Today in Physics 218: the classic conservation laws in electrodynamics
Today in Physics 218: the classic conservation laws in electrodynamics Poynting’s theorem Energy conservation in electrodynamics The Maxwell stress tensor (which gets rather messy) Momentum conservation in electrodynamics Electromagnetism on the sun, doing work on matter and emitting radiation. (TRACE satellite; Stanford U./Lockheed/NASA.) 23 January 2004
Physics 218, Spring 2004
1
Poynting’s theorem Suppose a collection of charges and currents lies entirely within a volume V. If they are released at some point in time, electromagnetic forces will begin to do work on them. Consider, for instance, the work done by the forces on charges and currents in an infinitesimal volume dτ during a time dt: 2
d Wmech.
v ⎛ ⎞ = F ⋅ dA = dq ⎜ E + × B ⎟ ⋅ vdt c ⎝ ⎠ = E ⋅ vdqdt i.e. B does no work. = E ⋅ Jdτ dt
Thus,
using dq = ρ dτ , J = ρ v
dWmech = E ⋅ Jdτ dt
∫
V
23 January 2004
Physics 218, Spring 2004
2
Poynting’s theorem (continued) We’ve seen this before, of course; it’s just another way to write P = VI. But we will learn something by elimination of J using Ampère’s law: 1 ∂E c J= —×B − : 4π 4π ∂t dWmech. 1 ∂E ⎞ ⎛ = dτ ⎜ cE ⋅ ( — × B ) − E ⋅ ⎟ . dt 4π ∂t ⎠ ⎝ V
∫
According to Product Rule #6,
E ⋅ ( — × B ) = −— ⋅ ( E × B ) + B ⋅ ( — × E ) 1 ∂B = −— ⋅ ( E × B ) − B ⋅ ∂t c 23 January 2004
Physics 218, Spring 2004
Use Faraday’s law: . 3
Poynting’s theorem (continued) The time-derivative terms in all of these expressions can be written as ∂E 1 ∂ 1 ∂ 2 E E⋅ = , (E ⋅ E) = ∂t 2 ∂t 2 ∂t ∂B 1 ∂ 2 and, similarly, B ⋅ = B . ∂t 2 ∂t Using the last three results in the expression for dW/dt gives us dWmech. 1 ∂B ∂E ⎞ ⎛ = − E⋅ dτ ⎜ −c— ⋅ ( E × B ) − B ⋅ ⎟ ∫ 4π ∂ ∂ dt t t ⎝ ⎠ V 1 =− 4π 23 January 2004
(
)
1 ∂ 2 ⎛ ⎞ dτ ⎜ c— ⋅ ( E × B ) + B + E2 ⎟ . 2 ∂t ⎝ ⎠ V
∫
Physics 218, Spring 2004
4
Poynting’s theorem (continued) V is bounded by surface S. Apply the divergence theorem to the first term in the integral, converting it to a surface integral over S: dWmech. 1 ∂B ∂E ⎞ ⎛ dτ ⎜ −c— ⋅ ( E × B ) − B ⋅ = − E⋅ ⎟ dt 4π ∂t ∂t ⎠ ⎝
∫
V
(
∂ 2 2 d B E τ + ∫ ∂t
c =− 4π
1 v∫ ( E × B ) ⋅ da − 8π
c =− 4π
1 d v∫ ( E × B ) ⋅ da − 8π dt
S
S
)
V
∫(
V
)
B2 + E2 dτ
. Poynting’s theorem
In the last step we have used the fact that time is the only variable that survives the integration. 23 January 2004
Physics 218, Spring 2004
5
Energy conservation in electrodynamics Poynting’s theorem of course expresses energy conservation. We have long known that the energy density stored in electric and magnetic fields – that is , the work required to assemble the charges and currents in the configuration they’re in – is 1 uEB = E2 + B2 . 8π
(
)
So the integral of u is evidently the energy within V stored in the E and B fields, and the term containing it just gives the rate at which the stored energy changes. 23 January 2004
Physics 218, Spring 2004
6
Energy conservation in electrodynamics (continued) The surface integral term contains a vector field that deserves its own name: c S= E×B . Poynting vector 4π
dWmech. dWEB d = − v S ⋅ da − uEB dτ = − v S ⋅ da − Thus, dt dt dt
∫
S
∫
V
∫
.
