Topics in Classical Electrodynamics

Finally, we would like to comment on the solutions of the Laplace equation
△ϕ = 0. It is not difficult to show that one cannot have an absolute minimum
or maximum in the region (in both directions, x and y) because for an
< 0
implying that in the other direction the second derivative must have an opposite
sign.
extremum to exist one requires ∂ϕ
∂x i
= 0 which results in ∂2 ϕ
∂x 2 i
> 0 or ∂2 ϕ
∂x 2 i
2 Electrodynamics
Here we will treat electrodynamics as a classical relativistic field theory. We
will rewrite the basic equations of electrodynamics in the manifestly Lorentzcovariant
form. We will also study the corresponding solutions.
2.1 Relativistic Particle in Electormagnetic Field
Let us first revisit some of the basics of special relativity written using tensor
notation. The Minkowski metric η µν that we will use has the signature
(+, −, −, −) and we will use the convention that the Latin indices run only
over the space coordinates (i.e. i, j, k... = 1, 2, 3), whereas the Greek indices
will include both time and space coordinates (i.e. µ, ν, σ, ρ... = 0, 1, 2, 3). Additionally,
in special relativity we will have to distinguish between 3-vectors
(those with only space components) and 4-vectors (having both space and
time components). The convention that we will use is that ⃗ A will denote a
3-vector, whereas A µ will denote a 4-vector. Using these definitions, we can
define the Lorentz invariant relativistic interval given by the expression
ds 2 = x µ x µ = c 2 dt 2 − ( dx i) 2
. (27)
The action for a relativistic particle has the following form
Rewriting (27), we get
S = −α
∫ b
a
√
∫ b
ds2 = −α ds .
a
ds =
√
dx
µ
dt
dx µ
dt dt2 =
√
dx
µ
dt
dx µ
dt
dt . (28)
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