[Bored chemist]

Does that mean that shaking a magnetic monopole about (if it existed) would emit radiation like shaking an electric charge does?

[paul cotter]

I can't answer that one, BC. The generation of waves, in simple everyday terms, arises from loops of the E and H fields mutually creating each other and thus propagating at c. A monopole would throw a spanner in the maths where divH is required to be =0. Maybe someone smarter could hazard a guess?

[Eternal Student]

Hi.

..one has to remember that the magnetic field is simply the electric field as seen from a different frame of reference...

     Well said,  I very nearly added that to the list of things @alancalverd has probably said to other people.

I never could do the maths, but I don't care.

    You've made a good argument without the Mathematics.

     LaTex still isn't working on the forum.    However, there's a good derivation for the e-m radiation that must arise in the far field due to an oscillating electric dipole  here:     

It's 24 minutes long and maybe nicer than working through the mathematics entirely on your own.   Sadly, I don't think this You Tuber went further and made a video for section 9.5 of Griffith's  which should be an oscillating magnetic dipole - but it's a reasonable template of how you would try and do the maths.   It's completely un-necessary to watch the video but there's at least two people in the world who might care about the maths.

I have looked up the derivation of the wave equation via the Laplacian of the E or H field in phasor form and it is completely symmetric: a time variation in either produces an em wave.

    Thank you.

- - - - - - - - -
    I think quite a few of us  (maybe 3 people and that's about half the forum) recognise that a rotating magnet must emit e-m radiation.

If the magnet falls over in a non-existent forest with no observers, does it matter?

   Hmm...  I've often wondered,  if I'm alone in the woods and my wife and daughters cannot hear me, am I still wrong?

A monopole would throw a spanner in the maths where divH is required to be =0.

    You've already given the smart answer.   Maxwells equations will not permit a magnetic monopole, so we can't use them and expect to get a sensible answer.   As far as I know,  quantum theories that do predict magnetic monopoles will have them behave much like an electric charge.

Best Wishes.

[alancalverd]

The force acting on the needle is electromagnetic.

It's certainly magnetic, but where is the E field? 

[alancalverd]

Pushing electrons to and fro in an antenna is cyclical (Particularly if the antenna is resonant.)

but doesn't involve any rotation.

[paul cotter]

∇xE=-μ partial time derivative of H. Thus a changing magnetic field give rise to an electric field with the property of circulation and that E field give rise to a H field with circulation and thus propagation.

[Bored chemist]

Pushing electrons to and fro in an antenna is cyclical (Particularly if the antenna is resonant.)

but doesn't involve any rotation.

And had anyone said it should?

[Jaaanosik]

Starting from the beginning, a rotating magnetic dipole  (or electric dipole) should generate e-m radiation.

I think not. You need to accelerate a charge to generate a photon. So it is entirely possible that a spinning neutron star could excite radiation from any gas cloud, plasma or passing ion, but a magnet in an infinite vacuum won't. 

What do you mean 'to accelerate a charge to generate a photon'?
How does it work?

[alancalverd]

So let's have a 1 tesla bar magnet spinning at 1000Hz in a vacuum. Can anyone please describe its electromagnetic emission spectrum? I just can't visualise it!

[Jaaanosik]

So let's have a 1 tesla bar magnet spinning at 1000Hz in a vacuum. Can anyone please describe its electromagnetic emission spectrum? I just can't visualise it!

I suspect the spectrum will be frame dependent.
It depends who is 'looking'.

[Bored chemist]

So let's have a 1 tesla bar magnet spinning at 1000Hz in a vacuum. Can anyone please describe its electromagnetic emission spectrum? I just can't visualise it!

Can you visualise the emission if you had an electric dipole rotating  at that rate?
(Say I have an insulating  horizontal disk with two metal balls on the perimeter on either end of a diameter and one is positively charged and the other negatively.) I rotate the disk at 1000 Hz.


Failing that, what about a (horizontal) dipole antenna being fed with 1000 Hz at the centre?

From the side, the disk looks like the dipole.
There's a slight complexity from the movement towards and away from you.

But if you consider two dipoles at right angles fed with two  signals (both at 1KHz) and pi/2 radians out of phase, the emission will look like the pair of charges on a turntable (or a rotating dipole).

I'm pretty sure that the emission pattern from a magnetic dipole would be the same (give or take a phase/ polarisation change)

[Jaaanosik]

...
The carrier of the EM force is the photon.
So, in this case, I know that a rotating magnet emits EM radiation. That force is periodic and thus there must be a component of the EM radiation with the same period.So it must have photons with energies corresponding to the frequency of rotation of the record player.


