# Frequency in Maxwell's equations

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$$\textbf{F} \cdot d\textbf{l}=q(\textbf{E}+\textbf{v} \times \textbf{B})\cdot \textbf{v} dt$$
If we denote $$q=\rho d \tau$$ and $$\rho \textbf{v}=\textbf{J}$$

$$\frac{dW}{dt}=\int_{V} (\textbf{E} \cdot \textbf{J}) d \tau.$$

From maxwell law's

$$\textbf{E} \cdot \textbf{J}=\frac{1}{\mu _0} \textbf{E} \cdot (\nabla \times \textbf{B}) - \epsilon _0 \textbf{E} \cdot \frac{\partial \textbf{E}}{\partial t}$$
we use the product rule:
$$\nabla \dot \cdot (\textbf{E} \times \textbf{B})=\textbf{B} \cdot (\nabla \times \textbf{E})-\textbf{E} \cdot (\nabla \times \textbf{B}).$$

$$\nabla \times \textbf{E}=-\frac{\partial \textbf{B}}{\partial t}$$
gives

$$\textbf{E} \cdot (\nabla \times \textbf{B})=-\frac{\partial \textbf{B}}{\partial t} - \nabla \cdot (\textbf{E} \times \textbf{B})$$

$$\textbf{B} \cdot \frac{\partial \textbf{B}}{\partial t}=\frac{1}{2} \frac{\partial}{\partial t}(B^2)$$

$$\textbf{E} \cdot \frac{\partial \textbf{E}}{\partial t}=\frac{1}{2} \frac{\partial}{\partial t}(B^2)$$

$$\textbf{E} \cdot \textbf{J}=\frac{1}{2} \frac{\partial}{\partial t}(\epsilon_0 E^2+\frac{1}{\mu_0} B^2) - \frac{1}{\mu_0} \nabla \cdot (\textbf{E} \times \textbf{B}).$$

The theory above is taken from Griffiths introduction to electrodynamics.

From there by rearranging:

$$\textbf{F} \cdot d \textbf{l}=\frac{q}{\rho}[-\frac{1}{2} \frac{\partial}{\partial t}(\epsilon_0 E^2+\frac{1}{\mu_0} B^2) - \frac{1}{\mu_0} \nabla \cdot (\textbf{E} \times \textbf{B})] dt$$

From a quantification perspective we could rewrite the last part to:
$$\textbf{F} \cdot d \textbf{l}=\frac{q}{\rho}[-\frac{1}{2} \frac{\partial}{\partial t}(\epsilon_0 E^2+\frac{1}{\mu_0} B^2) - \frac{1}{\mu_0}c \nabla \cdot ( \textbf{B}^2) ] dt$$

$$\frac{d W}{d t}=\frac{q}{\rho}[-\frac{1}{2} \frac{\partial}{\partial t}\int_{V}(\epsilon_0 E^2+\frac{1}{\mu_0} B^2) d \tau - \frac{1}{\mu_0}\int_{V}c \nabla \cdot ( \textbf{B}^2) \textbf{i}] d\tau$$

The first term is recognized as the energy density. If all the energy is sent out instead of stored in the system:

$$\frac{d W}{d t}= \frac{1}{\mu_0}\int_{V}c \nabla \cdot ( \textbf{B}^2) \textbf{i}] d\tau$$

From the divergence theorem:

$$\frac{d W}{d t}= \frac{c}{\mu_0}\int_{A} \textbf{B}^2 \cdot dA=\frac{c}{\mu_0} \textbf{B}^2 4 \pi r^2$$

If we denote this emitted energy as Em waves:

$$B=\textbf{B} \sin [2 \pi ft]$$

$$W=\frac{c}{\mu_0} \textbf{B}^2 \int_{0}^{T} \sin^2[2 \pi ft] 4 \pi r^2 dt$$

$$\sin^2x=\frac{1}{2}-\cos(2x)$$

$$W=\frac{c}{\mu_0} \textbf{B}^2 \int_{0}^{T}(\frac{1}{2}-\cos([4 \pi ft]) 4 \pi r^2 )dt=\frac{c}{\mu_0}\textbf{B}^2 \frac{1}{2} T 4 \pi r^2$$

