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dgh64
2010-Jul-31, 03:10 PM
So, imagine we have a very basic particle accelerator: a toroidal vacuum chamber surrounded by superconductor coils, and at one point a pair of charged plates with a small hole through them. A slow-moving proton (or electron, or other charged particle) is put in, and because of the superconductors it is forced to travel a circular path, part of which is between the charged plates. As it passes the plates, the electric field makes it move a little faster. Each time it goes around, it goes a little faster.

Now, here's the part I don't understand. Once you've overcome the inductance of the superconductors, they can maintain their field pretty much indefinitely without additional energy input. (aside from the cryo system to keep them cold, but that's not part of the system we're interested in) The charged plates, also, have no energy input -- once you've built up a charge on them, you can turn the switch off and they'll remain charged, assuming no electrons can jump off the negative plate and cross the gap.

So, where does the extra energy come from, to accelerate the proton?

grant hutchison
2010-Jul-31, 04:57 PM
Maybe I'm missing part of the set-up, but I don't see why the proton's average speed would increase with time.
The proton must decelerate once it's through the hole. In the vicinity of the charged plate, it exchanges potential for kinetic energy, and then (after passing through the hole) it trades kinetic for potential. There's no net energy gain.

Grant Hutchison

Strange
2010-Jul-31, 05:03 PM
What he said. Also, the energy to change the direction (velocity) of theprotons comes from the electromagnets. They may not lose energy through resistance (being superconducting) but they can lose energy in other ways. You can pulse the magnets to provide acceleration as well; that will also impart energy from the electromagnets.

dgh64
2010-Jul-31, 05:47 PM
The electromagnets are steady, not pulsed. Since the proton is only changing direction, not speed, there's no energy drain from the magnets, is there? I mean, you could do this experiment with some hypothetical extremely strong permanent magnets and forget the superconductors entirely.

The proton must decelerate once it's through the hole. In the vicinity of the charged plate, it exchanges potential for kinetic energy, and then (after passing through the hole) it trades kinetic for potential. There's no net energy gain.

Grant Hutchison

This is from my college physics class. The plates have to be very close together relative to their width. Then the electric field (and therefore force on the proton) is negligible outside the plates, but inside the gap the proton gets a "kick" each time it goes around. At least that was my understanding of it. But you're saying that there is a slight force pulling the proton back after it passes the holes?

Strange
2010-Jul-31, 06:11 PM
The electromagnets are steady, not pulsed.

OK. I just meant that was an alternative way of accelerating the protons.

Since the proton is only changing direction, not speed, there's no energy drain from the magnets, is there?

Yes, because a change in direction is a change in velocity (which is a vector: speed & direction) and therefore an acceleration.

I mean, you could do this experiment with some hypothetical extremely strong permanent magnets and forget the superconductors entirely.

And then the acceleration (change of direction) of the protons would (very slowly) demagnetize the magnets (I think).

grant hutchison
2010-Jul-31, 07:36 PM
This is from my college physics class. The plates have to be very close together relative to their width. Then the electric field (and therefore force on the proton) is negligible outside the plates, but inside the gap the proton gets a "kick" each time it goes around. At least that was my understanding of it. But you're saying that there is a slight force pulling the proton back after it passes the holes?I think "negligible" may not be entirely negligible. The proton is going to experience a slight deceleration over a relatively long distance as it approaches your pair of plates, a short burst of acceleration between the plates, and a slight deceleration over a relatively long distance on the other side. It seems to me that the combination will cancel out.

And, as Strange says, you're going to have to feed energy to the system to keep the proton moving in a circle, because accelerated charges radiate energy.

Grant Hutchison

IsaacKuo
2010-Aug-01, 12:30 AM
A slow-moving proton (or electron, or other charged particle) is put in, and because of the superconductors it is forced to travel a circular path, part of which is between the charged plates. As it passes the plates, the electric field makes it move a little faster. Each time it goes around, it goes a little faster.
The problem is that the proton will not go a little faster each time. The charged plates will produce an electric field, which is conservative. This means that a particle which travels in a closed loop will experience zero work from the electric field. Thus, the proton would end up with the same speed when it returns to the same spot in the loop.

The acceleration of the proton while between the plates is balanced by the deceleration of the proton while outside the plates. The magnitude of the deceleration is indeed smaller, but it acts over a longer distance.

The proton does experience some loss in speed due to cyclotronic radiation, but that's a secondary effect.

publius
2010-Aug-01, 12:34 AM
The energy always comes from the field, and to maintain that field, energy must be pumped in. That's the basic principle, but it can get sort of complex to follow.

