Why a pulsating charged sphere does not emit radiation?

[DaTario]

Hi All,

Consider an elastic sphere which is homogeneously charged and suferring an harmonic inflation and deflation.
Is it correct to explain the theoretical result that it does not emmit radiation, even though all of its charge elements are accelerating, as a systematic effect of destructive interferences occurring in the whole space?

Best wishes,
DaTario

 

[jambaugh]

Yes. Huygen-Fresnel principle at work. You could also point out that in all configurations of the sphere (i.e. all radii) the external E fields beyond the max radius are identical so no time dependence can manifest in the sequence of retarded fields for each configuration.

 

[tech99]

Hi All,

Consider an elastic sphere which is homogeneously charged and suferring an harmonic inflation and deflation.
Is it correct to explain the theoretical result that it does not emmit radiation, even though all of its charge elements are accelerating, as a systematic effect of destructive interferences occurring in the whole space?

Best wishes,
DaTario

The field lines of the charges point radially outwards. The acceleration is along this same direction, so no distortion of the field line occurs due to acceleration. In the case of a wire, the acceleration is at right angles to the field lines, so there is distortion and a transverse, or radiation, component is set up. No transverse component is set up with the sphere.
Had we been talking about sound waves, the sphere would radiate a longitudinal wave, but that does not exist for a free space EM wave.

 

[jambaugh]

Yes. Huygen-Fresnel principle at work. You could also point out that in all configurations of the sphere (i.e. all radii) the external E fields beyond the max radius are identical so no time dependence can manifest in the sequence of retarded fields for each configuration.

I am actually a bit wrong here with regard to the retarded fields. I'm wrong because the retarded field at a distant point will have contributions from the sphere at different times and thus it is not the simple unchanging field I mentioned. The answer should come out the same but the explanation is invalid.

 

[DaTario]

I am actually a bit wrong here with regard to the retarded fields. I'm wrong because the retarded field at a distant point will have contributions from the sphere at different times and thus it is not the simple unchanging field I mentioned. The answer should come out the same but the explanation is invalid.

When you say "a distant point" you are meaning any point which is beyond the max radius, isn´t it?

The field lines of the charges point radially outwards. The acceleration is along this same direction, so no distortion of the field line occurs due to acceleration.

The acceleration you are mentioning here is that of those charge elements located at the same radial line of the points you are studying the field, isn´t it? They seem not to be the only charge elements contributing to the field in the points of this radial straight line.

 

[tech99]

When you say "a distant point" you are meaning any point which is beyond the max radius, isn´t it?
The acceleration you are mentioning here is that of those charge elements located at the same radial line of the points you are studying the field, isn´t it? They seem not to be the only charge elements contributing to the field in the points of this radial straight line.

When you say "a distant point" you are meaning any point which is beyond the max radius, isn´t it?
The acceleration you are mentioning here is that of those charge elements located at the same radial line of the points you are studying the field, isn´t it? They seem not to be the only charge elements contributing to the field in the points of this radial straight line.

None of the points radiates. But if the acceleration were in a direction such that they did, then I agree they would form an array and the field at a distant point would be the vector sum of all of them.
Notice also that if they all radiated, they would be spherically uniform and so constitute an isotropic source, which we are taught does not exist.

 

[Vanadium 50]

This seems needlessly hard. A pulsating sphere has a constant dipole (and higher) moment - no radiation. Take advantage of the symmetry of the problem.

 

[DaTario]

This seems needlessly hard. A pulsating sphere has a constant dipole (and higher) moment - no radiation. Take advantage of the symmetry of the problem.

The theorem of the spherically symmetric distributions may be used, I guess, and it would yield a very simple explanation. The point I am trying to address is less the efficiency and simplicity of this explanation and more its correctness.

Best wishes,
DaTario

 

[DaTario]

None of the points radiates.

This statement doesn´t seem to be valid in general. Suppose a spherically symmetric distribution, centered at the origin. Consider the time dependent tranfomation acting on the points of this distribution:
$$
\left ( \begin{array}{ccc}
1+ 0.3 \sin (2t) & 0 & 0 \\
0 & 1+ 0.3 \sin (2t) & 0\\
0 & 0 & 1+ 0.3 \sin (2t) \\
\end{array} \right ).$$

The charge element located at the positive x-axis will emmit radiation to points at the y - axis.

 

[Vanadium 50]

is less the efficiency and simplicity of this explanation and more its correctness.

You are taking as the starting point "acceleration produces radiation" and trying to show how this is a special case. I am arguing that the better starting point is "a changing multipole produces radiation" and an accelerating point charge is one example of a changing multipole. in almost all cases this is much, much simpler.

 

[DaTario]

You are taking as the starting point "acceleration produces radiation" and trying to show how this is a special case. I am arguing that the better starting point is "a changing multipole produces radiation" and an accelerating point charge is one example of a changing multipole. in almost all cases this is much, much simpler.

Ok. Your argument seems to follow the same line of that hint in calculus books where the authors say: try to solve the tripple integrals figuring out the differential of highest order. I am correct to understand it this way?

 

[jambaugh]

I would add, that considering the ensemble of charges as a whole, when considering superposition (classical or not) it is improper to speak of a part radiating, or not radiating, except when the contributions of the whole are added up. When adding up the contributions one is doing math to solve for the logically necessary field. Once you've solved the math you can say definitively that the system does or does not radiate physically.

