Hao Mei

The University of South Dakota

Wednesday, Nov. 05, 2014

Vector and Scalar Potentials

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Maxwell Equations

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𝛻 · 𝐷 = ρ Coulomb’s Law

𝛻 × 𝐻 = 𝐽 +𝜕𝐷

𝜕𝑡 Ampere’s Law

𝛻 · 𝐵 = 0 Absence of free magnetic poles

𝛻 × 𝐸 +𝜕𝐵

𝜕𝑡= 0 Faraday’s Law

Potentials(Φ and 𝐴 )

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If a curl of a vector field (𝐹 ) vanishes(everywhere), then 𝐹 can be written as the gradient of a scalar potentials (Φ):

𝛻 x 𝐹 = 0 ⟺ 𝐹 = − 𝛻Φ

If a divergence of a vector field (𝐹 ) vanishes(everywhere),

then 𝐹 can be written as the curl of a vector potentials (𝐴 ):

𝛻 · 𝐹 = 0 ⟺ 𝐹 = 𝛻 x 𝐴

Potentials(Φ and 𝐴 )

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Since 𝛻 · 𝐵 = 0, we can still define 𝐵 in terms of a vector potential:

𝐵 = 𝛻 x 𝐴 (1)

Then the Faraday’s law can be written:

𝛻 × 𝐸 +𝜕𝐵

𝜕𝑡= 𝛻 × 𝐸 +

𝜕(𝛻 x 𝐴 )𝜕𝑡

= 𝛻 × (𝐸 +𝜕𝐴

𝜕𝑡)

= 0

Potentials(Φ and 𝐴 )

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Recall the definition of scalar potentials:

𝛻 x 𝐹 = 0 ⟺ 𝐹 = − 𝛻Φ

here we have

𝛻 × 𝐸 +𝜕𝐴

𝜕𝑡= 0

The vanishing curl means that we can define a scalar potential Φ satisfying:

−𝛻Φ = 𝐸 +𝜕𝐴

𝜕𝑡

or 𝐸 = −𝛻Φ −𝜕𝐴

𝜕𝑡 (2)

Maxwell equations in terms of Vector and Scalar Potentials

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Combining Equations (1) and (2),

𝐵 = 𝛻 x 𝐴 , 𝐸 = −𝛻Φ −𝜕𝐴

𝜕𝑡

These two equations, which is the definitions of 𝐵 and 𝐸 in

terms of Φ and 𝐴 , automatically satisfy the two homogeneous Maxwell equations.

This reduces the number of equations from 4 to 2.

Then the dynamic behavior of Φ and 𝐴 will be determined by the two inhomogeneous equations.

Maxwell equations in terms of Vector and Scalar Potentials

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At this stage we restrict our considerations to the vacuum form of the Maxwell equations.

Recall that ϵ0μ0 =1

𝑐2 , and 𝐻 =𝐵

μ0, 𝐷 = ϵ0𝐸.

Then the two inhomogeneous equations become

𝛻 · 𝐷 = ρ 𝛻 · 𝐸 =ρ

ϵ0

𝛻 · (−𝛻Φ −𝜕𝐴

𝜕𝑡) =

ρ

ϵ0

𝛻2Φ +𝜕(𝛻·𝐴 )

𝜕𝑡= −

ρ

ϵ0 (3)

Maxwell equations in terms of Vector and Scalar Potentials

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𝛻 × 𝐻 = 𝐽 +𝜕𝐷

𝜕𝑡, 𝐻 =

𝐵

μ0, 𝐷 = ϵ0𝐸

𝛻 ×𝐵

μ0= 𝐽 +

𝜕(ϵ0𝐸)

𝜕𝑡

𝛻 × 𝐵 = μ0𝐽 + ϵ0μ0𝜕𝐸

𝜕𝑡

𝛻 × 𝛻 x 𝐴 = μ0𝐽 + ϵ0μ0𝜕

𝜕𝑡(−𝛻Φ −

𝜕𝐴

𝜕𝑡)

