Discussion on Feynman lecture - E / B Field

[Sparky_]

Greetings,

In Feynman's lectures - vol. 2 either chap 19, 21 or 23 - I think 23 (I'm at work and don't have the lectures here), Feynman shows in a capacitor that a changing E-Field induces a B-field, then he shows that the B-field induces a new E-Field (he calculates the E-field) and then adds the 2 E-fields together. Well he says that "new" E-field creates a new B-field - he calculates that new B and adds the two B's together and so on.

He does 3 or so to see the pattern.

He ends up showing the E-field and B-field is this complicated expression involving a Bessel function.

I had under-grad electromagnetics - junior and senior level and this was never mentioned.

This Feynman lecture and results does not seem obvious to me -

I recall solving for the "E" field or the "B" field in class and it would be 1 step integration of derivation - box the answer and your finished.

Can you elaborate further on this?

Why was it not mentioned in electrodynamics (under-grad level at least)?

It seems so interestesting but I never would have known to go there, I would have stopped at the first calculation.

I find this lecture very interesting but do not feel comfortable with it (yet).

Can you all discuss it a little to help make it more obvious or clearer?

What other physical phenomina behave this way that are overlooked because we stop at the first step?

Thanks
-Sparky

 

[DrDu]

Although I do not know exactly what he was calculating this sounds like if he wanted to exemplify self-consistency. The recursive calculation of self-consistent quantities is quite popular on a computer although in cases like your example you will usually try to avoid a recursive solution and will try to solve for both B and E simultaneously. Take for example a free electromagnetic field of a free wave. You can argue that the change of the electric field will lead to a magnetic field and this in turn to an electric field and that is like a wave propagates. On the other hand, you can use some Fourier transformation and plug into the Maxwell equations and get directly a solution for the propagating wave.
Where recrusive methods are used in e.g. in quantum chemistry where you have to solve the Hartree-Fock equations. These are non-linear equations in all the electronic coordinates. You start from some trial wavefunctions and calculate new wavefunctions which you uses as new trial functions and so on.
Maybe the prototype of all these calculations is the Newton method for finding the zeros of a function f(x):
x_(i+1)=x_i-f(x_i)/f'(x_i)

 

[Bob S]

You might be talking about two closely related Maxwell equations (in free space without conductors):

Curl E = -μ₀ ∂H/∂t, and

Curl H = +ε₀ ∂E/∂t

So a changing magnetic field can produce an electric field, and a changing electric field can produce a magnetic field. The difference in the sign between the two equations, known as Lenz's Law, prevents additive regeneration (self-amplification) of either electric or magnetic fields. So for example, either E or H can be proportional to sin(ωt) or cos(ωt), and the other cos(ωt) or sin(ωt), but neither can be sinh(ωt) or cosh(ωt).

Bessel functions enter into solutions of electromagnetic equations in cylindrical geometry...?...

Bob S

 

[Sparky_]

Later this evening or within the next few days I will get the lecture out and see more about it.
It has bothered me for a while now.

Bob S - I think you are close (I just don't remember)

But that is one question I had was with each new field adding to the previous why didn't the solution "explode" additively.

 

[Bob S]

Bob S - I think you are close (I just don't remember)

But that is one question I had was with each new field adding to the previous why didn't the solution "explode" additively.

If you substitute one of the curl equations in my previous post into the other, you will get a second order differential equation that has a sine-like solution.

Curl Curl H + ε₀μ₀ ∂²H /∂t² = 0

If the second order equation had a sinh-like solution, the solution would be an exponential and would be unstable ("explode").

Bob S

 

[gnurf]

The Bessel functions show up in the solutions due to the cylindrical geometry of the waveguide. Begin with the curl equations in cylindrical coordinates ... [insert work here] ... and you get

\frac{1}{r} \frac{\delta}{\delta r}\left[r\frac{\delta \bar{\psi}}{\delta r}\right] + \frac{1}{r^2} \frac{\delta ^2 \bar{\psi}}{\delta \phi ^2} - \beta_c^2\bar{\psi} = 0

where \bar{\psi} represents \bar{E_z} or \bar{H_z} for TM and TE waves respectively.

Separation of variables:

\bar{\psi} = \bar{R}(r)\bar{\Phi}(\phi)e^{\mp j\beta_zz}

Plug that in, and you get two separate differential equations, one of them looks like this:

\bar{R}'' + \frac{1}{r} \bar{R}' + (\beta_c^2 - \frac{n^2}{r^2})\bar{R} = 0

which is a http://mathworld.wolfram.com/BesselDifferentialEquation.html" as solutions.