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Divergence with plane polars.



I'm not really sure where to start, I am so confused.
Reply 1
Avatar for nuodai
nuodai
17
Well r=x2+y2r=\sqrt{x^2+y^2} and tan⁡θ=yx\tan \theta = \dfrac{y}{x}, so if F(r,θ)=Frer+FθeθF(r, \theta) = F_r \mathbf{e}_r + F_{\theta}\mathbf{e}_{\theta} then you need to use the chain rule to find the corresponding expression for ∇⋅F\nabla \cdot \mathbf{F}.
Reply 2
Avatar for Zhen Lin
Zhen Lin
12
Original post by nuodai
Well r=x2+y2r=\sqrt{x^2+y^2} and tan⁡θ=yx\tan \theta = \dfrac{y}{x}, so if F(r,θ)=Frer+FθeθF(r, \theta) = F_r \mathbf{e}_r + F_{\theta}\mathbf{e}_{\theta} then you need to use the chain rule to find the corresponding expression for ∇⋅F\nabla \cdot \mathbf{F}.


It's probably easier to start from (x,y)=(rcos⁡θ,rsin⁡θ)(x, y) = (r \cos \theta, r \sin \theta). No funny implicit differentiation or inverse functions required here.
Reply 3
Avatar for refref
refref
OP
Thanks but I still don't know where to go. Do I change F(r,theta) in to a*i + b*j where a and b are some functions of r and theta? (by multiplying out F_r*e_r + F_theta*e_theta with e_r (costheta, sintheta), e_theta = (-sintheta, costheta)? Then find da/dx + db/dy? Or what?
Reply 4
Avatar for Zhen Lin
Zhen Lin
12
Well, the point is that you need to first express er\mathbf{e}_r and eθ\mathbf{e}_\theta in terms of the standard basis vectors ex\mathbf{e}_x and ey\mathbf{e}_y. You should know this by heart, if you don't, answer below:

Spoiler


Having done, this, you apply the standard formula ∇⋅(aex+bey)=∂a∂x+∂b∂y\displaystyle \nabla \cdot (a \mathbf{e}_x + b \mathbf{e}_y) = \frac{\partial a}{\partial x} + \frac{\partial b}{\partial y}, and then use the chain rule to rewrite it in terms of ∂∂r\displaystyle \frac{\partial}{\partial r} and ∂∂θ\displaystyle \frac{\partial}{\partial \theta}. (One way to do this is to calculate the Jacobian matrix ∂(x,y)∂(r,θ)\displaystyle \frac{\partial (x, y)}{\partial (r, \theta)} and invert it. This gives you the expressions you want in the form you need without fiddling with inverse functions.)
Reply 5
Avatar for refref
refref
OP
Thanks, it came out in the end (using the jacobian way). I'm assuming thats the way to find the general result with any curvilinear co-ordinates?

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