Download Energy in the Fields Linear Momentum

To review, in our original presentation of Maxwell’s
equations, ρall and J~all represented all charges, both “free”
and “bound”. Upon separating them, “free” from
“bound”, we have (dropping quadripole terms):
I For the electric field
I
I
I
I
I
~ called electric field
E
P~ called electric polarization is induced field
~ called electric displacement is field of “free charges”
D
~ = 0 E
~ + P~
D
For the magnetic field
I
I
I
I
Physics 504,
Spring 2011
Electricity
and
Magnetism
Shapiro
Poynting,
Energy and
Momentum in
the fields, Tµν
Poynting’s
Theorem
Energy in the Fields
The rate of work done by E&M fields on charged particles
is:
X
~ xj , t),
qj ~vj · E(~
~ called magnetic induction (unfortunately)
B
~ called magnetization is the induced field
M
~ called magnetic field
H
~ = 1B
~ −M
~
H
µ0
Harmonic
Fields
Then the two Maxwell equations with sources, Gauss for
~ and Ampère, get replaced by
E
~
~ · E
~ ×H
~ +H
~ ·∇
~ ×E
~ = −∇
~ E
~ ×H
~ −H
~ · ∂B
−∇
∂t
where we used Faraday’s law. Thus
Physics 504,
Spring 2011
Electricity
and
Magnetism
Shapiro
Poynting,
Energy and
Momentum in
the fields, Tµν
Poynting’s
Theorem
Dispersion
Harmonic
Fields
dP~mech
dt
=
X
j
Z
=
V
F~j =
X
~ xj ) + ~vj × B(~
~ xj )
qj E(~
j
~ + J~ × B,
~
ρE
provided no particles enter or leave the region V .
Let us postulate that the electromagnetic field has a
linear momentum density
1 ~
~ = 0 E
~ × B.
~
~g = 2 E
×H
c
~
~ ×H
~ − ∂D
∇
∂t
!
~
· E.
we have
∂u ~ ~ ~ ~
~ = 0.
+J ·E+∇· E×H
∂t
As this is true for any volume, we may interpret this
equation, integrated over some volume V with surface ∂V
as saying that the rate of increase in the energy in the
fields plus the energy of the charged particles plus the flux
of energy out of the volume is zero, that is, no energy is
created or destroyed. The flux of energy is then given by
the Poynting vector
Physics 504,
Spring 2011
Electricity
and
Magnetism
Shapiro
Poynting,
Energy and
Momentum in
the fields, Tµν
Poynting’s
Theorem
Dispersion
Harmonic
Fields
~=E
~ × H.
~
S
Let us assume the medium is linear without dispersion in
~ ∝H
~ and
electric and magnetic properties, that is B
~ ∝ E.
~ Then let us propose that the energy density of
D
the fields is
1 ~ ~
~ ·H
~ ,
u(~x, t) =
E·D+B
2
So as not to worry about such complications, let’s restrict
our discussion to the fundamental description, or
alternatively take our medium to be the vacuum, with the
fields interacting with distinct charged particles (mj , qj at
~xj (t)).
The mechanical
linear momentum in some region of space
P
is P~mech = j mj ~x˙j , so
Harmonic
Fields
Now by Ampère’s Law,
~
~
~ · E
~ ×H
~ +E
~ · ∂D + H
~ · ∂ B + J~ · E
~ = 0.
∇
∂t
∂t
Linear Momentum
Dispersion
This must be the rate of loss of energy U in the fields
themselves, so
Z
dU
~
−
J~ · E.
=
dt
V
~ =
J~ · E
~ · (V
~ ×W
~ )=W
~ · (∇
~ ×V
~)−V
~ · (∇
~ ×W
~ ),
∇
~ ×H
~ ·E
~ as
so we may rewrite ∇
Poynting’s
Theorem
V
~ ·D
~ = ρ
∇
~
∂
D
~ ×H
~ −
∇
= J~
∂t
From the product rule and the cyclic nature of the triple
product, we have for any vector fields
Shapiro
Poynting,
Energy and
Momentum in
the fields, Tµν
or if we describe it by current density,
Z
~
J~ · E.
