What is the vector quantity of e-m radiation produced by an accelerated charged particle?

[vitality00]

and also does a charge radiate when it decelerates as well?

[studiot]

1) Check out Poynting Vector

 

https://www.google.co.uk/search?hl=en-GB&source=hp&q=poynting+vector&gbv=2&oq=pointing+vect&gs_l=heirloom-hp.1.0.0i10l10.922.7203.0.10344.13.13.0.0.0.0.203.1360.8j4j1.13.0....0...1ac.1.34.heirloom-hp..0.13.1360.rmNrmHmqAjg

 

 

2) Yes, deceleration is acceleration with a preceding negtative sign

Edited October 2, 2014 by studiot

[swansont]

Bremsstrahlung

 

http://en.wikipedia.org/wiki/Bremsstrahlung

http://www.nt.ntnu.no/users/floban/KJ%20%203055/X%20%20Ray/Bremsstrahlung.htm

 

and also does a charge radiate when it decelerates as well?

 

Yes. The distribution depends on the angle between the motion and the acceleration

[I-try]

Vitality00.

Both Studiot and Swansont are correct regarding radiation resulting when an electron is accelerated. However, there is no measurable radiation resulting from the acceleration of an electron if the acceleration is induced by the gravitational effect.

Also, the vector quantity of e-m radiation would be dependent on the magnitude of the accelerating force instantaneously applied. The radiation referred to is always transmitted perpendicular to the direction of enforced acceleration and at the speed of light.

[swansont]

 

Both Studiot and Swansont are correct regarding radiation resulting when an electron is accelerated. However, there is no measurable radiation resulting from the acceleration of an electron if the acceleration is induced by the gravitational effect.

 

That doesn't sound right.

[studiot]

I don't know what happened to the original first question in the OP.

 

This was largely what I answered, and I considered the second about acceleration subsidiary.

 

This is in classical physics and I don't know if relativity was meant to be taken into account, but that needs to be done for bremstrallung and compton scattering etc.

[imatfaal]

Falling Charges. Charge will radiate under gravitation/acceleration - and there is no potential problem with continual radiation in orbit (which would be kinda difficult to explain away)

 

https://www.scribd.com/doc/100745033/Dewitt-1964

Cecile Morette DeWitt and Bryce S. DeWitt. Falling charges. Physics, 1:3–20, 1964.

 

If you think about it if gravitational acceleration did not cause a charge to emit it would be a problem with equivalance

[robinpike]

Bremsstrahlung

 

http://en.wikipedia.org/wiki/Bremsstrahlung

http://www.nt.ntnu.no/users/floban/KJ%20%203055/X%20%20Ray/Bremsstrahlung.htm

 

 

Yes. The distribution depends on the angle between the motion and the acceleration

 

And from the electron's point of view, I'm taking that the following is correct?

 

The electron is motionless when a nucleus goes whizzing by, causing the electron to move.

 

The closer the nucleus passes the electron, the greater the acceleration on the electron - and it is this 'greater acceleration' that equates to, in the reference frame of the nucleus, the statement: the distribution depends on the 'angle between the motion of the electron' and the acceleration?

Falling Charges. Charge will radiate under gravitation/acceleration...

 

So exactly how does that work?

 

For example, the electron accelerates in a gravitational field, heading towards the centre of mass of the object.

 

At what point, or points, does the electron emit radiation?

 

Looking at it from tiny increments, at each moment, the electron can be considered to be at rest when it receives a tiny acceleration from the gravitational field.

 

The next moment can then be considered, the electron again can be considered to be at rest when it receives another tiny acceleration from the gravitational field.

 

And so on.

 

So on that basis, which tiny acceleration triggers the emission of radiation?

 

 

In fact, I realise the same question can be asked for the example of the electron passing the nucleus.

 

Looking at the electron from moment to moment, at what point does the electron emit the radiation?

Edited October 3, 2014 by robinpike

[swansont]

 

And from the electron's point of view, I'm taking that the following is correct?

 

The electron is motionless when a nucleus goes whizzing by, causing the electron to move.

 

The closer the nucleus passes the electron, the greater the acceleration on the electron - and it is this 'greater acceleration' that equates to, in the reference frame of the nucleus, the statement: the distribution depends on the 'angle between the motion of the electron' and the acceleration?

