Rutherford's model

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  • #2
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Electrons circling in a wire don't radiate, as OP says. Rutherford atomic model had an electron circling the proton - not in a wire, but held in orbit by attraction of charges. This model was "rejected by physicists" as debajyoti datta says. Partly because it should radiate, lose energy, and dive into the proton, in about 10^-8 seconds. So debajyoti datta wonders why this conclusion holds in the atom model case when it doesn't hold in the wire case.

There are (I think) a number of possible answers, here's one. The Rutherford model has only one electron (obviously I'm talking of Hydrogen). So, in terms of tech99's answer, we don't have a static electric field pointing radially outwards everywhere, rather it only exists where the electron is at the current point in the orbit (in Rutherford incorrect model). So lines of electric field force are constantly changing direction and must radiate.
 
  • #3
First of all,thanks for picking up my doubt. I would add a little more doubt(s). We all know that accelerating charge produces transient radiation in the form of pulse and time varying current produces steady state harmonic variation radiation. these 2 forms viz.time varying current and accelerating charge are equivalent mathematically. I do not know LaTex,otherwise i would have written the mathematical relation. Now my doubt was, the electron circulates in a wire it's direction of velocity is constantly changing, owing to the fact that it experiences centripetal acceleration. But the acceleration vector and velocity vector are perpendicular to each other, I think that's why it does not radiate. Had there been any non 90 degree angle between acceleration vector and velocity vector then we could have got a transient radiation. Correct me please if I am wrong.
 
  • #4
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debajyoti datta: you, tech99, Vanadium 50, and most others at PF I suppose, know more about it than I do. Nevertheless I can give you some great advice! Read http://www.mathpages.com/home/kmath528/kmath528.htm

When done with that click the link at bottom ("Return to MathPages Main Menu") and read everything else Kevin Brown wrote. Finally: save that link and return to it often.
 
  • #5
I am just a learner, a newborn baby in terms of knowledge, how dare you compare me to others ? .....ha ha....but I am insanely curious. I've printed the page and bookmarked the link for future readings. Thanks for the link.
 
  • #6
tech99
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Electrons circling in a wire don't radiate, as OP says. Rutherford atomic model had an electron circling the proton - not in a wire, but held in orbit by attraction of charges. This model was "rejected by physicists" as debajyoti datta says. Partly because it should radiate, lose energy, and dive into the proton, in about 10^-8 seconds. So debajyoti datta wonders why this conclusion holds in the atom model case when it doesn't hold in the wire case.

There are (I think) a number of possible answers, here's one. The Rutherford model has only one electron (obviously I'm talking of Hydrogen). So, in terms of tech99's answer, we don't have a static electric field pointing radially outwards everywhere, rather it only exists where the electron is at the current point in the orbit (in Rutherford incorrect model). So lines of electric field force are constantly changing direction and must radiate.
Within a metal, we have a great number of free electrons, and they behave like a plasma and have a strong static field pointing everywhere out of the wire. Of course, we do not measure a static field from the wire because the field of the electrons is cancelled by the positive nuclei. But it is still there. And when an accelerating force is applied, the electrons are light and move but the nuclei are heavy and almost do not. So the new component of field, the transverse electric field associated with radiation, is not cancelled by the nuclei and we can observe it.
 
  • #7
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Couldn't this be settled by experiment??
 
  • #8
sophiecentaur
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Couldn't this be settled by experiment??
Wire loop antennae work very well.
 
  • #9
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Exactly, so it DOES radiate.
 
  • #10
sophiecentaur
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Exactly, so it DOES radiate.
Yes, it radiates at the appropriate frequency and not at optical frequencies, as in the inner orbitals of an H atom. QM is the clue here.
 
  • #11
jtbell
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Exactly, so it DOES radiate.
A loop antenna is driven by an alternating current, in which the electrons' average speed changes back and forth. A wire loop with a constant direct current, in which the electrons' average speed is constant, does not radiate.
 
  • #12
sophiecentaur
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A loop antenna is driven by an alternating current, in which the electrons' average speed changes back and forth. A wire loop with a constant direct current, in which the electrons' average speed is constant, does not radiate.
When DC passes round a loop, the current in every direction is equal so any radiated fields from individual electrons will cancel with all the rest.
 