Dispersion
Physics 504,
Spring 2011
Electricity
and
Magnetism
We have made assumptions which only fully hold for the
vacuum, as we assumed linearity and no dispersion (the
~ to E
~ independent of time).
ratio of D
Physics 504,
Spring 2011
Electricity
and
Magnetism
Shapiro
Poynting,
Energy and
Momentum in
the fields, Tµν
Poynting’s
Theorem
Then the total momentum inside the volume V changes
at the rate
Z
dP~Tot
~ + J~ × B
~ + 0 ∂ (E
~ × B).
~
ρE
=
dt
∂t
V
~ ~
Using
Maxwell’s laws to substitute 0 ∇ · E for ρ and
~ ×B
~ − ∂ E/∂t
~
~
0 c2 ∇
for J,
Dispersion
Harmonic
Fields
dP~Tot
dt
Z
~
~ ∇
~ · E)
~ + c2 (∇
~ × B)
~ ×B
~ +B
~ × ∂E
E(
∂t
V
∂ ~
~
−
B×E
∂t
Z
~
~
~
~
~ × B)
~ ×B
~ − ∂B × E
~
E(∇ · E) + c2 (∇
= 0
∂t
V
Z
~ × B)
~ ×B
~ −E
~× ∇
~ ×E
~ .
~ ∇
~ · E)
~ + c2 (∇
= 0
E(
= 0
V
Physics 504,
Spring 2011
Electricity
and
Magnetism
Shapiro
Poynting,
Energy and
Momentum in
the fields, Tµν
Poynting’s
Theorem
Dispersion
Harmonic
Fields
Physics 504,
Spring 2011
Electricity
and
Magnetism
~,
For any vector field V
~ × (∇
~ ×V
~
V
α
=
X
αβγ Vβ γµν
βγµν
∂
Vν
∂xµ
Physics 504,
Spring 2011
Electricity
and
Magnetism
Shapiro
1 ∂V 2 X ∂Vα
−
Vβ
2 ∂xα
∂xβ
β
X ∂
1
Vα Vβ − V 2 δαβ .
= −
∂xβ
2
Poynting,
Energy and
Momentum in
the fields, Tµν
=
Poynting’s
Theorem
Dispersion
Harmonic
Fields
β
Shapiro
Tµν
1~2
1~2
,
= 0 Eµ Eν − E
δµν + c2 Bµ Bν − B
δµν
2
2
which is called the Maxwell stress tensor. Then
!
XZ ∂
dP~Tot
=
Tµν .
dt
V ∂xν
ν
Poynting,
Energy and
Momentum in
the fields, Tµν
Poynting’s
Theorem
Dispersion
Harmonic
Fields
µ
So we see that the E terms in dP/dt may be written as
∂
1~2
0 Eα Eβ − E
δαβ .
∂xβ
2
By Gauss’s law,Pthe integral of this divergence over V is
the integral of β Tαβ n̂β over the surface ∂V of the
volune considered, so Tαβ n̂β is the flux of the α
component of momentum out of the surface.
The same may be done for the magnetic field, as the
~ ·B
~ is zero. Thus we define
missing ∇
Complex Fields, Dispersion
Physics 504,
Spring 2011
Electricity
and
Magnetism
~ = E
~ and B
~ = µH,
~ but
In linear media we can assume D
actually this statement is only good for the Fourier
transformed (in time) fields, because all media (other than
the vacuum) exhibit dispersion, that is, the permittivity and magnetic permeability µ depend on frequency. So we
need to deal with the Fourier transformed fields1
Z ∞
~ x, ω)e−iωt
~ x, t) =
dω E(~
E(~
−∞
Z ∞
~ x, t) =
~ x, ω)e−iωt
D(~
dω D(~
Shapiro
Poynting,
Energy and
Momentum in
the fields, Tµν
Poynting’s
Theorem
Dispersion
Harmonic
Fields
Note that the electric and magnetic fields in spacetime
are supposed to be real, not complex, valued. From
Z ∞
~ x, t) =
~ x, ω)e−iωt
E(~
dω E(~
−∞
Z ∞
Z ∞
~ ∗ (~x, t) =
~ ∗ (~x, ω)eiωt =
~ ∗ (~x, −ω)e−iωt
=E
dω E
dω E
−∞
−∞
~ x, ω) = E
~ ∗ (~x, −ω), and similarly for the
which tells us E(~
other fields. Thus the permittivity and permeability also
obey
(−ω) = ∗ (ω),
µ(−ω) = µ∗ (ω).