S
This means that the rate at which work is done on the charges and currents by the fields is balanced not just by the rate at which the stored energy decreases, but also by a new term – which, since it involves a flux integral over the surface bounding V, must be the rate at which the fields carry energy out of V. 23 January 2004
Physics 218, Spring 2004
7
Energy conservation in electrodynamics (continued) Put another way: if the stored energy decreases, not all of the resulting work goes into kinetic energy of the charges – some is radiated away. The Poynting vector tells us the energy per unit area and time that is radiated.
Rewrite the last result as d d umech. dτ = − ∫ — ⋅ Sdτ − ∫ uEB dτ ∫ dt dt V
∫
V
V
V
∂umech. ∂uEB dτ = − — ⋅ Sdτ − dτ ∂t ∂t
∫
V
∫
.
V
We may equate the integrands now, and obtain: 23 January 2004
Physics 218, Spring 2004
8
Energy conservation in electrodynamics (continued) ∂umech. ∂u = −— ⋅ S − EB , or ∂t ∂t ∂ ( umech. + uEB ) + — ⋅ S = 0 , ∂t an expression that invites comparison with the (chargecurrent) continuity equation, ∂ ρ +—⋅ J = 0 . ∂t Think of the Poynting vector S therefore as the “current density” of energy.
23 January 2004
Physics 218, Spring 2004
9
The Maxwell stress tensor We won’t use tensors very much in this class, and you won’t have to cope with this particular tensor after today. But the path to expressions for momentum conservation goes straight through the Maxwell stress tensor, and this tensor does turn out to be a valuable tool in more-advanced courses, so it won’t hurt to introduce it here. Return to the Lorentz force law, and make a new definition: 1 1 ⎛ ⎞ ⎛ ⎞ F = q⎜E + v×B⎟ ⇒ Fdτ = ρ ⎜ E + v × B ⎟ dτ , where c c ⎝ ⎠ ⎝ ⎠ V V
∫
∫
1 1 ⎛ ⎞ F= ρ ⎜ E + v × B ⎟ = ρ E + J × B c c ⎝ ⎠ is the force per unit volume exerted by E and B. 23 January 2004
Physics 218, Spring 2004
( J = ρv ) 10
The Maxwell stress tensor (continued) Using Gauss’s law and Ampère’s law, in the forms 1 c 1 ∂E , ρ= —⋅E , J = —×B − 4π 4π 4π ∂t we can eliminate the source terms, in favor of the fields:
1 1 ⎛ 1 ∂E ⎞ F= (— ◊ E) E + ⎜ — ¥ B − ⎟¥B . 4π 4π ⎝ c ∂t ⎠ The very last term can be put into a more useful form by noting that ∂B 1 ∂ 1 ∂E 1 Use Faraday’s ¥B+ E¥ ( E ¥ B) = law: c ∂t c ∂t c ∂t 1 ∂E = ¥ B − E ¥ (— ¥ E) . c ∂t 23 January 2004
Physics 218, Spring 2004
11
The Maxwell stress tensor (continued) Thus, 1 1 1 ⎛ ∂ F= ( — ◊ E ) E + ( — ¥ B ) ¥ B + ⎜ − ( E ¥ B ) − E ¥ ( — ¥ E ) ⎞⎟ 4π 4π 4π ⎝ ∂t ⎠ 1 1 1 ∂ = ⎡⎣( — ◊ E ) E − E ¥ ( — ¥ E ) ⎤⎦ − ⎡⎣ B ¥ ( — ¥ B ) ⎤⎦ − ( E ¥ B) . 4π 4π 4π c ∂t Since — ◊ B = 0, it changes nothing if we add ( — ◊ B ) B to the second square brackets to make the whole thing look more symmetrical: 1 ⎡⎣( — ◊ E ) E − E ¥ ( — ¥ E ) ⎤⎦ F= 4π 1 1 ∂ + ⎡⎣( — ◊ B ) B − B ¥ ( — ¥ B ) ⎤⎦ − ( E ¥ B) . 4π 4π c ∂t 23 January 2004
