Shan K said:
Will anyone help me to under stand the covariant and contravariant vector ? And can anyone show me the derivation of
dr=(dr/dx)dx + (dr/dy)dy + (dr/dz)dz
The last, at least, is just the "chain rule" from Calculus. Suppose r is a function of x, y, and z and they, themselves, depend on the parameter t. We can think of r as, perhaps, the radius of curvature of a surface defined by x, y, and z, and (x(t), y(t), z(t)) as the trajectory of an object moving over that surface. Of course, then we can think of r(t)= r(x(t), y(t), z(t)) as giving the
radius of curvature at each point along that trajectory.
By the chain rule,
[tex]\frac{dr}{dt}= \frac{\partial r}{\partial x}\frac{dx}{dt}+ \frac{\partial r}{\partial y}\frac{dyz}{dt}+ \frac{\partial r}{\partial z}\frac{dz}{dt}[/tex]
In terms or "differentials" we can write
[tex]dr= \left(\frac{\partial r}{\partial x}\frac{dx}{dt}+ \frac{\partial r}{\partial y}\frac{dy}{dt}+ \frac{\partial r}{\partial z}\frac{dz}{dt}\right)dt= \frac{\partial r}{\partial x}\frac{dx}{dt}dt+ \frac{\partial r}{\partial y}\frac{dy}{dt}dt+ \frac{\partial r}{\partial z}\frac{dz}{dt}dt[/tex]
[tex]dr= \frac{\partial r}{\partial x}dx+ \frac{\partial r}{\partial y}dy+ \frac{\partial r}{\partial z}dz[/tex]