Physics 504,
Spring 2011
Electricity
and
Magnetism
Shapiro
Poynting,
Energy and
Momentum in
the fields, Tµν
Poynting’s
Theorem
Dispersion
Harmonic
Fields
The power transferred to charged particles includes an
integral of
ZZ
~
−i(ω−ω 0 )t
~ · ∂D =
~ ∗ (ω 0 )(−iω(ω)) · E(ω)e
~
E
,
dωdω 0 E
∂t
~
where we took the complex conjugate expression for E(t),
but alternatively we could have taken the complex
~ and interchanged ω and ω 0 ,
conjugate expression for D
−∞
and we define linear permittivity as
~ x, ω), and similarly
D(~x, ω) = (ω)E(~
~ x, ω). The inverse fourier transform
B(~x, ω) = µ(ω)H(~
does not like multiplication — if (ω) is not constant, we
~ x, t) proportional to E(~
~ x, t).
do not have D(~
1
The Maxwell Stress Tensor
Note the 2π inconsistency with 6.33, and slide 8 of lecture 1.
~
~ · ∂D =
E
∂t
ZZ
Physics 504,
Spring 2011
Electricity
and
Magnetism
0
−i(ω−ω )t
~ ∗ (ω 0 )(iω 0 ∗ (ω 0 )) · E(ω)e
~
,
dωdω 0 E
Shapiro
or averaging the two expressions
~
~ ∂D = i
E·
∂t
2
ZZ
0 ~∗
0
0 ∗
0
−i(ω−ω 0 )t
~
(ω )−ω(ω)]e
dωdω E (ω )·E(ω)[ω
We are often interested in the situation where the fields
are dominately near a given frequency, and if we ignore
the rapid oscillations in this expression which come from
one ω positive and one negative, we may assume the ω’s
differ by an amount for which a first order variation of is
enough to consider, so
ω 0 ∗ (ω 0 ) ≈ ω∗ (ω) + (ω 0 − ω)d(ω∗ (ω))/dω, and
[iω 0 ∗ (ω 0 ) − iω(ω)] = 2ωIm (ω) − i(ω − ω 0 )
d
(ω∗ (ω))
dω
,
Poynting,
Energy and
Momentum in
the fields, Tµν
Poynting’s
Theorem
Dispersion
Harmonic
Fields
~ · ∂ D/∂t,
~
Inserting this back into E
the −i(ω − ω 0 ) can be
interpreted as a time derivative, so
ZZ
~
0
~ · ∂D =
~ ∗ (ω 0 ) · E(ω)ωIm
~
E
(ω)e−i(ω−ω )t
dωdω 0 E
∂t
ZZ
∂ 1
d ∗ −i(ω−ω0 )t
~ ∗ (ω 0 ) · E(ω)
~
+
.
dωdω 0 E
ω (ω) e
∂t 2
dω
If were pure real, the first term would not be present,
and if were constant, the d[ω∗ (ω)]/dω would be ,
consistent with the u = 12 E 2 we had in our previous
consideration. There is, of course, a similar result from
~ · dB/dt
~
the H
term. More generally, we can think of the
first term as energy lost to the motion of bound charges
~ which goes
within the molecules, not included in J~ · E,
into heating the medium, while the second term describes
the energy density in the macroscopic electromagnetic
fields.
