The Plane Electromagnetic Wave in Vacuum

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SUMMARY

The discussion centers on the representation of a plane electromagnetic wave in vacuum, described by the equation E = E₀ exp[i(kx - ωt)]. The phase velocity, denoted as c, is defined by the relationship c = ω/k. Participants clarify that while the values of ω and k change under Lorentz transformation, the phase velocity c remains constant. The conversation also touches on deriving the formulas for the relativistic Doppler Effect from these transformations, emphasizing a deeper understanding of the underlying physics.

PREREQUISITES
  • Understanding of electromagnetic wave equations
  • Familiarity with Lorentz transformations
  • Knowledge of phase velocity and its significance
  • Basic concepts of the relativistic Doppler Effect
NEXT STEPS
  • Study the derivation of the relativistic Doppler Effect formulas
  • Explore the implications of Lorentz transformations on wave properties
  • Investigate the mathematical representation of electromagnetic waves in different inertial frames
  • Learn about the significance of phase velocity in various physical contexts
USEFUL FOR

Physicists, students of electromagnetism, and anyone interested in the mathematical foundations of wave mechanics and relativistic effects.

Anamitra
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A plane electromagnetic wave[traveling in the x-direction in an inertial frame]in vacuum is usually represented by an equation of the form:

{E}{=}{E_{0}}{exp{[}{i}{(}{k}{x}{-}{\omega}{t}{)}{]}}

The wave velocity[phase velocity] is given by:

{c}{=}{\frac{\omega}{k}}

We can perform Lorentz transformation on t and x[and y,z] to some other inertial frame moving in the x-x' direction.

The value {\frac{\omega}{k}} changes after transformation.But the value of c [wave velocity=phase velocity]should not change.
How does one explain this?
[If one considers a wave packet in vacuum each component should have the same speed=c]
 
Last edited:
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Anamitra said:
The value {\frac{\omega}{k}} changes after transformation.
No it doesn't. You must have made a mistake in your math.
 
{t}{=}{\gamma}{(}{t^{'}}{+}{\frac{vx^{'}}{c^{2}}{)}
{x}{=}{\gamma}{(}{x^{'}}{+}{vt^{'}}{)}

{{k}{x}{-}{\omega}{t}}{=}{x^{'}}{\gamma}{(}{k}{-}{v}{/}{c^{2}}{)}{-}{{t}^{'}}{\gamma}{(}{\omega}{-}{kv}{)}

{\frac{{\omega}^{'}}{k^{'}}}{=}{\frac{{\omega}{-}{kv}}{{k}{-}{{v}{\omega}{/}{c^{2}}}
 
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Yes, you almost have it. Now just make the substitution:
\omega=ck
And simplify.
 
Dalespam is of course correct. Thanks for that.

In fact we can get the formulas for relativistic Doppler Effect from the transformations of "omega" and "k" in the formulas of thread #3, without any reference to the position/location of the source[there is a small typing error in the second last formula of thread #3]
 
Anamitra said:
In fact we can get the formulas for relativistic Doppler Effect from the transformations of "omega" and "k" in the formulas of thread #3, without any reference to the position/location of the source
Yes, that is a very useful formula. I am glad you know how to derive it. I think that is a deeper level of understanding than just using the formula.
 

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