Re. "but a magnet in an infinite vacuum won't. "
If the magnet falls over in a non-existent forest with no observers, does it matter?

I contend that it will still emit EM radiation.
Imagine that we somehow have a magnet in an infinite empty universe.
We set it spinning (We know it is spinning, because an ant who happens to be standing on it notices the centrifugal effect).

After a while, and at some distance from the magnet, we magically call a compass needle into being.

Does the needle have to "wait" for EM radiation from the magnet to reach it, or is that changing field already there?

I can't see how it would so I think the magnet must have been emitting EM radiation all along.

But... if there's EM radiation, then there are photons.
And, if there are photons, the universe isn't empty.
So the solution may be that you can't have a rotating magnet in an empty universe.

All seems a bit esoteric..
But let's ignore nearly the whole of the universe and consider some hydrogen atoms.
Some of them have the magnetic dipoles of the electron and the proton aligned parallel, and in others it's antiparallel.
And if one happens to flip from the first state to the second, it emits a photon of about 21 cm wavelength.

That photon crosses space and is picked up many years later by a detector here on earth.

But we have only been constructing such detectors for about 100 years.
So, for a source more than 100 light years away, the detector had not been built when the photon was emitted.

I think that's close enough to " we magically call a compass needle into being." for the analogy to work.

An electron- with a magnetic diploe moment- was flipped and sent out EM radiation. It did so in a universe in which the detector did not exist.
You may say that, without the proton of the hydrogen atom, the energy of the photon would be undefined. Which is a fair point
But, at that point, I think you need to be able to do Laplace transforms and, as I said, I can't do the maths.

Apparently there are virtual photons that enable fields, wave propagation, potential gradients, ...
... and there are photons, energy wave packets, energy quanta changes, ...

The hydrogen 21cm photon is the quanta change.
The hydrogen atom as a proton-electron system has a binding energy.
When the 21cm photon flies away then the electron is bound to the proton with stronger force.
The amount of energy that left the system means that exactly this much more energy is going to be required for electron to fly away from proton when compared to the electron state prior to the flip and prior to the 21cm emission.
The same logic why quantum number 1 requires more energy to break electron free than higher quantum number.

Photons are tricky.

[Eternal Student]

Hi.

What do you mean 'to accelerate a charge to generate a photon'?
How does it work?

    I'm not alancalverd....   but one way you could do this is with a device called a cyclotron.    Put a charged particle like a proton in to a cyclotron with some initial velocity (and remember to switch the magnetic field generator on).   Then e-m radiation will be emitted.

So let's have a 1 tesla bar magnet spinning at 1000Hz in a vacuum. Can anyone please describe its electromagnetic emission spectrum? I just can't visualise it!

       I guess you couldn't find Griffith's introduction to electrodynamics then.   I don't have a copy of it either, which is a bit shameful, I may get one for Christmas if I'm lucky.
    Page 150 and 151   of this  online reference will suffice:
https://www.damtp.cam.ac.uk/user/tong/em/el5.pdf

   That describes the far-field with quite arbitrary motion allocated to the magentic dipole  m(t).     Very close to the magnetic dipole many of the approximations used during the derivation would be poor,  so the near-field can be quite different.

   Is LaTex working yet?   I'm going to try it and post to see how it looks....


If that's not working we'll be down to words.
Late Editing:   A subsequent post was written and then removed.  It'll take an hour to cover stuff without too many formulae, check it all through and try to avoid having too many errors  etc.

Best wishes but will be back shortly.

[Jaaanosik]

Quote from: Jaaanosik on Yesterday at 20:37:01

    What do you mean 'to accelerate a charge to generate a photon'?
    How does it work?

    I'm not alancalverd....   but one way you could do this is with a device called a cyclotron.    Put a charged particle like a proton in to a cyclotron with some initial velocity (and remember to switch the magnetic field generator on).   Then e-m radiation will be emitted.

Adding energy to a charged particle, acceleration in a straight line, does not generate a photon.
Changing trajectory direction of previously accelerated charged particle is the causality of the photon emission.
https://en.wikipedia.org/wiki/Bremsstrahlung
This is exactly 'flip of the charged particle spin', similar to 21cm hydrogen line spin-flip.
The magnetic field is only the tool that is causing the change of charged particle trajectory direction.
The charged particle spin-flip itself is the cause of the photon emission.
The charged particle is a system, photon (quanta energy change) is leaving the charged particle meaning the charged particle is losing energy, 'breaking'.

[Eternal Student]

Hi.

Adding energy to a charged particle, acceleration in a straight line, does not generate a photon.