Physics 504,
Spring 2011
Electricity
and
Magnetism
Shapiro
Poynting,
Energy and
Momentum in
the fields, Tµν
Poynting’s
Theorem
Dispersion
Harmonic
Fields
Harmonic Fields
We are often interested in considering fields oscillating
with a given frequency, so
~ x, t) = Re
E(~
i
h
~ x)e−iωt = 1 E(~
~ ∗ (~x)eiωt .
~ x)e−iωt + E
E(~
2
~ x, t), the dot product
If we have another such field, say J(~
is
~ x, t) · E(~
~ x, t)
J(~
i
1 h ~
~ x)e−iωt + E
~ ∗ (~x)eiωt
=
J(~x)e−iωt + J~∗ (~x)eiωt · E(~
4
h
i
1
~ x) + J(~
~ x) · E(~
~ x)e−2iωt .
=
Re J~∗ (~x) · E(~
2
Physics 504,
Spring 2011
Electricity
and
Magnetism
Shapiro
Poynting,
Energy and
Momentum in
the fields, Tµν
Poynting’s
Theorem
Dispersion
Harmonic
Fields
The second term is rapidly oscillating, so can generally be
ignored, and the average product is just half the product
of the two harmonic (fourier transformed) fields, one
complex conjugated.
~ = 1E
~ ×H
~ ∗ and
We define the complex Poynting vector S
2
the harmonic densities of electric and magnetic energy,
~ ·D
~ ∗ and wm = 1 B
~ ·H
~ ∗ , and thus find (using
we = 14 E
4
Gauss’s divergence law)
Z
I
Z
1
~ · n̂ = 0,
~ + 2iω (we − wm ) +
J~∗ · E
S
2 V
V
∂V
which, as a complex equation, contains energy flow. If the
medium may be taken as pure lossless dielectrics and
magnetics, with real and µ, the real part of this
equation says
Z
I
1
~ +
~ · n̂ = 0,
Re J~∗ · E
Re S
2 V
∂V
which says that the power transferred to the charges and
that flowing out of the region, on a time averaged basis, is
zero. But there is also an oscillation of energy between
the electric and magnetic fields. For example, with a pure
~ and H
~ in phase, S is real,
plane wave in vacuum, with E
J~ vanishes, and the imaginary part tells us the electric
and magnetic energy densities are equal.
Physics 504,
Spring 2011
Electricity
and
Magnetism
Shapiro
Poynting,
Energy and
Momentum in
the fields, Tµν
Poynting’s
Theorem
Dispersion
Harmonic
Fields
Fourier transforming Maxwell’s equations, for the
Harmonic fields, for which ∂/∂t becomes −iω, we see
~ ·B
~ =0
∇
~
~
∇·D =ρ
~ ×E
~ − iω B
~ =0
∇
~
~
~ = J~
∇ × H + iω D
Physics 504,
Spring 2011
Electricity
and
Magnetism
Shapiro
Poynting,
Energy and
Momentum in
the fields, Tµν
Poynting’s
Theorem
The current induced will also be harmonic, so the power
Dispersion
lost to current is
Harmonic
Fields
Z
1
3 ~∗ ~
d xJ · E
2
Z
1
~· ∇
~ ×H
~ ∗ − iω D
~∗
=
d3 xE
2
Z
i
h
1
~ ×E
~ · H ∗ − iω E
~ · E
~ ×H
~∗ + ∇
~ ·D
~∗
=
d3 x −∇
2
Z
i
h
1
~ · H ∗ − iω E
~ · E
~ ×H
~ ∗ + iω B
~ ·D
~∗
=
d3 x −∇
2
Rotational and PCT properties
Let us briefly describe how our fields behave under
rotational symmetry, reflection, charge conjugation and
time reversal.
From mechanics, forces are vectors, so as a charge
~ E
~ is a vector under rotations,
experiences a force q E,
unchanged under time reversal, reverses in sign under
~ → −E,
~ as a
charge conjugation, and under parity E
proper vector should.2
Charge and charge density are proper scalars, reversing
under charge conjugation (by definition).