Physics 218, Spring 2004
12
The Maxwell stress tensor (continued) We can “simplify” the terms in square brackets if we reintroduce tensor notation. Consider the three Cartesian directions to be represented by x, y, z = 1, 2, 3; for example, ⎡ Ex ⎤ ⎡ E1 ⎤ ⎡ x ⎤ ⎡ r1 ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ r = ⎢ y ⎥ = ⎢ r2 ⎥ , or E = ⎢Ey ⎥ = ⎢ E2 ⎥ . ⎢ E ⎥ ⎢E ⎥ ⎢⎣ z ⎥⎦ ⎢⎣ r3 ⎥⎦ ⎣ z⎦ ⎣ 3⎦ Then the x-component of the first term in [] can be written as
⎛ ∂E1 ∂E2 ∂E3 ⎞ ⎡⎣( — ◊ E ) E − E ¥ ( — ¥ E ) ⎤⎦ = E1 ⎜ + + ⎟ 1 ∂r3 ⎠ ⎝ ∂r1 ∂r2 ⎛ ∂E2 ∂E1 − E2 ⎜ − ⎝ ∂r1 ∂r2 23 January 2004
Physics 218, Spring 2004
⎛ ∂E1 ∂E3 ⎞ ⎞ − ⎟ ⎟ + E3 ⎜ r r ∂ ∂ 1 ⎠ ⎠ ⎝ 3 13
The Maxwell stress tensor (continued) It will pay to multiply this out and regroup: ∂E3 ∂E1 ∂E2 + E1 + E1 [...]1 = E1 ∂r1 ∂r2 ∂r3 ∂E3 ∂E2 ∂E1 ∂E1 − E2 + E2 + E3 − E3 ∂r1 ∂r2 ∂r3 ∂r1 ∂ ∂ 1 ∂ 2 = ( E1 ) + ( E1E2 ) + ( E1E3 ) ∂r2 ∂r3 2 ∂r1 1 ∂ 2 2 1 ∂ − ( E2 ) − ( E3 ) 2 ∂r1 2 ∂r1 ∂ ∂ ∂ 1 ∂ 2 E12 + E22 + E32 = ( E1 ) + ( E1E2 ) + ( E1E3 ) − ∂r1 ∂r2 ∂r3 2 ∂r1
(
23 January 2004
Physics 218, Spring 2004
) 14
The Maxwell stress tensor (continued)
[...]1
(
∂ ∂ ∂ 1 ∂ 2 E12 + E22 + E32 = ( E1 ) + ( E1E2 ) + ( E1E3 ) − ∂r1 ∂r2 ∂r3 2 ∂r1
)
⎡ 3 ∂ ⎤ 1 ∂ 3 ∂ ⎛ 1 2 ⎞ E1 E j ⎥ − E2 = E E E =⎢ − δ1 j ⎟ , ⎜ 1 j ⎢ j =1 ∂rj ⎥ 2 ∂r1 2 ∂rj ⎝ ⎠ j = 1 ⎣ ⎦ where we have reintroduced the Kronecker delta, ⎧ 1, i = j δ ij = ⎨ ⎩0, i ≠ j
∑
(
)
∑
This is for component 1; for component i, we’d get 3 ∂ ⎛ 1 2 ⎞ ⎡⎣( — ◊ E ) E − E ¥ ( — ¥ E ) ⎤⎦ = ⎜ Ei E j − E δ ij ⎟ . i ∂r 2 ⎠ j =1 j ⎝
∑
23 January 2004
Physics 218, Spring 2004
15
The Maxwell stress tensor (continued) Similarly,
3
∂ ⎛ 1 2 ⎞ ⎡⎣( — ◊ B ) B − B ¥ ( — ¥ B ) ⎤⎦ = ⎜ Bi B j − B δ ij ⎟ . i ∂r 2 ⎠ j =1 j ⎝ Let us now define a nine-component object,
∑
(
)
Maxwell stress 1 2 ⎛ ⎞ 2 ⎜ Ei E j + Bi B j − E + B δ ij ⎟ , tensor 2 ⎝ ⎠ with which we can re-write the ith component of both of the ugly terms in []: 1 ⎡⎣( — ◊ E ) E − E ¥ ( — ¥ E ) ⎤⎦ i 4π
1 Tij = 4π
3 ∂ 1 Tij + ⎡⎣( — ◊ B ) B − B ¥ ( — ¥ B ) ⎤⎦ = ∑ i ∂r 4π j =1 j 23 January 2004
Physics 218, Spring 2004
. 16
The Maxwell stress tensor (continued) The best way to envision the sum on the right-hand side is as matrix multiplication: ⎡T11 T21 T31 ⎤ 3 ⎡ ∂ ∂ ∂ ∂ ⎤⎢ ⎥ T T T T = ∑ ∂rj ij ⎢ ∂r1 ∂r2 ∂r3 ⎥ ⎢ 12 22 32 ⎥ . ⎣ ⎦ ⎢T j =1 ⎣ 13 T23 T33 ⎥⎦ The result is represented by a three-component object; that is, a vector. We can use Ia vector-algebra-like symbolism for this operation, by using T to denote the second-rank tensor whose nine components are Tij . The sum is represented thus: 3 I ∂ ∑ ∂rj Tij ↔ — ◊ T , j =1 since it now looks so much like a divergence. 23 January 2004
Physics 218, Spring 2004
17
The Maxwell stress tensor (continued) Finally, the force per unit volume becomes, in terms of the Maxwell stress tensor, 1 F= ⎡⎣( — ◊ E ) E − E ¥ ( — ¥ E ) ⎤⎦ 4π 1 1 ∂ + ⎡⎣( — ◊ B ) B − B ¥ ( — ¥ B ) ⎤⎦ − ( E ¥ B) 4π 4π c ∂t I ∂ = — ◊ T − g EB , ∂t Momentum 1 1 g ≡ E ¥ B = S . where EB density 4π c c2
23 January 2004
Physics 218, Spring 2004
18
Momentum conservation in electrodynamics The total force on the charges within volume V can be found by integrating F : I ∂ dpmech. ⎛ ⎞ Use divergence = ∫ F dτ = ∫ ⎜ — ◊ T − g EB ⎟ dτ theorem: dt ∂t ⎝ ⎠ V
=v ∫ S
V
I d T ◊ da − dt
∫ g EBdτ
,
V
We can now identify pEB ≡
∫ g EBdτ as the total momentum
V
stored in the fields E and B, so that g EB =
1 4π c
E ¥ B is the
momentum density in the fields. 23 January 2004
Physics 218, Spring 2004
19
Momentum conservation in electrodynamics (continued) Rearranging slightly, we have I d Momentum conservation + = d p p T ◊ a . ( mech. EB ) in electrodynamics dt
v∫ S
Interpretation: Changes in mechanical momentum (mvs for all the charges) and momentum stored in fields within V are I caused by the pressure and shear T distributed over the boundary surface S, and vice versa. I The terms on the diagonal of T , 1 ⎛ 2 2 1 2 2 ⎞ + − + E B E B ⎜ i ⎟ , i 4π ⎝ 2 ⎠ represent pressures in the E and B fields.
( )
(
23 January 2004
Physics 218, Spring 2004
)
20
Momentum conservation in electrodynamics (continued) The off-diagonal terms,
(
1 Ei E j + Bi B j 4π
)
,
represent shear in the fields – the kind of stress that causes strain in a direction different than the stress. One can also write the momentum-conservation relation in differential form: I ∂ dpmech. ∂ ⎛ ⎞ g mech. dτ = ⎜ — ◊ T − g EB ⎟ dτ , or = dt ∂t ∂t ⎝ ⎠
∫
V
∫
V
I ∂ ( g mech. + g EB ) − — ◊ T = 0 , ∂t where g mech.is the mechanical (mv-type) momentum per unit volume for the charges within V. 23 January 2004
Physics 218, Spring 2004
21
Summary Electric and magnetic fields can store energy and momentum: 1 energy/volume uEB = E2 + B2 , 8π 1 g EB ≡ E¥B , momentum/volume 4π c or, for that matter, even angular momentum: 1 angular momentum/ LEB = r ¥ g EB = r ¥ ( E ¥ B) , volume 4π c and can transport energy (and thus all the other quantities): c energy/area/time S= E×B . 4π Electromagnetic waves are the normal method of transport.
(
23 January 2004
)
Physics 218, Spring 2004
22
Physics 218, Spring 2004
1
Poynting’s theorem Suppose a collection of charges and currents lies entirely within a volume V. If they are released at some point in time, electromagnetic forces will begin to do work on them. Consider, for instance, the work done by the forces on charges and currents in an infinitesimal volume dτ during a time dt: 2
d Wmech.
v ⎛ ⎞ = F ⋅ dA = dq ⎜ E + × B ⎟ ⋅ vdt c ⎝ ⎠ = E ⋅ vdqdt i.e. B does no work. = E ⋅ Jdτ dt
Thus,
using dq = ρ dτ , J = ρ v
dWmech = E ⋅ Jdτ dt
∫
V
23 January 2004
Physics 218, Spring 2004
2
Poynting’s theorem (continued) We’ve seen this before, of course; it’s just another way to write P = VI. But we will learn something by elimination of J using Ampère’s law: 1 ∂E c J= —×B − : 4π 4π ∂t dWmech. 1 ∂E ⎞ ⎛ = dτ ⎜ cE ⋅ ( — × B ) − E ⋅ ⎟ . dt 4π ∂t ⎠ ⎝ V
∫
According to Product Rule #6,
E ⋅ ( — × B ) = −— ⋅ ( E × B ) + B ⋅ ( — × E ) 1 ∂B = −— ⋅ ( E × B ) − B ⋅ ∂t c 23 January 2004
Physics 218, Spring 2004
Use Faraday’s law: . 3
Poynting’s theorem (continued) The time-derivative terms in all of these expressions can be written as ∂E 1 ∂ 1 ∂ 2 E E⋅ = , (E ⋅ E) = ∂t 2 ∂t 2 ∂t ∂B 1 ∂ 2 and, similarly, B ⋅ = B . ∂t 2 ∂t Using the last three results in the expression for dW/dt gives us dWmech. 1 ∂B ∂E ⎞ ⎛ = − E⋅ dτ ⎜ −c— ⋅ ( E × B ) − B ⋅ ⎟ ∫ 4π ∂ ∂ dt t t ⎝ ⎠ V 1 =− 4π 23 January 2004
(
)
1 ∂ 2 ⎛ ⎞ dτ ⎜ c— ⋅ ( E × B ) + B + E2 ⎟ . 2 ∂t ⎝ ⎠ V
∫
Physics 218, Spring 2004
4
Poynting’s theorem (continued) V is bounded by surface S. Apply the divergence theorem to the first term in the integral, converting it to a surface integral over S: dWmech. 1 ∂B ∂E ⎞ ⎛ dτ ⎜ −c— ⋅ ( E × B ) − B ⋅ = − E⋅ ⎟ dt 4π ∂t ∂t ⎠ ⎝
∫
V
(
∂ 2 2 d B E τ + ∫ ∂t
c =− 4π
1 v∫ ( E × B ) ⋅ da − 8π
c =− 4π
1 d v∫ ( E × B ) ⋅ da − 8π dt
S
S
)
V
∫(
V
)
B2 + E2 dτ
. Poynting’s theorem
In the last step we have used the fact that time is the only variable that survives the integration. 23 January 2004
Physics 218, Spring 2004
5
Energy conservation in electrodynamics Poynting’s theorem of course expresses energy conservation. We have long known that the energy density stored in electric and magnetic fields – that is , the work required to assemble the charges and currents in the configuration they’re in – is 1 uEB = E2 + B2 . 8π
(
)
So the integral of u is evidently the energy within V stored in the E and B fields, and the term containing it just gives the rate at which the stored energy changes. 23 January 2004
Physics 218, Spring 2004
6
Energy conservation in electrodynamics (continued) The surface integral term contains a vector field that deserves its own name: c S= E×B . Poynting vector 4π
dWmech. dWEB d = − v S ⋅ da − uEB dτ = − v S ⋅ da − Thus, dt dt dt
∫
S
∫
V
∫
.
S
This means that the rate at which work is done on the charges and currents by the fields is balanced not just by the rate at which the stored energy decreases, but also by a new term – which, since it involves a flux integral over the surface bounding V, must be the rate at which the fields carry energy out of V. 23 January 2004
Physics 218, Spring 2004
7
Energy conservation in electrodynamics (continued) Put another way: if the stored energy decreases, not all of the resulting work goes into kinetic energy of the charges – some is radiated away. The Poynting vector tells us the energy per unit area and time that is radiated.
Rewrite the last result as d d umech. dτ = − ∫ — ⋅ Sdτ − ∫ uEB dτ ∫ dt dt V
∫
V
V
V
∂umech. ∂uEB dτ = − — ⋅ Sdτ − dτ ∂t ∂t
∫
V
∫
.
V
We may equate the integrands now, and obtain: 23 January 2004
Physics 218, Spring 2004
8
Energy conservation in electrodynamics (continued) ∂umech. ∂u = −— ⋅ S − EB , or ∂t ∂t ∂ ( umech. + uEB ) + — ⋅ S = 0 , ∂t an expression that invites comparison with the (chargecurrent) continuity equation, ∂ ρ +—⋅ J = 0 . ∂t Think of the Poynting vector S therefore as the “current density” of energy.
23 January 2004
Physics 218, Spring 2004
9
The Maxwell stress tensor We won’t use tensors very much in this class, and you won’t have to cope with this particular tensor after today. But the path to expressions for momentum conservation goes straight through the Maxwell stress tensor, and this tensor does turn out to be a valuable tool in more-advanced courses, so it won’t hurt to introduce it here. Return to the Lorentz force law, and make a new definition: 1 1 ⎛ ⎞ ⎛ ⎞ F = q⎜E + v×B⎟ ⇒ Fdτ = ρ ⎜ E + v × B ⎟ dτ , where c c ⎝ ⎠ ⎝ ⎠ V V
∫
∫
1 1 ⎛ ⎞ F= ρ ⎜ E + v × B ⎟ = ρ E + J × B c c ⎝ ⎠ is the force per unit volume exerted by E and B. 23 January 2004
Physics 218, Spring 2004
( J = ρv ) 10
The Maxwell stress tensor (continued) Using Gauss’s law and Ampère’s law, in the forms 1 c 1 ∂E , ρ= —⋅E , J = —×B − 4π 4π 4π ∂t we can eliminate the source terms, in favor of the fields:
1 1 ⎛ 1 ∂E ⎞ F= (— ◊ E) E + ⎜ — ¥ B − ⎟¥B . 4π 4π ⎝ c ∂t ⎠ The very last term can be put into a more useful form by noting that ∂B 1 ∂ 1 ∂E 1 Use Faraday’s ¥B+ E¥ ( E ¥ B) = law: c ∂t c ∂t c ∂t 1 ∂E = ¥ B − E ¥ (— ¥ E) . c ∂t 23 January 2004
Physics 218, Spring 2004
11
The Maxwell stress tensor (continued) Thus, 1 1 1 ⎛ ∂ F= ( — ◊ E ) E + ( — ¥ B ) ¥ B + ⎜ − ( E ¥ B ) − E ¥ ( — ¥ E ) ⎞⎟ 4π 4π 4π ⎝ ∂t ⎠ 1 1 1 ∂ = ⎡⎣( — ◊ E ) E − E ¥ ( — ¥ E ) ⎤⎦ − ⎡⎣ B ¥ ( — ¥ B ) ⎤⎦ − ( E ¥ B) . 4π 4π 4π c ∂t Since — ◊ B = 0, it changes nothing if we add ( — ◊ B ) B to the second square brackets to make the whole thing look more symmetrical: 1 ⎡⎣( — ◊ E ) E − E ¥ ( — ¥ E ) ⎤⎦ F= 4π 1 1 ∂ + ⎡⎣( — ◊ B ) B − B ¥ ( — ¥ B ) ⎤⎦ − ( E ¥ B) . 4π 4π c ∂t 23 January 2004
