How Do You Redefine Grad, Div, and Curl in a New Coordinate System?

[NoobixCube]

hi all,
Simple questions..

I am dealing with the del operator (grad, div curl) in one coord system, but say I parametrise my system into another one. How then do I redefine the grad, div, and curl operators.

Any links would be really helpful.

 

[HallsofIvy]

Use the chain rule. For example, the grad of a function in Cartesian coordinates is given by f_x\vec{i}+ f_y\vec{j}.

In terms of polar coordinates,
\frac{df}{dx}= \frac{df}{dr}\frac{dr}{dx}+ \frac{df}{d\theta}\frac{d\theta}{dx}

Of course r= (x^2+ y^2)^{1/2} so dr/dx= (1/2)(x^2+ y^2)^{-1/2}(2x)= x/(x^2+ y^2)^{-1/2}= r cos(\theta)/r= cos(\theta)

and \theta= arctan(y/x) so d\theta/dx= (1/(1+ y^2/x^2))(-y/x^2) and d\theta/dx= -y/(x^2+ y^2)= -r sin(\theta)/r^2= (-1/r) sin(\theta)

That is, df/dx= cos(theta) df/dr- (1/r) sin(\theta) df/d\thetaand you can do the same thing for df/dy.

 

[NoobixCube]

Use the chain rule. For example, the grad of a function in Cartesian coordinates is given by f_x\vec{i}+ f_y\vec{j}.

In terms of polar coordinates,
\frac{df}{dx}= \frac{df}{dr}\frac{dr}{dx}+ \frac{df}{d\theta}\frac{d\theta}{dx}

Of course r= (x^2+ y^2)^{1/2} so dr/dx= (1/2)(x^2+ y^2)^{-1/2}(2x)= x/(x^2+ y^2)^{-1/2}= r cos(\theta)/r= cos(\theta)

and \theta= arctan(y/x) so d\theta/dx= (1/(1+ y^2/x^2))(-y/x^2) and d\theta/dx= -y/(x^2+ y^2)= -r sin(\theta)/r^2= (-1/r) sin(\theta)

That is, df/dx= cos(theta) df/dr- (1/r) sin(\theta) df/d\theta and you can do the same thing for df/dy.

Hey, thanks a lot for the reply!