Mathematical Tools for Physics

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

(47)