 

 

From the perspective of electron, it is going to start moving well before it gets to the target nucleus (nuclei) for that source of Bremsstrahlung (e.g. with an x-ray machine). But the nucleus won't accelerate much at all, owing to the mass difference; since the target is usually a macroscopic collection in a lattice, that effectively increases the mass of the target. So that's really moot. I think you need to view this from an inertial reference frame, so you can unambiguously assign the acceleration and velocity vectors. That's what the formula assumes.

 

Looking at the electron from moment to moment, at what point does the electron emit the radiation?

 

I think you'd have to look at what's happening to its electric field as a result of the acceleration, and what wave would be emitted as a result of any radiation, combined with whether energy and momentum could be conserved were a photon to be emitted. I'm not sure if that's sufficient to give the answer; it's an interesting question but I've never worked this out.

[MigL]

I'm not sure I understand you correctly, RobinPike. Are you applying Zeno's paradox to electron motion and radiation ?

[I-try]

Imatfaal.

Thanks for the link you provided in post number 7. Whilst I was unable to understand the mathematics provided by the authors, I was able to understand some of the their argument about which I will not be providing comment. However, I am in agreement with their findings that the radiation resulting from acceleration of an electron by the gravitational effect is too feeble to be measurable. In that regard, perhaps I should have also underlined the word measurable in post number 4.

[imatfaal]

Imatfaal.

Thanks for the link you provided in post number 7. Whilst I was unable to understand the mathematics provided by the authors, I was able to understand some of the their argument about which I will not be providing comment. However, I am in agreement with their findings that the radiation resulting from acceleration of an electron by the gravitational effect is too feeble to be measurable. In that regard, perhaps I should have also underlined the word measurable in post number 4.

 

That article was from half a century ago - I wonder if in those years we have been able to measure it, or at least we could measure it in certain circumstances. My consternation was due to the implication that no radiation is emitted - and the, admittedly subtle but important difference, between 'too weak to be measurED' and 'too weak to be measureABLE' .

 

I wonder if - for instance - the Rossi-Hall experiment could be so refined that a significant difference could be measured between the results as predicted without the influence of this additional factor and the experimental observations; the difference being that which would be predicted by the additional retarding influence.

I suppose the number of muons reaching the surface should be reduced - the emission of radiation will rob them of KE and decelerate them; thus they will be less dilated from the earth's reference frame (or alternatively there will be less contraction of the atmosphere from their frame) and they will simply cover the distance at a slower rate. Thus the time as viewed from surface or the distance as viewed from muon will both be changed as will the speed from both perspectives - all these factors will mean that more muons decay in the atmosphere than would otherwise be expected.

[studiot]

 

That article was from half a century ago - I wonder if in those years we have been able to measure it, or at least we could measure it in certain circumstances. My consternation was due to the implication that no radiation is emitted - and the, admittedly subtle but important difference, between 'too weak to be measurED' and 'too weak to be measureABLE' .

 

That difference is too subtle for me, please explain.

 

Here is a quote from Clemmow

 

 

It must also be noted that the repulsive electromagnetic force between two protons at rest exceeds the attractive gravitational force between them by a factor of 10³⁶ and that the proton mass is 1836 times the electron mass.

 

 

So the experimental implications of excluding electric accelerations on electrons are that we must measure to 1 part in 10³⁹.

[imatfaal]

 

That difference is too subtle for me, please explain.

 

Here is a quote from Clemmow

 

 

 

So the experimental implications of excluding electric accelerations on electrons are that we must measure to 1 part in 10³⁹.

 

The explanation was in the previous sentence - may be "too weak to have been measured yet" would be more obvious.

 

We must be able to measure with extreme accuracy undoubtedly - why though are you comparing with electromagnetic force? Sure the force between two protons is dominated by the EM - but EM is cancelled out whereas Gravity is always additive; ie we are not talking about an electron being gravitationally and em'ly attracted to a proton, rather any charged particle being attracted to a massive body that is close to neutral (ie compared to its enormous mass its charge is close to balanced)

[studiot]

I was comparing the relative strengths of the accelerating forces because almost everywhere we look and find charges being accelerated, they are accelerated (principally) by electromagnetic means.

