# From Poynting vector to EM wave

1. Dec 15, 2013

### simoncks

After reading Griffiths' book, I find the explanation from poynting vector to sinusodial EM wave pretty blurred.

EM wave has two requirements.
1. E / B = c
2. E and B are sinusodial

How could we ensure the Poynyig vector ultimately will produce the form satisfying the two points mentioned? Or there is other form of field energy transfer? Like non-sinusodial wave?

2. Dec 15, 2013

### mikeph

I don't understand the question. How do you believe a Poynting vector (which is a property of an EM field) is used to produce an EM wave?

The free space EM wave equation comes straight from Maxwell's equations.

3. Dec 15, 2013

### simoncks

My question is -
The EM wave must come from something, which then propagates in free space. But, the Poynting theorem only tells the flow of energy which is EM wave, while Maxwell's equations (in most books) explain how EM wave propagates.
Is there any proof / linkage between the propagation and the origin of EM wave?

4. Dec 15, 2013

### WannabeNewton

What does your question even have to do with Poynting's theorem and Poynting vectors? Maxwell's equations describe the propagation of EM waves in free space and in materials. In simple cases you can understand the production of EM waves as coming from the "kinks" in the electric field due to time varying dynamics of a charged system possessing acceleration: http://web.mit.edu/jbelcher/www/java/plane/plane.pdf

5. Dec 15, 2013

### simoncks

If there is a decelerating charge, there should be Poynting vector. If I take the limit where distance is infinity, should I get a sinusoidal EM wave?
For all cases where there is Poynting vector, should the limit of all cases tends to sinusoidal EM wave only?
I mean is there a theorem to guarantee it?

6. Dec 15, 2013

### mikeph

I still don't understand the questions, you need to describe them in more detail, for example:

"For all cases where there is Poynting vector, should the limit of all cases tends to sinusoidal EM wave only?"

Should what limit? What cases? The Poynting vector is ExH, that's it. Poynting's theorem is an energy conservation equation.

7. Dec 15, 2013

### Staff: Mentor

If I understand what you are asking then the answer is "no". You can have non sinusoidal EM waves with perfectly valid Poynting vectors. The Poynting vector is not related in any way to a sinusoidal EM wave.

You seem to have an idea that there is some requirement for sinusoidal waves in EM. There is not. They are a mathematical convenience, not a physical necessity.

8. Dec 15, 2013

### simoncks

To mikeph
I mean if I take the any Poynting vector a limit where distance from source tends to infinity, should I always get sinusoidal EM wave? It seems not the case by DaleSpam.

To DaleSpam
So, is there any other form of EM wave in nature? Like a Gaussian distribution pulse or something like that? If I want to have a mathematical model about the generation of EM wave, should I still take Poynting vector the limit infinity distance from source? I could not take the limit for E or B fields, since their surface flux with infinitely far sheet (with same solid angle) is basically zero.

9. Dec 15, 2013

### simoncks

Sorry, I think the flux thing has nothing to do with our discussion...

10. Dec 15, 2013

### WannabeNewton

Maxwell's equations in vacuum reduce to two decoupled wave equations, one for the electric field and one for the magnetic field. The solutions to these are superpositions of plane waves; in other words, arbitrary solutions to said equations can be Fourier decomposed in terms of plane waves. Plane waves are sinusoidal but a general Fourier decomposition in terms of plane waves need not be sinusoidal. An example of such a Fourier decomposition, in particular for an electric field solving the vacuum Maxwell equations, is $\vec{E}(\vec{x},t) = \int_{-\infty}^{\infty} E_k e^{i(\vec{k}\cdot x - \omega t)}d^{3}k$ where $\vec{k}$ is the wave 3-vector. You seem to be very confused about the basic formalism of EM.

EDIT: How much of Griffiths did you read?

Last edited: Dec 15, 2013
11. Dec 16, 2013

### simoncks

The EM wave in free space could be directly formulated from Maxwell's equations. Simply solve -
2E = μ0ε0t2E
2B = μ0ε0t2B

My question might not be precise enough, sorry for that. Please let me explain in a more detailed manner. (I am not used to the symbol typing, so the equations might be difficult to read)

Let E be the electric field produced by a distribution of charge, and B be that of magnetic field. By fourier transform, E and B could be expressed by a linear combination of ei(kr-ωt).
El = ƩClei(klr-ωlt)
Bm = ƩCmei(kmr-ωmt)

If I simply plugged the fourier-ed E and B into the two equations above, it must be valid since it must satisfy Maxwell's equations. But, the plane wave should propagate with E and B (same frequency) perpendicular to each other, in the direction E X B (due to my limited knowledge, it should be the case).
1. Do we just match E and B with same frequency, and say they would form a plane wave? Does it ensure the perpendicular requirement?
2. The solution of plane waves shows E / B = c. Is it guaranteed by the matching? i.e. Cl / Cm = c for all l = m?
3. How about the cross terms? i.e. l ≠ m

EM wave 'has' energy, and the propagating direction of EM wave is -
Ei X Bi
where i is the i th term after fourier transform

Meanwhile, the Poynting vector is given by -
E X B / μ
which represents the energy flow

4. Could the EM wave's energy density (by counting number of EM wave in a unit volume) and propagation related to Poynting vector? Because EM wave represents the energy 'escaping' from the source.
5. If they are related, how could we extract EM wave from Poynting vector?

