Download Poynting Paradox

Jim Wiss
July 29, 2009
Resistor Poynting Power Paradox
Abstract
We discuss the use of Poynting’s theorem to obtain the power converted into heat for a
cylindrical resistor with a static, uniform current within the resistor and zero current and Efield outside of the resistor. We show that Poynting’s theorem works for cylindrical
bounding surfaces contained within the resistor but fails for surfaces which enclose the
resistor. The problem is that our model E-field has a non-zero curl in the φˆ direction at the
resistor surface. We show how Poynting’s theorem can be resurrected by adding a singular,
artificial time variation in the magnetic field at the resistor surface. We conclude with a
discussion of a similar paradox when one uses the Maxwell stress tensor to calculate the
pressure on the resistor surface.
Creating the Paradox
Consider applying Poynting’s theorem to a simple cylindrical resistor of radius a , length
h a with uniform current density J = J 0 zˆ and conductivity. σ ,
2a
h
J 0 zˆ
σ
∂u EM
+ E i J = 0 . Integrating over volume
∂t
and using the divergence theorem gives us the integral form
∂U EM
+ ∫ E i J dτ = 0 where S is any surface which “encloses” V and we dropped
∫ S ida +
∂t
S⊆ V
V
∂U EM / ∂t contribution since the fields are static in our model. The Poynting vector is given
by
sJ 2 sˆ
E × B 1 ⎡ J ⎤ ⎡ μ sJ φˆ ⎤
− 0 for s < a
= ⎢ zˆ ⎥ × ⎢
S=
⎥ = 2σ
μ ⎣ σ ⎦E ⎣ 2 ⎦B
μ0
0 for s > a
The differential form Poynting’s theorem is ∇i S +
0
0
0
0
which disappears outside of the resistor since E disappears. If we apply the integral form
to a bounding surface of length h and a radius R within the resistor ( R < a ) we get:
Page 1 of 4
∫ S ida +
S⊆ V
⎡ RJ 2 sˆ
⎤
∂U EM
+ ∫ E i J dτ = ⎢ − 0 i( 2π Rh sˆ ) ⎥
∂t
V
⎣ 2σ
⎦ ∫
(
S i da
S⊆ V
−π R hJ
2
σ
2
0
+
⎡⎛ J zˆ
⎤
⎞
π R2h ⎥
+ [ 0] ∂U EM + ⎢⎜ 0 i J 0 zˆ ⎟
τ
∂t
⎠E i J
⎣⎝ σ
⎦ ∫ E i J dτ
)
V
π R hJ
= 0 for R < a
σ
2
2
0
The paradox comes about when we use an enclosing bounding surface with R > a .
2
2
∫ E i J dτ is π a hJ 0 / σ but now ∫ S ida =0 since S ( s > a ) = 0 which violates our integral
S⊆ V
V
version of Poynting’s theorem. What went wrong?
Removing the Paradox
The problem is that our static resistor model violates Faraday’s law which is necessary to
prove Poynting’s theorem.
Here is a plot of the electrical field as a function of s in our model.
J0
Ez
σ
J
E = zˆ 0 ⎡⎣1 − Θ ( s − a ) ⎤⎦
σ
s
a
1 if s > a
where we introduce the Heaviside function Θ ( s − a ) =
. The derivative of the
0 if s < a
Heaviside function is a delta function implying a non-zero curl for the electric field.
J
∂E
∂ ⎧J
∂B
⎫
∇ × E = −φˆ z = −φˆ ⎨ 0 ⎡⎣1 − Θ ( s − a ) ⎤⎦ ⎬ = φˆ 0 δ ( s − a ) = −
σ
∂s
∂s ⎩ σ
∂t
⎭
This means our model is not correct for a static current resistor. We can still check
Poynting’s theorem, however, by including the singular ∂U EM / ∂t term due to ∂B / ∂t once
the volume includes s = a which would be true when R > a .
2
∂U EM
∂u
∂u ∂ ⎪⎧ Bφ ⎪⎫ Bφ ( s ) ∂Bφ ⎡ J 0 s ⎤ ⎡ J 0
⎤
= ∫ [ 2π hsds ]dτ ;
= ⎨
=⎢
− δ ( s − a )⎥
⎬=
⎥
⎢
μ0 ∂t ⎣ 2 ⎦ B ⎣ σ
∂t
∂t
∂t ∂t ⎪⎩ 2μ0 ⎪⎭
⎦ ∂Bφ
∂t
μ
(1.1)
0
⎡ J
⎤
sJ
J2
∂U EM
∂u
∂u
δ (s − a) →
δ ( s − a )⎥ = − 0 π a 2 h
=−
= ∫ 2π hsds = 2π h ∫ s 2 ds ⎢ −
σ
2σ
∂t
∂t
∂t
⎣ 2σ
⎦ ∂u
2
0
2
0
∂t
Let us write the integral form of Poynting’s theorem for the case where R > a :
⎡ J 02π a 2 h ⎤
⎡ π a 2 hJ 02 ⎤
∂U EM
τ
S
da
E
J
d
0
0
i
i
+
+
=
→
+
−
+
=0
[ ]∫ S i da ⎢ σ ⎥
∫
∫
⎢
⎥
σ
∂t
U
∂
V
EM
⎣
⎦
⎣
⎦ ∫ E i J dτ
∂t
V
Page 2 of 4
The Poynting vector surface integral disappears for an enclosing bounding cylinder ( R > a ) ,
and the rate of change of the E&M model cancels the E i J integral to restore Poynting’s
theorem .Thus Poynting’s Theorem is still valid once we compensate for the Faraday’s law
violation by adding an artificial, singular ∂Bφ / ∂t at the resistor surface.
