Earnshaw's theorem

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Earnshaw's theorem states that a collection of point charges cannot be maintained in a stable stationary equilibrium configuration solely by the electrostatic interaction of the charges. This was first proved by Samuel Earnshaw in 1842. It is usually referenced to magnetic fields, but originally applied to electrostatic fields, and, in fact, applies to any classical inverse-square law force or combination of forces (such as magnetic, electric, and gravitational fields).

1 Explanation
2 Proofs for magnetic dipoles
3 References
4 External links

[edit] Explanation

This follows from Gauss's law. Intuitively, for a particle to be in a stable equilibrium, small perturbations ("pushes") on the particle in any direction should not break the equilibrium; the particle should "fall back" to its previous position. This means that the force field lines around the particle's equilibrium position should all point inwards, towards that position. (If any of the field lines pointed outwards, then if the particle moved along those field lines, it would leave its equilibrium, which would therefore not be stable.) Because all of the surrounding field lines point towards the equilibrium point, the divergence of the field at that point must be nonzero (in fact, negative). However, Gauss's Law says this is impossible: the force acting on an object F(x) (as a function of position) due to a combination of inverse-square law forces (forces deriving from a potential which satisfies Laplace's equation) will always be divergenceless (·F = 0) in free space. Therefore, there is no point in empty space where the force due to the field points inward from all directions, and a stable equilibria of particles cannot exist. There are no local minima or maxima of the field potential in free space, only saddle points.

This theorem also states that there is no possible static configuration of ferromagnets which can stably levitate an object against gravity, even when the magnetic forces are stronger than the gravitational forces. There are, however, several exceptions to the rule's assumptions which allow magnetic levitation.

Earnshaw’s Theorem, in addition to the fact that configurations of classical charged particles orbiting one another are also unstable due to electromagnetic radiation, pointed the way to quantum mechanical explanations of the structure of the atom.

[edit] Proofs for magnetic dipoles

Introduction

While a more general proof may be possible, three specific cases are considered here. The first case is a magnetic dipole of constant magnitude that has a fixed (unchanging) orientation. The second and third cases are magnetic dipoles where the orientation changes to remain aligned either parallel or anti-parallel to the field lines of the external magnetic field. In paramagnetic and diamagnetic materials the dipoles are aligned parallel and anti-parallel to the field lines, respectively.

Background

The proofs considered here are based on the following principles.

The energy U of a magnetic dipole M in an external magnetic field B is given by

$U = -\mathbf{M}\cdot\mathbf{B} = -(M_x B_x + M_y B_y + M_z B_z)$

The dipole will only be stably levitated at points where the energy has a minimum. The energy can only have a minimum at points where the Laplacian of the energy is greater than zero. That is, where

$\nabla^2 U = {\partial^2 U \over \partial x} + {\partial^2 U \over \partial y} + {\partial^2 U \over \partial z} > 0$

Finally, because both the divergence and the curl of a magnetic field are zero (in the absence of current or a changing electric field), the Laplacians of the individual components of a magnetic field are zero. That is

$\nabla^2 B_x = 0, \nabla^2 B_y = 0, \nabla^2 B_z = 0$

This is proved at the very end of this article as it is central to understanding the overall proof.

Summary of Proofs

For a magnetic dipole of fixed orientation (and constant magnitude) the energy will be given by

$U = -\mathbf{M}\cdot\mathbf{B} = -(M_x B_x + M_y B_y + M_z B_z)$

where $M x$ , $M y$ and $M z$ are constant. In this case the Laplacian of the energy is always zero

$\nabla^2 U = 0$

so the dipole can have neither an energy minimum or an energy maximum. That is, there is no point in free space where the dipole is either stable in all directions or unstable in all directions.

Magnetic dipoles aligned parallel or anti-parallel to an external field with the magnitude of the dipole proportional to the external field will correspond to paramagnetic and diamagnetic materials respectively. In these cases the energy will be given by

$U = -\mathbf{M}\cdot\mathbf{B} = -k\mathbf{B}\cdot\mathbf{B} = -k (B_x^2 + B_y^2 + B_z^2)$

Where k is constant greater than zero for paramagnetic materials and less than zero for diamagnetic materials.

In this case, it will be shown that

$\nabla^2 (B_x^2 + B_y^2 + B_z^2) \geq 0$

which, combined with the constant k, shows that paramagnetic materials can have energy maxima but not energy minima and diamagnetic materials can have energy minima but not energy maxima. That is, paramagnetic materials can be unstable in all directions but not stable in all directions and diamagnetic materials can be stable in all directions but not unstable in all directions. Of course, both materials can have saddle points.

Finally, the magnetic dipole of a ferromagnetic material (a permanent magnet) that is aligned parallel or anti-parallel to a magnetic field will be given by

$\mathbf{M} = k{\mathbf{B} \over |\mathbf{B}|}$

so the energy will be given by

$U = -\mathbf{M}\cdot\mathbf{B} = -k{\mathbf{B} \over |\mathbf{B}|}\cdot\mathbf{B} = -k{(B_x^2 + B_y^2 + B_z^2) \over (B_x^2 + B_y^2 + B_z^2)^{1/2}} = -k(B_x^2 + B_y^2 + B_z^2)^{1/2}$

but this is just the square root of the energy for the paramagnetic and diamagnetic case discussed above and, since the square root function is monotonically increasing, any minimum or maximum in the paramagnetic and diamagnetic case will be a minimum or maximum here as well.