So before we can attribute any measurement of radiation induced by gravitational acceleration of a charged particle, we have to eliminate electromagnetic effects.

 

I do not know of any measurement system now or in the foreseeable future that is accurate to 1 part in 10⁴⁰.

[I-try]

I was comparing the relative strengths of the accelerating forces because almost everywhere we look and find charges being accelerated, they are accelerated (principally) by electromagnetic means.

So before we can attribute any measurement of radiation induced by gravitational acceleration of a charged particle, we have to eliminate electromagnetic effects.

 

I do not know of any measurement system now or in the foreseeable future that is accurate to 1 part in 10⁴⁰.

I am in complete agreement with the information supplied by Studiot in the quote above. In that regard, perhaps those able to do so should examine the concepts and mathematics provided by the authors of the paper located in the link provided by Imatfaal in post number 7.

If the mathematics cannot be faulted and the concepts drawn are in conformity to the mathematics, then that is an example of the astounding precision of mathematics and the commendable ability of the authors of that paper.

[imatfaal]

I was comparing the relative strengths of the accelerating forces because almost everywhere we look and find charges being accelerated, they are accelerated (principally) by electromagnetic means.

So before we can attribute any measurement of radiation induced by gravitational acceleration of a charged particle, we have to eliminate electromagnetic effects.

 

I do not know of any measurement system now or in the foreseeable future that is accurate to 1 part in 10⁴⁰.

 

 

I make the difference about 10^12 for an electron at distance of geostationary orbit - widely dependant on what the actual net charge of mother earth actually is. Still far too subtle at today's technology status

[robinpike]

I'm not sure I understand you correctly, RobinPike. Are you applying Zeno's paradox to electron motion and radiation ?

 

No that was not the intention.

 

If I am understanding this correctly, when a free electron is accelerated, it will emit radiation.

 

The query is around at what point does the electron emit the radiation? Does it emit a 'sizeable' photon after a certain period of time under acceleration? Or does it emit a 'continuous stream' of very weak photons of the smallest energy?

 

If the observation is the former, then there is the question as to how the electron is 'remembering' the amount of acceleration it has recently received, in order to emit the photon equivalent to that 'cumulative acceleration'?

 

I know that in the atom, the radiation is emitted as a single photon, corresponding to the difference in the start and next orbital the electron moves to.

 

But here the query is around the acceleration of a free electron, another example being that of an electron spiralling in a magnetic field.

Edited October 6, 2014 by robinpike

[studiot]

 

If I am understanding this correctly, when a free electron is accelerated, it will emit radiation.

 

The query is around at what point does the electron emit the radiation? Does it emit a 'sizeable' photon after a certain period of time under acceleration? Or does it emit a 'continuous stream' of very weak photons of the smallest energy?

 

If the observation is the former, then there is the question as to how the electron is 'remembering' the amount of acceleration it has recently received, in order to emit the photon equivalent to that 'cumulative acceleration'?

 

I know that in the atom, the radiation is emitted as a single photon, corresponding to the difference in the start and next orbital the electron moves to.

 

But here the query is around the acceleration of a free electron, another example being that of an electron spiralling in a magnetic field.

 

 

 

This is where you need to clearly distinguish between the predictions of QM and classical electrodynamic wave theory (EWT).

 

It is classical EWT that predicts the electromagnetic emission from accelerating charged particles. As such there is no quantisation, the waves are freely emitted. E and H fields at are generated at right angles to each other such that

Where E and H are both proportional to the acceleration (a) and the charge (e) and inversely proportional to the distance from the charge ®

[math]E,H \propto \frac{{ae\sin (\varphi )}}{r}[/math]

 

phi is the angle made between the direction r is measured in and the direction of acceleration.

 

According to this model, the electric and magnetic fields are always perpendicular to the direction of propagation and move with velocity c.

Further because of the sine term they vary in intensity from zero in the direction of propagation to a maximum at right angles to it.

 

I stress again there is not quantisation of these fields.

 

The above is OK for charge speeds small compared to c but needs to be modified for speeds a significant fraction of c.

This leads to an apparent energy and mass gain and quantisation effects.

 

You might also find the following non mathematical explanation as to why a single accelerating charge should classically radiate EM waves.

