Download Electromagnetic Angular Momentum

CURIOSITIES IN THEORETICAL PHYSICS – V
Electromagnetic Angular Momentum
T. PADMANABHAN
IUCAA, Pune 411007
([email protected])
You would have certainly learnt that
electromagnetic field possesses energy and
momentum. The usual expressions for energy
per unit volume (U) and momentum per unit
volume (P) are
U=
1
1
( E 2 + B 2 ); P =
(E × B)
8π
4 πc
(1)
What is not stressed adequately in text
books is that electromagnetic field − and pretty
simple ones at that − also possesses angular
momentum. Just as the electromagnetic field
can exchange its energy and momentum with
charged particles, it can also exchange its
angular momentum with a system of charged
particles often leading to rather surprising
results. In this installment, we shall explore one
such example.
One simple configuration in which
Physics Education • January − March 2007
exchange of angular momentum occurs is
shown in Figure 1, discussed in volume II of
Feynman Lectures in Physics.1 A plastic disk,
located in the x – y plane, is free to rotate about
the vertical z-axis. On the disk is embedded a
thin metallic ring of radius a carrying a
uniformly distributed charge Q. Along the zaxis, there is a thin long current carrying
solenoid producing a magnetic field B
contributing a total flux Φ. This initial
configuration is completely static with a
magnetic field B confined within the solenoid
and an electric field E produced by the charge
located on the ring. Let us suppose that the
current source is disconnected leading the
magnetic field to die down. The change in the
magnetic flux will lead to an electric field
which will act tangential to the ring of charge
thereby giving it a torque. Once the magnetic
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field has died down, this torque would have
resulted in the disk spinning about the z-axis
with a finite angular momentum.
Feynman, as usual, makes a song and dance
about this problem but it is pretty much
obvious that the angular momentum in the
initial field is what appears as the mechanical
angular momentum of the rotating disk in the
The angular momentum of the final rotating
disk is easy to compute. The rate of change of
angular momentum dL/dt due to the torque
acting on the ring of charge is along the z-axis
and so we only need to compute its magnitude.
This is given by
dL
Q
Q ∂Φ
E. dl = −
= aQE =
2π
2 πc ∂ t
dt
∫
(2)
Here E is the tangential electric field
generated due to the changing magnetic field
and the last equality follows from the
Faraday’s law. Integrating this equation and
noting that the initial angular momentum of the
disc and the final magnetic flux are zero, we
get
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final stage. What is really important and
interesting is to work this out and explicitly
verify that the angular momentum is conserved
(which Feynman doesn't do!). Let us see what
is involved. (There is large literature on this
problem not all of which is illuminating; one
place to start the search is from ref.[2].)
L=
Q
Φ initial
2πc
(3)
It is interesting that the final angular
momentum depends only on the total flux and
not on other configurational details.
We now need to show that the initial static
electromagnetic configuration had this amount
of stored angular momentum. I will first do this
in a rather unconventional manner and then
indicate the connection with the more familiar
approach. To do this, let us recall that the
canonical momentum of a charge q located in a
magnetic field is given by p – (q/c)A where A
is the vector potential related to the magnetic
field by B =−∇×A and p is the usual
momentum. This suggests that one can
Physics Education • January − March 2007
associate with charges located in a magnetic
field, a momentum (q/c)A. For a distribution of
charge, with a charge density ρ, the field
momentum per unit volume will be (1/c)ρA.
Hence, to a charge distribution located in a
region of vector potential A, we can attribute
an angular momentum
LA =
∫
1
d 3 xρ(x)[x × A (x)]
c
(4)
In our problem, the charge distribution is
confined to a ring of radius a and there is
negligible magnetic field at the location of the
charge. But the vector potential will exist
outside the solenoid and the above expression
can be non zero. To compute this, let us use a
cylindrical coordinate system with (r,θ,z) as the
coordinates. We will choose a gauge in which
the vector potential has only the tangential
component; that is, only Ap is non zero. Using
∫
A.dl = Φ
(5)
where Φ is the total magnetic flux, we get
2πrAp = Φ for a line integral of A around any
circle. Hence Ap = Φ/(2πr). This can be written
in a nice vectorial form as
A=
Φ
( z$ × r )
2 πr 2
(6)
where ẑ is the unit vector in the z-direction.
When we substitute this expression in Eq.(4)
and calculate the angular momentum, the
integral gets contribution only from a circle of
radius a. Using further the identity,
r × ( z$ × r ) = z$ r 2 we get the result that
LA =
Q
Φ initial z$
2πc
(7)
which is exactly the final angular momentum
which we computed in Eq.(3). Rather nice!
This elementary derivation, as well as the
expression for electromagnetic angular
Physics Education • January − March 2007
momentum in Eq.(4) raises several intriguing
issues. On the positive side, it makes vector
potential a very tangible quantity, something
which we learnt from relativity and quantum
mechanics but could never be clearly
demonstrated within the context of classical
electromagnetism. In the process, it also gives
a physical meaning to the field momentum
(q/c)A which is somewhat mysterious in
conventional approaches. On the flip side, one
should note that A, by very definition, is gauge
dependent and one would have preferred a
definition
of
electromagnetic
angular
momentum which is properly gauge invariant.