    I'm fairly sure it would generate some photons.
The link that you provided gives a formula for the power radiated when acceleration is parallel to the velocity of the charged particle. 

When any charged particle (such as an electron, a proton, or an ion) accelerates, energy is radiated in the form of electromagnetic waves.
  -  Taken from a slightly different Wikipedia page.   https://en.wikipedia.org/wiki/Larmor_formula

    The only minor issue that will arise is whether all of this oscillation in E and B fields would always be recognisable as "a photon", a discrete packet of energy.   Classically, it's an oscillation in the E and B fields but it should be a continous emission rather than something that occurrs in bursts at discrete places and times.

Best Wishes.

[paul cotter]

Hi, ES. I took a chance at using "del" and "mu", hoping the forum software did not mangle them, as it does with other symbols. I promptly logged out and then logged in again to see if I still had a meaningful post and it was correct. When I joined this forum originally and complained that my keyboard did not have math capability you said to use LaTex. I had no clue what you were talking about and I should have asked- to me a latex is an emulsion produced as a secretion by various plants!

[Jaaanosik]

Hi.

Adding energy to a charged particle, acceleration in a straight line, does not generate a photon.

    I'm fairly sure it would generate some photons.
The link that you provided gives a formula for the power radiated when acceleration is parallel to the velocity of the charged particle. 

When any charged particle (such as an electron, a proton, or an ion) accelerates, energy is radiated in the form of electromagnetic waves.
  -  Taken from a slightly different Wikipedia page.   https://en.wikipedia.org/wiki/Larmor_formula

    The only minor issue that will arise is whether all of this oscillation in E and B fields would always be recognisable as "a photon", a discrete packet of energy.   Classically, it's an oscillation in the E and B fields but it should be a continous emission rather than something that occurrs in bursts at discrete places and times.

Best Wishes.

This comes back to 'who is looking'.

What is a straight line acceleration in one inertial frame is a curved acceleration in other moving frames.

The rest frame of the plates, in the middle, an electron accelerates in a straight line from the middle of the blue plate, cathode, to the middle of the red plate, anode.
The K'1 frame in which the plates are moving to the left,
and K'2 frame in which the plates are moving to the right,
the end-result is curved trajectory accelerations in K'1, and K'2 frames.

[hamdani yusuf]

This comes back to 'who is looking'.

Interesting.
Is a stationary charged particle release photons when seen by an inertially moving observer?
What about a stationary magnet?

[Eternal Student]

Hi.

you said to use LaTex. I had no clue what you were talking about and I should have asked-

     It's a markup language especially intended for mathematical symbols.
You may already have noticed that if you want to underline something in this forum,     a set of tags appears around what you were trying to underline:

Code: [Select]

To  [u]underline[/u]  you enclose the start and end tags   [u]  and [/u]  around the thing you want to underline.
   That displays like this:
To  underline  you enclose the start and end tags     and   around the thing you want to underline.
You can see how the tags have been removed and would just become a long underline  around  the word  "and".

You could also use tags that mark the start and end of a block of  LaTeX  code   (The capital letters in the acromym LaTeX, are supposed to be as shown,  some capitals here and there although I often write LaTEX).

Code: [Select]

the  [tex]   and  [/tex] tags mark the start and end of the LaTeX code block.

[tex] \sum \limits_{i=1}^n  {3i}   [/tex]

Inside of the LaTeX code block  you can use code that signifies or "marks up" the symbols you want.
The bit of code above would have produced  a Sigma symbol for a sum over an index  i  from 1 to n  of terms 3i.
   
It would have looked like this:



At the moment, whoever last edited, or updated the forum software may not have realised how LaTeX is supposed to work.   I don't honestly know what happend.    The buttons that appear along the top while you're creating a forum post all work.... so  it just so happens you can press a button to get precisely what I've shown above....   You can get the sum from i =1 to n    of terms  3i.     However,  if you change any of that,  for example to display the sum from i =1 to 2n  then it displays  nothing.... well  a numerical code linking to a  gif image that the computer never generates.

Code: [Select]

This should be the sum from  i=1 to 2n  of terms 3i.

[tex] \sum \limits_{i=1}^{2n}  {3i}   [/tex]

This is what it produces:


If you can see what I mean... they may have thought.... "ahh... I have to get all the buttons to work".....  instead of  "ahh.... I'm supposed to support the full use of LaTEX markup language".
    In the old days we had full LaTeX support,  now we have 8 buttons for just 8 things you can display.

Best Wishes.

[paul cotter]

I have to admit to being virtually computer illiterate.