Mathematical Methods for Physics and Engineering : A Comprehensive Guide

26.23 EXERCISES

Writing out the covariant derivative, we obtain
(
dti
ds

+Γijktj

duk

ds

)

ei= 0.

But, sincetj=duj/ds, it follows that the equation satisfied by a geodesic is

d^2 ui

ds^2

+Γijk

duj

ds

duk

ds

=0. (26.101)

Find the equations satisfied by a geodesic (straight line) in cylindrical polar coordinates.

From (26.83), the only non-zero Christoffel symbols are Γ^122 =−ρand Γ^212 =Γ^221 =1/ρ.
Thus the required geodesic equations are

d^2 u^1

ds^2

+Γ^122

du^2

ds

du^2

ds

=0 ⇒

d^2 ρ

ds^2

−ρ

(

dφ

ds

) 2

=0,

d^2 u^2

ds^2

+2Γ^212

du^1

ds

du^2

ds

=0 ⇒

d^2 φ

ds^2

+

2

ρ

dρ

ds

dφ

ds

=0,

d^2 u^3

ds^2

=0 ⇒

d^2 z

ds^2

=0.

26.23 Exercises

26.1 Use the basic definition of a Cartesian tensor to show the following.

(a) That for any general, but fixed,φ,

(u 1 ,u 2 )=(x 1 cosφ−x 2 sinφ, x 1 sinφ+x 2 cosφ)

are the components of a first-order tensor in two dimensions.

(b) That

(

x^22 x 1 x 2

x 1 x 2 x^21

)

is not a tensor of order 2. To establish that a single element does not

transform correctly is sufficient.

26.2 The components of two vectors,AandB, and a second-order tensor,T, are given
in one coordinate system by

A=





1

0

0



, B=





0

1

0



, T=





2

√

√^30

340

002



.

In a second coordinate system, obtained from the first by rotation, the components

ofAandBare

A′=

1

2





√

3

0

1



, B′=^1

2





− 1

√^0

3



.

Find the components ofTin this new coordinate system and hence evaluate,

with a minimum of calculation,

TijTji,TkiTjkTij,TikTmnTniTkm.