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Radius increases and decreases periodically (as a pulse).And so does the charges on the surface of sphere.

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I can't get what is gonna happen.the EM waves are produced perpendicularly to motion of the charges,but here where is the perpendicular ? Will the EM waves get produced into the space of sphere?

Also, can two EM waves intersect each other or not? [as Electric fields lines can't intersect.]

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What are the multipole moments of this charge distribution? How do they vary with time? –  Jerry Schirmer Mar 29 '13 at 15:44
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The story goes that Bethe (I think...?) said "Find the answer first, then do the integral." :) In this case you can find the solution without doing any calculation at all, just using symmetry and two of Maxwell's equations. Make sure you understand the reason for the answer before you try to do any fancy integral over dipoles. What do you mean by intersect when you say electric fields can't intersect? You can add two electric fields just fine. Not sure what else you could mean by that. –  Michael Brown Mar 29 '13 at 16:18
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The other contributors so far have tried to give hints without giving away the answer, so I'm going to do the same. This problem is actually a physical description of a hydrogen atom in a superposition of the 1s and 2s states. There is a time-varying charge density in this superposition, and it is pretty much exactly what the problem here describes. –  Marty Green Mar 29 '13 at 19:36
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@MichaelBrown The questioner is probably referring to electric field lines not intersecting. Thus, I would counsel for a thorough review of the fundamentals of electrostatics before grappling with a problem of E&M radiation –  kleingordon Mar 29 '13 at 20:24
    
@exploringnet Can you tell me what the electric field of a static charged sphere is? Which of Maxwell's equations do you use to determine this? –  Michael Brown Mar 30 '13 at 8:51

2 Answers 2

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The electric field due a uniformly charged sphere with its radius increasing at a constant rate at any point outside the sphere is the same as if the radius were not changing at all. In this scenario, you can still invoke the radial symmetry argument and use the Gauss's law along with invariance of electric charge to obtain the electric field which turns out to be the same as that of a point charge at the centre of the sphere.

Any change in the rate of increase of radius of the sphere does not lead to a change in the electric field at any point farther from the sphere than the maximum radius the sphere can attain while it is pulsating. So the electric field at all points whose distance from the centre of the sphere is larger that the maximum radius that can be attained by the pulsating sphere is constant over time. So beyond the maximum attainable radius there is no electromagnetic radiation.

However, I think there will be some electromagnetic radiation in the region between the minimum attainable radius and the maximum attainable radius. This could be in the form of standing waves confined to that region.

At any point whose distance from the centre of the sphere is less than the minimum attainable radius, I think you can again apply the above mentioned logic and conclude that there is to electromagnetic radiation in that region.

So, to conclude, I feel there should be no net electromagnetic radiation emanating from the sphere.

P.S. I have not done a detailed study of electromagnetic radiation. So, I'm not sure if the above analysis is correct. I'd be glad if somebody were to post the correct analysis of the scenario in question.

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You basically got it: the E-field is independent of radius, so there is no change in E-field. By symmetry, there is no B-field tangential to the sphere. Thus the Poynting vector (hence radiation) is zero. –  Chris Gerig Apr 2 '13 at 1:10
    
@Marty NR-Nonradiation condition by Marengo "The simplest example is provided by a pulsating (or even collapsing) spherically symmetric source." –  Helder Velez Jun 28 '13 at 1:16

RADIATION FROM A PULSATING ELECTRICALLY CHARGED SPHERE

Here is a modest approach to find an answer to this problem which I hope will generate some discussion in the right direction, and someone will enrich and extend.

There is no point in rediscovering the wheel, so we can use some well known results from classical electromagnetism. It is known that any electrically charged particle subjected to acceleration radiates EM waves, the power, $P$, of which is given by the equation

$P=\frac{2}{3}\frac{q^2}{c^3}a^2$ --------------------(1)

(see QUANTUM PHYSICS Stephen Gasiorowicz, p25), where

$q$ is the electric charge of the accelerated particle

$c$ the speed of light

$a$ the magnitude of the acceleration of the charged particle (some would like to see it written as $|\vec{\bf a}|$.)

Since all particles move with the pulsating sphere, they move in a synchronised way, they have the same amplitude and frequency of oscillation. Although all particles pulsate the same way, we can write independently the radial displacement of the $k^{th}$ oscillating charged particle on the sphere, with equilibrium radius $R$, as

$r_k=R+A\sin(\omega t)$ where $A$ is the amplitude of oscillation and $\omega=2\pi f$, so that the magnitude of the acceleration is $a=A\omega^2\sin(\omega t)$. Therefore equation (1) becomes

$P_k(t)=\frac{2}{3}\frac{\omega^2A^2q_k^2}{c^3}\sin^2(\omega t)$. --------------------(2)

As there are $N$ particles on the sphere all doing the same thing the above can be summed up for the total power (power is a scalar quantity)

$P_T(t)=\frac{2}{3}\frac{\omega^2A^2}{c^3}\sin^2(\omega t)\Sigma_{k}q_k^2 $. --------------------(3)

The intensity of the emitted radiation at some large distance $r>>R$ from the pulsating sphere will be

$I_T(r,t)=\frac{P_T(t)}{4\pi r^2}$--------------------(4)

From equation (4) we can find the average intensity of emitted radiation over one period of the oscillating charges

$<I_T(r)>=\frac{1}{T}\int_0^TI_T(t)dt$--------------------(5)

or

$<I_T(r)>=\frac{1}{8}\frac{A^2}{\pi c^3r^2}\omega^4 \Sigma_{k}{q^2_k}.$. --------------------(6)

In order to generalise equation (6) to account for a uniform continuous distribution of charges, one must take into account the pulsation of the area element $\Delta A$ of the sphere. Although the electric charge on the area element remains fixed, the electric charge density pulsates. Also, the interactions between neighbouring electric charges are ignored, assuming that the charges are fixed on the surface and are constraint to move only along the radius of the sphere. At the centre of the sphere there might be intense radiation power as all charges emit towards that point too.

I will be happy to be shown errors in the above analysis, other than the ones I have already mentioned, and I hope someone will see how to generalise this to a continuous uniform distribution.

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Unfortunately, this line of reasoning is not correct, even though it is tempting. The Poynting flux is computed from the net E&M fields, accounting for all interference from the fields emitted from all the antennae. As the charge distribution on the sphere approaches one that is continuous and perfectly spherically symmetric, the net flux will go to zero, as indicated in @Charuhas' answer. A finite number of separated charges will produce a flux that is slightly positive but with a much smaller integrated power than indicated in your formula, which has neglected field interference. –  kleingordon Apr 1 '13 at 5:14

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