{\partial^2 \mathbf{A}}{\partial t^2} = \mathbf{0} \] which is the wave equation for \(\mathbf{A}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electromagnetism
Electromagnetism is a fundamental branch of physics that explores the interactions between electric and magnetic fields. One of the most crucial sets of equations in electromagnetism is Maxwell's Equations. These describe how electric and magnetic fields propagate and interact in space. The four equations are:

1. **Gauss's law for magnetism**: \(abla \cdot \mathbf{B} = 0\)
2. **Gauss's law for electricity**: \(abla \cdot \mathbf{E} = 0\)
3. **Faraday's law of induction**: \(abla \times \mathbf{E} + \frac{\partial \mathbf{B}}{\partial t} = 0\)
4. **Ampère's circuital law** (with Maxwell's correction): \(abla \times \mathbf{B} - \frac{1}{c^{2}} \frac{\partial \mathbf{E}}{\partial t} = 0\)

In free space, meaning there are no charges or currents affecting the fields, these equations simplify to the forms shown above. These simplified equations help describe how the electric and magnetic fields (\(\mathbf{E}\) and \(\mathbf{B}\)) behave and propagate as waves. Understanding these forms is critical for grasping the foundational principles of electromagnetism.
Vector Calculus
Vector calculus is a branch of mathematics focused on differentiation and integration of vector fields. This is crucial in physics, particularly in electromagnetism, for describing fields like \(\mathbf{E}\) (electric field) and \(\mathbf{B}\) (magnetic field). In this context, specific vector calculus identities and operations play a pivotal role.

Key operations include:

- **Divergence** (\(abla \cdot \mathbf{v}\)): Measures the magnitude of a source or sink at a given point in a vector field.
- **Curl** (\(abla \times \mathbf{v}\)): Describes the rotation or twisting force in a vector field.
- **Gradient** (\(abla f\)): Represents the rate and direction of change in a scalar field.

For Maxwell's Equations, these operations allow us to express fields in compact mathematical forms. For example, using the identity \(abla \cdot (abla \times \mathbf{A}) = 0\), we show that the magnetic field \(\mathbf{B} = abla \times \mathbf{A}\) satisfies Gauss's law for magnetism, \(abla \cdot \mathbf{B} = 0\). Mastery of vector calculus is essential for solving problems and deriving deeper results, like the wave equations for electromagnetic fields.
Wave Equations
Wave equations describe how waves propagate through different media. In electromagnetism, these equations explain how electric and magnetic fields travel as waves in free space.

When applying the Lorenz gauge condition, we derive the wave equations for the scalar potential \(\phi\) and the vector potential \(\mathbf{A}\). They are:

1. **Scalar Potential (\(\phi\)) Wave Equation**: \(abla^{2} \phi - \frac{1}{c^{2}} \frac{\partial^{2} \phi}{\partial t^{2}} = 0\)
2. **Vector Potential (\(\mathbf{A}\)) Wave Equation**: \(abla^{2} \mathbf{A} - \frac{1}{c^{2}} \frac{\partial^{2} \mathbf{A}}{\partial t^{2}} = 0\)

These equations reveal that both \(\phi\) and \(\mathbf{A}\) propagate as wave phenomena at the speed of light, \(c\). These wave equations are fundamental in predicting how electromagnetic waves behave, including radio waves, microwaves, and light itself. Understanding these equations and their solutions are critical for exploring advanced topics in electromagnetism, such as wave optics and electromagnetic radiation.
Lorenz Gauge
The Lorenz gauge is a condition used in electromagnetism to simplify the equations for the scalar potential \(\phi\) and the vector potential \(\mathbf{A}\). This gauge condition is described by:

\(abla \cdot \mathbf{A} + \frac{1}{c^{2}} \frac{\partial \phi}{\partial t} = 0\)
This choice of gauge helps decouple the equations governing \(\phi\) and \(\mathbf{A}\), making it easier to solve the wave equations for these potentials.

By imposing the Lorenz gauge, we ensure that the divergence of \(\mathbf{A}\) plus the time derivative term involving \(\phi\) balances out. This simplifies the Maxwell's equations in free space and helps derive:

- The wave equation for the scalar potential, \(\phi\)
- The wave equation for the vector potential, \(\mathbf{A}\)

This gauge choice is particularly useful in theoretical and computational electromagnetics, as it streamlines analytical and numerical methods for solving the behavior of electromagnetic fields. Mastering the use of the Lorenz gauge provides students with powerful tools to tackle more complex problems in electromagnetism.

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Most popular questions from this chapter

(a) Simplify $$ \nabla \times \mathbf{a}(\nabla \cdot \mathbf{a})+\mathbf{a} \times[\nabla \times(\nabla \times \mathbf{a})]+\mathbf{a} \times \nabla^{2} \mathbf{a} $$ (b) By explicitly writing out the terms in Cartesian coordinates prove that $$ [\mathbf{c} \cdot(\mathbf{b} \cdot \nabla)-\mathbf{b} \cdot(\mathbf{c} \cdot \nabla)] \mathbf{a}=(\nabla \times \mathbf{a}) \cdot(\mathbf{b} \times \mathbf{c}) $$ (c) Prove that \(\mathbf{a} \times(\nabla \times \mathbf{a})=\nabla\left(\frac{1}{2} a^{2}\right)-(\mathbf{a} \cdot \nabla) \mathbf{a}\).

