Download Today in Physics 218: gauge transformations

Today in Physics 218: gauge transformations
More updates as we move from
quasistatics to dynamics:
‰ #3: potentials
‰ For better use of potentials:
gauge transformations
‰ The Coulomb and Lorentz
gauges
‰ #4: force, energy, and
momentum in
electrodynamics
The spectre of the Brocken. Photo by
Galen Rowell.
21 January 2004
Physics 218, Spring 2004
1
Update #3: potentials
In electrodynamics the divergence of B is still zero, so
according to the Helmholtz theorem and its corollaries (#2, in
this case), we can still define a magnetic vector potential as
B = —× A .
However, the curl of E isn’t zero; in fact it hasn’t been since
we started magnetoquasistatics. What does this imply for the
electric potential? Note that Faraday’s law can be put in a
suggestive form:
1 ∂
—× E = −
( — × A ) , or
c ∂t
1 ∂A ⎞
⎛
—×⎜ E +
⎟=0 .
c ∂t ⎠
⎝
21 January 2004
Physics 218, Spring 2004
2
Potentials (continued)
Thus Corollary #1 to the Helmholtz theorem allows us to
define a scalar potential for that last bracketed term:
1 ∂A
E+
= −— V
c ∂t
1 ∂A
⇒ E = −— V −
c ∂t
so we can still use the scalar electric potential in
electrodynamics, but now both the scalar and the vector
potential must be used to determine E.
21 January 2004
Physics 218, Spring 2004
3
“Reference points” for potentials
Our usual reference point for the scalar potential in
electrostatics is V → 0 at r → ∞. For the vector potential in
magnetostatics we imposed the condition — ⋅ A = 0.
‰ These reference points arise from exploitation of the builtin ambiguities in the static potentials: one can add any
gradient to A and any constant to V, and still get the same
fields.
‰ So we decided to add whatever was necessary to make the
second-order differential equations in A and V look like
Poisson’s equation (i.e. easy to solve).
In electrodynamics these choices no longer produce that last
result:
21 January 2004
Physics 218, Spring 2004
4
“Reference points” for potentials (continued)
For instance, Gauss’s law gives us
1 ∂A ⎞
⎛
— ⋅ E = 4πρ ⇒ — ⋅ ⎜ −—V −
⎟ = 4πρ
c ∂t ⎠
⎝
1 ∂
2
⇒∇ V+
— ⋅ A = −4πρ ,
c ∂t
which with — ⋅ A = 0 still leaves us with a Poisson equation,
but Ampère’s law gives
4π
1 ∂
1 ∂2 A
— × ( — × A) =
J−
—V − 2
c
c ∂t
c ∂t 2
— ( — ⋅ A) − ∇2 A =
or
21 January 2004
(P.R. #11)
⎛ 2
1 ∂2 A ⎞
1 ∂V ⎞
4π
⎛
J .
⎜⎜ ∇ A − 2
⎟ − —⎜—⋅ A +
⎟=−
2 ⎟
c ∂t ⎠
c
⎝
c ∂t ⎠
⎝
Physics 218, Spring 2004
5
“Reference points” for potentials (continued)
This latter equation does not of course reduce to a Poisson
equation with any of the reference conditions we have
imposed. Thus we must look harder to use the built-in
ambiguity of the potentials to make the differential equations
simpler. The general way to do this, which we will cover next
time, is called a gauge transformation.
21 January 2004
Physics 218, Spring 2004
6
Gauge transformations
In electro- and magentostatics, we showed that we could
always choose our conventional reference points,
V → 0 as r → ∞
—⋅ A = 0
without placing any peculiar constraints on E or B. Now we
have two, more complicated equations to simplify, and a
more general approach is more fruitful.
Consider performing a transformation on A and V: add a
vector to A and a scalar to V, giving new potential functions:
A′ = A + α
21 January 2004
V′ = V + β
Physics 218, Spring 2004
7
Gauge transformations (continued)
Now, we can’t add just any old thing to the potentials; we
need for the fields arising from the new potentials to be the
same as those from the old:
B′ = B
— × A′ = — × A + — × α
⇓
— × α = 0 , or
α = —λ ′ ,
E′ = E
—V ′ = —V + —β
1 ∂A′
1 ∂A
− E′ −
= −E −
+ —β
c ∂t
c ∂t
1 ∂
1 ∂α
′
—β =
A
A
−
=
−
(
)
c ∂t
c ∂t
where λ ′ is a scalar function of r and t.