You might look at an ion drive system. Particles are ionized and electric fields are use to accelerate the heavy ions, throwing mass out and thus producing thrust. Now, for the system to work, it must remain neutral, and thus throw off electrons to keep the thruster from gaining a net charge and thus cancelling out the thrust (the positive ions would be progressively attracted to the thruster frame and that would kill exhaust velocity).

If we reduce it all to circuit theory, any time we have power being expended, we're going to have some VI afoot there, and the plate supply system is going to be flowing a current. Although it may get complicated in the details, in the big picture there, what do we have? We have a net neutral exhaust stream, and a net neutral thruster. As far as the power supply circuits are concerned, there can only be a circulation, not seperation of charge.

-Richard

grant hutchison
2010-Aug-01, 12:36 AM
And we can really forget about the cyclotron as part of the problem. Either the proton violates conservation of energy with a single pass through the plates, or it doesn't. Bringing it around for multiple passes makes no difference to the problem.

Grant Hutchison

dgh64
2010-Aug-01, 01:41 PM
Thank you for all your replies. This answers my question to my satisfaction. I was hoping to have finally invented an unlimited energy source...

Ken G
2010-Aug-01, 02:50 PM
I think "negligible" may not be entirely negligible. The proton is going to experience a slight deceleration over a relatively long distance as it approaches your pair of plates, a short burst of acceleration between the plates, and a slight deceleration over a relatively long distance on the other side. It seems to me that the combination will cancel out.This is the key point, and the astute OP question brings up an important issue not often raised in simple physics-- the importance of a boundary condition "at infinity". In physics class, when you want to treat parallel plates, for simplicity you make them "infinitely wide", meaning, much wider than they are separated from each other. But this also means they have to be much wider than the distance the particle is from the plates at all times. The difference can be seen in the global behavior of the voltage-- if the plates are truly infinite, then they divide all space into a +V and a -V region. But even so, there are 4 regions of interest, not 3-- the +V region, the -V region, the transition between the plates, and the reverse transition "at infinity". To pass the proton through twice, you have to "pass through infinity", where the kinetic energy it got is lost.

Since there really is no place called "infinity", what actually happens is what Grant said above, but it can still be spoken of in idealized terms if we mindfully treat infinity. Push infinity "off the page" (out of sight, out of mind), and you will encounter many types of problems and paradoxes, and that's what the OP question asks us to resolve.

JohnD
2010-Aug-03, 04:50 PM

The consensus from the above seems to be that a proton confined in a circle by magnets cannot be accelerated to higher and higher energies.
See Grant and Isaac's words for example.

I thought that was the whole point of the LHC, that the protons and other particles therein could be raised to very high energies.

John

grant hutchison
2010-Aug-03, 05:00 PM
I thought that was the whole point of the LHC, that the protons and other particles therein could be raised to very high energies.And that's why high-energy accelerators need to be big: to reduce the acceleration required to keep fast particles in "orbit". The diameter of the accelerator sets a limit on how much energy you can pump into the particles before they start to radiate it away as fast as you add it.

Grant Hutchison

Ken G
2010-Aug-03, 05:09 PM
I thought that was the whole point of the LHC, that the protons and other particles therein could be raised to very high energies.
I think the key thing you are missing is actually the time-dependence of the electric field. If you have a static field, it fills all space with a set voltage, and then conservation of energy says that if you return to the same place, you have the same kinetic energy. LHC wouldn't work. So instead, what they do is invoke a time-dependent electric field, so there is no global voltage and they can continue to do work on the particle.

dgavin
2010-Aug-03, 08:09 PM
Ok this is just a pure guess on my part, but considering superconductors have also been used to produce magentogravitational effects, this might be a case where the kick comes from the MG field superconductos can put out.

As to the conservation of energy, I'd tentivley suspect the energy is lost out of the electrical field traveling through the superconductor, via the MG field that produces. It would probably take a very sensitive instrument to dedect the loss of the current out of the superconductor however.

Ken G
2010-Aug-04, 12:15 AM
It would probably take a very sensitive instrument to dedect the loss of the current out of the superconductor however.The power comes from the time-varying electric fields, and it's not hard to see this energy-- indeed, it represents a non-negligible fraction of the energy usage of the entire city of Geneva when the beam is on!

JohnD
2010-Aug-04, 05:42 PM
I think the key thing you are missing is actually the time-dependence of the electric field. If you have a static field, it fills all space with a set voltage, and then conservation of energy says that if you return to the same place, you have the same kinetic energy. LHC wouldn't work. So instead, what they do is invoke a time-dependent electric field, so there is no global voltage and they can continue to do work on the particle.