In a Cartesian system, \(A\) and \(B\) are the points \((0,0,-1)\) and \((0,0,1)\) respectively. In a new coordinate system a general point \(P\) is given by \(\left(u_{1}, u_{2}, u_{3}\right)\) with \(u_{1}=\frac{1}{2}\left(r_{1}+r_{2}\right), u_{2}=\frac{1}{2}\left(r_{1}-r_{2}\right), u_{3}=\phi ;\) here \(r_{1}\) and \(r_{2}\) are the distances \(A P\) and \(B P\) and \(\phi\) is the angle between the plane \(A B P\) and \(y=0\). (a) Express \(z\) and the perpendicular distance \(\rho\) from \(P\) to the \(z\)-axis in terms of \(u_{1}, u_{2}, u_{3}\) (b) Evaluate \(\partial x / \partial u_{i}, \partial y / \partial u_{i}, \partial z / \partial u_{i}\), for \(i=1,2,3\). (c) Find the Cartesian components of \(\hat{\mathbf{u}}_{j}\) and hence show that the new coordinates are mutually orthogonal. Evaluate the scale factors and the infinitesimal volume element in the new coordinate system. (d) Determine and sketch the forms of the surfaces \(u_{i}=\) constant. (e) Find the most general function \(f\) of \(u_{1}\) only that satisfies \(\nabla^{2} f=0\)

(a) Using the parameterization \(x=u \cos \phi, y=u \sin \phi, z=u \cot \Omega\), find the sloping surface area of a right circular cone of semi-angle \(\Omega\) whose base has radius \(a\). Verify that it is equal to \(\frac{1}{2} \times\) perimeter of the base \(\times\) slope height. (b) Using the same parameterization as in (a) for \(x\) and \(y\), and an appropriate choice for \(z\), find the surface area between the planes \(z=0\) and \(z=Z\) of the paraboloid of revolution \(z=\alpha\left(x^{2}+y^{2}\right)\)

If two systems of coordinates with a common origin \(O\) are rotating with respect to each other, the measured accelerations differ in the two systems. Denoting by \(\mathbf{r}\) and \(\mathbf{r}^{\prime}\) position vectors in frames \(O X Y Z\) and \(O X^{\prime} Y^{\prime} Z^{\prime}\) respectively, the connection between the two is $$ \ddot{\mathbf{r}}^{\prime}=\ddot{\mathbf{r}}+\dot{\omega} \times \mathbf{r}+2 \omega \times \dot{\mathbf{r}}+\omega \times(\omega \times \mathbf{r}) $$ where \(\omega\) is the angular velocity vector of the rotation of \(O X Y Z\) with respect to \(O X^{\prime} Y^{\prime} Z^{\prime}\) (taken as fixed). The third term on the RHS is known as the Coriolis acceleration, whilst the final term gives rise to a centrifugal force. Consider the application of this result to the firing of a shell of mass \(m\) from a stationary ship on the steadily rotating earth, working to the first order in \(\omega\left(=7.3 \times 10^{-5} \mathrm{rad} \mathrm{s}^{-1}\right)\). If the shell is fired with velocity \(\mathbf{v}\) at time \(t=0\) and only reaches a height that is small compared to the radius of the earth, show that its acceleration, as recorded on the ship, is given approximately by $$ \ddot{\mathbf{r}}=\mathbf{g}-2 \omega \times(\mathbf{v}+\mathbf{g} t) $$ where \(m \mathbf{g}\) is the weight of the shell measured on the ship's deck. The shell is fired at another stationary ship (a distance \(\mathbf{s}\) away) and \(\mathbf{v}\) is such that the shell would have hit its target had there been no Coriolis effect. (a) Show that without the Coriolis effect the time of flight of the shell would have been \(\tau=-2 \mathbf{g} \cdot \mathbf{v} / g^{2}\) (b) Show further that when the shell actually hits the sea it is off target by approximately $$ \frac{2 \tau}{g^{2}}[(\mathbf{g} \times \omega) \cdot \mathbf{v}](\mathbf{g} \tau+\mathbf{v})-(\omega \times \mathbf{v}) \tau^{2}-\frac{1}{3}(\omega \times \mathbf{g}) \tau^{3} $$ (c) Estimate the order of magnitude \(\Delta\) of this miss for a shell for which \(v=300\) \(\mathrm{m} \mathrm{s}^{-1}\), firing close to its maximum range ( \(\mathbf{v}\) makes an angle of \(\pi / 4\) with the vertical) in a northerly direction, whilst the ship is stationed at latitude \(45^{\circ}\) North.

For a description in spherical polar coordinates with axial symmetry of the flow of a very viscous fluid, the components of the velocity field \(\mathbf{u}\) are given in terms of the stream function \(\psi\) by $$ u_{r}=\frac{1}{r^{2} \sin \theta} \frac{\partial \psi}{\partial \theta}, \quad u_{\theta}=\frac{-1}{r \sin \theta} \frac{\partial \psi}{\partial r} $$ Find an explicit expression for the differential operator \(E\) defined by $$ E \psi=-(r \sin \theta)(\nabla \times \mathbf{u})_{\phi} $$ The stream function satisfies the equation of motion \(E^{2} \psi=0\) and, for the flow of a fluid past a sphere, takes the form \(\psi(r, \theta)=f(r) \sin ^{2} \theta\). Show that \(f(r)\) satisfies the (ordinary) differential equation $$ r^{4} f^{(4)}-4 r^{2} f^{\prime \prime}+8 r f^{\prime}-8 f=0 $$
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