12—Tensors 387
in terms of the different coordinate functions. Call the two sets of coordinatesxiandyi. Each of them defines
a set of basis vectors such that a given velocity is expressed as
~v=~ei
dxi
dt
=~e′j
dyi
dt
(45)
What you need is an expression for~e′j in terms of~eiat each point. To do this, take a particular path for the
particle — along they^1 -direction (y^2 &y^3 constant). The right hand side is then
~e′ 1
dy^1
dt
Divide bydy^1 /dtto obtain
~e′ 1 =~ei
dxi
dt
/
dy^1
dt
But this quotient is just
~e′ 1 =~ei
∂xi
∂y^1
∣
∣
∣
∣
y^2 ,y^3
And in general
~e′j=~ei
∂xi
∂yj
(46)
Do a similar calculation for the reciprocal vectors
gradφ=~ei
∂φ
∂xi
=~e′j
∂φ
∂yj
Take the case for whichφ=yk, then
∂φ
∂yj
=
∂yk
∂yj
=δjk
which gives
~e′k=~ei
∂yk
∂xi