From another approach of the theory in Griffiths:

Again we look at the energy sent out and assume that no energy is stored

$$\frac{d W}{d t}= \frac{1}{\mu_0}\int_{V}c \nabla \cdot ( \textbf{B}^2) \textbf{i}] d\tau$$

Now instead of using the divergence theorem there could perhaps be possible to do this volume integral above and then equate it to:

$$\frac{c}{\mu_0}\textbf{B}^2 \frac{1}{2} T 4 \pi r^2$$

And then solve for $$T=\frac{1}{f}$$

What I have done so far:

We assume that B goes radially outwards and use spherical coordinates to solve the divergence theorem:

The divergence in spherical coordinates:

$$\frac{1}{r^2} \frac{\partial r^2 A_r}{\partial r} + \frac{1}{r sin \theta}\frac{\partial}{\partial \theta}A_{\theta}sin_{\theta} + \frac{1}{r sin \theta} \frac{\partial A_{\varphi}}{ \partial \varphi}$$

Since we only have radial divergence we can obtain:

$$\frac{d W}{d t}= \frac{1}{\mu_0}\int_{V}c \nabla \cdot ( \textbf{B}^2) \textbf{i}] d\tau=\frac{c}{\mu_0}\int_{V} \frac{1}{r^2} \frac{d}{dr}(r^2B^2)4 \pi r^2 dr=\frac{4 \pi c}{\mu_0}\int_{0}^r d (r^2B^2)=\frac{4 \pi c}{\mu_0} (r^2B^2)$$

Which we see is the same result as we ended up with above as:

$$W=\frac{c}{\mu_0}\textbf{B}^2 \frac{1}{2} T 4 \pi r^2$$

But what if we instead tried to integrate

$$\frac{c}{\mu_0}\int_{V} \frac{1}{r^2} \frac{d}{dr}(r^2B^2)4 \pi r^2 dr$$

by rewriting:

$$B=\textbf{B} \sin [2 \pi ft]=\textbf{B} \sin [2 \pi f\frac{r}{c}]$$

Thus

$$\frac{c}{\mu_0}\int_{V} \frac{1}{r^2} \frac{d}{dr}(r^2\textbf{B}^2 \sin^2 [2 \pi f\frac{r}{c}])4 \pi r^2 dr$$

Then we have a r dependency for the integral. I have tried and I ended up with a longer expression. The better news was that if I equated it with

$$\frac{c}{\mu_0}\textbf{B}^2 \frac{1}{2} T 4 \pi r^2$$

I ended up with an equation that only had R and f as variables since B is the same for the two divergence theorems versions from the derivations above. I guess I could initially determine R from a setup point of view. But unfortunately it did not give any solutions. It seems that since there are 2 Work equations. Equating thoose two could give a solution of frequency. But I have not found a solution that makes sense. Anyone that wants to try to find a solution for the frequency with this setup?

Here is my attempt attached just in case anyone want to take a look (the answer gave me 0 answers as far as I could tell):   • etotheipi

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Is there a question?

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Is there a question?
Sorry. I guess my qestion would be: I did end up with an equation where R, the radius, and f, the frequency, are variables. But it did not yield any solutions. (look at the last attachment part for this). The problem for me is that I can't seem to see where there is an error in the calculations. I use maxwells equations to the best of my knowledge first. Then I use the formula for an EM wave on sinus form. And then I rearrange an emission of EM waves from an electrons surface with the divergence theorem.

What I am looking for is a value for the frequency, f. But my last equation did not yield any answer when I used an equation solver like this one:

https://www.symbolab.com/solver/equation-calculator/\frac{\frac{4\cdot \pi \cdot x}{3\cdot 10^{8}}+2-\left(4\cdot \pi \cdot x\cdot \frac{1}{3\cdot 10^{8}}\right)^{2}\cdot y^{2}}{8\cdot\pi\cdot x\cdot\frac{1}{3\cdot 10^{8}}\cdot y-1}=\frac{1}{\tan\left(4\cdot\pi\cdot x\cdot\frac{1}{3\cdot 10^{8}}\cdot y\right)}

Sorry for the long link. So I want to solve for frequency. Anyone who can get it done? Or can point out why I am failing to solve for f?

