8.022 (E&M) – Lecture 19
Topics:
The missing term in Maxwell’s equation
Displacement current: what it is, why it’s useful
The complete Maxwell’s equations
And their solution in vacuum: EM waves
Maxwell’s equations so far
⎧ ∇ i E = 4πρ
⎪
⎪∇ i B = 0
⎪
⎨ ∇ × E = − 1 ∂B
⎪
c ∂t
⎪
4π
⎪∇ × B =
J
c
⎩
Gauss’s law: relates E and charge density (ρ)
Magnetic field lines are always closed!
Faraday’s law: change in B flux creates e.m.f. (E)
Ampere’s law: relates B and its sources (J)
Is this set of equations completely consistent?
Not quite…
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8.022 – Lecture 19
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Maxwell’s equations so far (2)
⎧ ∇ i E = 4πρ
⎪
⎪ ∇ i B = 0
⎪
⎨ ∇ × E = − 1 ∂B
⎪
c ∂
t
⎪
4π ⎪∇ × B =
J
c
⎩
Is this set of equations consistent? Not quite…
Take the divergence of Ampere’s law
4π ∂ ρ
⎛ 4π ⎞ 4π
∇ i⎜
J⎟=
∇iJ = −
(using continuity equation)
c
c ∂t
⎝ c
⎠
∇ i∇ × B = 0
(∇ i∇ × v is ALWAYS 0!)
Ampere’s law works only when dρ/dt=0 which works in most cases but
not always: Ampere’s law is incomplete!
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Fixing the inconsistency
Since ∇i∇ × v ≡ 0 we need to add some term to the right hand side
to that its divergence will be identically 0
Generalized Ampere’s law: ∇ × B =
4π
J+F
c
What is F? We know that its divergence must be =0:
∂ρ
⎛ 4π
⎞
∇ i⎜
J + F ⎟ = 0 ⇒ ∇ i ( cF ) = − 4π ∇ i J = 4π
∂t
c
⎝
⎠
⇒ ∇ i ( cF ) = 4π
∂ρ
Similar to Gauss's law!
∂t
Take time derivative of Gauss’s law:
⎛ ∂E
∂
∂ρ
∂
∇ i E = 4π
⇒
∇ i E = ∇ i⎜
∂t
∂t
∂t
⎝ ∂t
(
)
⇒ F =
(
)
⎞
⎟ = ∇ i ( cF ) time and space derivatives commute
⎠
1 ∂E
c ∂t
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Displacement currents
Generalized Ampere’s equation
∇× B =
4π
1 ∂E
J+
c
c ∂t
This can also be written as:
∇× B =
4π
(J + Jd )
c
With Jd = displacement current (density):
Jd =
1 ∂E
4π ∂t
What is the Jd?
Not a real current: does not describe charges flowing through
some region
But it acts like a real current: whenever we have changing E
field, we can treat its effect as if due to as a real current Jd
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What is a displacement current?
Consider a current flowing in a circuit and charging a capacitor C
S
S’
C
Standard integral Ampere’s law:
∫ B idl
=
C
4π
c
I encl =
4π
c
∫S J ida
Let’s choose the path C and the surface S as in the drawing above:
It all makes sense!
Now choose the same path C but the surface S’ (ok by Stokes…)
No standard current J through the surface (no charge crosses C!)
But there is a flux of displacement current Jd through the plates!
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What is a displacement current? (2)
We can use the generalized Ampere’s Law:
4π
∫C B idl = c ( I encl + I d )
with I d =
∫S
'
J d ida =
1
4π
∫S
∂E
1 ∂
ida =
4π ∂t
∂t
'
∫S
'
E id a =
1 ∂Φ E
4π ∂t
The displacement current is related to the change over time of the
flux of the electric field.
In the example above, the electric field E is the one produced in between
the plates of the capacitor C
E
S’
C
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What is a displacement current? (3)
The electric field E:
Points in the same direction of the current (+x)
At a given instant in time: E = 4π Q xˆ
A
The flux of E will then be: ΦE = 4π Q (yes, Gauss's law!)
∂ΦE
∂Q
=4π I
= 4π
∂t
∂t
Where I is the current that is charging the capacitor
The rate of the change if this flux is:
Comparing this with results in the previous page:
Id =
∫S
'
J id a =
∫S J d id a
=I
generalized Ampere’s Law is valid no matter what surface we use
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The importance of
displacement currents
When we examined the following circuit:
we said the same current I was flowing in each circuit element.
How is it possible? No current flows through the plates of a capacitor!
Displacement currents fix this inconsistency!
Displacement current “continues” the “real” current across the capacitors
ensuring the validity of Kirchoff’s laws.
