Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Chatroom created by @pcr for discussing this: http://chat.stackexchange.com/rooms/2824/direction-of-rotation-of-proton-in-magnetic-field

Here's a small paradoxical question I was asked a long time ago (and have been asked twice since). I think I do know the answer, but I though it would be fun to ask it here.

Let's take a proton and fling it into a magnetic field coming out of the plane of the paper/screen ($\vec{B}=B_0\odot$) Now, looking from above the plane, the proton goes clockwise.

Alright. lets take our right hand and find the direction of the dipole moment. It's a proton going clockwise, so it's a clockwise current. This is a downwards ($\vec{M}=M_0\otimes$) dipole moment.

But, potential energy of a dipole is $U=-\vec{M}\cdot\vec{B}$. If they are antiparallel, then the dot product is negative, so we get $U=-M_0B_0(\odot\cdot\otimes)=M_0B_0$, and is positive. Compare that with the case where $\vec{M}\parallel\vec{B}$, we get a negative value of potential energy.

As we all know, a system tends to reduce its potential energy. Then why, in this case, does a proton deliberately choose the direction of rotation with the maximum potential energy?

Reason for bounty

Multiple things. The bountybox does not provide the ability to overlap reasons, unfortunately ;)

  • I have multiple conflicting answers, and while each one is individually convincing, when brought together the whole situation becomes a jumble
  • I need more people to take a look at this, upvote answers they agree with, comment, and/or add more answers.
  • The answers could be clearer
  • It would help if the answers explained the paradox for various levels of understanding.
share|improve this question
    
I personally have two, simple explanations for this. What I want to see is how deep one can go into this.. And having been asked this question thrice, I guess it's appropriate for this site as it may be useful to future visitors. –  Manishearth Mar 15 '12 at 9:54
    
Atleast, I don't think so. If you can get another explanation from this, then please post it as an answer :). I have a feeling that this paradox can be explained in multiple ways, and some pretty deep. –  Manishearth Mar 15 '12 at 10:56
    
Don't ask me if I've thought of something or not. I already [think I] know the answer. I've posted this here to (a) Let others have fun; (b) see how deep one can go; (c) see how many explanations this has [most probably they'll all boil down to the same point]; (d) Put a [probably] common confusion/paradox on the site. If you feel you have an explanation, just post it as an answer. –  Manishearth Mar 15 '12 at 11:01
    
This question actually sounds familiar, I wonder if it's been asked here before? –  David Z Mar 15 '12 at 18:34
    
@DavidZaslavsky: I asked it on chat once. No response. –  Manishearth Mar 16 '12 at 0:41
show 1 more comment

5 Answers 5

up vote 3 down vote accepted
+100

The potential energy in this case should be $U=+\vec{m}.\vec{B}$, hence the potential energy is minimized, as it should be. Here is the explanation:

Let’s look at the derivation of interaction energy between magnetic dipole and magnetic field carefully. The dipole energy $U=-\vec{m}.\vec{B}$ is derived using principle of virtual work with an assumption that the dipole moment is constant, and thus it’s self energy is permanent. However, if the dipole moment is allowed to change as in this case, the self energy of the dipole is no longer permanent. We can imagine it as the $\frac{1}{2}LI^2$ energy for the case of a current loop, if we change its dipole moment, its internal energy will also change. So for the case like this, the self energy can be extracted into mechanical energy. If we take into account the additional work to change the self energy in the principle of virtual work we will end up with $U=+\vec{m}.\vec{B}$. We can always calculate the work $\int \tau d\theta$ to change the orientation of the dipole. However in this case the dipole moment is not permanent, so its magnitude will be be different for different orientations. Thus the work calculation will be messy, but there is an easy way to do that. We can use some sort of "battery" to keep the dipole moment constant and calculate the work using $−\vec{m}.\vec{B}$. At the end of the process we put back all the energy given/stolen by the battery, which means I changed back the dipole moment to the value it should have been if the battery was not there. In other words I already get rid of all the influences of the battery. The work done by the battery turns out to be $(−2\vec{m}.\vec{B})$, then we get

$U=-\vec{m}.\vec{B}-(-2\vec{m}.\vec{B})=+\vec{m}.\vec{B}$

We can also get the same $U=+\vec{m}.\vec{B}$ if we calculate the total electromagnetic field energy, some details of the derivation is in my blog:

http://emitabsorb.wordpress.com/2011/08/21/m-b-or-m-b/

Magnetic field does no work on a proton, then how do we define potential energy?

