Geodesics

TENSORS
components of a second-order tensor T are
δT ij
duk
≡ T ij ; k
,
δt
dt
δT ij
duk
≡ T ij; k
,
δt
dt
k
δTij
du
≡ Tij; k
.
δt
dt
The derivative of T along the curve r(t) may then be written in terms of, for
example, its contravariant components as
dT
δT ij
duk
=
ei ⊗ ej = T ij ; k
ei ⊗ ej .
dt
δt
dt
26.22 Geodesics
As an example of the use of the absolute derivative, we conclude this chapter
with a brief discussion of geodesics. A geodesic in real three-dimensional space
is a straight line, which has two equivalent defining properties. Firstly, it is the
curve of shortest length between two points and, secondly, it is the curve whose
tangent vector always points in the same direction (along the line). Although
in this chapter we have considered explicitly only our familiar three-dimensional
space, much of the mathematical formalism developed can be generalised to more
abstract spaces of higher dimensionality in which the familiar ideas of Euclidean
geometry are no longer valid. It is often of interest to find geodesic curves in
such spaces by using the defining properties of straight lines in Euclidean space.
We shall not consider these more complicated spaces explicitly but will determine the equation that a geodesic in Euclidean three-dimensional space (i.e.
a straight line) must satisfy, deriving it in a sufficiently general way that our
method may be applied with little modification to finding the equations satisfied
by geodesics in more abstract spaces.
Let us consider a curve r(s), parameterised by the arc length s from some point
on the curve, and choose as our defining property for a geodesic that its tangent
vector t = dr/ds always points in the same direction everywhere on the curve, i.e.
dt
= 0.
(26.100)
ds
Alternatively, we could exploit the property that the distance between two points
is a minimum along a geodesic and use the calculus of variations (see chapter 22);
this would lead to the same final result (26.101).
If we now introduce an arbitrary coordinate system ui with basis vectors ei ,
i = 1, 2, 3, then we may write t = ti ei , and from (26.99) we find
dt
duk
= ti ; k
ei = 0.
ds
ds
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