It should be noted, however, there are no known configurations of permanent magnets that stably levitate so there may be other reasons not discussed here why it is not possible to maintain permanent magnets in orientations anti-parallel to magnetic fields (at least not without rotational motion - see Levitron).

Detailed Proofs

Earnshaw's Theorem was originally formulated for electrostatics (point charges) to show that there is no stable configuration of a collection of point charges. The proofs presented here for individual dipoles should be generalizable to collections of magnetics dipoles because they are formulated in terms of energy which is additive. A rigorous treatment of this topic, however, is currently beyond the scope of this article.

Fixed Orientation Magnetic Dipole

It will be proved that at all points in free space

$\nabla \cdot (\nabla U) = \nabla^2 U = {\partial^2 U \over \partial x} + {\partial^2 U \over \partial y} + {\partial^2 U \over \partial z} = 0$

The energy U of the magnetic dipole M in the external magnetic field B is given by

$U = -\mathbf{M}\cdot\mathbf{B} = -(M_x B_x + M_y B_y + M_z B_z)$

The Laplacian will be

$\nabla^2 U = -[ {\partial^2 (M_x B_x + M_y B_y + M_z B_z) \over \partial x} + {\partial^2 (M_x B_x + M_y B_y + M_z B_z) \over \partial y} + {\partial^2 (M_x B_x + M_y B_y + M_z B_z) \over \partial z} ]$

Expanding and rearranging the terms (and noting that the dipole M is constant) we have

$\nabla^2 U = -( M_x({\partial^2 B_x \over \partial x} + {\partial^2 B_x \over \partial y} + {\partial^2 B_x \over \partial z}) + M_y({\partial^2 B_y \over \partial x} + {\partial^2 B_y \over \partial y} + {\partial^2 B_y \over \partial z}) + M_z({\partial^2 B_z \over \partial x} + {\partial^2 B_z \over \partial y} + {\partial^2 B_z \over \partial z}) )$

$\nabla^2 U = -(M_x \nabla^2 B_x + M_y \nabla^2 B_y + M_z \nabla^2 B_z)$

but the Laplacians of the individual components of a magnetic field are zero in free space (not counting electromagnetic radiation) so

$\nabla^2 U = -(M_x 0 + M_y 0 + M_z 0) = 0$

which completes the proof.

Magnetic Dipole Aligned with External Field Lines

The case of a paramagnetic or diamagnetic dipole is considered first. The energy is given by

$U = -k (B_x^2 + B_y^2 + B_z^2)$

Expanding and rearranging terms,

but since the Laplacian of each individual component of the magnetic field is zero

$\nabla^2 (B_x^2 + B_y^2 + B_z^2) = 2[ | \nabla B_x |^2 + | \nabla B_y |^2 + | \nabla B_z |^2 ]$

and since the square of a magnitude is always positive

$\nabla^2 (B_x^2 + B_y^2 + B_z^2) \geq 0$

As discussed above, this means that the Laplacian of the energy of a paramagnetic material can never be positive (no stable levitation) and the Laplacian of the energy of a diamagnetic material can never be negative (no instability in all directions).

Further, because the energy for a dipole of fixed magnitude aligned with the external field will be the square root of the energy above, the same analysis applies.

Laplacian of Individual Components of a Magnetic Field

It is proved here that the Laplacian of each individual component of a magnetic field is zero. This shows the need to invoke the properties of magnetic fields that the divergence of a magnetic field is always zero and the curl of a magnetic field is zero in free space (that is, in the absence of current or a changing electric field). See Maxwell's equations for a more detailed discussion of these properties of magnetic fields.

Consider the Laplacian of the x component of the magnetic field

$\nabla^2 B_x = {\partial^2 B_x \over \partial x} + {\partial^2 B_x \over \partial y} + {\partial^2 B_x \over \partial z} = {\partial \over \partial x} {\partial \over \partial x} B_x + {\partial \over \partial y} {\partial \over \partial y} B_x + {\partial \over \partial z} {\partial \over \partial z} B_x$

Because the curl of B is zero,

${\partial B_x \over \partial y} = {\partial B_y \over \partial x}$

and

${\partial B_x \over \partial z} = {\partial B_z \over \partial x}$

so we have

$\nabla^2 B_x = {\partial \over \partial x} {\partial \over \partial x} B_x + {\partial \over \partial y} {\partial \over \partial x} B_y + {\partial \over \partial z} {\partial \over \partial x} B_z$

but since $B x$ is continuous the order of differentiation doesn't matter giving

$\nabla^2 B_x = {\partial \over \partial x}( {\partial B_x \over \partial x} + {\partial B_y \over \partial y} + {\partial B_z \over \partial z} ) = {\partial \over \partial x}(\nabla \cdot \mathbf{B})$

The divergence of B is constant (zero, in fact) so

$\nabla^2 B_x = {\partial \over \partial x}(\nabla \cdot \mathbf{B} = 0) = 0$

The Laplacian of the y component of the magnetic field $B y$ field and the Laplacian of the z component of the magnetic field $B z$ can be calculated analogously.

[edit] References

Earnshaw, S., On the nature of the molecular forces which regulate the constitution of the luminiferous ether., 1842, Trans. Camb. Phil. Soc., 7, pp 97-112.

[edit] External links

http://www.hfml.kun.nl/levitation-possible.html - A discussion of Earnshaw's theorem and its consequences for levitation, along with several ways to levitate with electromagnetic fields
Biography and other information about Samuel Earnshaw

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Categories: Electrostatics | Levitation | Physics theorems