21 January 2004
Physics 218, Spring 2004
8
Gauge transformations (continued)
Combine those last two results:
1 ∂
—β = −
—λ ′
c ∂t
1 ∂λ ′ ⎞
⎛
—⎜ β +
⎟=0
c ∂t ⎠
⎝
and integrate the second one over volume, applying the
fundamental theorem of calculus:
The integration “constant”
1 ∂λ ′
β+
= f (t )
f is not a function of
c ∂t
position
21 January 2004
Physics 218, Spring 2004
9
Gauge transformations (continued)
We can combine the integration “constant” f with λ ′
by defining
t
λ = λ ′ − c ∫ f ( t ′ ) dt
0
1 ∂λ ′
1 ⎛ ∂λ
1 ∂λ
⎞
+ f (t ) = − ⎜
+ cf ( t ) ⎟ + f ( t ) = −
Then, β = −
c ∂t
c ⎝ ∂t
c ∂t
⎠
α = —λ ′ = —λ ,
Thus for any scalar function λ = λ ( r , t ) , the transformation
1 ∂λ
Gauge
V′ = V −
A′ = A + — λ
transformation
c ∂t
makes new potentials but leaves the fields E and B
unchanged.
21 January 2004
Physics 218, Spring 2004
10
Gauge transformations (continued)
This sort of operation on potentials is called a gauge
transformation, and a particular choice of λ is called a gauge
condition.
‰ Clever choices of λ can simplify one or the other of the
second-order differential equations for the potentials.
‰ The solution of these simpler equations for the
transformed potentials gives the same fields as the
solutions to the untransformed, complicated equations
For instance, to simplify ∇ 2V + 1 ∂ — ⋅ A = −4πρ ,
c ∂t
we could pick λ such that
—⋅ A = 0
21 January 2004
Physics 218, Spring 2004
Coulomb gauge
11
Gauge transformations (continued)
which, as we showed in PHY 217 (see
http://www.pas.rochester.edu/~dmw/phy217/Lectures/Lect_28b.pdf )
we can always do; that is, there always exists a function λ that
we can add to A to give an A’ with zero divergence. With this
gauge condition,
∇ 2V ′ = −4πρ ,
just like in electrostatics (hence the name).
‰ Coulomb gauge only does a lot of good in
magnetoquasistatics, because otherwise the time
derivative of A gets big enough that you have to
remember that
1 ∂A
.
E = −— V −
c ∂t
21 January 2004
Physics 218, Spring 2004
12
Lorentz gauge
It’s hard to compute A in Coulomb gauge. On the other hand,
we could choose λ such that
Lorentz
1 ∂V
—⋅ A+
=0 ,
gauge
c ∂t
for which the second-order PDEs we saw several pages back
become
1 ∂
2
∇ V+
— ⋅ A = −4πρ
c ∂t
2
⇒ ∇ V−
21 January 2004
1 ∂ 2V
c
2
∂t
2
= −4πρ
Physics 218, Spring 2004
13
Lorentz gauge (continued)
and
⎛ 2
1 ∂2 A ⎞
1 ∂V ⎞
4π
⎛
J
⎜⎜ ∇ A − 2
⎟ − —⎜—⋅ A +
⎟=−
2 ⎟
c ∂t ⎠
c
⎝
c ∂t ⎠
⎝
1 ∂2 A
4π
⇒ ∇ A− 2
=−
J
2
c
c ∂t
2
‰ Not utterly simple, but at least V and A are separately
determined, and the four equations are very similar to one
another.
‰ In fact, the four second-order PDEs here are
(inhomogeneous) wave equations, the solution of which
will concern us for the bulk of this semester.
21 January 2004
Physics 218, Spring 2004
14
Lorentz gauge (continued)
Can one always use the Lorentz gauge? I think so, because:
1 ∂V ′
=0
— ⋅ A′ +
c ∂t
1 ∂⎛
1 ∂λ ⎞
— ⋅ ( A + —λ ) +
⎜V −
⎟=0
c ∂t ⎝
c ∂t ⎠
1 ∂2λ
1 ∂V ⎞
⎛
∇ λ − 2 2 = −⎜—⋅ A +
⎟ = g (r ,t)
c ∂t ⎠
⎝
c ∂t
That is, the Lorentz gauge condition λ always obeys an
inhomogenous wave equation, just as do the potentials in
Lorentz gauge. In MTH 281 you proved the existence of
solutions to such equations; we’ll demonstrate such solutions
this semester.