Usually one asks the question before starting the calculation. What question prompted the calculation?

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Usually one asks the question before starting the calculation. What question prompted the calculation?
Sorry. Same as in my last post really. I was looking for a equation that would solve the frequency in the EM wave equation:

$$B=\textbf{B} \sin [2 \pi ft]$$

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Is there a question?
Let me repeat - is there a question?

Here's what I wrote six months ago:

Seriously, learn to use TeX. Posting microscopic pictures is disrespectful of the time and effort people have put in trying to help you. Use units, like Zz said. Making people guess is disrespectful of the time and effort people have put in trying to help you. Finally, explainnwhat the heck you are trying to do rather than just dumping a big-wall-O-calculations. Doing otherwise is disrespectful of the time and effort people have put in trying to help you.
Making us guess what problem you are trying to solve is extremely disrespectful.

• etotheipi
etotheipi
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$$B=\textbf{B} \sin [2 \pi ft]$$
Maybe I'm out of the loop, but what exactly does this equation mean? It doesn't make sense to me, for starters, the RHS is a vector and the LHS a scalar...

Last edited:
• fisher garry
PeterDonis
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We assume that B goes radially outwards
You can't. It's impossible to have a purely radial magnetic field consistent with Maxwell's Equations, because of ##\nabla \cdot \mathbf{B} = 0##.

I rearrange an emission of EM waves from an electrons surface
It looks like you are trying to make the outgoing EM wave spherically symmetric. It's impossible to have a spherically symmetric EM wave consistent with Maxwell's Equations; the lowest order EM radiation is dipole.

So as far as I can tell, you found an equation with no solution because you made assumptions that are inconsistent.

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It looks like you are trying to make the outgoing EM wave spherically symmetric.
Sure would be nice to be told what the target here is.

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Let me repeat - is there a question?

Here's what I wrote six months ago:

Making us guess what problem you are trying to solve is extremely disrespectful.
Sorry. I will write it out. Ok I could ask: Show me the values for the frequency f that is given in the equation I ended up with in my first post. Here is one attempt:

I did ponder on solving the equation with the charge radius. Since I have been told that the charge radius is where electrons emit light:

$$\frac{4 \pi f \frac{1}{c}+2-[4\pi f \frac{1}{c}]^2R^2}{8 \pi f \frac{1}{c} R - 1}= \frac{1}{tan[4 \pi f \frac{1}{c} R]}$$
$$R=2.82 \cdot 10^{-15}$$

$$\frac{4 \pi f \frac{1}{3 \cdot 10^8}+2-[4\pi f \frac{1}{3 \cdot 10^8}]^2(2.82 \cdot 10^{-15})^2}{8 \pi f \frac{1}{3 \cdot 10^8} 2.82 \cdot 10^{-15} - 1}= \frac{1}{tan[4 \pi f \frac{1}{3 \cdot 10^8} 2.82 \cdot 10^{-15}]}$$

$$\frac {4.2 \cdot 10^{-8} f+2-418.2 f \cdot 10^{-38}}{23.6\cdot 10^{-23}f - 1}= \frac{1}{tan(11.8 f \cdot 10^{-23})}$$
approx:
$$\frac {4.2 \cdot 10^{-8} f+2}{23.6\cdot 10^{-23} f- 1}= \frac{1}{tan(11.8 f \cdot 10^{-23})}$$

$$4.2 \cdot 10^{-8} f+2= \frac{ 23.6\cdot 10^{-23} f- 1 }{tan(11.8 f \cdot 10^{-23})}$$

Is there a solution to this?

Gold Member
You can't. It's impossible to have a purely radial magnetic field consistent with Maxwell's Equations, because of ##\nabla \cdot \mathbf{B} = 0##.

It looks like you are trying to make the outgoing EM wave spherically symmetric. It's impossible to have a spherically symmetric EM wave consistent with Maxwell's Equations; the lowest order EM radiation is dipole.

So as far as I can tell, you found an equation with no solution because you made assumptions that are inconsistent.
Ok. Thanks for feedback :) If I could ask: I have a setup here with EM waves coming out of an object I would believe. I want to use the divergence theorem to find a solution for the frequency. How would the direction of the EM waves be in order to be able to use this setup? Thanks :)

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Show me the values for the frequency f that is given in the equation I ended up with in my first post.
Stop making us guess. Just stop it.