Yes the total kinetic energy of the system is conserved, but we can separate it into parts. For example we can lump together the kinetic energy due to $v_x$ & $v_y$ and give it a name say $U_1$. The change in $U_1$ will affect the particle’s movement in $z$ direction, thus we can say that $U_1$ is the potential energy for $z$ direction. In this case we would like to know the tendency of the proton’s angular revolution velocity to align or counter-align with the magnetic field, so we lumped together part of the kinetic energy and magnetic field energy. As how it is derived, this energy can be written as $\tau=-dU/d\theta$. Thus if the lumped energy is not minimum, there will be torque perpendicular to $\vec{B}$.

So why don’t we also use $U=+\vec{m}.\vec{B}$ for the case of permanent dipole since it is the actual total energy with self energy already included in it?

Yes it is true that the right total energy is $U=+\vec{m}.\vec{B}$. But in this case the potential energy, the one that tends to minimize itself is $U=-\vec{m}.\vec{B}$. The part of energy that can minimize itself is the one that can be written as $F=-\nabla U$, that is to say the force will tend to any particle affected by the potential to the place where $U$ is lower. For example, consider a system of an earth and a moon orbiting it. Then suddenly the earth becomes twice as big as before with the same mass. We know that the self gravitational potential energy of the earth is changed, but it leaves no effect on the moon. So in this case the self gravitational potential energy of the earth is not a potential energy for the moon.

Now the only problem remaining in the permanent dipole case is that besides the $–\vec{m}.\vec{B}$ part which can change back and forth with mechanical energy, the remaining $+2\vec{m}.\vec{B}$(part of it is from the dipole’s self energy, and the remaining is from the self energy of the constant $\vec{B}$ field provider) part also changes mysteriously and which means the energy is not conserved. To save the principle conservation of energy we can always invent a new kind of energy so that the $+2\vec{m}.\vec{B}$ is not missing or being created freely but instead it is just changing its form between electromagnetic energy and this new energy. But I think it is not necessary, because what I was doing is not to protect the principle of conservation of energy, but instead to protect the field energy interpretation. Actually the field energy is also derived using the principle of virtual work in the first place, but in this case the increase in the total field energy is not equal to the decrease in mechanical energy. Thus for the case of permanent dipole I think the field energy interpretation is no longer valid. If we stick to the definition $F=-\nabla U$ , these difficulties would never occurs.

share|improve this answer
    
Here is my comment from chat: "Chris' answer makes sense, but he doesn't address the real problem here. The fact that the proton radiates doesn't dictate the fact that it goes clockwise. We already know that the proton goes clockwise from basic EM law (or if you want, Lenz law). And now, we're trying to force the -m.B paradigm into our system -> which definitely won't work because we're not dealing with a permanent magnet." Radiation damping is necessary for 'full' description of the system, but that's for different question. –  pcr Mar 18 '12 at 14:56
    
@Emitabsorb I think the battery argument works because in magnetostatics, we're dealing with a conservative system? Though it is no longer the case with the radiation included -> but again it's not the main issue, we can still explain the situation in the realm of magnetostatics –  pcr Mar 18 '12 at 15:07
    
No we cannot, the problem is that you are trying to use a quasistatic approach without taking into account the full dynamics. And this view is ensured by my Berkeley colleagues. –  Chris Gerig Mar 18 '12 at 22:52
    
Do you mind if we chat? this comment thread is way too long. . Here chat.stackexchange.com/rooms/2824/… . I can't agree with you and your Berkeley colleagues without really understanding why I'm mistaken. –  pcr Mar 18 '12 at 23:09
    
Yeah, this needs to be taken to the chat room. In the meantime I'll clean up the comments here. –  David Z Mar 19 '12 at 7:55
show 5 more comments

I’m not an expert in electromagnetism, but if one of my students (general physics, BSc level) asked, I'd say the following: the rotating proton indeed generates an magnetic field, and it behaves like a magnetic dipole for these purposes (and at large enough distances). However, that only concerns the field created by this proton, and its interaction of this proton with other particles at long distance. The interaction energy between magnetic field and proton cannot be described as $-\vec M\cdot \vec B$, because that expression is the energy of a point dipole, which our system is not. It's like you would be calculating the torque of something that's not an rigid solid, it just doesn't apply.

share|improve this answer
add comment

1) The answer of F'x is incorrect, there is definitely a magnetic dipole moment (and there is no self-energy here of its moment with its own magnetic field), and it indeed points downward. However, the proton accelerates in its cyclotron movement, inducing radiation and hence depleting the energy, reconciling this paradox.

2) Alexey's answer is flawed because his Lagrangian does not take into consideration the full dynamics. He also argues that his Lagrangian system is not conserved (but it is in his post).