  • #13
jtbell
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My memory could be playing tricks on me, but I seem to remember once seeing a derivation for the classical radiation produced by N charges moving at constant speed around a circular path, equally spaced. As N increases, the amplitude of the radiation decreases.
 
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  • #14
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I am insanely curious.
That's exactly why I sent you to Kevin Brown. If you were just a student trying to pass a course I wouldn't recommend him - too distracting, and iconoclastic.

KB is (was?) a smart and knowledgeable guy, with (at least) a PhD and plenty of teaching experience, who thought deeply about physics (and other) issues and tells it like he sees it. Goes along with the establishment view only when it makes sense (which, of course, it usually does). Has a nose for interesting questions, writes good and short essays. As far as I know, he gave up around 1994. I picture him hiding out in a log cabin in Montana, disgusted with modern physics, the whole modern world; but have no idea if that's true. If I were more ambitious, I'd copy his website and be ready to mirror it when it disappears (assuming he didn't object).

When DC passes round a loop, the current in every direction is equal so any radiated fields from individual electrons will cancel with all the rest.
Undoubtedly right, especially for a brief answer (the best kind), but there are also plenty of interesting details I think

My memory could be playing tricks on me, but I seem to remember once seeing a derivation for the classicalradiation produced by N charges moving at constant speed around a circular path, equally spaced. As N increases, the amplitude of the radiation decreases.
Interesting detail
 
  • #15
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So remove the positive lattice ie electrons in a constant current circle ... radiate or not?
 
  • #16
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So remove the positive lattice ie electrons in a constant current circle ... radiate or not?
How would you get the electrons to cooperate? Answer may depend on that detail. Also, how closely spaced they are (as jtbell indicates); and how closely you look for the radiation

By the way, are you aware that a recent study shows that questioning fine points can be bad for your health? Evidently it causes a decrease in financial status leading to physical health issues. Just thought I'd let you know
 
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  • #17
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I wouldn't ask then nicely. ....big magnetic field.

They'll go round.

too late on that other issue.
 
  • #18
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Good spelling, on the other hand, never hurt anybody
 
  • #19
jtbell
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My memory could be playing tricks on me, but I seem to remember once seeing a derivation for the classical radiation produced by N charges moving at constant speed around a circular path, equally spaced. As N increases, the amplitude of the radiation decreases.
Now that I've had a couple of cups of coffee, I think my memory must indeed be playing tricks on me. Synchrotron radiation is a fact of life at circular particle accelerators, especially with electrons, and they don't work by sending individual electrons around one at a time!

I think a more relevant factor for the DC-current-carrying loop is the electron drift speed that corresponds to the current. This is typically on the order of millimeters per second, producing a correspondingly small centripetal acceleration (depending of course on the radius of the loop). And the radiated power varies as the cube of the acceleration IIRC.
 
  • #20
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I know LHC sends protons around in packets; guess they're all that way? So the leading and trailing edges of those packets would still have changing fields, producing the synchrotron (what a cool word!) radiation. The dense part of the packet may emit no radiation, in line with sophiecentaur's comment.

Nevertheless I wouldn't be surprised if you can have (essentially) continuous flow and still get radiation with the right conditions (as houlahound wonders). Perhaps, higher speed combined with tighter loop, for greater centripetal acceleration? However in chip design if the corner of a CPU's bus (for instance) isn't rounded off, the electrons will tunnel through and fly away - so that's pretty extreme acceleration. Yet I don't think they have a problem with radiation emission.

It seems to me electromagnetism is still not well understood by those who understand it.

It's amazing that some physicists can tell you exactly how many child universe generations are necessary to produce intelligent life, assuming they reside on 10-dimensional multi-trillion light-year colliding branes (although it's still unknown for 11-dimensional branes, and a massive infusion of taxpayer's money is necessary to research the question). But they can't tell you how a D-cell battery works.

{EDIT] I'm not thinking of anyone posting at PF! Incredibly brilliant genius can't be bothered with such mundane activity
 
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  • #21
sophiecentaur
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The dense part of the packet may emit no radiation, in line with sophiecentaur's comment.
My logic would suggest that a sector of moving charges around the curve would all radiate more or less in phase. It would be the diametrically opposite charges in a continuous ring that would cancel.
 