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Maxwell’s equations (complete!)
cgs
⎧ ∇ i E = 4πρ
⎪
⎪∇ i B = 0
⎪
⎨∇ × E = − 1 ∂B
⎪
c ∂t
⎪
⎪ ∇ × B = 4π J + 1 ∂ E
c
c ∂t
⎩
ρ
⎧
SI
⎪∇ i E = ε
0
⎪
⎪∇ i B = 0
⎪
Typo in Purcell
⎨
Eq 15 ch 9
∂B
⎪∇ × E = −
∂t
⎪
⎪
∂E
⎪ ∇ × B = µ 0 J + µ 0ε 0
∂t
⎩
Generalized Ampere’s law
NB: when Maxwell introduced the term dE/dt in the generalized
Ampere’s law, his arguments were based purely on symmetry
Yes, he was a theorist!
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Maxwell’s equations: integral form
⎧ Φ = E i da = 4π Q enc
( Gauss's law )
⎪ E ∫S
⎪Φ B = 0
( M agnetic field line are clo sed)
⎪
⎪
1 ∂Φ B
⎨ emf = ∫ E i dl = −
(Faraday's law )
c ∂t
⎪
C
⎪
⎪ B i d l = 4π ( I + I d )
( Generalized Ampere's law )
⎪⎩ C∫
c
where the curren ts I and Id are defined as I= ∫ J i d a and
S
Id =
1 ∂Φ E ( S )
4π
∂t
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3 good reasons to remember
Maxwell’s equations
1)
They compactly and beautifully summarize all the E&M we learned so far!
2)
You will see them on T shirts for the rest of your life at MIT:
better to get familiar with them ASAP!
3) On the first day of 8.03 next semester you will be asked to write them
down on a piece of paper to check what you learned in your first semester
at MIT: save your honor (and mine)
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Displacement current: application
Consider the following RC circuit:
As C charges up, Id flows
Id induces B inside the plates
Assuming cylindrical plates of radius a
Calculate B inside the plates
1) F i nd E(t):
E (t ) = 4 πσ =
πa2
Jd =
2) D i sp l . current density:
3) Rem em ber that I (t ) =
4 π Q (t )
1 ∂E
1 ∂ E (t )
1 ∂ Q (t ) I (t )
=
=
=
π a 2 ∂t
πa 2
4π ∂t
4π ∂t
V b − t / RC
e
R
4) Magnet i c f ield i ns i de the p l ate (Am pere's l aw ):
G. Sciolla – MIT
8.022 – Lecture 19
∫ B idl
C
=
4π
c
⇒ B (r ) =
∫C J d id a
2 rV b − t / RC
e
13
ca 2R
Maxwell equations in vacuum
What happens when we write Maxwell’s equations in vacuum?
Vacuum: no sources, ρ=0 and J=0
⎧ ∇ i E = 4πρ
⎪
⎪∇ i B = 0
⎪
⎨ ∇ × E = − 1 ∂B
⎪
c ∂t
⎪
⎪ ∇ × B = 4π J + 1 ∂ E
c
c ∂t
⎩
⎧∇ i E = 0
⎪
⎪ ∇ i B = 0
⎪
1 ∂B
⎨∇ × E = −
⎪
c ∂t
⎪
⎪∇ × B = 1 ∂E
c ∂t
⎩
Except for a – sign, these equations are exquisitely symmetric!
Consequence: an electric filed E varying in time will create a magnetic
filed B; a B varying in time creates a E: E and B are intimately related!
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8.022 – Lecture 19
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Maxwell equations in vacuum:
solution
⎧∇ i E = 0
⎪
⎪∇ i B = 0
Separate E and B in equations
⎪
How?
⎨∇ × E = − 1 ∂B
c ∂t
Take the curl of equations (3) and (4) ⎪
⎪
Use other equations as needed
⎪∇ × B = 1 ∂E
c ∂t
⎩
How to solve these equations?
Uncouple them!
