"Showing E&B Obey Wave Equation w/ Maxwell's Curl

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SUMMARY

This discussion focuses on demonstrating that the electric field \(\bf{E}\) and magnetic field \(\bf{B}\) obey the wave equation using Maxwell's equations. The key equations involved are \(\nabla \times \bf{E} = -\frac{1}{c}\frac{\partial\bf{B}}{\partial t}\) and \(\nabla \times \bf{B} = \frac{1}{c}\frac{\partial\bf{E}}{\partial t}\). By applying the curl operator and utilizing the identity \(\nabla\times\left(\nabla\times A\right) = \nabla\left(\nabla\cdot A\right) - \nabla^2 A\), the discussion concludes that both fields satisfy the wave equations \(\nabla^2\bf{E} = \frac{1}{c^2}\frac{\partial^2\bf{E}}{\partial t^2}\) and \(\nabla^2\bf{B} = \frac{1}{c^2}\frac{\partial^2\bf{B}}{\partial t^2}\).

PREREQUISITES
  • Understanding of vector calculus, specifically curl and divergence operations.
  • Familiarity with Maxwell's equations in electromagnetism.
  • Knowledge of wave equations and their mathematical representation.
  • Proficiency in using mathematical identities involving differential operators.
NEXT STEPS
  • Study the derivation of wave equations from Maxwell's equations in detail.
  • Learn about the physical significance of electromagnetic waves and their propagation.
  • Explore the applications of curl and divergence in vector calculus.
  • Investigate the implications of boundary conditions on electromagnetic wave solutions.
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Students and professionals in physics, particularly those focusing on electromagnetism, as well as mathematicians interested in vector calculus applications in physical contexts.

knowlewj01
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Homework Statement



This question is closely related to physics but it's in a maths assignment paper i have so here it is:

By taking curls of the following equations:

\nabla \times \bf{E} = -\frac{1}{c}\frac{\partial\bf{B}}{\partial t}

\nabla \times \bf{B} = \frac{1}{c}\frac{\partial\bf{E}}{\partial t}

show that both E and B obey the wave equation with speed c, that is:

\nabla^2\bf{E} = \frac{1}{c^2}\frac{\partial^2\bf{E}}{\partial t^2}

and

\nabla^2\bf{B} = \frac{1}{c^2}\frac{\partial^2\bf{B}}{\partial t^2}


Homework Equations



\nabla\cdot\bf{E} = 0
\nabla\cdot\bf{B} = 0
\nabla \times \bf{E} = -\frac{1}{c}\frac{\partial\bf{B}}{\partial t}
\nabla \times \bf{B} = \frac{1}{c}\frac{\partial\bf{E}}{\partial t}


The Attempt at a Solution



I don't really know how to start this, so i took a wild guess. I started by looking at the first equation with \nabla \times \bf{E} in it. and did the cross product, needless to say i got some horrible expression.

given that the divergence of both vector fields are 0, we should be able to write these vector fields as the curl of another vector field, such that:

\bf{E} = \nabla \times \bf{A}

iff

\nabla\cdot\bf{E} = 0

i think this is the starting point for the question. But i hit a brick wall at this point, anyone know what is next? thanks
 
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i have got a bit further with this, not sure how to finish it off though.

take curls of the equations:

\nabla \times \left(\nabla\times\bf{E}\right) = -\frac{1}{c}\frac{\partial}{\partial t}\left(\nabla\times\bf{B}\right) [1]

\nabla \times \left(\nabla\times\bf{B}\right) = \frac{1}{c}\frac{\partial}{\partial t}\left(\nabla\times\bf{E}\right) [2]

Use the identity: \nabla\times\left(\nabla\times A\right) = \nabla\left(\nabla\cdot A\right) - \nabla^2 A

and since \nabla\cdot \bf{E} = \nabla\cdot \bf{B} = 0

eqn's [1] and [2] become:

\frac{1}{c}\frac{\partial}{\partial t}\left(\nabla\times\bf{B}\right) = \nabla^2\bf{E} [3]
\frac{1}{c}\frac{\partial}{\partial t}\left(\nabla\times\bf{E}\right) = -\nabla^2\bf{B} [4]

is this correct? i don't know wether i should curl something that is already operated by a d/dt also, where should the second time derivative come into it?
 
Last edited:

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