~ to be a proper
Velocity is a proper vector, so for q~v × B
~ must be a pseudovector, whose
vector as well, B
components are unchanged under parity, because the cross
product (and αβγ ) are multiplied by −1 under parity. Of
~ odd under charge conjugation.
course we also have B
Physics 504,
Spring 2011
Electricity
and
Magnetism
Shapiro
Poynting,
Energy and
Momentum in
the fields, Tµν
Poynting’s
Theorem
Dispersion
Harmonic
Fields
2
under reflection in a mirror, the component perpendicular to the
mirror reverses, but parity includes a rotation of 180◦ about that
axis, reversing the other two components as well.
Physics 504,
Spring 2011
Electricity
and
Magnetism
Under time-reversal, forces and acceleration are
~
unchanged (∝ d2 /dt2 ) but velocity changes sign, so E
~ are even under time-reversal, but B
~ (and
(and P~ and D)
~ and H)
~ are odd under time-reversal, that is, the get
M
multiplied by −1.
Maxwell’s equations, the Lorentz force law, and all the
other formulae we have written are consistent with
symmetry under rotations and P,C, and T reflections.
That means the laws themselves are invariant under these
symmetries.
Energy in Harmonic Fields
Shapiro
Poynting,
Energy and
Momentum in
the fields, Tµν
Poynting’s
Theorem
Dispersion
Harmonic
Fields
Magnetic Monopoles
~ and B
~ almost
In vacuum, Maxwell’s equations treat E
!
~
E
~
identically. Then if we consider a doublet D =
~ the
cB
two Gauss’ laws say
~ ·D
~ = 0,
∇
~
~ × D + iσ2 1 ∂ D = 0,
∇
c ∂t
0 1
where iσ2 =
. This not only lets us write 4 laws
−1 0
as 2, but shows that the equations would be unchanged
by a rotation in this two dimensional space. But this
symmetry is broken by our having observed electric
charges and currents but no magnetic charges (magnetic
monopoles). But maybe we just haven’t found them yet,
and we should add them in.
Physics 504,
Spring 2011
Electricity
and
Magnetism
Shapiro
Poynting,
Energy and
Momentum in
the fields, Tµν
Poynting’s
Theorem
Dispersion
Harmonic
Fields
Maxwell with Monopoles
Physics 504,
Spring 2011
Electricity
and
Magnetism
Then Maxwell’s equations become
~
~ ×H
~ = ∂ D + J~e
Shapiro
∇
∂t
Poynting,
~
Energy and
~ ·B
~ = ρm ,
~ ×E
~ = ∂ B + J~m
∇
−∇
Momentum in
∂t
the fields, Tµν
Poynting’s
From these equations and our previous symmetry
Theorem
Dispersion
properties, we see that ρe is a scalar, ρm is a pseudoscalar,
Harmonic
Fields
J~e a proper vector and J~m a pseudovector. Define the
doublets3
!
!
!
!
1
1
1
1
2 ~
2 ~
µ02 ρe
µ02 J~e
~ = µ01 D
~ = 01 E
R=
B
J~ =
H
1
1
~
~
µ02 H
02 B
02 J~m
02 ρm
~ ·D
~ = ρe ,
∇
Physics 504,
Spring 2011
Electricity
and
Magnetism
Shapiro
Thus we see that the equations are invariant under a
simultaneous rotation of these doublets in the two
dimensional space, in particular
~ →E
~ cos ξ + Z0 H
~ sin ξ.
E
~ is a proper vector and H
~ is
But this is very pecular, as E
a pseudovector. So such an object would not have a well
defined parity.
Poynting,
Energy and
Momentum in
the fields, Tµν
Poynting’s
Theorem
Dispersion
Harmonic
Fields
so Maxwell’s equations become
~ ·B
~ = R,
∇
~
~ × H = ∂ B + J~ .
iσ2 ∇
∂t
√
√
3
The 0 and µ0 are due to the unfortunate SI units we are
p
using. Let Z0 = µ0 /0 .