Start from (3):
(
)
(
)
Left: ∇ × ∇ × E = ∇ ∇ i E − ∇ 2 E = −∇ 2 E
(1)
(2)
(3)
(4)
( since ∇ i E = 0 in vacuum )
⎛ 1 ∂B ⎞
1 ∂
1 ∂2E
Right: ∇ × ⎜ −
∇×B = − 2
( using (4))
⎟=−
c ∂t
c ∂t 2
⎝ c ∂t ⎠
1 ∂2E
G. Sciollac –2 MIT
∂t 2
⇒ ∇2E =
8.022 – Lecture 19
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Maxwell equations in vacuum:
solution
Now repeat the procedure starting from
(
)
(
∇×B =
)
Left: ∇ × ∇ × B = ∇ ∇ i B − ∇ 2 B = −∇ 2 B
1 ∂E
c ∂t
(4)
( since ∇ i B = 0 in vacuum )
⎛ 1 ∂E ⎞ 1 ∂
1 ∂2B
Right: ∇ × ⎜
( using (3))
=
∇
×
=
E
−
⎟
c 2 ∂t 2
⎝ c ∂t ⎠ c ∂t
⇒ ∇2B =
1 ∂2B
c 2 ∂t 2
This is a special case of a known equation: the wave equation:
1 ∂2 f
where f = f ( x ± vt )
v 2 ∂t 2
where f is any function that has well-behaved derivatives
∇2 f =
NB: we are restricting ourselves to the 1D case; extension to 3D next lecture
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∇2 f =
1 ∂2 f
v 2 ∂t 2
Solution of wave equation: prove
Prove that f = f ( x ± vt ) is a solution of the wave equation
Just calculate time and space
derivatives.
2
2
2
Keep in mind that ∇ 2 = ∂ 2 + ∂ 2 + ∂ 2
∂x
∂y
∂z
u = x ± vt
Define
2
∂ f ( x ± vt ) ∂ f ∂ f
∂f
∂ 2 f ( x ± vt )
2 ∂ f
v
=
= ±v
⇒
=
∂u ∂t
∂t
∂u
∂t 2
∂u 2
2
2
∂ f ( x ± vt ) ∂ f ∂ f
∂f
∂ f ( x ± vt ) ∂ f
=
=
⇒
=
∂u ∂x ∂u
∂x
∂t 2
∂u 2
∂2 f
∂2 f
1
Plug the above results into the equation ⇒
= 2 v2
⇒ identity!
2
v
∂u
∂u 2
As we wanted to prove!
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∇2 f =
1 ∂2 f
v 2 ∂t 2
Wave equation solution
What is a function such as f = f ( x − vt ) ?
f(x)
Assume v=1 cm/s
At time t=0:
Position of the max: x0
x0
At time t=1 s:
The peak still occurs when the argument of f is x0
But since the time is not 0
the function will be shifted in x by “vt”=1 cm
Position of the max: x1=x0+1
f(x)
x1= x0+1
f = f ( x − vt ) represents a wave traveling in the +x direction with velocity v
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∇2 f =
1 ∂2 f
v 2 ∂t 2
Wave equation solution
What is a function such as f = f ( x + vt ) ?
f(x)
t=0
Assume v=1 cm/s
At time t=0:
Position of the max: x0
x0
At time t=1 s:
The peak still occurs when the argument of f is x0
But since the time is not 0
the function will be shifted in x by “vt”=1 cm
Position of the max: x1=x0-1
f(x)
t=1
x1= x0-1
f = f ( x + vt ) represents a wave traveling in the -x direction with velocity v
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EM waves
Wave equation: ∇ 2 f =
Solution:
1 ∂2 f
v 2 ∂t 2
f = f ( x ± vt )
Any function of argument x ± vt
These solution represent waves traveling with velocity v
x − vt
x + vt
represents a wave traveling in the +x direction
represents a wave traveling in the –x direction
Maxwell’s equation:
∇2 f =
1 ∂2 f
c 2 ∂t 2
Same equation! Only difference: v=c
Solution: EM waves traveling with speed of light
The light IS an EM wave!!!
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EM waves in SI
This same result looks much more interesting in SI.
Maxwell’s equations in SI:
∂2 f
∇ 2 f = µ 0ε 0
∂t 2
where ε0 is the permittivity of free space
and µ0 is the permeability of free space
Maxwell’s equations tell us what the velocity of an EM wave is:
v = 1 / µ 0ε 0
ε0 and µ0 can be measured
we can predict velocity of EM waves:
v=2.998 108 m/s2 which is the speed of light!
Maxwell was the first to realize that E&M equations were leading to a wave
equation that was propagating at the speed of light: light is an EM wave!
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8.022 – Lecture 19
How to measure c (demo A4)
Experimental setup: a neon laser beam is sent into a beam splitter.
Part of it is reflected and part of it is refracted first and then reflected
by a mirror.
laser
Ch2
Ch1
• Difference in path between the 2 beams:
~ 17.15 m x 2 = 34.3 meters
• Measure the delay of channel 2 wrt channel 1
on the scope: 116 ns
v=34.3 m / 116 ns = 2.96 108 m/s
Beam splitter
Ch 1: longer path
Ch 2: shorter path
Mirror
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– MIT
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Summary and outlook
Today:
Complete Maxwell’s equations
The missing term leads to displacement currents
Solution of Maxwell’s equations in vacuum
Wave equation
light is an EM radiation
Next time:
Properties of EM radiation
Polarization and scattering of light
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