10.2 Review of E&MHomework group: just me n leek42@uw.edu
Realistically, all of the E&M review comes from https://peppyhare.github.io/griffiths-em/
10.1.1 Maxwell Stress TensorThe Maxwell stress tensor encompasses the way that electromagnetic fields an exert forces/stresses in 3D space:
p = momentum
\vec p = \text{ momentum }
p = momentum
The basic force laws of E&M are
d p d t = q ( E + v × B )
\dv{\vec p}{t} = q \left( \vec E + \vec v \cross \vec B \right)
d t d p = q ( E + v × B )
On an element of volume, the change in momentum is
d P d t = ∫ V ( ρ E + j × B ) d V
\dv{P}{t} = \int_V (\rho E + j \cross B) \dd V
d t d P = ∫ V ( ρ E + j × B ) d V
The maxwell equations give the source terms as
ρ = ϵ 0 ∇ ⋅ E j = 1 μ 0 ( ∇ × B − μ 0 ϵ 0 ∂ E ∂ t )
\rho = \epsilon _0 \div \vec E \qquad \vec j = \frac{1}{\mu_0} \left( \curl \vec B - \mu_0 \epsilon_0 \pdv{E}{t} \right)
ρ = ϵ 0 ∇ ⋅ E j = μ 0 1 ( ∇ × B − μ 0 ϵ 0 ∂ t ∂ E )
ρ E + j × B = E ϵ 0 ( ∇ ⋅ E ) − B μ 0 × ( ∇ × B − μ 0 ϵ 0 ∂ E ∂ t )
\rho \vec E + \vec j \cross \vec B = \vec E \epsilon_0 (\div E) - \frac{B}{\mu_0} \cross ( \curl B - \mu_0 \epsilon_0 \pdv{E}{t} )
ρ E + j × B = E ϵ 0 ( ∇ ⋅ E ) − μ 0 B × ( ∇ × B − μ 0 ϵ 0 ∂ t ∂ E )
Using the chain rule ∂ ∂ t ( E × B ) = ∂ E ∂ t × B + E × ∂ B ∂ t \pdv{}{t} (E \cross B) = \pdv{E}{t} \cross B + E \cross \pdv{B}{t} ∂ t ∂ ( E × B ) = ∂ t ∂ E × B + E × ∂ t ∂ B B ( ∇ ⋅ B ) = 0 B( \div B) = 0 B ( ∇ ⋅ B ) = 0
ρ E + j × B = ϵ 0 E ( ∇ ⋅ E ) − B × ( ∇ × B ) μ 0 + ϵ 0 [ E × ∂ B ∂ t − ∂ ∂ t ( E × B ) ]
\rho E + j \cross B = \epsilon_0 E( \div E) - \frac{B \cross (\curl B)}{\mu_0} + \epsilon_0 \left[ E \cross \pdv{B}{t} - \pdv{}{t} (E \cross B) \right]
ρ E + j × B = ϵ 0 E ( ∇ ⋅ E ) − μ 0 B × ( ∇ × B ) + ϵ 0 [ E × ∂ t ∂ B − ∂ t ∂ ( E × B ) ]
From Faraday, ∂ B ∂ t → ∇ × E \pdv{B}{t} \rightarrow \curl E ∂ t ∂ B → ∇ × E
= ϵ 0 E ( ∇ ⋅ E ) + B ( ∇ ⋅ B ) μ 0 − B × ( ∇ × B ) μ 0 − ϵ 0 E × ( ∇ × E ) − ϵ 0 ∂ ∂ t ( E × B )
= \epsilon_0 E(\div E) + \frac{B(\div B)}{\mu_0} - \frac{B \cross (\curl B)}{\mu_0} - \epsilon_0 E \cross (\curl E) - \epsilon_0 \pdv{}{t} (E \cross B)
= ϵ 0 E ( ∇ ⋅ E ) + μ 0 B ( ∇ ⋅ B ) − μ 0 B × ( ∇ × B ) − ϵ 0 E × ( ∇ × E ) − ϵ 0 ∂ t ∂ ( E × B )
Finally we've got everything in place to use the identity
1 2 ∇ ( B 2 ) = 1 2 ∇ ( B ⋅ B ) = ( B ⋅ ∇ ) B + B × ( ∇ × B )
\frac{1}{2} \grad (B^2) = \frac{1}{2} \grad (\vec B \cdot \vec B) = (\vec B \cdot \grad) \vec B + \vec B \cross (\curl \vec B)
2 1 ∇ ( B 2 ) = 2 1 ∇ ( B ⋅ B ) = ( B ⋅ ∇ ) B + B × ( ∇ × B )
All together,
ρ E + j × B = ϵ 0 [ E ( ∇ ⋅ E ) + ( E ⋅ ∇ ) E − 1 2 ∇ ( E 2 ) ] + 1 μ 0 [ B ( ∇ ⋅ B ) + ( B ⋅ ∇ ) B − 1 2 ∇ ( B 2 ) ] − ϵ 0 ∂ ∂ t ( E × B )
\rho E + j \cross B = \epsilon_0 \left[ E ( \div E ) + (\vec E \cdot \grad) E - \frac{1}{2} \grad (E^2) \right] + \frac{1}{\mu_0} \left[B (\div B) + (\vec B \cdot \grad) B - \frac{1}{2} \grad (B^2) \right] - \epsilon_0 \pdv{}{t} ( \vec E \cross \vec B)
ρ E + j × B = ϵ 0 [ E ( ∇ ⋅ E ) + ( E ⋅ ∇ ) E − 2 1 ∇ ( E 2 ) ] + μ 0 1 [ B ( ∇ ⋅ B ) + ( B ⋅ ∇ ) B − 2 1 ∇ ( B 2 ) ] − ϵ 0 ∂ t ∂ ( E × B )
Integrating over a volume,
d P d t + ϵ 0 ∂ ∂ t ∫ E × B d V
\dv{P}{t} + \epsilon_0 \pdv{}{t} \int \vec E \cross \vec B \dd V
d t d P + ϵ 0 ∂ t ∂ ∫ E × B d V
= ∫ ϵ 0 [ E ( ∇ ⋅ E ) + ( E ⋅ ∇ ) E − 1 2 ∇ ( E 2 ) ] + 1 μ 0 [ B ( ∇ ⋅ B ) + ( B ⋅ ∇ ) B − 1 2 ∇ ( B 2 ) ] d V
\qquad = \int \epsilon_0 \left[ E(\div E) + (E \cdot \grad) E - \frac{1}{2} \grad(E^2) \right] + \frac{1}{\mu_0} \left[ B(\div B) + (B \cdot \grad) B - \frac{1}{2} \grad (B^2) \right] \dd V
= ∫ ϵ 0 [ E ( ∇ ⋅ E ) + ( E ⋅ ∇ ) E − 2 1 ∇ ( E 2 ) ] + μ 0 1 [ B ( ∇ ⋅ B ) + ( B ⋅ ∇ ) B − 2 1 ∇ ( B 2 ) ] d V
To simplify, we use another identity
E ( ∇ ⋅ E ) + ( E ⋅ ∇ ) E − 1 2 ∇ ( E 2 ) = ∇ ⋅ ( E E − 1 2 1 E 2 )
E (\div E) + (E \cdot \grad) E - \frac{1}{2} \grad (E^2) = \div (\vec E \vec E - \frac{1}{2} \vec 1 E^2)
E ( ∇ ⋅ E ) + ( E ⋅ ∇ ) E − 2 1 ∇ ( E 2 ) = ∇ ⋅ ( E E − 2 1 1 E 2 )
And the same is true for B \vec B B
T = ϵ 0 E E + B B μ 0 − 1 2 1 ( ϵ 0 E 2 + B 2 μ 0 )
\vec T = \epsilon_0 \vec E \vec E + \frac{\vec B \vec B}{\mu_0} - \frac{1}{2} \vec 1 \left( \epsilon_0 E^2 + \frac{B^2}{\mu_0} \right)
T = ϵ 0 E E + μ 0 B B − 2 1 1 ( ϵ 0 E 2 + μ 0 B 2 )
That lets us re-cast the momentum change in a volume as a stress tensor on a surface
d p d t + ϵ 0 d d t ∫ ( E × B ) d V = ∫ ∇ ⋅ T d V = ∫ T ⋅ n ^ d a
\dv{\vec p}{t} + \epsilon_0 \dv{}{t} \int ( \vec E \cross \vec B) \dd V = \int \div \vec T \dd V = \int \vec T \cdot \vu{n} \dd \vec a
d t d p + ϵ 0 d t d ∫ ( E × B ) d V = ∫ ∇ ⋅ T d V = ∫ T ⋅ n ^ d a
Example: Application to FRC (Field-Reversed Configuration)
In a FRC confinement experiment, a bias field B ∞ B_{\infty} B ∞
T = B B μ 0 − 1 2 I ( B 2 μ 0 )
\vec T = \frac{\vec B \vec B}{\mu_0} - \frac{1}{2} \vec I \left( \frac{B^2}{\mu_0} \right)
T = μ 0 B B − 2 1 I ( μ 0 B 2 )
If we want to confine a plasma in the center of the FRC, we want to calculate the force on our plasma (electron fluid).
F = ∫ T ⋅ n ^ d a
\vec F = \int \vec T \cdot \vu n \dd a
F = ∫ T ⋅ n ^ d a
At the left end of the configuration where we have B = B z z ^ \vec B = B_z \vu z B = B z z ^ B 1 = B r , B 2 = B z , B 3 = B θ B_1 = B_r, B_2 = B_z, B_3 = B_\theta B 1 = B r , B 2 = B z , B 3 = B θ
T = [ − 1 2 B 0 2 μ 0 0 0 0 B 0 2 μ 0 − 1 2 B 0 2 μ 0 0 0 0 − 1 2 B 0 2 μ 0 ]
\vec T = \begin{bmatrix} - \frac{1}{2} \frac{B_0 ^2}{\mu_0} & 0 & 0 \\ 0 & \frac{B_0 ^2}{\mu_0} - \frac{1}{2} \frac{B_0 ^2}{\mu_0} & 0 \\ 0 & 0 & - \frac{1}{2} \frac{B_0 ^2}{\mu_0} \end{bmatrix}
T = ⎣ ⎢ ⎢ ⎢ ⎡ − 2 1 μ 0 B 0 2 0 0 0 μ 0 B 0 2 − 2 1 μ 0 B 0 2 0 0 0 − 2 1 μ 0 B 0 2 ⎦ ⎥ ⎥ ⎥ ⎤
T ⋅ n ^ = [ 0 − 1 2 B 0 2 μ 0 z ^ 0 ]
\vec T \cdot \vu n = \begin{bmatrix} 0 \\ -\frac{1}{2} \frac{B_0 ^2}{\mu_0} \vu z \\ 0 \end{bmatrix}
T ⋅ n ^ = ⎣ ⎢ ⎢ ⎡ 0 − 2 1 μ 0 B 0 2 z ^ 0 ⎦ ⎥ ⎥ ⎤
Now what about at the sides? There, B r ≠ 0 B_r \neq 0 B r = 0
T ⋅ n ^ = [ − 1 2 B 2 μ 0 + B r 2 μ 0 B r B z μ 0 0 B r B z μ 0 − 1 2 B 2 μ 0 + B z 2 μ 0 0 0 0 − 1 2 B 2 μ 0 ] [ 1 0 0 ] = [ − 1 2 B 2 μ 0 + B r 2 μ 0 B r B z μ 0 0 ]
\vec T \cdot \vu n = \begin{bmatrix} - \frac{1}{2} \frac{B^2}{\mu_0} + \frac{B_r ^2}{\mu_0} & \frac{B_r B_z}{\mu_0} & 0 \\ \frac{B_r B_z}{\mu_0} & - \frac{1}{2} \frac{B^2}{\mu_0} + \frac{B_z ^2}{\mu_0} & 0 \\ 0 & 0 & - \frac{1}{2} \frac{B^2}{\mu_0} \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} - \frac{1}{2} \frac{B^2}{\mu_0} + \frac{B_r ^2}{\mu_0} \\ \frac{B_r B_z}{\mu_0} \\ 0 \end{bmatrix}
T ⋅ n ^ = ⎣ ⎢ ⎢ ⎢ ⎡ − 2 1 μ 0 B 2 + μ 0 B r 2 μ 0 B r B z 0 μ 0 B r B z − 2 1 μ 0 B 2 + μ 0 B z 2 0 0 0 − 2 1 μ 0 B 2 ⎦ ⎥ ⎥ ⎥ ⎤ ⎣ ⎢ ⎡ 1 0 0 ⎦ ⎥ ⎤ = ⎣ ⎢ ⎢ ⎡ − 2 1 μ 0 B 2 + μ 0 B r 2 μ 0 B r B z 0 ⎦ ⎥ ⎥ ⎤
The total force is the sum
F = ∫ T ⋅ n ^ d a = ( − 1 2 B 0 2 μ 0 + 1 2 B 0 2 μ 0 ) π a 2 z ^ + contrib. from sides
\vec F = \int \vec T \cdot \vu n \dd a = \left( - \frac{1}{2} \frac{B_0 ^2}{\mu_0} + \frac{1}{2} \frac{B_{0} ^2}{\mu_0} \right) \pi a ^2 \vu z + \text{ contrib. from sides}