= μ0𝐽 − ϵ0μ0𝜕

𝜕𝑡(𝛻Φ +

𝜕𝐴

𝜕𝑡)

Maxwell equations in terms of Vector and Scalar Potentials

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Using the identity( in Jackson cover):

𝛻 × 𝛻 x 𝐴 = 𝛻 𝛻·𝐴 − 𝛻2𝐴

𝛻2𝐴 −1

𝑐2

𝜕2𝐴

𝜕𝑡2 − 𝛻(𝛻·𝐴 +1

𝑐2

𝜕Φ

𝜕𝑡) = −μ0𝐽 (4)

The four first order coupled differential equations (Maxwell equations) reduce to two second order differential equations, but they are still coupled.

𝛻2Φ +𝜕(𝛻 · 𝐴 )

𝜕𝑡= −

ρ

ϵ0

𝛻2𝐴 −1

𝑐2

𝜕2𝐴

𝜕𝑡2 − 𝛻(𝛻·𝐴 +1

𝑐2

𝜕Φ

𝜕𝑡) = −μ0𝐽

Gauge Transformation

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Since 𝐵 = 𝛻 x 𝐴 , the vector potential is arbitrary to the extent that the gradient of some scalar function Ʌ can be added.

𝐵 is unchanged by the transformation:

𝐴 𝐴′ = 𝐴 + 𝛻Ʌ

• For 𝐸 to remain unchanged as well, we require

ΦΦ′ = Φ −𝜕Ʌ

𝜕𝑡

Quick check: 𝐸 = −𝛻Φ −𝜕𝐴

𝜕𝑡

−𝛻Φ′ −𝜕𝐴′

𝜕𝑡= −𝛻 Φ −

𝜕𝐴

𝜕𝑡−

𝜕 𝐴 + 𝛻Ʌ

𝜕𝑡

= −𝛻Φ +𝜕

𝜕𝑡𝛻Ʌ −

𝜕𝐴

𝜕𝑡−

𝜕

𝜕𝑡𝛻Ʌ

= 𝐸

Lorenz condition and wave equations

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We can use gauge freedom to specify useful conditions on

Φ, 𝐴 .

Until now only 𝛻 x 𝐴 has been specified; the choice of 𝛻·𝐴 is still arbitrary. Imposing the so-called Lorenz condition:

𝛻·𝐴 +1

𝑐2

𝜕Φ

𝜕𝑡= 0

Applying the Lorenz condition, it will decouples Eqs.(3) and (4), and results in a considerable simplification:

Lorenz condition and wave equations

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For Equation (3): 𝛻2Φ +𝜕(𝛻·𝐴 )

𝜕𝑡= −

ρ

ϵ0

applying the Lorenz condition:

𝛻·𝐴 +1

𝑐2

𝜕Φ

𝜕𝑡= 0

So, 𝛻2Φ −𝜕(

1

𝑐2𝜕Φ

𝜕𝑡)

𝜕𝑡= −

ρ

ϵ0

𝛻2Φ −1

𝑐2

𝜕2Φ

𝜕𝑡2 = −ρ

ϵ0 (5)

Lorenz condition and wave equations

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For Equation (4): 𝛻2𝐴 −1

𝑐2

𝜕2𝐴

𝜕𝑡2 − 𝛻(𝛻·𝐴 +1

𝑐2

𝜕Φ

𝜕𝑡) = −μ0𝐽

applying the Lorenz condition:

𝛻·𝐴 +1

𝑐2

𝜕Φ

𝜕𝑡= 0

So, 𝛻2𝐴 −1

𝑐2

𝜕2𝐴

𝜕𝑡2 − 𝛻(0) = −μ0𝐽

𝛻2𝐴 −1

𝑐2

𝜕2𝐴

𝜕𝑡2 = −μ0𝐽 (6)

Equations (5) and (6), form a set of equations equivalent in all respects to the Maxwell Equations in vacuum. Thus, the Lorentz condition makes and satisfies inhomogeneous wave equations of similar forms.

Questions and Discussion

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Thanks!