Page 79 - Introduction to Continuum Mechanics

64 Curvilinear Coordinates

Again, from Fig. 2D.4b, we have

therefore,

that is,

From Fig. 2D.4a, it is clear that de<j> = d<t>(-e r'), therefore,

Substituting Eq,(2D3.4a) into Eq.(2D3.2), we have

We are now in a position to obtain the components of Vjf,Vv, div v, curl v and div T in
spherical coordinates.

(i)Components ofVf
Let (r,0,0) be a scalar field. By the definition of the gradient of/, we have,

i.e.,

From calculus, the total derivative of/is

Comparing Eq. (2D3.7) with Eq. (2D3.8), we obtain

(ii) Components ofVv
Let the vector field v be represented as:

Letting T=Vv, we have

Now by definition of the components of tensor T in spherical coordinates