Ah! So that is where the energy goes. It uses energy to change a field, yes?
John

Ken G
2010-Aug-04, 06:41 PM
Ah! So that is where the energy goes. It uses energy to change a field, yes?
It uses energy if a field is doing work on the particle. So you set the fields up to do that work, and you find it takes energy to set up fields like that.

heldervelez
2010-Aug-05, 06:32 PM
The energy always comes from the field, and to maintain that field, energy must be pumped in. That's the basic principle, ...

"..must be pumped in." And drained from where ? (I'm trying to avoid a free lunch)

An isolated system of one charged particle and his field, that expands at 'c' speed'.

To 'source' the field we must 'drain' the energy from somewhere, obviously the particle.

That's the basic principle, in your words, that justifies the Evanescence model ( Shrinking matter ) that was mentioned BF here (http://www.bautforum.com/showthread.php/83475-Other-way-to-look-at-the-Universe-s-Expansion?p=1420081#post1420081)

publius
2010-Aug-06, 09:43 PM
"..must be pumped in." And drained from where ? (I'm trying to avoid a free lunch)

An isolated system of one charged particle and his field, that expands at 'c' speed'.

To 'source' the field we must 'drain' the energy from somewhere, obviously the particle.

That's the basic principle, in your words, that justifies the Evanescence model ( Shrinking matter ) that was mentioned BF here (http://www.bautforum.com/showthread.php/83475-Other-way-to-look-at-the-Universe-s-Expansion?p=1420081#post1420081)

The energy to establish the field comes from mechanical work done on charged particles to establish the field. A simple example: matter is generally bulk neutral, but made of equal and opposite charged particles. The net external electric field is zero. Now, seperate that charge and establish a external field. That requires work be done on those particles. When the charges come back together, the field dimishes again, and that energy is returned.

This can all get very mathematically tedious in the the details, but any time there is field energy being converted back and forth to what we will call "mechanical energy", there will be a non-zero Poynting vector flux.

Consider the simple, most elementary example of a a charged particle being accelerated in a static electric field from EM 101. Guess what? We no longer have an actual static field! The field of the accelerating particle is moving, accelerating, and a moving charge makes a magnetic field. We indeed have a dynamic E x B afoot there. The field is now dynamic, moving energy (and momentum) around through space. That sort of complicates the simple static picture we first consider doesn't it? :) But yet the simple qV energy picture will give the correct answers for the results of that dynamic process.

Indeed, if we draw a closed surface around our accelerating particle, the flux of ExH over the surface will give us the rate at which field energy is being transferred to the particle's mechanical energy. Things get complicated in the full picture, though. An accelerating particle radiates, so there is some energy being transferred that doesn't go to mechanical energy, but to the radiation energy. And there will be some "reactive energy", going to the magnetic field that is established by the particle as it begins to move. That energy is not mechanical, but stays within the EM field realm.

-Richard

heldervelez
2010-Aug-07, 09:59 AM
As I expected Magnetic field is an artifact.
from Hans de Vries web page (give a try to his good book at http://physics-quest.org)
The simplest, and the full derivation of Magnetism as a Relativistic side effect of ElectroStatics
http://physics-quest.org/Magnetism_from_ElectroStatics_and_SR.pdf

Mr publius, you are evading the most simple exercise of one particle, bringing more than one into question.
The particle at all time radiates the same way, but the radiated field interferes with itself (self-interaction) as the particle moves in relation to the exterior field (its own radiated field or others). This gives us what we call a dynamic field.
The particle does not have two regimes, it is unaware of its movement, at least in principle. It does not see at a distance other particle moving and say to itself: hey I will have to switch (in its own past, to be more precise) to the dynamic/H regime,.. hoho the electrostatic regime is no longer valid,.. hoho please help me... ;)
The facts that we both are reporting are the same but the interpretations diverge.

When the particle stops moving, or starts moving, accelerating, ones must say 'in relation to what'.
I say 'in relation to the exterior field'. This way we can considere the simple scenario of a single particle, accelerating or stationary.

Now the basic fact: the particle radiates at all time, and always the same way.
If, as we all know, if the stationary particle stop radiating at any time, by any moment, ... (someone try an answer, pls)

Try to answer to my original question,pls :
one particle and its field that spreads away at 'c' speed.
1) What is the source of the field, -- my answer : the particle
2) what is sourced into the field, -- my answer : energy (say it electromagnetic and/or gravitational)
3) and what must me drained from the particle. -- my answer : energy
consider one stationary particle to avoid any 'mechanical' energy, accelerating, etc ...

If my answers are not correct then one more question : How can 'source' without 'drain' ?

Ken G
2010-Aug-09, 03:57 AM
helderverez, if you are having trouble getting an answer, it may be for the reason I cannot answer either: the things you said that make sense sound just like standard physics, and just like what publius said. The rest doesn't make sense. So what is there to answer?

