The Energy in an Electromagnetic Wave and the shape of the Wave

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SUMMARY

The energy of an electromagnetic (EM) wave is directly proportional to the square of its amplitude, a fundamental principle in electromagnetism. The shape of an EM wave is sinusoidal, and the wave equation is linear, meaning that any sum of sinusoidal solutions remains a valid solution. The shape of the EM wave can change when it travels through different media, depending on the dispersion relation. This discussion emphasizes the importance of Maxwell's equations in understanding the energy density of electromagnetic fields.

PREREQUISITES
  • Understanding of electromagnetic wave properties
  • Familiarity with Maxwell's equations
  • Knowledge of wave equations and linearity
  • Concept of dispersion relations in different media
NEXT STEPS
  • Study the relationship between amplitude and energy in electromagnetic waves
  • Explore the implications of Maxwell's equations on electromagnetic fields
  • Research dispersion relations and their effects on wave propagation
  • Learn about energy density calculations in electromagnetic fields
USEFUL FOR

Students of physics, electrical engineers, and researchers interested in the properties of electromagnetic waves and their applications in various media.

fizzyfiz
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no need of equations
 
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fizzyfiz said:
I met a statement that the energy of EM wave is proportional to the amplitude square
Don't keep us in the dark: tell us where you encountered this statement. In a dark alley or in broad daylight ?

the shape of EM wave is sinusoidal
The wave equation is linear: any sum of sinusoidals is also a solution.

Does shape of EM wave change when it travels through medium
Depends on the dispersion relation.

fizzyfiz said:
no need of equations
Totally disagree
 
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BvU said:
The wave equation is linear: any sum of sinusoidals is also a solution.
Any sum of sinusoidals that are themselves solutions is also a solution, and these sums are not necessarily sinusoidal

BvU already knows this of course; I'm just adding footnotes for other people who may be reading the thread.
 
Though you forbid me to use the adequate language, which are formulae, I answer, because maybe it can help you to understand the issue. No matter which specific electromagnetic field you have (of course it must obey the Maxwell equations for the given charge-current distribution) the energy density of the electromagnetic field in a vacuum is (in SI units)
$$u_{\text{em}}(t,\vec{x})=\frac{\epsilon_0}{2} \vec{E}^2(t,\vec{x}) + \frac{1}{2 \mu_0} \vec{B}^2(t,\vec{x}).$$
 

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