 

[DaTario]

I would add, that considering the ensemble of charges as a whole, when considering superposition (classical or not) it is improper to speak of a part radiating, or not radiating, except when the contributions of the whole are added up. When adding up the contributions one is doing math to solve for the logically necessary field. Once you've solved the math you can say definitively that the system does or does not radiate physically.

But don´t you think it is completely justifiable to say that there is interference when there are at least two parts emitting radiation to a common region of space?

 

[Vanadium 50]

Ok. Your argument seems to follow the same line of that hint in calculus books where the authors say: try to solve the tripple integrals figuring out the differential of highest order. I am correct to understand it this way?

I think I am saying the opposite. I am saying if you work with multipoles as the fundamental entity, rather than infinitesmal charges, the calculations get easier. You can avoid the triple integrals.

 

[jambaugh]

But don´t you think it is completely justifiable to say that there is interference when there are at least two parts emitting radiation to a common region of space?

Yes, but I believe that in the examples you have in mind, "is radiating" doesn't change when one is considering either part alone and the whole. But consider the example of a mirror. You can solve for the resulting field by noting that the incident em-wave induces effective currents in the conducting plane as it passes through. Then the conducting plane emits a resulting wave which has a component which adds to and interferes completely with the transmitted wave.

In short you can say that a.) the wave passes through the mirror but also that the mirror emits a counter field, or b.) the wave is reflected. The observed fields it the same and my point is what's empirically meaningful is the prediction of what the observed fields will be, not the way you go about adding up the pieces.

It's a philosophical quibble to be sure and more an issue when considering the quantum cases. But it's an important point for avoiding arguments over which of two different ways of adding up the pieces to get the same result is "what really happens".

 

[DaTario]

I think I am saying the opposite. I am saying if you work with multipoles as the fundamental entity, rather than infinitesmal charges, the calculations get easier. You can avoid the triple integrals.

I understand the way you took my words, and I appologize for not being clear. In fact I used the tripple integrals metaphorically. According to your suggestion it is more interesting to catch a collective entity (which is the multipole) than the punctual charge element. In my metaphor, when calculating the volume of a cylinder it is more convenient to integrate the volumes of disks of infinitesimal heigth (collective entity) than by using the dV = dx dy dz (punctual element), although both methods are correct.

Thank you,

DaTario

 

[rude man]

The field lines of the charges point radially outwards. The acceleration is along this same direction, so no distortion of the field line occurs due to acceleration. In the case of a wire, the acceleration is at right angles to the field lines, so there is distortion and a transverse, or radiation, component is set up. No transverse component is set up with the sphere.
Had we been talking about sound waves, the sphere would radiate a longitudinal wave, but that does not exist for a free space EM wave.

Feynman in his Lectures on Physics gives the formula (21.1) for the total E field produced by a moving charge. There are three terms: the first is the (time-retarded) Coulomb formula, the 2nd has the 1st time derivative of the (time-retarded) unit vector in the r direction (direction of Q to P) and the 3rd has the 2nd time derivative of the same unit vector. Since neither time derivative has a non-zero value for charge oscillations solely along r there is no time-varying E field.

 

[tech99]

Feynman in his Lectures on Physics gives the formula (21.1) for the total E field produced by a moving charge. There are three terms: the first is the (time-retarded) Coulomb formula, the 2nd has the 1st time derivative of the (time-retarded) unit vector in the r direction (direction of Q to P) and the 3rd has the 2nd time derivative of the same unit vector. Since neither time derivative has a non-zero value for charge oscillations solely along r there is no time-varying E field.

Why does the Coulomb field have to be retarded? I understood that for the induction field there was zero net flow of energy, so the propagation delay was zero.

 

[rude man]

Why does the Coulomb field have to be retarded? I understood that for the induction field there was zero net flow of energy, so the propagation delay was zero.

Nothing moves faster than the speed of light. The consequence of this is that the effective distance r' between an observation of E and its occurrence is not the distance at the time of occurrence but at an earlier distance corresponding to where the charge was r'/c earlier. So r' is called the "retardation distance" to distinguish it from the simultaneous distance r.

Cf. http://feynmanlectures.caltech.edu/I_28.html
and http://feynmanlectures.caltech.edu/II_21.html

 

[tech99]

Nothing moves faster than the speed of light. The consequence of this is that the effective distance r' between an observation of E and its occurrence is not the distance at the time of occurrence but at an earlier distance corresponding to where the charge was r'/c earlier. So r' is called the "retardation distance" to distinguish it from the simultaneous distance r.

Cf. http://feynmanlectures.caltech.edu/I_28.html
and http://feynmanlectures.caltech.edu/II_21.html

I do not think a pure magnetic field has a defined velocity. Similarly with a radial electric field. The energy flows out and back in equal measure. This is because there is no net transfer of energy or information. So far as I know, we do not observe a delay between the plates of a capacitor, for instance.

 

[DaTario]

Just to check a point: if the point charge Q oscillates sinusoidally between points A and B, on the y-axis, one is to observe a radiation field at point P located at z-axis (with the z component of point P larger than the y component of point B).
Is it correct?

OBS: the restriction in the parenthesis is to make sense when the point charge is part of the pulsating sphere.

Best wishes.