PHYS 532 Lecture 11 Page 1

Conservation of Momentum

Consider a situation with free (conduction) currents only, no polarization or magnetization currents. We start by writing the electromagnetic force exerted on the particles in volume V. Then using Maxwell’s equations to eliminate ρ and J in favor of fields:

F = (X6.37)

We try to do to this equation what we did for the energy equation, namely express all the terms as either a divergence or a time derivative. Switching to index notation, the first term on the right becomes

(X6.38)

– !ij k " # E) k where εijk is the completely antisymmetric 3rd rank tensor.

Digression on the completely antisymmetric tensor. This tensor is defined as follows:

(X6.39)

It is useful for representing a cross product or curl: (X6.40)

(X6.41)

The following mathematical identity is very useful:

(X6.42) PHYS 532 Lecture 11 Page 2

Returning to the problem of “simplifying” (X6.38), we use Faraday's law to write

!(E E ) !(E E ) !B ( E) E = i j 1 j j k !" i - - #ijk Ej !xj 2 !xi !t

(X6.43)

Now let's work on the second term of (X6.37).

$B m $Bm [(! " B) " B] i = #ijk # jlm Bk = #kij # jlm Bk $x l $x l Now use the identity (X6.42) to write

!B i !B k = B k – Bk !x k !x i

However, ∂Bk/∂xk=0 because ∇ · B = 0, so that we have now

! 1 !"B " B i = BkBi - #ik BjBj (X6.44) !xk 2

Substituting (X6.43) and (X6.44) in (X6.37) gives

E E 1 B B F = d 3 x E E k k + B B k k E B (X6.45) i V o i j ij o i j ij ( o ijk j k ) xi 2 µo 2µo t

whi ch can also be written

PHYS 532 Lecture 11 Page 3

dP where mech = rate at which EM field transfers momentum to particles, per unit volume, and dt Poynting flux Pfield = momentum density in the EM field = εo E × B = c2

To see whether it is reasonable that momentum-density/Poynting-flux = c-2, consider a beam of photons with a number density N of photons, each with energy . The energy flux is .

The momentum density is . The ratio of momentum density to energy flux is k /ωc = c-2.

The “Maxwell stress tensor” is defined by

B B !o 2 i j 1 2 Tij = !oEiE j – " ijE + – " ij B (6.120) 2 µ o 2µ o = − flux of electromagnetic momentum

(Momentum flux is a tensor: Tij = flux of i-component of momentum transferred in j-direction.)

The momentum-conservation law including the electromagnetic field can be written

(6.122)

Electromagnetic momentum transferred into V through boundary where

PHYS 532 Lecture 11 Page 4

! Physical meaning of o E 2 : 2 " ij

Force and transmission direction are the same

! Similar stress tensors are used in theories of o E2 = “electric pressure” 2 continuum mechanics - elastic media and fluids. The non-diagonal terms are called 2 B = “magnetic pressure” “shear stress” 2µ o

For the εοEiEj term, consider the case where E||x

A −x force transmitted in the + direction is a tension acting along the electric field lines. Thus

2 εoE = electric tension (force per unit area) 2 B /µo = magnetic tension (force per unit area)

2 The term εοEiEj represents a tension εοE per unit area acting along the direction of E. 2 Similarly BiBj/µo represents a tension B /µo per unit area acting along B. PHYS 532 Lecture 11 Page 5

Applications of the Maxwell stress tensor

(A) Force between plates in a parallel-plate capacitor:

Q 2 Electric tension along field lines A !o Attractive force !oA ! Q 2 Electric pressure – A o Repulsive force 2 !oA ! Q 2 Net force A o Attractive 2 !oA

The same result can be obtained by integrating ρE through a plate, with ρ = εo∇⋅E. This gives 1 Q2 F = QE = QEinside = 2 2!oA

(B) Force on a planar current sheet (magnetic pressure):

Answer: B2 B2 1 ! 2 2 2 µo µo to the right.

The same result can be obtained by integrating J×B across the current layer, with J = ∇×B/µo. PHYS 532 Lecture 11 Page 6

(C) Magnetic tension in a planar current sheet:

Which way is the magnetic force on the current sheet? Answer: to the left.

It’s due to the magnetic tension on the field lines crossing the sheet. Note that a component of B normal to the current sheet is needed to transmit this force.

The homework (problem 3) asks you to work this out quantitatively for a specific case.

2 It’s OK to visualize magnetic field lines as rubber bands, with a tension B /µo per unit area. Just remember to also include the effect of magnetic pressure, if it has a gradient.

A configuration like the above example is appropriate for a magnetospheric tail. There is a 2 magnetic pressure gradient force [ - ∇( B /2µo )] toward the current sheet from both sides (top & bottom), which is balanced in equilibrium by a maximum of plasma pressure in the sheet. There is, in addition, a magnetic tension force to the left (toward the planet), which is balanced in equilibrium by an opposing gradient of magnetic + plasma pressure.

In a vacuum magnetic field configuration (e.g. a dipole field), there are large magnetic tension and pressure forces, but they cancel one another identically, because their sum is J×B, and J=0 in a vacuum.