PHY 641: EM-1

Problems and Solutions

Gulmammad Mammadov

Physics Department, Syracuse University,Syracuse, NY 13244-1130, USA

Email: gmammado@syr.edu or gulmammad@gmail.com

January 31, 2008

Problem 1

Let ∗Fµν be the field dual to Fµν with

∗Fµν =12εµνλρF

λρ (1)

Calculate ∗Fij and ∗F4i = i∗F0i in terms of Ei and Bi. What is ∗FµνFµν in

terms of E and B?

Solution:To solve the problem we will use the metric signature (+−−−) and ε0123 = −1.In this convention the field-strength tensor is

Fµν =

0 −Ex −Ey −EzEx 0 −Bz ByEy Bz 0 −BxEz −By Bx 0

(2)

Where the index µ = 0, 1, 2, 3 or (t, x, y, z) labels the rows, and the index ν thecolumns.Using (1), (2) and antisymmetric property of both Fµν and εµνλρ we can easilycompute components of ∗Fµν .

∗F12 =12

(ε1230F 30 + ε1203F03) = ε1230F

30 = Ez

∗F13 =12

(ε1320F 20 + ε1302F02) = −ε1230F 20 = −Ey

1

∗F23 =12

(ε2310F 10 + ε2301F01) = −ε3210F 10 = Ex

∗F01 =12

(ε0123F 23 + ε0132F32) = ε0123F

23 = Bx

∗F02 =12

(ε0213F 13 + ε0231F31) = −ε0123F 13 = By

∗F03 =12

(ε0312F 12 + ε0321F21) = −ε3012F 12 = Bz

Therefore the dual field-strength tensor is

∗Fµν =

0 Bx By Bz−Bx 0 Ez −Ey−By −Ez 0 Ex−Bz Ey −Ex 0

(3)

And if i < j

∗FµνFµν = 2(∗F0lF

0l + ∗FijFij) = −2(BxEx +ByEy +BzBz

+BxEx +ByEy +BzBz) = −4(BxEx +ByEy +BzBz) = −4E ·B

∗FµνFµν = −4E ·B (4)

Note that the left hand side of (4) being tensorial expression is coordinate in-dependent, and so is the right hand side. As it can be seen from the right handside of (4), ∗FµνF

µν is a pseudoscalar (changes sign under a parity inversion)quantity. ∗FµνF

µν is also one of two (another FµνFµν) fundamental invariantsof the electromagnetic field.

Problem 2

If there are elementary magnetic monopoles, then there will be a magneticcurrent, JMµ, to which ∂λ

∗Fλµ is proportional. Find this equation for ∂λ∗Fλµusing its similarity to ∂λFλµ in the presence of an electric current. Prove fromthis equation that ∂µJMµ = 0.

Solution:

Using (2) we can write

Fλµ = ηλσFσρηρµ

≡

1 0 0 00 −1 0 00 0 −1 00 0 0 −1

0 −Ex −Ey −EzEx 0 −Bz ByEy Bz 0 −BxEz −By Bx 0

1 0 0 00 −1 0 00 0 −1 00 0 0 −1

2

=

0 Ex Ey Ez−Ex 0 Bz −By−Ey −Bz 0 Bx−Ez By −Bx 0

which satisfies

∂λFλµ =4πcJµ. (5)

On the other hand we have already computed ∗Fλµ which is given by

∗Fλµ =

0 Bx By Bz−Bx 0 Ez −Ey−By −Ez 0 Ex−Bz Ey −Ex 0

. (6)

If we compare ∗Fλµ to Fλµ, we see that they are related by B → E. Thus weexpect that if there are magnetic monopoles in the nature, ∗Fλµ will satisfy thesame equation as one for Fλµ,

∂λ∗Fλµ =

4πcJMµ (7)

where JMµ is a magnetic current.For the second part of the problem we use antisymmetry of ∗Fλµ and the

symmetric property of partial derivatives:

∂µ∂λ∗Fλµ = ∂λ∂µ

∗Fµλ = −∂λ∂µ∗Fλµ = −∂µ∂λ∗Fλµ

If we put left and right hand sides together, we get

2∂µ∂λ∗Fλµ = 0

which implies∂µJMµ = 0.

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