A Related Stress Tensor Paradox
We start with the wrong approach that leads to a paradox.
We use the stress tensor to calculate the pressure on the
resistor surface assuming static E and B fields. We will
calculate the pressure using a infinitesimal cylinder with end
plane area A which extends just within and just outside of
the resistor centered at ( x y ) = ( a 0 ) . The force due to the
ε 0 J 02
ε 0 J 02
Fx = T A − T A = −
( −A ) = 2 A → Poutward = 2
2
2σ
2σ
<
xx
ε 0 Ez2
a
E
E
currents within the cylinder will be
>
xx
y
B
A
x
( wrong!)
We only considered the electrical stress tensor since it is the only tensor contribution which
is discontinuous across the resistor boundary. I believe this answer is wrong for the
following reasons.
1. The pressure in this wrong treatment is evidently due to the electrical field because it
is obtained from the electrical stress tensor and involves σ which only affects the
electric field. The conductivity σ is irrelevant to the magnetic field.
2. The electric field is parallel to the resistor surface and would not contribute to a force
normal to the cylindrical surface.
3. There are no free charges anywhere for the electrical fields to act on.
4. One could expect a magnetic pressure contribution of the form:
⎛ B + B< ⎞
P = K ×⎜ >
⎟ where we use the average of the field on either side of the
⎝ 2 ⎠
boundary to get the magnetic field not due to the surface current K . But there is only
a finite J and no surface current K . This is reinforced by noting that B .is
continuous across the resistor surface.
5. Two parallel magnetic currents would attract implying a negative outward pressure,
but our answer suggests a current repulsion.
The foregoing suggests that the correct answer must be P = 0 on the surface.
Removing the Stress Tensor Paradox
We will calculate the pressure on the surface using the differential form of the stress tensor
equation:
Page 3 of 4
a +δ
∂T elect
∂S
∂S
→ f xsingular = xx − ε 0 μ0 x ; Poutward = limδ →0 ∫ f xsingular dx
∂t
∂x
∂t
a −δ
In order to affect the pressure on the resistor surface we are looking for singular parts of
the force density since the surface force is the integral of force density over the infinitesimal
cylinder. We use Txxelect since this the only singular part of ∇iT . The magnetic stress tensor
is continuous. We next calculate the singular parts of the electric stress tensor and the rate
of change of the field momentum. The stress tensor discontinuity is due to the discontinuity
in Ez which is discussed in Eq. (1.1)
f = ∇iT − ε 0 μ0
2
J ∂ ⎧ J0
∂Txxelect
∂Ez
∂ ⎧ε E 2 ⎫
⎫ ε0 J0
x=s
⎡
⎤
1
x
a
= − ⎨ 0 z ⎬ = −ε 0 Ez
⎯⎯⎯
→ −ε 0 0
−
Θ
−
=
(
)
⎨
⎦⎬ σ 2 δ ( x − a)
σ ∂x ⎩ σ ⎣
∂x
∂x ⎩ 2 ⎭
∂x
⎭
The field momentum time derivative is due to the singular ∂B / ∂t which is also discussed in
Eq. (1.1)
∂ E×B
ε 0 J 02
∂S x
⎡ J 0 ⎤ ∂By
⎡ J0 ⎤ ⎡ J0
⎤
y =φ
x
ε 0 μ0
= ε0
= −ε 0 ⎢ ⎥
⎯⎯⎯
→ −ε 0 ⎢ ⎥ ⎢ − δ ( x − a ) ⎥ = 2 δ ( x − a )
σ
∂t
∂t
⎣ σ ⎦ z ∂t
⎣σ ⎦⎣ σ
⎦ ∂By
(
)
∂t
Thus the singular f x terms cancel:
∂Txxelect
∂S
ε J2
ε J2
− ε 0 μ0 x = 0 20 δ ( x − a ) − 0 20 δ ( x − a ) = 0 → Poutward = 0
∂x
∂t
σ
σ
We thus confirm that there is no pressure on the cylindrical surface of the resistor.
f xsingular =
Page 4 of 4