Quantization of Charge
Dirac noticed that a point charge and a point monopole,
both at rest, gives a momentum density and an angular
momentum density in the electromagnetic fields. Consider
a monopole at the origin, ρm = gδ(~x), and a point charge
ρe = qδ(~x − (0, 0, D)) on the z axis a distance D away.
Physics 504,
Spring 2011
Electricity
and
Magnetism
Shapiro
Poynting,
Energy and
Momentum in
the fields, Tµν
Poynting’s
Theorem
We have magnetic and electric fields
But the angular momentum density, ~r × ~g , has a
component everywhere in the −z direction, so its integral
will not vanish.
~ =
L
=
Dispersion
~ =
H
g êr
,
4πµ0 r2
~ =
E
q êr0
.
4π0 r0 2
Harmonic
Fields
B
r’
The momentum density in the fields
~ × H,
~ which points in the
is ~g = c12 E
aximuthal direction, ∝ êφ in spherical coordinates. As the magnitude is
independent of φ, when we integrate
this vector in φ, we will get zero, so
the total momentum vanishes.
g
r
E
e2
h2 0
= 4 =
4π0
e µ0
h0 c
e2
g0
4π
g0
4π
Z
~ × ~r)
g
~r × (E
~ × H)
~ =
d3 r ~r × (E
d3 r
2
4πµ0 c
r3
Z
Z
~ − ~r(~r · E)
~
r2 E
g0
~ · ∇)
~ ~r
=
d3 r
d3 r (E
r3
4π
r
I
Z
~
r
~
r
~ ·E
~
~ −
∇
n̂ · E
r
∂V
V r
Shapiro
Poynting,
Energy and
Momentum in
the fields, Tµν
Poynting’s
Theorem
Dispersion
Harmonic
Fields
~ = − qg êz .
L
4π
There are several interesting things about this result. One
is that it is independent of how far apart the two objects
are. But even more crucial, if we accept from quantum
mechanics the requirement that angular momentum is
quantized in usits of ~/2, we see that if just one monopole
of magnetic charge g exists anywhere, all purely electric
charges must be
qn = nh/g
where n is an integer. Furthermore, as
there are electrons, the smallest nonzero monopole has a
“charge” at least h/e, and the Coulomb force between two
such charges would be
Z
Taking the volume to be a large sphere, the first term
~ is constant/r2 , so the integral is
vanishes
because n̂ · E
R
∝ dΩn̂ = 0. In the second term
~ ·E
~ = qδ 3 (~r − (0, 0, D)/0 so overall
∇
L density
Charge Quantization
g2
4πµ0
=
1
c2
Physics 504,
Spring 2011
Electricity
and
Magnetism
2
=
137
2
2
∼ 4700
times as big as the electric force between two electrons at
the same separation.
Physics 504,
Spring 2011
Electricity
and
Magnetism
Shapiro
Poynting,
Energy and
Momentum in
the fields, Tµν
Poynting’s
Theorem
Dispersion
Harmonic
Fields
Vector Potential and Monopoles
The ability to define vector and scalar potentials to
represent the electromagnetic fields depended on the two
sourceless Maxwell equations. If we have monopoles,
these conditions don’t apply whereever a monopole exists,
~ is not divergenceless everywhere. Even if ∇
~ ·B
~
so that B
~ cannot be
fails to vanish at only one point, it means B
written as a curl throughout space. Poincaré’s Lemma
tells us it is possible on a contractible domain, which is
not true for a sphere surrounding the monopole. One can
~ consistently everywhere other than on a “Dirac
define A
~ is
string” extending from the monopole to infinity, but A
not defined on the string.
Physics 504,
Spring 2011
Electricity
and
Magnetism
Shapiro
Poynting,
Energy and
Momentum in
the fields, Tµν
Poynting’s
Theorem
Dispersion
Harmonic
Fields