Potential Energy of a Dipole

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Homework Statement


The problem is to show that the potential energy of a dipole (whether electric or magnetic) is given by ##U = -\vec{m}\cdot\vec{B}## in the magnetic case and ##U = -\vec{p}\cdot\vec{E}## in the electric case assuming the respective magnitudes of the dipole moment remain fixed. Since the argument is identical in both cases, I will stick with the magnetic dipole throughout.

Homework Equations


Torque on a magnetic dipole: ##\vec{N} = \vec{m} \times \vec{B}##
Force on a magnetic dipole: ##\vec{F} = \nabla(\vec{m}\cdot\vec{B})##

The Attempt at a Solution


First of all, the potential energy of a dipole is the work required to bring it in from infinity and place it in its final position. The question now is how should the dipole be brought in. I propose to move the dipole in from infinity when it already points in its final direction. Thus, I will not have to do any rotational work and the only work that is left is translational. From here the argument is straight-forward: [tex]U = -\int\vec{F}\cdot\vec{dl} = -\int_\infty^r \nabla(\vec{m}\cdot\vec{B})\cdot\vec{dl} = -\vec{m}\cdot\vec{B}|_\infty^r = -\vec{m}\cdot\vec{B}[/tex] where I have used the fundamental theorem for gradients and the assumption that the energy at ##\infty## is zero. However, virtually all of the websites I have seen derive the equation via bringing the dipole so that it is perpendicular to the field and thus, do no translational work. Then they calculate only the rotational work. (See for example http://hyperphysics.phy-astr.gsu.edu/hbase/electric/diptor.html and http://www.chem.ox.ac.uk/teaching/Physics%20for%20CHemists/Electricity/Dipoles.html [Broken]).
Is there anything flawed with my argument that causes everyone to apply the rotational one? As far as I can see, my argument is simpler and I can detect no problems with it.
Any comments/corrections will be greatly appreciated.
 
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Answers and Replies

  • #3
ehild
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That is the "translational" potential energy of an aligned dipole in inhomogeneous magnetic field . But the dipole can rotate, and in a lot of practical cases the question is how the energy of the dipole changes during its rotation in a homogeneous magnetic field.
 
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That is the "translational" potential energy of an aligned dipole in inhomogeneous magnetic field . But the dipole can rotate, and in a lot of practical cases the question is how the energy of the dipole changes during its rotation in a homogeneous magnetic field.
But isn't the formula identical in both cases, namely: ##U = -\vec{m}\cdot\vec{B}## ?
 
  • #5
ehild
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What would be your integral in uniform magnetic field?

The dipole can both translate and rotate. In constant field, the magnetic field has no force on the dipole, so no work is done if it translates. But there is torque on the dipole if it is not aligned to the field. In stable position it is aligned in the direction of the field. You need to do work against the torque of the field if you want to change the direction of the dipole. How do you determine the potential energy of the dipole if it is not aligned to B?
 
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What would be your integral in uniform magnetic field?

The dipole can both translate and rotate. In constant field, the magnetic field has no force on the dipole, so no work is done if it translates. But there is torque on the dipole if it is not aligned to the field. In stable position it is aligned in the direction of the field. You need to do work against the torque of the field if you want to change the direction of the dipole. How do you determine the potential energy of the dipole if it is not aligned to B?
Well, suppose that the dipole starts out at an angle ##\phi_0## with the uniform magnetic field and its final angle is ##\phi##. The torque exerted on it would be ##\vec{N} = \vec{m} \times \vec{B}## and its magnitude ##N = mB\sin\phi'##. Then, the work required to rotate the dipole to its final position: ##U = \int_{\phi_0}^{\phi}mB\sin\phi'\ d\phi' = mB(-\cos\phi')|_{\phi_0}^{\phi} = mB(\cos\phi_0 - \cos\phi)##. If, for convenience, we choose ##U = 0## when ##\phi_0 = \pi/2## (since the magnitude of the potential doesn't matter, only its difference), we get ##U = -mB\cos\phi = -\vec{m}\cdot\vec{B}## just as we got with the translational approach.
My original integral indeed vanishes, but so is this one if we start with the dipole aligned already in its final position. So is it a coincidence that both of them come out the same and in general, translational and rotational potential energies are not the same? Or are both approaches valid and general?
 
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ehild
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The translational an rotational energies are different things. Translation and rotation are different degrees of freedom. The kinetic energies are also different.
 
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  • #8
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The translational an rotational energies are different things. Translation and rotation are different degrees of freedom. The kinetic energies are also different.
I'm afraid that I still don't fully get it. The question asks about the total potential energy of a dipole in a magnetic field, and I get the exact same formula for both a purely rotational work and purely translational. Are both of them valid and give the correct general formula for the total potential energy?
By the way, thank you for the replies, they are appreciated!
 
  • #10
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I'm afraid that I still don't fully get it. The question asks about the total potential energy of a dipole in a magnetic field, and I get the exact same formula for both a purely rotational work and purely translational. Are both of them valid and give the correct general formula for the total potential energy?
By the way, thank you for the replies, they are appreciated!
Yes, both are valid and give the same result. Note that a constant offset for the energy does not have a physical relevance (unless we go to special relativity), only energy differences are important.
 
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