PHYSICS 603 - Winter/Spring Semester 2020 - ODU

Classical Mechanics - Lecture notes 4/23/20

This derivation addresses several things we’ve learned over the last 2-3 weeks: Lagran-gian formalism for continuous degrees of freedom, electromagnetic interaction in Lagran-gian formulation, and relativity. Consider a system which has a pre-determined, given charge density and current density distribution, expressed by the 4 components of the charge current-density four-vector

everywhere in space and time. We consider the 4 components of

the 4-vector electromagnetic potential as the 4 continuous, inde-

pendent degrees of freedom, all dependent on space-time. (NOTE: In the following we assume the Lorentz gauge, i.e. ).

In the following, we also use the electromagnetic field (2-)tensor

where and

is the usual 4-dimensional gradient. In other words, can be

considered a function of the space-time derivatives of our continuous degrees of free-dom. Following our formalism, we can now define a Lagrangian density

. Here, the first part is the “potential energy” due to the

interaction between the current and the field (analog to the term 𝑞�⃗� ∙ 𝐴 − 𝑞Φ in the La-grangian for a single charge interacting with a given electromagnetic field) and the sec-ond part is just equal to the negative of the electromagnetic field energy density, 𝑤 =!!"𝐸+⃗ " + #

"$!𝐵+⃗ ".

First, we need to re-express the Lagrangian density in terms of only the fields and their derivatives :

. We use the

metric tensor to raise the indices of A and lower those of as needed:

jµ (xµ ) = cρ,!j( )(ct, !x)

Aµ (xµ ) = Φc,!A

⎛

⎝⎜

⎞

⎠⎟(ct,!x)

∂µAµ = 0

F µν =∂ µAν −∂νAµ =

0 −Ex / c −Ey / c −Ez / c

Ex / c 0 −Bz ByEy / c Bz 0 −BxEz / c −By Bx 0

⎛

⎝

⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟

∂ µ = gµα∂α

∂α =∂∂ct, ∂∂x, ∂∂ y, ∂∂ z

⎛

⎝⎜

⎞

⎠⎟ F µν

ℓ(Aµ ,∂νAµ ) = − jµA

µ −14µo

F µνFµν

Aµ

∂νAµ

ℓ(Aµ ,∂νAµ ) = − jµA

µ −14µo

F µνFµν = − jµAµ −

14µo

∂ µAν −∂νAµ( ) ∂µAν −∂νAµ( )

€

∂

PHYSICS 603 - Winter/Spring Semester 2020 - ODU

.

Straightforward multiplication of the last two brackets gives

where we made use of the fact that the metric tensor is symmetric and its own inverse: . (The last line also uses the fact that one can rename any indices

over which we are summing.) We are now ready to calculate the various derivatives re-quired by Eq. (13.23) in Goldstein; to avoid any confusion, I will be using brand new in-dices a, b:

and

where we have again used the fact that the metric tensor is symmetric and that we can re-name any indices over which we are summing. Now we are ready to write down the full equation of motion for the component :

Here, we used the Lorentz-gauge condition (together with the fact that derivatives can be interchanged) to cancel the second term in the bottom row. The final result relates two 1-forms to each other; we can easily apply the inverse of

on both sides and end up with . But this is equivalent to Maxwell’s equa-

tions for the 4-vector potential, see Eq. (7.67a) on page 297 in Goldstein! So we were able to derive Maxwell’s equations for the electromagnetic field from our Lagrangian density.

ℓ(Aµ ,∂νAµ ) = − jµA

µ −14µo

gµρ∂ρAν − gνρ∂ρA

µ( ) gνσ∂µAσ − gµσ∂νAσ( )

gµρgνσ∂ρAν∂µA

σ − gµρgµσ∂ρAν∂νA

σ − gνρgνσ∂ρAµ∂µA

σ + gνρgµσ∂ρAµ∂νA

σ( )= gµρgνσ∂ρA

ν∂µAσ − g ρσ∂ρA

ν∂νAσ − g ρσ∂ρA

µ∂µAσ + gνρgµσ∂ρA

µ∂νAσ( )

= gµρgνσ∂ρAν∂µA

σ −∂ρAν∂νA

ρ −∂ρAµ∂µA

ρ + gνρgµσ∂ρAµ∂νA

σ( )= 2gµρgνσ∂ρA

ν∂µAσ − 2∂ρA

ν∂νAρ( )

gµρgµσ = gρµgµσ = δρσ

∂ℓ∂Aα

= − jα

∂ℓ∂(∂βA

α )= −

14µo

2gµβgασ∂µAσ + 2g βρgνα∂ρA

ν − 2∂αAβ − 2∂αA

β( ) =−g βρgαν∂ρA

ν +∂αAβ

µo

Aα

∂β∂ℓ

∂(∂βAα )−∂ℓ∂Aα

=1µo

−g βρgαν∂β∂ρAν +∂β∂αA

β( )+ jα = 0

=1µo

−gαν∂ρ∂ρA

ν +∂α∂βAβ( )+ jα ⇒ gαν∂

ρ∂ρAν =∂ ρ∂ρAα = µo jα

gαν∂ ρ∂ρA

ν = µo jν

PHYSICS 603 - Winter/Spring Semester 2020 - ODU

Here is a screenshot from today’s whiteboard: