• I
Lets say you have a "magnetic field" (call it B) and lets say it has a z direction, all over the space. Now let's kick an electron into the scene perpendicular to z.
It will feel a force F=e v B to the axis z.
Given a velocity v = v0 this won't change. We have an electron running in a circle for eternity and the energy of the particle is conserved!
But we didn't applied the full set of Maxwell equations. There is an interaction, otherwise the electron would follow a straight line.
The electron will radiate and get the energy from the magnetic field.

Is this correct?

Related Other Physics Topics News on Phys.org
Simon Bridge
Homework Helper
When you are examining established physics closely like this, it is a good idea to be as precise as you can muster.
Lets say you have a "magnetic field" (call it B) and lets say it has a z direction, all over the space.
"has a z direction" is not very precise ... what do you mean? Do you mean that, say, ##\vec B = B\hat k## where ##\hat k = (0,0,1)^t## in cartesian coordinates?

Aside: Is 'a "magnetic field"' different from a magnetic field? (ie: what do the quotes signify?)

Now let's kick an electron into the scene perpendicular to z.
Similarly "kick an electron..." is not precise ... you mean an electron is present and has velocity ##\vec v## such that ##\vec v\cdot\vec B = 0##?
It will feel a force F=e v B to the axis z.
I don't know what you mean by "to the axis" and the equation is not strictly correct.
The resultant force will be perpendicular to both the velocity and the magnetic field, and is given by: ##\vec F = -e(\vec v \times \vec B)##.
This is all in the non-relativistic case.

Given a velocity v = v0 this won't change. We have an electron running in a circle for eternity and the energy of the particle is conserved!
But we didn't applied the full set of Maxwell equations. There is an interaction, otherwise the electron would follow a straight line.
The electron will radiate and get the energy from the magnetic field.
This is a reasonable statement of the usual paradox for orbital motion of an electron when you don't have quantum mechanics.
If you like you can have a go working out how much radiation to expect in different situations.
ie. see: http://www.astro.utu.fi/~cflynn/astroII/l4.html
Where are you going with this?

Omega0 and sophiecentaur
Sorry about my sloppyly formulated problem.
Do you mean that, say, ##\vec B = B\hat k## where ##\hat k = (0,0,1)^t## in cartesian coordinates?
Exactly.
Aside: Is 'a "magnetic field"' different from a magnetic field? (ie: what do the quotes signify?)
You would have been needed to be a mind reader to understand the quotes, sorry again. I meant that in non-relativistic QM the ##\vec B## is substituted in the Hamiltonian, ## \hat{H} = \frac{1}{2m} (\hat{\vec{p}} - q \vec{A}(t,\hat{\vec{x}}))^2 + q \phi (t,\hat{\vec{x}}) ##
As Feynman said, not cited: ##\vec A## decides, not ##\vec B## if you are going to quantum mechanics.
The resultant force will be perpendicular to both the velocity and the magnetic field, and is given by: ##\vec F = -e(\vec v \times \vec B)##.
Exactly this.
This is all in the non-relativistic case.
This is a reasonable statement of the usual paradox for orbital motion of an electron when you don't have quantum mechanics.
If you like you can have a go working out how much radiation to expect in different situations.
ie. see: http://www.astro.utu.fi/~cflynn/astroII/l4.html
Where are you going with this?
Thanks for the link. The basic idea of my post was to clarify a point. In the discussion about "What is a field"
I got the answer for the magnetic field: "charged particles follow curved paths that don't change their energy because the force is always perpendicular to the direction of motion."
I think this answer is not correct under respect of the STR because of radiation. The more, if only ##\vec A## plays a role for the Aharonov-Bohm-effect the question might be if ##\vec B## shouldn't be replaced in our thinking by ##\vec A##.
I think it is not correct that the neutron is not influenced by ##\vec A## because of the spin which was said in the thread.
In this sence I love the GRT: Every sort of energy has an influence on the space-time structure.