3) They key response to Emitabsorb's answer:
$U=-m\cdot B$ is the interaction potential from a moment in a field; it only includes the work done in establishing an a priori permanent magnetic moment in the field. It does not include the work done in creating the magnetic moment and keeping it permanent, however. If we include this extra work, the total energy of the system becomes $U=+\bar{m}\cdot \bar{B}$. But you still want to minimize the interaction-potential only (this is explained in Jackson's Classical EM textbook, pg190 + pg214). HOWEVER, we assumed here that you first had a moment $m$ independent of any magnetic fields, and then brought it into $B$. This simply doesn't apply to our scenario... if you write down the Lagrangian for a proton in a magnetic field, there is no interaction potential $\pm m\cdot B$ that appears (because there is no a priori fixed moment); it is simply $\mathcal{L}=\frac{1}{2}m_0v^2-\frac{e}{c}\bar{v}\cdot \bar{A}$. This paragraph, coupled with my acceleration-remark, resolves the paradox!!!


My acceleration-remark:
In effect, you're trying to consider a quasistatic solution to this problem, without taking into consideration the full dynamics. The proton circles around and produces cyclotron radiation, damping its motion (i.e. radiation damping). You can form a Lagrangian to handle this, and minimum potential energy is achieved.

share|improve this answer
add comment

Here are my original resolutions to the paradox.

Looking at the other answer, this is most probably wrong, but I'll post it for completeness' sake. Also, it will emphasise what I suspected would happen: multiple resolutions to one paradox.

  • Magnetic field does no work on a proton, thus there is no PE and the whole discussion is moot. Note that this may be wrong as the proton itself influences the local magnetic field.

  • A proton doing UCM cannot be called a current loop. In a current loop, at any given instant, we have moving charges on every side. This gives us a couple. There is no such couple here.

share|improve this answer
add comment

The argument of minimal potential energy is not applicable here, as it works only for conservative forces. Consider lagrangian $L=T(\dot q_i)-U(q_i)$ with $q_i$ being some generalised coordinates. One assumes normally, that any motion causes dissipative forces to diminish $\dot q_i$ with time, and hence the term $T(\dot q_i)$. Lagrange equations then read off $\dfrac{\partial U}{\partial q_i}=0$ and consideration of how the system approaches that point will show that it is a local mininum. By the way, the principle is particularly nice for being versatile with respect to the choice of $q_i$.

Now consider the proton. Its lagrangian function is $L=\dfrac{m v^2}{2}+\dfrac{e}{c}{\bf A}{\bf v}$. Potential energy here may be either:

1) Considered zero. Then any state is the minimal energy state.

2) Taken to be $-\dfrac{e}{c}{\bf A}{\bf v}$. Then the system is not conservative and the argument doesn't apply.

Hence, the energy argument $U\rightarrow U_\min$ doesn't say anything about the state where the system shall be.

If you switch to magnetic momentum formulation, you use lagrangian $L=\dfrac{m v^2}{2}+{\bf m}{\bf H}$. Implicit assumption about ${\bf m}$ being fixed makes one expect that after enough disipation the system would reach ${\bf m}{\bf H}=-|{\bf m}{\bf H}|$. Instead, as seen from lagrangian, the system would just trivially reach the state ${\bf v}=0$ and hence ${\bf m}=0$.

To conclude, the paradox stems from applying $U\rightarrow U_\min$ and assuming ${\bf m}$ constant for dipole discription.

share|improve this answer
    
I think this is slightly mudddling the point, and is an attempted restatement of my answer. Your Lagrangian does not take into consideration radiation damping (in this case being cyclotron radiation), which is what my post is all about (and this is equivalent to "nonconservative force"). –  Chris Gerig Mar 20 '12 at 20:40
    
Indeed, you are making use of your Lagrangian and showing the the principle of minimum (total) energy does not hold here... But that is clear because the system is not closed, there is radiation (as I have stated). The real principle going on here is the minimum potential energy principle which holds for ALL systems, and that is what my post addresses. –  Chris Gerig Mar 20 '12 at 20:50
    
Thank you, Chris, but: 1) Radiation damping is a dissipative force and hence cannot be used in lagrangian, 2) Were there no radiation damping, the paradox would still hold, possibly in a modified form, involving statistics –  Alexey Bobrick Mar 20 '12 at 20:54
    
Chris, your post is great, I don't claim that you are wrong, or not original, but I just give a simple rigorous explanation of what is going on. Besides, the minimal potential energy principle doesn't hold for "ALL" systems, you are wrong here. –  Alexey Bobrick Mar 20 '12 at 20:57
    
And actually, I take it back, your argument is simply wrong because $-A\cdot v$ IS a generalized conservative potential energy. The fact is, you have not considered radiation. –  Chris Gerig Mar 20 '12 at 20:57
show 5 more comments

protected by Qmechanic Jan 23 '13 at 17:15

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.