Transformation of Operators

  • Thread starter NoobixCube
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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.
 

Answers and Replies

  • #2
HallsofIvy
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Use the chain rule. For example, the grad of a function in Cartesian coordinates is given by [itex]f_x\vec{i}+ f_y\vec{j}[/itex].

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

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

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

That is, [itex]df/dx= cos(theta) df/dr- (1/r) sin(\theta) df/d\theta[itex] and you can do the same thing for df/dy.
 
  • #3
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Use the chain rule. For example, the grad of a function in Cartesian coordinates is given by [itex]f_x\vec{i}+ f_y\vec{j}[/itex].

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

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

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

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

Hey, thanks a lot for the reply!
 

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