What I wanted to state: A physical field is only a field if there is at least one particle which interacts with this field and energy exchange is possible.

Is this correct?

Simon Bridge
Homework Helper
Thanks for the link. The basic idea of my post was to clarify a point. In the discussion about "What is a field"
What I wanted to state: A physical field is only a field if there is at least one particle which interacts with this field and energy exchange is possible.
Nope: fields are considered to exist whether or not there is anything to interact with it.
The word "field" is a technical term in mathematics, which is applied to a number of different concepts in physics - all of which have the same intrinsic math.
For instance, in QFT a particle is thought of as a perturbation of the field which, depending on the symmetry of the perturbation, may be a real particle. like an electron, or a virtual particle (ie force-mediating bosons). With this sort of field, it is possible in maths (ie by definition) to have a field with no perturbations and thus no particles.

I think fields like gravity and the classical electric field as you are taught in high school are less subtle. The field is basically a diagram that tells you the magnitude and direction of the force on a unit-quantity test-particle at a particular position if there were one there. But there isn't one there ... and it is usually important that it is not there because that would change the field in question everywhere else. So a force field is the mathematical result of mathematically removing all subject particles (apart from the source of the field you want to study).

ie we may define a gravitational field as ##\vec F/m## where ##\vec F=-(GMm/r^3)\vec r## where ##\vec r## is the position of m in relation to M.
So it's the force you would have if m were present but it isn't - or, put in a more engineeringly way, the force that would be on m if m=1unit which it doesn't in general.

This is philosophically tricky to nail down, but easy to imagine as a bunch of imaginary lines or arrows - so physicists do this kind of thing all the time and try not to talk to philosophers much if they can help it.
Asking if the field is there when there is no particle to experience it is like that koan about the tree falling in the forest...

Simon Bridge
Homework Helper
Per the rest - electron radiation etc. I think the link I gave you approaches this: "how can the kinetic energy remain unchanged if the electron is radiating"?
I think that's the point right?

So: do the particles in a cyclotron lose kinetic energy by radiating?

You have been looking at this with an eye on the quantum mechanics ... it is important not to mix up models: the usual high school descriptions of the electromagnetic field are incomplete and are replaced by the quantum description so you cannot treat them as exact equivalents.

Is the cyclotron an example of a magnetic field only existing if there is a particle to interact with it?
Well, that must be inconclusive: the source for the field is a big electromagnet and the particles it is composed of must be interacting electromagnetically ... ie there are always going to be those particles present. Clearly the field is independent of the test particle orbiting between the magnetic poles though. Just try prise the poles apart and see.

The advantage of the field description is that it removes the need to talk about what gives rise to the field, which simplifies the maths considerably.
That's why we use it.

tech99
Gold Member
Lets say you have a "magnetic field" (call it B) and lets say it has a z direction, all over the space. Now let's kick an electron into the scene perpendicular to z.
It will feel a force F=e v B to the axis z.
Given a velocity v = v0 this won't change. We have an electron running in a circle for eternity and the energy of the particle is conserved!
But we didn't applied the full set of Maxwell equations. There is an interaction, otherwise the electron would follow a straight line.
The electron will radiate and get the energy from the magnetic field.

Is this correct?
Sorry I am just am amateur scientist. This is clearly a complicated question, as several have described, due to the interaction of the fields. But my understanding is that it will radiate energy, as happens in a cyclotron, and the energy will come from the kinetic energy of the electron, so it will slow down. The "kinetic energy" of the electron may be regarded as being mainly stored in its own magnetic field, which arises due to its velocity, and gives the electron an increase in inertia (or effective mass).