It is, of course, possible to write down
another expression for the electromagnetic
angular
momentum
which
is
more
conventional. Using the definition of the field
momentum density, one can define an angular
momentum by
LA =
=
∫
1
d 3xρ(x)[x × A (x)] LEM
c
∫
1
d 3x[x × (E × B )
4 πc
(8)
which just replaces the momentum density
ρA/c in Eq.(4) by E×B/4πc. It is trivial to
verify that, as momentum densities, these two
expressions are, in general, unequal. But what
is relevant, as far as our computation goes, is
the integral over the whole space of these two
expressions. If these two expressions differ by
terms which vanish when integrated over
whole space, then we have an equivalent,
gauge invariant, definition of field angular
momentum.
It turns out that this is indeed the case in
any static configuration if we choose to
describe the magnetic field in a gauge which
satisfies ∇.A = 0. One can then show that
1
1
(E × B) α =
(E × (∇ × B )) α
4π
4π
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= ρA α + ∂ β + ∂ βV βα
(9)
where Vβα is a complicated second rank tensor
built out of field variables, ∂α is a condensed
notation for the partial derivative ∂/∂xα and we
are using the convention that Greek indices are
summed over 1,2,3. While one can provide a
proof of Eq.(9) using vector identities (you
should try it out!), it is a lot faster and neater to
use four dimensional notation and special
relativity to get this result. Let me briefly
outline this for those who are familiar with the
four dimensional notation.
We begin with the expression for the
momentum density of the electromagnetic field
in terms of the stress tensor Tab of the
electromagnetic field. The T00 component of
this tensor is proportional to the energy density
of the electromagnetic field while the T0α is
proportional to the momentum density Pα.
More precisely,
1
T0α =
(E × B ) α = cP α
4π
(10)
On the other hand, the electromagnetic
stress tensor can be written in terms of the four
dimensional field tensor Fab (with the
convention that Latin letters range over 0,1,2,3)
in the form T0α = − 1(1 / 4 π) F αβ F0β . We will
manipulate this expression using the facts that
(i) the configuration is static and (ii) the vector
potential satisfies the gauge condition
∇ ⋅ A = ∂ α A α = 0 to prove Eq.(9). Using the
definition of the field tensor in terms of the
four vector potential, Fij = ∂ i A j − ∂ i A j , we can
write
T0α = −
=−
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1 αβ
1 α β β α
F F0β = −
( ∂ A − ∂ A ) F0β
4π
4π
∂ β F0β
1 α β
1 β
( ∂ A ) F0β +
∂ ( F0β A α ) − A α
4π
4π
4π
=−
1
1 β
∂ ( F0β A α )
( − ∂ α A β ∂ β Aα ) +
4π
4π
− Aα
∂ β F0β
4π
(11)
To arrive at the second line we have used
the chain rule for differentiation and to obtain
the third line we have used ∂0Aβ=0 since the
configuration is time independent. We next use
the result ∂βF0β = −∇.E = −4πρ in the last term
and another application of chain rule in the first
term, using the gauge condition ∇.A = ∂αAα =
0. This gives
1
T0α = ρA α +
∂ β [ A0 ∂ α Aβ − A α ∂ β A0 ] (12)
4π
We thus find that
cP α = ρA α + ∂ βV βα ;
V βα ≡
1
[ A0 ∂ α Aβ − A α ∂ β A0 ]
4π
(13)
which proves the equivalence between the two
expressions for electromagnetic momentum
density, when used in integrals over all space,
provided the second term vanishes sufficiently
fast. For the case we are discussing, this is
indeed true.
Now that everything appears to be well
settled, I will leave you with two questions to
worry about. First, let us take one more look at
the final configuration. The solenoid has no
current and it produces no magnetic field. But
we do have a ring of charge which is rotating
around, which constitutes a current loop!
Surely this current produces a magnetic field.
We also have the electric field due to the
charged ring itself. Together, won’t this
configuration have some field angular
momentum? Then our angular momentum
balance seems to go for a toss! Second, let us
note that nowhere in the problem we really
worried about the radius l of the solenoid in the
initial configuration. So the results should hold
Physics Education • January − March 2007
even if we take the radius to be infinitesimally
small with a correspondingly large magnetic
field, keeping the flux πBl2 constant. But now
the magnetic field is confined to the z-axis.
(Mathematically, it is given by Bz(x,y,z) =
Φδ(x)δ(y).) The electric field on the z-axis is
along the z-axis as well − which means that
E×B vanishes and there is no angular
momentum according to Eq.(8) while Eq.(4),
of course gives the same angular momentum as
before! What is going on?
Physics Education • January − March 2007
References
1. Feynman, R.P., Feynman Lectures in Physics,
Volume II; section 17-4, McGraw-Hill, New York
(1970).
2. J.M. Agqirregabiria, A.Hernandez, (1981)
Eur.J.Phys., 2, 168.
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