Question

Maxwell's equations relating the electric field \(E\)and magnetic field \(H\)as they vary with time in a region containing no charge and no current can be stated as follows:

\(divE = 0\;\;\;divH = 0\)

\(curlE = - \frac{1}{c}\frac{{\partial H}}{{\partial t}}\;\;\;curlH = \frac{1}{c}\frac{{\partial E}}{{\partial t}}\)

where\(c\)is the speed of light. Use these equations to prove the following:

(a) \(\nabla \times (\nabla \times E) = - \frac{1}{{{c^2}}}\frac{{{\partial ^2}E}}{{\partial {t^2}}}\)

(b) \(\nabla \times (\nabla \times H) = - \frac{1}{{{c^2}}}\frac{{{\partial ^2}H}}{{\partial {t^2}}}\)

(c) \({\nabla ^2}E = \frac{1}{{{c^2}}}\frac{{{\partial ^2}E}}{{\partial {t^2}}}\;\;\;\) (Hint: Use Exercise 27.)

(d) \({\nabla ^2}H = \frac{1}{{{c^2}}}\frac{{{\partial ^2}H}}{{\partial {t^2}}}\)

Short answer

It is proved that,

(a)\(\nabla \times (\nabla \times E) = - \frac{1}{{{c^2}}}\frac{{{\partial ^2}E}}{{\partial {t^2}}}\)

(b)\(\nabla \times (\nabla \times H) = - \frac{1}{{{c^2}}}\frac{{{\partial ^2}H}}{{\partial {t^2}}}\)

(c)\({\nabla ^2}E = \frac{1}{{{c^2}}}\frac{{{\partial ^2}E}}{{\partial {t^2}}}\;\;\;\)

(d)\({\nabla ^2}H = \frac{1}{{{c^2}}}\frac{{{\partial ^2}H}}{{\partial {t^2}}}\)

Step-by-step solution

To Find the equation (a)

We need to prove

\(\nabla \times (\nabla \times E) = - \frac{1}{{{c^2}}}\frac{{{\partial ^2}E}}{{\partial {t^2}}}\)

We are given

\(\begin{array}{l}{\mathop{\rm div}\nolimits} E = 0\\{\mathop{\rm div}\nolimits} H = 0\\{\mathop{\rm curl}\nolimits} E = - \frac{1}{c}\frac{{\partial H}}{{\partial t}}\\{\mathop{\rm curl}\nolimits} H = \frac{1}{c}\frac{{\partial E}}{{\partial t}}\end{array}\)

Starting with the equation

\(\begin{array}{c}\nabla \times E = {\mathop{\rm curl}\nolimits} E\\ = - \frac{1}{c}\frac{{\partial H}}{{\partial t}}\end{array}\)

Apply curl once again;

\(\begin{array}{c}\nabla \times (\nabla \times E) = \nabla \times \left( { - \frac{1}{c}\frac{{\partial H}}{{\partial t}}} \right)\\ = - \frac{1}{c}\nabla \times \left( {\frac{{\partial H}}{{\partial t}}} \right)\end{array}\)

We can change the order of curl and differentiation with respect to\(t\);

\(\nabla \times (\nabla \times E) = - \frac{1}{c}\frac{\partial }{{\partial t}}(\underbrace {\nabla \times H}_{{\rm{curl }}{\bf{H}}})\)

Now use the Maxwell equation

\({\mathop{\rm curl}\nolimits} H = \frac{1}{c}\frac{{\partial E}}{{\partial t}}\)

to get

\(\begin{array}{c}\nabla \times (\nabla \times E) = - \frac{1}{c}\frac{\partial }{{\partial t}}\left( {\frac{1}{c}\frac{{\partial E}}{{\partial t}}} \right)\\ = - \frac{1}{{{c^2}}}\frac{{{\partial ^2}E}}{{\partial {t^2}}}\end{array}\)

To Find the equation (b)

We need to prove

\(\nabla \times (\nabla \times H) = - \frac{1}{{{c^2}}}\frac{{{\partial ^2}H}}{{\partial {t^2}}}\)

We are given

\(\begin{array}{l}{\mathop{\rm div}\nolimits} E = 0\\{\mathop{\rm div}\nolimits} H = 0\\{\mathop{\rm curl}\nolimits} E = - \frac{1}{c}\frac{{\partial H}}{{\partial t}}\\{\mathop{\rm curl}\nolimits} H = \frac{1}{c}\frac{{\partial E}}{{\partial t}}\end{array}\)

Starting with the equation

\(\begin{array}{c}\nabla \times H = {\mathop{\rm curl}\nolimits} E\\ = \frac{1}{c}\frac{{\partial E}}{{\partial t}}\end{array}\)