F = ∫ T ⋅ n ^ d a = ( − 2 1 μ 0 B 0 2 + 2 1 μ 0 B 0 2 ) π a 2 z ^ + contrib. from sides
By symmetry, the r r r
F z = π a 2 2 μ 0 ( B ∞ 2 − B 0 2 ) + 2 π a μ 0 ∫ B r B z d z
F_z = \frac{\pi a^2}{2 \mu_0} (B_{\infty} ^2 - B_0 ^2) + \frac{2 \pi a}{\mu_0} \int B_r B_z \dd z
F z = 2 μ 0 π a 2 ( B ∞ 2 − B 0 2 ) + μ 0 2 π a ∫ B r B z d z
The magnetic flux at the open ends must go somewhere
Φ = B z π a 2 ( 1 − x s 2 )
\Phi = B_z \pi a^2 (1 - x_s ^2)
Φ = B z π a 2 ( 1 − x s 2 )
2 π a B r d z = − d Φ
2 \pi a B_r \dd z = - \dd \Phi
2 π a B r d z = − d Φ
F z = 1 2 μ 0 π a 2 ( Φ ∞ 2 − Φ 0 2 ) − 1 μ 0 ∫ Φ 0 Φ ∞ Φ π a 2 ( 1 − x s 2 ) d Φ
F_z = \frac{1}{2 \mu_0 \pi a^2} ( \Phi_{\infty} ^2 - \Phi_0 ^2 ) - \frac{1}{\mu_0 } \int_{\Phi_0} ^{\Phi_{\infty}} \frac{\Phi}{\pi a^2 ( 1 - x_s ^2)} \dd \Phi
F z = 2 μ 0 π a 2 1 ( Φ ∞ 2 − Φ 0 2 ) − μ 0 1 ∫ Φ 0 Φ ∞ π a 2 ( 1 − x s 2 ) Φ d Φ
Assume x s x_s x s
F z = 1 2 μ 0 π a 2 ( Φ ∞ 2 − Φ 0 2 ) − Φ ∞ 2 − Φ 0 2 2 μ 0 π a 2 ( 1 − x s 2 )
F_z = \frac{1}{2 \mu_0 \pi a^2} ( \Phi_{\infty} ^2 - \Phi_0 ^2 ) - \frac{ \Phi_{\infty} ^2 - \Phi_0 ^2}{2 \mu_0 \pi a ^2 (1 - x_s ^2)}
F z = 2 μ 0 π a 2 1 ( Φ ∞ 2 − Φ 0 2 ) − 2 μ 0 π a 2 ( 1 − x s 2 ) Φ ∞ 2 − Φ 0 2
= π a 2 2 μ 0 ( B ∞ 2 − B 0 2 ) ( 1 − 1 1 − x s 2 )
= \frac{\pi a^2}{2 \mu_0} ( B_\infty ^2 - B_0 ^2) \left( 1 - \frac{1}{1 - x_s ^2} \right)
= 2 μ 0 π a 2 ( B ∞ 2 − B 0 2 ) ( 1 − 1 − x s 2 1 )
= π a 2 2 μ 0 ( B 0 2 − B ∞ 2 ) x s 2 1 − x s 2
= \frac{\pi a^2}{2 \mu_0 } ( B_0 ^2 - B_{\infty} ^2 ) \frac{x_s ^2}{1 - x_s ^2}
= 2 μ 0 π a 2 ( B 0 2 − B ∞ 2 ) 1 − x s 2 x s 2
= A f r c 2 μ 0 ( B 0 2 − B ∞ 2 ) ( 1 1 − x s 2 )
= \frac{A_{frc}}{2 \mu_0} ( B_0 ^2 - B_{\infty} ^2 ) \left( \frac{1}{1 - x_s ^2} \right)
= 2 μ 0 A f r c ( B 0 2 − B ∞ 2 ) ( 1 − x s 2 1 )
Where A f r c = π a 2 x s 2 A_{frc} = \pi a^2 x_s^2 A f r c = π a 2 x s 2