2
21 January 2004
Physics 218, Spring 2004
15
Update #4: force, energy, and momentum in
electrodynamics
The Lorentz force law,
1
⎛
⎞
F = q⎜E + v×B⎟
c
⎝
⎠
hasn’t changed since we first learned it, but can be used to
illuminate the relation of the potentials to mechanics. To wit:
1 ∂A 1
⎛
⎞
F = q ⎜ −—V −
+ v× —× A⎟
c ∂t c
⎝
⎠
According to product rule #4, written for v and A,
— (v ⋅ A) = v × ( — × A) + A × ( — × v ) + (v ⋅ — ) A + ( A ⋅ — ) v
0
0
= v × (— × A) + (v ⋅ — ) A
because v doesn’t depend
explicitly on r
21 January 2004
Physics 218, Spring 2004
16
Force, energy, and momentum in electrodynamics
(continued)
so
1
⎛ 1 ∂A 1
⎛
⎞⎞
+ (v ⋅ — ) A + — ⎜V − v ⋅ A ⎟ ⎟
F = −q ⎜
c
⎝
⎠⎠
⎝ c ∂t c
It will be useful to introduce total time derivatives:
dA ∂A ∂A dx ∂A dy ∂A dz
=
+
+
+
dt
∂t ∂x dt ∂y dt ∂z dt
∂A ⎛
∂
∂
∂ ⎞
∂A
=
+ ⎜ vx
+ vy
+ vz ⎟ A =
+ (v ⋅ — ) A
∂t ⎝ ∂x
∂y
∂z ⎠
∂t
so
21 January 2004
1
⎛ 1 dA
⎛
⎞⎞
F = −q ⎜
+ — ⎜V − v ⋅ A ⎟ ⎟
c
⎝
⎠⎠
⎝ c dt
Physics 218, Spring 2004
17
Force, energy, and momentum in electrodynamics
(continued)
But F = dp dt , where p is the momentum of the point charge
q, so
dp
1
⎛ 1 dA
⎛
⎞⎞
= −q ⎜
+ — ⎜V − v ⋅ A ⎟ ⎟
dt
c
⎝
⎠⎠
⎝ c dt
q ⎞
d⎛
1
⎛ ⎡
⎤⎞
p + A ⎟ = −— ⎜ q ⎢V − v ⋅ A ⎥ ⎟
⎜
dt ⎝
c ⎠
c
⎦⎠
⎝ ⎣
d
This has the form F = pcanonical = −—U , where the canonical
dt
momentum is
q
Since
v = dr/dt, r
Remember
pcanonical = p + A ,
c
isLagrangian
the conjugate
and
and the related potential energy is
canonical
Hamiltonian
1
⎛
⎞
coordinate
dynamics?
U = q ⎜V − v ⋅ A ⎟ .
c
⎝
⎠
21 January 2004
Physics 218, Spring 2004
18
Force, energy, and momentum in electrodynamics
(continued)
In quantum mechanics one normally uses the Hamiltonian
formulation of dynamics; this last result represents the easiest
way to incorporate electrodynamics into quantum mechanics.
From correspondence to classical mechanics, the Hamiltonian
H = x canonical ⋅ pcanonical − L
gives rise to the quantum-mechanical equations of motion,
Hψ = Eψ
,
as well as the classical ones,
∂H
∂pcanonical
21 January 2004
= x canonical
,
∂H
∂xcanonical
Physics 218, Spring 2004
= p canonical
.
19
Force, energy, and momentum in electrodynamics
(continued)
The classical Lagrangian for a charge q in an electromagnetic
field is therefore
q
1
1
2
2
L = mx canonical − U = mv − qV + v ⋅ A ,
2
2
c
so the classical Hamiltonian is
q
q
1
2
2
H = x canonical ⋅ pcanonical − L = mv + v ⋅ A − mv + qV − v ⋅ A
c
2
c
2
2
p
q
1
1
2
= mv + qV =
+ qV =
pcanonical − A + qV .
2
2m
2m
c
In quantum mechanics, pcanonical → −i =— .
21 January 2004
Physics 218, Spring 2004
20