Apply curl once again;

\(\begin{array}{c}\nabla \times (\nabla \times H) = \nabla \times \left( {\frac{1}{c}\frac{{\partial E}}{{\partial t}}} \right)\\ = \frac{1}{c}\nabla \times \left( {\frac{{\partial E}}{{\partial t}}} \right)\end{array}\)

We can change the order of curl and differentiation with respect to\(t\);

\(\nabla \times (\nabla \times H) = \frac{1}{c}\frac{\partial }{{\partial t}}(\underbrace {\nabla \times E}_{{\rm{curl E}}})\)

Now use the Maxwell equation

\({\mathop{\rm curl}\nolimits} E = - \frac{1}{c}\frac{{\partial H}}{{\partial t}}\)

to get

\(\begin{array}{c}\nabla \times (\nabla \times H) = \frac{1}{c}\frac{\partial }{{\partial t}}\left( { - \frac{1}{c}\frac{{\partial H}}{{\partial t}}} \right)\\ = - \frac{1}{{{c^2}}}\frac{{{\partial ^2}H}}{{\partial {t^2}}}\end{array}\)

To Find the equation (c)

According to Q29, We have

\({\mathop{\rm curl}\nolimits} ({\mathop{\rm curl}\nolimits} E) = {\mathop{\rm grad}\nolimits} ({\mathop{\rm div}\nolimits} E) - {\nabla ^2}E\)

Therefore,

\({\nabla ^2}E = {\mathop{\rm grad}\nolimits} ({\mathop{\rm div}\nolimits} E) - {\mathop{\rm curl}\nolimits} ({\mathop{\rm curl}\nolimits} E)\)

Substitute\({\mathop{\rm div}\nolimits} E = 0\), To get

\(\begin{array}{l}{\nabla ^2}E = {\mathop{\rm grad}\nolimits} (0) - {\mathop{\rm curl}\nolimits} ({\mathop{\rm curl}\nolimits} E)\\{\nabla ^2}E = - {\mathop{\rm curl}\nolimits} ({\mathop{\rm curl}\nolimits} E)\end{array}\)

Recall that\({\mathop{\rm curl}\nolimits} F = \nabla \times F\)

Therefore,

\({\nabla ^2}E = - \nabla \times (\nabla \times E)\)

From part(a), we know that\(\nabla \times (\nabla \times E) = - \frac{1}{{{c^2}}}\frac{{{\partial ^2}E}}{{\partial {t^2}}}\)

Therefore,

\(\begin{array}{c}{\nabla ^2}E = - \left( { - \frac{1}{{{c^2}}}\frac{{{\partial ^2}E}}{{\partial {t^2}}}} \right)\\ = \frac{1}{{{c^2}}}\frac{{{\partial ^2}E}}{{\partial {t^2}}}\end{array}\)

Hence proved

To Find the equation (d)

According to\({\bf{Q29}}\), We have

\({\mathop{\rm curl}\nolimits} ({\mathop{\rm curl}\nolimits} H) = {\mathop{\rm grad}\nolimits} ({\mathop{\rm div}\nolimits} H) - {\nabla ^2}H\)

Therefore,

\({\nabla ^2}H = {\mathop{\rm grad}\nolimits} ({\mathop{\rm div}\nolimits} H) - {\mathop{\rm curl}\nolimits} ({\mathop{\rm curl}\nolimits} H)\)

Substitute\({\mathop{\rm div}\nolimits} H = 0\), To get

\(\begin{array}{l}{\nabla ^2}H = {\mathop{\rm grad}\nolimits} (0) - {\mathop{\rm curl}\nolimits} ({\mathop{\rm curl}\nolimits} H)\\{\nabla ^2}H = - {\mathop{\rm curl}\nolimits} ({\mathop{\rm curl}\nolimits} H)\end{array}\)

Recall that\({\mathop{\rm curl}\nolimits} F = \nabla \times F\)

Therefore,

\({\nabla ^2}H = - \nabla \times (\nabla \times H)\)

From part\(({\bf{b}})\), we know that\(\nabla \times (\nabla \times H) = - \frac{1}{{{c^2}}}\frac{{{\partial ^2}H}}{{\partial {t^2}}}\)Therefore,

\(\begin{array}{c}{\nabla ^2}H = - \left( { - \frac{1}{{{c^2}}}\frac{{{\partial ^2}H}}{{\partial {t^2}}}} \right)\\ = \frac{1}{{{c^2}}}\frac{{{\partial ^2}H}}{{\partial {t^2}}}\end{array}\)

Hence proved