What is the problem you are trying to solve? Not "here's this equation". For example, "A sphere with charge Q is pulsating with frequency f. What is the frequency of the radiation?" (This is Peter's guess, I think)

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Stop making us guess. Just stop it.

What is the problem you are trying to solve? Not "here's this equation". For example, "A sphere with charge Q is pulsating with frequency f. What is the frequency of the radiation?" (This is Peter's guess, I think)  This is taken from introduction to electrodynamics by Griffiths which the rearrangment of the maxwell equations are taken from. I will rewrite what it says there if it is ok:

Suppose we have some charge and current configuration which at time t, produces fields E and . In the next instant, dt, the charges move around a bit. How much work dW is done by the electromagentic forces acting on these charges in the interval dt? According to the Lorentz force law, the work done on a charge is.

So I guess it could be any charge contribution. For example an electron. Peter Donis did use the relation that ##\nabla \cdot \mathbf{B} = 0##. But at the same time electrons emit EM waves from deacceleration for example in hydrogen molecule. So would there not then be a setup for using the divergence theorem then if some EM waves are coming out of the electron? And if so how would the spatial arrangment of the EM waves be?

PeterDonis
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This is taken from introduction to electrodynamics by Griffiths
That question is about work done by EM fields on charges. Is that the question you are trying to answer?

If it is, pretty much everything else you have written in this thread up to now is irrelevant.

• Gold Member
Maybe I'm out of the loop, but what exactly does this equation mean? It doesn't make sense to me, for starters, the RHS is a vector and the LHS a scalar...
$$\frac{d^2B}{dx^2}c^2=\frac{d^2B}{d t^2}$$

$$\textbf{B}\frac{d^2 sin[2 \pi f \frac{x}{c}]}{dx^2}c^2=\textbf{B}\frac{d^2sin[2 \pi f t]}{d t^2}$$

$$-4 \pi^2 f^2 \frac{1}{c^2}\textbf{B}sin[2 \pi f \frac{x}{c}] c^2=-4 \pi^2 f^2\textbf{B} sin[2 \pi f t]$$

So it seems to be a solution? As for what solution one should use of the wave equation. I dont know?

I did once try to prove how there could be multiple solution to this wave equation. But I believe I failed. It could perhaps have helped to identify what solutions that are valid to the wave equation. And which one that could be used.

I would perhaps say that it could add up from a quantification perspective. Since it was assumed that the EM waves where going outwards. One could say that they move with time t as variable given as:

##B=\textbf{B} \sin [2 \pi ft]##

But one could also say that one dont need vector notation because of symmetry so that one instead could use

##B=\textbf{B} \sin [2 \pi f \frac{r}{c}]##

Or?

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But at the same time electrons emit EM waves from deacceleration for example in hydrogen molecule.
No they don't.

Are we back to the personal theory of yours?

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That question is about work done by EM fields on charges. Is that the question you are trying to answer?

If it is, pretty much everything else you have written in this thread up to now is irrelevant.
Where does it fail in the first post. And why does it fail?

PeterDonis
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at the same time electrons emit EM waves from deacceleration for example in hydrogen molecule
A hydrogen molecule is a quantum object. This is the classical physics forum, and this thread is about classical EM, Maxwell's Equations. If you want to ask a question about the hydrogen molecule, you need to start a new thread in the quantum physics forum--and for any such thread, everything else you have written in this thread is irrelevant.

PeterDonis
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Where does it fail in the first post. And why does it fail?
I have already told you where your first post fails: you are making inconsistent assumptions, so of course you get nonsense answers. That is true regardless of what question you are actually trying to answer.

If you want to know why everything you have written is irrelevant to the question about work done by EM fields on charges, that should be obvious: because nothing you have written has anything to do with work done by EM fields on charges. It's all about some inconsistent assumptions about radiation.

• etotheipi
PeterDonis
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@fisher garry I am closing this thread because there is nothing here that can serve as a useful basis for further discussion. I would recommend taking some time to think carefully about what you actually want to know so you can formulate a better question if you want to start a new thread.

• etotheipi