Per the rest - electron radiation etc. I think the link I gave you approaches this: "how can the kinetic energy remain unchanged if the electron is radiating"?
I think that's the point right?
So: do the particles in a cyclotron lose kinetic energy by radiating?
The particle neccessarily radiates and the energy needs to come from somewhere. I wrote that this energy comes from the magnetic field, this may be on the first view misleading or call it wrong. Let's say it comes from the "kinetic energy".
This time I explain why I use quotes. In nature there is only one kind of energy known, whatever this is. It is simply nice to differentiate between those energies to help the lerner to talk about a "transform of energy forms into each other". I think this is a pretty old picture coming from mechanics and thermodynamics.
You have been looking at this with an eye on the quantum mechanics ... it is important not to mix up models: the usual high school descriptions of the electromagnetic field are incomplete and are replaced by the quantum description so you cannot treat them as exact equivalents.
Correct. I wanted to say that with respect of the model of non-relativistic QM the classical picture of the ##\vec B## seems to be ruled out with respect of the Aharonov-Bohm-effect. The picture of the field ##\vec B## is surely not really wrong, classical electrodynamics - but it's the same for classical mechanics. As you said, the usual high school description is not complete.
Now going to someone constructing a bridge to recommend relavistictic mechanics or for the metal relavistic condesed matter physics or going to someone constructing an electronic circuit to explain that he should use ##\vec A##. This is ridiculous. So I agree - the classical field picture is useful (as well as classical mechanics).
Is the cyclotron an example of a magnetic field only existing if there is a particle to interact with it?
Well, that must be inconclusive: the source for the field is a big electromagnet and the particles it is composed of must be interacting electromagnetically ... ie there are always going to be those particles present.
That's it. We are not able to contruct an electromagnetic field without "particles". I personally don't like this particle image coming from classical mechanics, I would prefer a thermodynamical point of view. If you do it this way the following makes sense:
Clearly the field is independent of the test particle orbiting between the magnetic poles though. Just try prise the poles apart and see.
This is a thermodynamical point of view. You have a mighty reservoir of a charged gas building that field. A test particle plays no role, it will run around and run around and radiate - but the energy comes from the ##\dot{\vec{B}}##. The charged test particle looses energy and is pumped up again.
In short: Those classical fields, as well as the force field in classical mechanics is nothing without the idea of "particles" they are made of and so naturally they interact with.

Correct?
The advantage of the field description is that it removes the need to talk about what gives rise to the field, which simplifies the maths considerably.
That's why we use it.
I understand that it is comfortable to avoid the creation of the field "by particles" but this makes it in my eyes very obscure. I would say that there is nothing but fields and what we call "particles" is nothing but measurable eigenstates of this field.
In my eyes it is not something like "Is there a forest when you don't here a tree falling?" but "Is there a forest when there is no tree?".

Simon Bridge
Homework Helper
I personally don't like this particle image coming from classical mechanics, I would prefer a thermodynamical point of view.
Nature does not care what you personally like or do not like.

I understand that it is comfortable to avoid the creation of the field "by particles" but this makes it in my eyes very obscure.
Find an EM field that does not have source charges?
However - a field may be present without particles, other than the source partcles, interacting with it.
The self-interaction is usually considered a different issue to the interaction used to define the field in the first place: which was the subject of your original question.

I don't know what you mean by "obscure" here - there is nothing obscure in the model: it is all in the open. Perhaps you mean "abstract"? This would be true - unfortunately the abstract models work the best so we are forced to use them.
I would say that there is nothing but fields and what we call "particles" is nothing but measurable eigenstates of this field.
It can be confusing keeping track of all the models ... but I believe your original question has been answered.

Delta2
Homework Helper
Gold Member
But we didn't applied the full set of Maxwell equations. There is an interaction, otherwise the electron would follow a straight line.
The electron will radiate and get the energy from the magnetic field.

Is this correct?
Treating the problem classically, yes the electron will radiate, but can not get energy from the magnetic field, magnetic field can not do work on matter(you can read many threads in physics forums about this subject).

So the electron will radiate and keep losing kinetic energy (kinetic energy transforms to radiating EM wave energy), its trajectory will be a spiral and it will stop at some future time when it has lost all of his kinetic energy.