EM wave/radiation from which equations?

  • Thread starter xxxyyy
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  • #1
xxxyyy
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Hi there,
in Feynmann's lectures, (Vol 2 eq (21.1) Heaviside-Feynman forumla)
https://www.feynmanlectures.caltech.edu/II_21.html

it is said that radiation comes from the last term, the one with acceleration in it. This formula is a particular case of the Jefimenko equations, the ones derived from Lienard-Wiechert potentials.
These potentials are particular solutions of the inhomogeneus wave equations for the potentials (in Lorenz gauge). I should add the general solution of the homogeneus equestions, i.e. wave solutions, to get the true general solution.

Does wave solutions for the potentials translate into wave solutions for the fields?

And where does exactly the wave phenomenon come from? from the homogeneus solution or from the second time derivative Feynman was talking about in his (21.1) formula?
Thank you!
 
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  • #2
Delta2
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Does wave solutions for the potentials translate into wave solutions for the fields?
Absolutely yes. If you have solutions ##V,\vec{A}## for the scalar potential and vector potential that satisfies the homogeneous or inhomogeneous wave equation, then the fields defined by $$\vec{E}=-\nabla V-\frac{\partial \vec{A}}{\partial t}$$ and $$\vec{B}=\nabla\times\vec{A}$$ satisfy the wave equation as well. To prove it is a not so hard vector calculus excercise, you will utilize Maxwell's equation (in differential form) and some vector calculus identities and the Lorentz gauge condition.
 
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Set ##\phi = 0## such that ##\mathbf{E} = - \dfrac{\partial \mathbf{A}}{\partial t}## and ##\mathbf{B} = \nabla \times \mathbf{A}##. Set also ##\nabla \cdot \mathbf{A} = 0## then from Maxwell\begin{align*}
\Delta \mathbf{A} - \frac{\partial^2 \mathbf{A}}{\partial t^2} &= 0 \ \ \ (1) \\
\end{align*}Take the curl of ##(1)##,\begin{align*}
\Delta ( \nabla \times \mathbf{A} )- \frac{\partial^2}{\partial t^2}\left( \nabla \times \mathbf{A} \right) &= 0 \\

\Delta \mathbf{B} - \frac{\partial^2 \mathbf{B}}{\partial t^2} &= 0

\end{align*}Instead take the time derivative of ##(1)##,\begin{align*}
\Delta \left( \frac{\partial \mathbf{A}}{\partial t}\right) - \frac{\partial^2}{\partial t^2} \left( \frac{\partial \mathbf{A}}{\partial t} \right) &= 0\\

\Delta \mathbf{E} - \frac{\partial^2 \mathbf{E}}{\partial t^2} &= 0

\end{align*}The wave equations holds on ##\mathbf{E}## and ##\mathbf{B}##. Let ##\xi## be any component of either of these vectors, i.e. ##\Delta \xi - \dfrac{\partial^2 \xi}{\partial t^2} = 0## and let ##\xi(\boldsymbol{x},t) = f(p)## where ##p = k_1 x + k_2 y + k_3 z + k_4 t##. Then ##\dfrac{\partial^2 \xi}{\partial x_i^2} = k_i^2 \dfrac{d^2 f}{dp^2}## hence ##k_1^2 + k_2^2 + k_3^2 - k_4^2 = 0##. You may simply let ##k_4 = \pm 1## so that ##\hat{\mathbf{k}} = (k_1, k_2, k_3)## form the components of a unit vector. Then ##p_{\pm} = \hat{\mathbf{k}} \cdot \boldsymbol{x} \pm t## and for each component the general solution is ##\xi(\boldsymbol{x}, t) = f(p_+) + g(p_{-})##
 
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  • #4
xxxyyy
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Does wave phenomena arise from the third term in Feynman equation (21.1)?
At the end of 21-3 paragraph, Feyman boldy says that the complete theory of (classical) EM is in that box... but why doesn't he include the solutions of the homogeneous equations (the wave solutions)? Is really everything, that's a wave, in that third term of (21.1)?
I have here a (very respectable) book that includes them and it even says that, for bounded charge distributions, the Lienard-Wiechert potential are negligible far away from the sources, because they fall of faster then 1/r.
I'm confused.
 

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