Physics 218, Spring 2004
12
The Maxwell stress tensor (continued) We can “simplify” the terms in square brackets if we reintroduce tensor notation. Consider the three Cartesian directions to be represented by x, y, z = 1, 2, 3; for example, ⎡ Ex ⎤ ⎡ E1 ⎤ ⎡ x ⎤ ⎡ r1 ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ r = ⎢ y ⎥ = ⎢ r2 ⎥ , or E = ⎢Ey ⎥ = ⎢ E2 ⎥ . ⎢ E ⎥ ⎢E ⎥ ⎢⎣ z ⎥⎦ ⎢⎣ r3 ⎥⎦ ⎣ z⎦ ⎣ 3⎦ Then the x-component of the first term in [] can be written as
⎛ ∂E1 ∂E2 ∂E3 ⎞ ⎡⎣( — ◊ E ) E − E ¥ ( — ¥ E ) ⎤⎦ = E1 ⎜ + + ⎟ 1 ∂r3 ⎠ ⎝ ∂r1 ∂r2 ⎛ ∂E2 ∂E1 − E2 ⎜ − ⎝ ∂r1 ∂r2 23 January 2004
Physics 218, Spring 2004
⎛ ∂E1 ∂E3 ⎞ ⎞ − ⎟ ⎟ + E3 ⎜ r r ∂ ∂ 1 ⎠ ⎠ ⎝ 3 13
The Maxwell stress tensor (continued) It will pay to multiply this out and regroup: ∂E3 ∂E1 ∂E2 + E1 + E1 [...]1 = E1 ∂r1 ∂r2 ∂r3 ∂E3 ∂E2 ∂E1 ∂E1 − E2 + E2 + E3 − E3 ∂r1 ∂r2 ∂r3 ∂r1 ∂ ∂ 1 ∂ 2 = ( E1 ) + ( E1E2 ) + ( E1E3 ) ∂r2 ∂r3 2 ∂r1 1 ∂ 2 2 1 ∂ − ( E2 ) − ( E3 ) 2 ∂r1 2 ∂r1 ∂ ∂ ∂ 1 ∂ 2 E12 + E22 + E32 = ( E1 ) + ( E1E2 ) + ( E1E3 ) − ∂r1 ∂r2 ∂r3 2 ∂r1
(
23 January 2004
Physics 218, Spring 2004
) 14
The Maxwell stress tensor (continued)
[...]1
(
∂ ∂ ∂ 1 ∂ 2 E12 + E22 + E32 = ( E1 ) + ( E1E2 ) + ( E1E3 ) − ∂r1 ∂r2 ∂r3 2 ∂r1
)
⎡ 3 ∂ ⎤ 1 ∂ 3 ∂ ⎛ 1 2 ⎞ E1 E j ⎥ − E2 = E E E =⎢ − δ1 j ⎟ , ⎜ 1 j ⎢ j =1 ∂rj ⎥ 2 ∂r1 2 ∂rj ⎝ ⎠ j = 1 ⎣ ⎦ where we have reintroduced the Kronecker delta, ⎧ 1, i = j δ ij = ⎨ ⎩0, i ≠ j
∑
(
)
∑
This is for component 1; for component i, we’d get 3 ∂ ⎛ 1 2 ⎞ ⎡⎣( — ◊ E ) E − E ¥ ( — ¥ E ) ⎤⎦ = ⎜ Ei E j − E δ ij ⎟ . i ∂r 2 ⎠ j =1 j ⎝
∑
23 January 2004
Physics 218, Spring 2004
15
The Maxwell stress tensor (continued) Similarly,
3
∂ ⎛ 1 2 ⎞ ⎡⎣( — ◊ B ) B − B ¥ ( — ¥ B ) ⎤⎦ = ⎜ Bi B j − B δ ij ⎟ . i ∂r 2 ⎠ j =1 j ⎝ Let us now define a nine-component object,
∑
(
)
Maxwell stress 1 2 ⎛ ⎞ 2 ⎜ Ei E j + Bi B j − E + B δ ij ⎟ , tensor 2 ⎝ ⎠ with which we can re-write the ith component of both of the ugly terms in []: 1 ⎡⎣( — ◊ E ) E − E ¥ ( — ¥ E ) ⎤⎦ i 4π
1 Tij = 4π
3 ∂ 1 Tij + ⎡⎣( — ◊ B ) B − B ¥ ( — ¥ B ) ⎤⎦ = ∑ i ∂r 4π j =1 j 23 January 2004
Physics 218, Spring 2004
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The Maxwell stress tensor (continued) The best way to envision the sum on the right-hand side is as matrix multiplication: ⎡T11 T21 T31 ⎤ 3 ⎡ ∂ ∂ ∂ ∂ ⎤⎢ ⎥ T T T T = ∑ ∂rj ij ⎢ ∂r1 ∂r2 ∂r3 ⎥ ⎢ 12 22 32 ⎥ . ⎣ ⎦ ⎢T j =1 ⎣ 13 T23 T33 ⎥⎦ The result is represented by a three-component object; that is, a vector. We can use Ia vector-algebra-like symbolism for this operation, by using T to denote the second-rank tensor whose nine components are Tij . The sum is represented thus: 3 I ∂ ∑ ∂rj Tij ↔ — ◊ T , j =1 since it now looks so much like a divergence. 23 January 2004
