Show that a longitudinal wave is electrostatic

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SUMMARY

This discussion centers on demonstrating that all longitudinal waves are electrostatic by applying Faraday's law of electromagnetic induction. The key equation referenced is Faraday's law, expressed as \(\frac{\partial \vec{B}}{\partial t} = - \nabla \times \vec{E}\). Participants explore the relationship between electric and magnetic fields in longitudinal electromagnetic waves, specifically using the equation \(\vec{B} = \vec{B_0} \sin{[i(kx-\omega t)]}\) and its implications for the curl of the electric field. The conversation highlights the complexity of deriving meaningful results from these equations.

PREREQUISITES
  • Understanding of Faraday's law of electromagnetic induction
  • Familiarity with electromagnetic wave equations
  • Knowledge of vector calculus, particularly curl operations
  • Basic concepts of longitudinal and transverse waves
NEXT STEPS
  • Study the derivation of electromagnetic wave equations from Maxwell's equations
  • Explore the implications of longitudinal versus transverse wave propagation
  • Learn about the real and imaginary components of complex numbers in physics
  • Investigate applications of Faraday's law in various electromagnetic contexts
USEFUL FOR

Students of physics, particularly those studying electromagnetism, as well as educators and researchers interested in the properties of electromagnetic waves and their mathematical descriptions.

Logarythmic
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Homework Statement


Show that all longitudinal waves must be electrostatic by using Faraday's law.


Homework Equations


Faraday's law:

\frac{\partial \vec{B}}{\partial t} = - \nabla \times \vec{E}


The Attempt at a Solution


Where should I start??
 
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A good place to start would be to write down any equations for the electric and magnetic fields (or auxiliary field H) for longitudinal EM waves. Then calculate dB/dt (or dH/dt)...what do you get?
 
I tried with

\vec{B} = B_0 \sin{[i(kx-\omega t)]}

so

\nabla \times \vec{E} = i \omega B_0 \cos{[i(kx-\omega t)]}

But that doesn't really help me.
 
Logarythmic said:
I tried with

\vec{B} = B_0 \sin{[i(kx-\omega t)]}

so

\nabla \times \vec{E} = i \omega B_0 \cos{[i(kx-\omega t)]}

But that doesn't really help me.

Don't you mean:

\vec{B} = \vec{B_0} \sin{[i(kx-\omega t)]}

and

\nabla \times \vec{E} = \Re[i \omega \vec{B_0} \cos{[i(kx-\omega t)]}]


...what is the real part of a purely imaginary number?:wink:
 
Sometimes I feel so smart that I don't know what to do with myself. ;) Thanks!
 

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