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Transforming divergence from cartesian to cylindrical coordinates

  • Thread starter jaejoon89
  • Start date
1. Homework Statement

Compute the divergence in cylindrical coordinates by transforming the expression for divergence in cartestian coordinates.

2. Homework Equations

F = F_x i + F_y j + F_z k
div F = ∂F_x/∂x + ∂F_y/∂y + ∂F_z/∂z .......... (divergence in Cartesian coordinates)

I need to transform this into

divF = (1/rho)(∂(rho*F_rho)/∂rho) + (1/rho)(∂F_theta/∂theta) + ∂F_z/∂z ...... (divergence in cylindrical coordinates)

3. The Attempt at a Solution

Using the chain rule,
∂F_x/∂x = (∂F_x/∂rho)(∂rho/∂x) + (∂F_x/∂theta)(∂theta/∂x) + (∂F_x/∂z)(∂z/∂x)
∂F_y/∂y = (∂F_y/∂rho)(∂rho/∂y) + (∂F_y/∂theta)(∂theta/∂y) + (∂F_y/∂z)(∂z/∂y)
∂F_z/∂z = (∂F_z/∂rho)(∂rho/∂z) + (∂F_z/∂theta)(∂theta/∂z) + (∂F_z/∂z)(∂z/∂z)

∂rho/∂x = x/∂ = costheta
∂theta/∂x = -y/rho^2 = -sintheta/rho
∂z/∂x = 0
∂rho/∂y = y/∂ = sintheta
etc. (these are the transformational equations)

Then I try inputing this into the cartesian definition for divergence and obtain
divF = [(∂F_x/∂rho)costheta + (∂F_x/∂theta)(-sintheta/rho)] + [(∂F_y/∂rho)sintheta + (∂F_y/∂theta)(costheta/rho)] + ∂F_z/∂z

But how does that simplify to the expression in cylindrical coordinates?
 
any 1 here plzzzzzzzzzzzzzz. solve this!
i also need divergence in spherical!
 

HallsofIvy

Science Advisor
Homework Helper
41,683
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1. Homework Statement

Compute the divergence in cylindrical coordinates by transforming the expression for divergence in cartestian coordinates.

2. Homework Equations

F = F_x i + F_y j + F_z k
div F = ∂F_x/∂x + ∂F_y/∂y + ∂F_z/∂z .......... (divergence in Cartesian coordinates)

I need to transform this into

divF = (1/rho)(∂(rho*F_rho)/∂rho) + (1/rho)(∂F_theta/∂theta) + ∂F_z/∂z ...... (divergence in cylindrical coordinates)

3. The Attempt at a Solution

Using the chain rule,
∂F_x/∂x = (∂F_x/∂rho)(∂rho/∂x) + (∂F_x/∂theta)(∂theta/∂x) + (∂F_x/∂z)(∂z/∂x)
∂F_y/∂y = (∂F_y/∂rho)(∂rho/∂y) + (∂F_y/∂theta)(∂theta/∂y) + (∂F_y/∂z)(∂z/∂y)
∂F_z/∂z = (∂F_z/∂rho)(∂rho/∂z) + (∂F_z/∂theta)(∂theta/∂z) + (∂F_z/∂z)(∂z/∂z)

∂rho/∂x = x/∂ = costheta[/math]
∂theta/∂x = -y/rho^2 = -sintheta/rho
∂z/∂x = 0
∂rho/∂y = y/∂ = sintheta
etc. (these are the transformational equations)

Then I try inputing this into the cartesian definition for divergence and obtain
divF = [(∂F_x/∂rho)costheta + (∂F_x/∂theta)(-sintheta/rho)] + [(∂F_y/∂rho)sintheta + (∂F_y/∂theta)(costheta/rho)] + ∂F_z/∂z

But how does that simplify to the expression in cylindrical coordinates?
You haven't said anything about what [itex]F_\rho[/itex] and [itex]F_\theta[/itex] are in terms of [itex]F_x[/itex] and [itex]F_y[/itex].
 
lets assume that fx=f_rho*cos(theta)-f_theta*sin(theta)
fy=f_rho*sin(theta)+f_theta*cos(theta)
 

tiny-tim

Science Advisor
Homework Helper
25,790
246
oooh, this is all virtually unreadable :rolleyes:

can everybody please use the usual symbols? … :smile:
 
ok
fx=fρ*cosφ-fφ*sinφ
fy=fρ*sinφ+fφ*cosφ
 
Nay 1 here pleaseeeeeeeeeee solve this
i need it badly!
 

gabbagabbahey

Homework Helper
Gold Member
5,001
6
lets assume that fx=f_rho*cos(theta)-f_theta*sin(theta)
fy=f_rho*sin(theta)+f_theta*cos(theta)
Okay, then
[tex]\frac{\partial f_x}{\partial\rho}=\frac{\partial}{\partial\rho}(f_\rho\cos\theta-f_\theta\sin\theta)=\frac{\partial f_\rho}{\partial\rho}\cos\theta-\frac{\partial f_\theta}{\partial\rho}\sin\theta[/tex]

Right?
Then I try inputing this into the cartesian definition for divergence and obtain
divF = [(∂F_x/∂rho)costheta + (∂F_x/∂theta)(-sintheta/rho)] + [(∂F_y/∂rho)sintheta + (∂F_y/∂theta)(costheta/rho)] + ∂F_z/∂z

But how does that simplify to the expression in cylindrical coordinates?
Calculate [tex]\frac{\partial f_y}{\partial\rho}[/tex], [tex]\frac{\partial f_x}{\partial\theta}[/tex] and [tex]\frac{\partial f_y}{\partial\theta}[/tex] the same way and substitute them all into this expression.
 

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