My idea is to take the Poynting vector far away from the source. Since EM wave represents the same thing as that of Poynting vector, the 'motion' of energy (defer by a constant c), they should agree with each other at distance far away. But I am not sure whether my concepts of Poynting vector and EM wave are wrong.

I scan through the whole book two years ago. I am sorry for raising stupid questions.

Last edited: Dec 16, 2013
12. Dec 17, 2013

### Staff: Mentor

Yes. In fact, since no EM wave can possibly be infinite in duration there actually is no example of a pure sinusoidal wave in nature.

I don't understand your question. The mathematical model is Maxwell's equations. You can derive the Poynting vector as well as the fields from Maxwell's equations.

13. Dec 17, 2013

### Staff: Mentor

I am not sure that I understand your concern, but if you decompose E and B onto an arbitrary set of basis functions there is no guarantee that the individual components will satisfy Maxwell's equations.

For example, if you have a given E and B field then you can solve for the sources required to create that field. It is possible to choose a basis such that those sorces are non-zero for every element in the basis. Such a basis element cannot represent a wave in free space, by definition. However, as long as all of the basis sources sum to zero for a given expansion, then the basis is still a valid expansion of a wave in free space.

14. Dec 18, 2013

### simoncks

Most existing theorems only describe EM wave propagating, but how are they generated? What most textbooks say - 'they come from radiation source', which I am not satisfied without the explicit theorem for generation of EM wave

15. Dec 19, 2013

### Staff: Mentor

Again, that is Maxwells equations. The "propagating" equation you are talking about is only for vacuum, I.e. with currents and charges removed. If you put the current and charge distributions back into Maxwells equations then you can describe the generation of EM waves.

16. Dec 19, 2013

### Staff: Mentor

Any intermediate-level (or above) E&M textbook e.g. Griffiths derives the radiation from at least a dipole antenna (one of the simplest antennas to analyze). A Google search on something like "dipole antenna radiation" should lead you to something useful.

Beyond that, there are whole textbooks on antenna theory alone.

17. Dec 20, 2013

### simoncks

Thank you very much. The links and replies are useful!

18. Dec 20, 2013

### vanhees71

It's a pretty important issue, introducing the technique of Green's functions in relativistic field theory (which is utmost important in the modern quantum field theories).

Let's take the simplest case of a located charge-current distribution $\rho(t,\vec{x}),\vec{j}(t,\vec{x})$. The homogeneous Maxwell equations lead to the introduction of the scalar and the vector potential,
$$\vec{E}=-\frac{1}{c} \partial_t vec{A}-\vec{\nabla} \Phi, \quad \vec{B}=\vec{\nabla} \times \vec{A}.$$
You are free to "choose a gauge", because $\vec{A}$ is only determined up to the gradient of a scalar field. The remainder of the derivation becomes most simple, choosing the Lorenz gauge, imposing the condition
$$\frac{1}{c} \partial_t \Phi+\vec{\nabla} \cdot \vec{A}=0$$
on the potentials.

Plugging this into the inhomogeneous Maxwell equations, leads to
$$\Box \Phi=\rho, \quad \Box \vec{A}=\frac{1}{c} \vec{j} \quad \text{with} \quad \Box=\frac{1}{c^2} \partial_t^2-\Delta.$$
Now the scalar potential and the components of the vector potential separate thanks to the Lorenz-gauge condition, and we need to bother only with the solution of the inhomogeneous wave equation. We also need to take into account the proper boundary conditions. What we want is to describe waves going out from the charge-current distribution, and the fields most depend only on these distributions at times in the past of the present time, which leads to the socalled retarded potentials.

Instead of solving the pretty formal equation for the Green's function,
$$\Box G(t,\vec{x};t',\vec{x}')=-\delta(t-t') \delta^{(3)}(\vec{x}-\vec{x}'),$$
we can use a short cut by making a physical argument, knowing that electromagnetic-wave signals move with the universal velocity of light. This leads to the assumption that the scalar potential is given by Coulomb's Law, but for a charge distribution, taken at the "retarded times" for each point of the charge distribution, i.e.,
$$\Phi(x)=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' \frac{\rho(t-|\vec{r}-\vec{r}'|/c,\vec{r}')}{4 \pi |\vec{r}-\vec{r}'|}.$$
Correspondingly you find
$$\vec{A}(x)=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' \frac{\vec{j}(t-|\vec{r}-\vec{r}'|/c,\vec{r}')}{4 \pi c |\vec{r}-\vec{r}'|}.$$
You can check by applying the d'Alembert operator to these expressions that these integrals indeed solve the inhomogeneous wave equations. In this check you must make use of the continuity equation, i.e., the conservation of electric charge, which is a consistency condition for the Maxwell equations, ensuring gauge invariance:
$$\partial_t \rho + \vec{\nabla} \cdot \vec{j}=0.$$

The Poynting vector gives the energy flow of the electromagnetic field and is not directly needed in the derivation of the fields from the charge-current distribution, but you can calculate the energy flow using it, as soon as you have found the fields.