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The Maxwell stress tensor (continued) Finally, the force per unit volume becomes, in terms of the Maxwell stress tensor, 1 F= ⎡⎣( — ◊ E ) E − E ¥ ( — ¥ E ) ⎤⎦ 4π 1 1 ∂ + ⎡⎣( — ◊ B ) B − B ¥ ( — ¥ B ) ⎤⎦ − ( E ¥ B) 4π 4π c ∂t I ∂ = — ◊ T − g EB , ∂t Momentum 1 1 g ≡ E ¥ B = S . where EB density 4π c c2
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Momentum conservation in electrodynamics The total force on the charges within volume V can be found by integrating F : I ∂ dpmech. ⎛ ⎞ Use divergence = ∫ F dτ = ∫ ⎜ — ◊ T − g EB ⎟ dτ theorem: dt ∂t ⎝ ⎠ V
=v ∫ S
V
I d T ◊ da − dt
∫ g EBdτ
,
V
We can now identify pEB ≡
∫ g EBdτ as the total momentum
V
stored in the fields E and B, so that g EB =
1 4π c
E ¥ B is the
momentum density in the fields. 23 January 2004
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Momentum conservation in electrodynamics (continued) Rearranging slightly, we have I d Momentum conservation + = d p p T ◊ a . ( mech. EB ) in electrodynamics dt
v∫ S
Interpretation: Changes in mechanical momentum (mvs for all the charges) and momentum stored in fields within V are I caused by the pressure and shear T distributed over the boundary surface S, and vice versa. I The terms on the diagonal of T , 1 ⎛ 2 2 1 2 2 ⎞ + − + E B E B ⎜ i ⎟ , i 4π ⎝ 2 ⎠ represent pressures in the E and B fields.
( )
(
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Momentum conservation in electrodynamics (continued) The off-diagonal terms,
(
1 Ei E j + Bi B j 4π
)
,
represent shear in the fields – the kind of stress that causes strain in a direction different than the stress. One can also write the momentum-conservation relation in differential form: I ∂ dpmech. ∂ ⎛ ⎞ g mech. dτ = ⎜ — ◊ T − g EB ⎟ dτ , or = dt ∂t ∂t ⎝ ⎠
∫
V
∫
V
I ∂ ( g mech. + g EB ) − — ◊ T = 0 , ∂t where g mech.is the mechanical (mv-type) momentum per unit volume for the charges within V. 23 January 2004
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Summary Electric and magnetic fields can store energy and momentum: 1 energy/volume uEB = E2 + B2 , 8π 1 g EB ≡ E¥B , momentum/volume 4π c or, for that matter, even angular momentum: 1 angular momentum/ LEB = r ¥ g EB = r ¥ ( E ¥ B) , volume 4π c and can transport energy (and thus all the other quantities): c energy/area/time S= E×B . 4π Electromagnetic waves are the normal method of transport.
(
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