 

Consider a single charged particle travelling between A and B.

The faster it goes the greater the steady current it is equivalent to.

So if it is accelerating then that current is not steady but increasing.

The magnetic induction is proportional to the rate of change of current.

So as the particle accelerates, the equivalent current increases, which in turn causes a changing magnetic field.

A changing magnetic field gives rise to a (changing) electric field.

A changing electric field gives rise to a changing magnetic field.

 

And so the wave is born.

Edited October 6, 2014 by studiot

[robinpike]

 

 

 

 

...the waves are freely emitted...

 

I stress again there is not quantisation of these fields.

 

...You might also find the following non mathematical explanation as to why a single accelerating charge should classically radiate EM waves.

 

Consider a single charged particle travelling between A and B.

The faster it goes the greater the steady current it is equivalent to.

So if it is accelerating then that current is not steady but increasing.

The magnetic induction is proportional to the rate of change of current.

So as the particle accelerates, the equivalent current increases, which in turn causes a changing magnetic field.

A changing magnetic field gives rise to a (changing) electric field.

A changing electric field gives rise to a changing magnetic field.

 

And so the wave is born.

 

Thanks, so if I am understanding this correctly, multiple electrons undergoing the same acceleration (under exactly the same conditions) emit equivalent photons and the same number of those photons?

 

And the rate of production of the photons and their wavelengths is related to how quickly the change in the magnetic and electric fields occurs due to the accelerating electron.

 

So it is not the electron 'that remembers' from moment to moment what acceleration it has undergone - it is the increase in the magnetic and electric fields that contains the 'preparation of the creation of the photon' from moment to moment?

 

If that is a valid way of looking at this, it resolves my query - thanks.

[studiot]

Classical EM theory is not quantised so any mention of photons is outside the scope.

 

Discretisation of charge is admissible in classical theory since ions were included.

The quantity called electric current' is carried by unspecified charge carriers and is considered as continuous stream of such carriers.

 

This is equivalent to saying that the fluid elements are small enough to be considerd point masses for the purpose of fluid mechanics, so the mechanics of fluids is the mechanics of a continuum.

 

Classical electrical theory including electric and magnetic fields and EM waves is also a branch of continuum physics.

 

Waves are, of course, the bridge between QM and classical physics because they show mathematically how discretised phenonmena can arise naturally in a continuum.

 

I don't think it is possible to discuss energy flows without some more sophisticated mathematics.

Edited October 7, 2014 by studiot

[swansont]

 

Thanks, so if I am understanding this correctly, multiple electrons undergoing the same acceleration (under exactly the same conditions) emit equivalent photons and the same number of those photons?

 

And the rate of production of the photons and their wavelengths is related to how quickly the change in the magnetic and electric fields occurs due to the accelerating electron.

 

So it is not the electron 'that remembers' from moment to moment what acceleration it has undergone - it is the increase in the magnetic and electric fields that contains the 'preparation of the creation of the photon' from moment to moment?

 

If that is a valid way of looking at this, it resolves my query - thanks.

 

Since you're basically anthropomorphizing the electron, the approach is probably not on solid ground.

 

We know that driving an electron at some frequency will produce radiation at that frequency, so I suspect that for an arbitrary acceleration, the radiation will be at whatever Fourier components are present.

[imatfaal]

 

Since you're basically anthropomorphizing the electron, the approach is probably not on solid ground.

 

We know that driving an electron at some frequency will produce radiation at that frequency, so I suspect that for an arbitrary acceleration, the radiation will be at whatever Fourier components are present.

 

I thought cyclotron radiation was at a frequency the same (the centre of the spectrum) as the orbital frequency

 

edit - sorry I thought you said half that frequency; I'm going mad

[swansont]

 

I thought cyclotron radiation was at a frequency the same (the centre of the spectrum) as the orbital frequency

 

edit - sorry I thought you said half that frequency; I'm going mad

 

I was thinking of radios, but cyclotron works, too.

[Bluemoon]

I'm not sure that I really understand this matter.

Does this all mean that if someone measures the weight of an electron in the 'rocket experiment', that its weight will be measured to be

the conventional rest mass if the rocket is static on the launchpad but

weigh more when the rocket is acelerating at 1g in Deep space?