  • #22
sophiecentaur
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the electrons will tunnel through and fly away
This isn't like a sharp bend on a race track. The electrons are only moving at a snail's pace.
It is interesting that you would choose such a modest frequency and low power example.
It's amazing that some physicists can tell you exactly how many child universe generations are necessary to produce intelligent life, assuming they reside on 10-dimensional multi-trillion light-year colliding branes (although it's still unknown for 11-dimensional branes, and a massive infusion of taxpayer's money is necessary to research the question). But they can't tell you how a D-cell battery works.
Actually, the two cutting edge examples of such speculations are probably a good deal more fuzzy than the present understanding of the operation of a chemical cell. It's just that they are presented differently in the press.
 
  • #23
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My logic would suggest that a sector of moving charges around the curve would all radiate more or less in phase. It would be the diametrically opposite charges in a continuous ring that would cancel.
I see, could have guessed that. Brings up a couple ideas.

Undoubtedly cyclic accelerator designers have thought of this idea, and already use it as much as may be practical: Couldn't they could "synchronize" their packets so two of them radiate out of phase, and cancel? Perhaps avoid packets entirely, use a continuous flux? One obvious problem, the radius at LHC is 8.6 miles; would it work at all at such a distance? Could you get around it perhaps with a two-ring system, a second loop running next to the first? Or (with two sharp bends - that's a problem) make most of the track two close "stretches", like a race track. LHC does use two rings, closely spaced, for the two opposing beams (clockwise and counter). But they can't cancel because they're not coherent? In general interference usually depends on coherence, is that necessary also in this case?

Obviously I'm not going to come up with anything they haven't thought of, so the general question is, are such considerations part of cyclic accelerator design?

Back to ordinary electric current, I don't think there will be any radiation no matter how big the radius is? I'm surprised they can cancel correctly if it's very big? Obviously no great care need be taken to ensure that when the radiation meets they cancel; randomness takes care of that. That's also why any arbitrary shape of the loop (square for instance) should have cancellation also. I wonder if, with a smaller circuit, is it possible to calibrate it so exactly that they don't cancel? Maybe create a new kind of antenna that way?

Again, if I can think of it, physicists and electrical engineers have already thought of it, so the question is, do any devices exist which take advantage of that possibility?

This isn't like a sharp bend on a race track. The electrons are only moving at a snail's pace.
It is interesting that you would choose such a modest frequency and low power example.
Actually I know they are moving slowly but still they are being accelerated sharply. Anyway - suppose we used a high potential, can you get a sharp bend to produce radiation - I mean without a spark?

By the way it's not so interesting I'd choose a poor example, I'm no physicist and EM, in particular, has always been confusing (not as much as the dynamics of a top, though!) My degree is in functional analysis so re. QM I'm dangerous (as in, "a little knowledge is a dangerous thing"). You may notice at times I sound like an expert. That's because I googled some specific topic, may have spent an hour on it. Anything off the top of my head is liable to betray ignorance. Nevertheless I know history of physics better than most physicists.

I'm hitting you with a bunch of random questions here, sorry! Any thoughts welcome.

Actually, the two cutting edge examples of such speculations are probably a good deal more fuzzy than the present understanding of the operation of a chemical cell. It's just that they are presented differently in the press.
Yes, but that's with the cooperation of the superstars involved, and the acquiescence of everyone else. Of course I'm being sarcastic (not good) but sometimes can't help it, such speculation should not be allowed in physics at all - the nuthouse is the proper place for it. But let's not get into that.

=========================
Could somebody please explain me the problem 11.17 in Griffith's Electrodynamics 3rd edition ?
Try http://www.physicspages.com/2015/02/13/radiation-reaction-a-few-examples/
 
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  • #25
sophiecentaur
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Back to ordinary electric current, I don't think there will be any radiation no matter how big the radius is? I'm surprised they can cancel correctly if it's very big?
However big or whatever shape, there will always be a complete loop path for the current. Assuming you have DC, the effect of one part of the wire will always be cancelled by the rest. If you have a step change or AC, you can expect radiation, in any case.
 

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