Strange
2010-Aug-09, 09:21 AM
Now the basic fact: the particle radiates at all time, and always the same way.
If, as we all know, if the stationary particle stop radiating at any time, by any moment, ... (someone try an answer, pls)

Try to answer to my original question,pls :
one particle and its field that spreads away at 'c' speed.
1) What is the source of the field, -- my answer : the particle
2) what is sourced into the field, -- my answer : energy (say it electromagnetic and/or gravitational)
3) and what must me drained from the particle. -- my answer : energy
consider one stationary particle to avoid any 'mechanical' energy, accelerating, etc ...

If my answers are not correct then one more question : How can 'source' without 'drain' ?

I'm not quite sure what you are asking either. The original question was about particles being accelerated, which clearly requires an input of energy.

A stationary particle (as you seems to agree) will not lose any energy so does not require any energy source. So where is the problem?

heldervelez
2010-Aug-10, 10:14 PM
I'm not quite sure what you are asking either. The original question was about particles being accelerated, which clearly requires an input of energy.

A stationary particle (as you seems to agree) will not lose any energy so does not require any energy source. So where is the problem?

Not so: a stationary particle will lose energy (it radiates energy away) even if it is not accelerating as one can find in this very recent study (Aug/2010) :
On the Definition of Radiation by a System of Charges by K. McDonald of Princeton Edu
quoting "..then definition B is preferable, even though it implies that
fields of the individual charges.."
Even without that paper one can say that electrostatic field has energy, that this energy goes away at 'c' speed.
It is not even necessary to have a Poynting vector <> 0 to have energy going away, as I found today in a paper about physics of 'plasmas'.
Anyway: if energy is sourced to the field, because it expands, then it is drained from the particle.
E=mc^2 leaves me with the sense that the particle itself feeds the field.
As no one has measured this situation we are tempted to say that it does not exist. But it seems to me an absolute impossibility of a directly measure of it.

publius
2010-Aug-10, 11:26 PM
Not so: a stationary particle will lose energy (it radiates energy away) even if it is not accelerating as one can find in this very recent study (Aug/2010) :
On the Definition of Radiation by a System of Charges by K. McDonald of Princeton Edu
quoting "..then definition B is preferable, even though it implies that
fields of the individual charges.."

The divergence between what that paper is saying and what you think it is saying is a chasm as wide as the Grand Canyon.

What is he talking about. A system of *oscillating charges*, which means they are of course accelerating (and jerking) and he is fretting over the definition of "radiation field" of that system of charges. His A and B choices are whether to define radiation as the surface integral of the Poynting vector of the total (time-varying, dynamic) field, or just that of what is defined as the "radiation field", picking terms out of the field expressions.

He may have a point of some sort, but in my limited understanding of these things, we can somewhat clumsily boil it down to circuit theory at least for the power source that is driving the antenna. The Poynting flux of those terms dubbed the "radiation field" can be made to look and treated as a "resistive load" to that source, and the Poynting flux of the rest of the field is a "reactive load" to that source.

The latter is the flow of energy through space over time that has to do with the energy of the E and B fields (or at least the non-radiation part, which splitting off is what the author is fretting over), while the radiation is energy "thrown away to infinity" that will never return to the source. When you pump a field, you have electric and magnetic field energy which plays the role of capacitors and inductors, while the radiation part looks resistive.

-Richard

Geo Kaplan
2010-Aug-10, 11:38 PM
Not so: a stationary particle will lose energy (it radiates energy away) even if it is not accelerating as one can find in this very recent study (Aug/2010) :

As publius notes, you completely misunderstand the paper you've referenced. I recommend that you study, e.g., Haus's classic paper on radiation (http://dx.doi.org/10.1119%2F1.14729), which improves on earlier work by Goedecke and others. A stationary particle will never radiate. A constant velocity particle may radiate, but not in free space (see Cerenkov). Haus gives the general requirement for radiation: The current produced by the moving charge(s) must have Fourier components with the velocity of light.

heldervelez
2010-Sep-18, 09:14 AM
A stationary particle will lose energy (it radiates energy away) even if it is not accelerating as I show at (ATM) Growing-fields-imply-Evanescent-Matter (http://www.bautforum.com/showthread.php/107793-Growing-fields-imply-Evanescent-Matter).

Coincidently when I was trying to find Goedeck's original paper (and found it in McDonald's site) you made the very first reference to Goedeck in this forum.

pzkpfw
2010-Sep-18, 11:11 AM
A stationary particle will lose energy (it radiates energy away) even if it is not accelerating as I show at (ATM) Growing-fields-imply-Evanescent-Matter (http://www.bautforum.com/showthread.php/107793-Growing-fields-imply-Evanescent-Matter).

ATM claims are to be kept in the ATM sub-forum, please.