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Hello: I have 3 components of the Navier equation. One of them is this:

$\mu\left(\dfrac{\partial v_x}{\partial y} + \dfrac{\partial v_y}{\partial x}\right) = \tau_{xy} = \tau_{yx}$

Can anyone help me convert this to cylindrical coordinates?

Hint: The answer is: $\mu\left[r\dfrac{\partial}{\partial r}\left(\dfrac{v_\theta}{r}\right) + \dfrac{1}{r}\left(\dfrac{\partial v_r}{\partial \theta}\right)\right]$

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    $\begingroup$ What have you tried? Should just be a change of variables, I assume you've at least got as far as $x=f(r,\theta,z)$, $y=$, $z=$ and so on. Don't forget the chain rule for the derivatives... If you can show specifically where you're getting hung up, you're much more likely to get an answer. Also, I cleaned up the math in your post. Have a look and make sure it's still as intended, and try and use the math markup on future posts. $\endgroup$ – Kyle Jul 10 '13 at 17:19
  • $\begingroup$ I am not sure how to relate the partial derivatives to each other. Does d/dx = (d/dx)(dx/dr) + (d/dy)(dy/dr)? $\endgroup$ – Ben Jul 10 '13 at 17:21
  • $\begingroup$ @Kyle Could you help show me the first couple of steps to get me started? I am not sure if my equation relating d/dx to cylindrical coordinates is even right. $\endgroup$ – Ben Jul 10 '13 at 17:31
  • $\begingroup$ This is purely a math question and doesn't have any direct relation to physics aside from the equation being manipulated. $\endgroup$ – tpg2114 Jul 11 '13 at 2:51
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For first order partial derivatives you can just use the chain-rule:

$\frac{\partial}{\partial x} = \frac{\partial r}{\partial x}\frac{\partial}{\partial r}+\frac{\partial \theta}{\partial x}\frac{\partial}{\partial\theta}+\frac{\partial z}{\partial x}\frac{\partial}{\partial z}$

Now in polar coördinates the last term drops out since z and x are independent variables. By using the transformation-rules for polar-coördinates the partial derivatives can be calculated.

$r = \sqrt{x^2+y^2}$ yields $\frac{\partial r}{\partial x} = \frac{x}{r}$, or upon using $x = r\cos\theta$: $\frac{\partial r}{\partial x} = \cos \theta$.

The same steps can be made for the second term.

$\theta = \arctan(\frac{y}{x})$ yields (with $y = r\sin\theta$): $\frac{\partial\theta}{\partial x} = -\sin\theta$. This gives:

$\frac{\partial}{\partial x} = \cos\theta\frac{\partial}{\partial r}-\sin\theta\frac{\partial}{\partial\theta}$

For $\frac{\partial}{\partial y}$ you should do the same steps.

Now you also need to transform your velocity using the transformations (remember $v_x = \dot{x} = ...)$!

These definitions you'll probably need:

$v_r = \dot{r}$ and $v_\theta = r\dot{\theta}$.

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I'll do a quick example going in the other direction (transforming into cartesian coordinates) since that's the case where it's easiest to follow what's going on; from there, you should be able to figure out how to proceed to transform into cylindrical coords. The chain rule for partial derivatives of a real-valued function $f(\vec{u})$ is:

$\dfrac{\partial f}{\partial x_i} = \nabla f\cdot\dfrac{\partial \vec{u}}{\partial x_i}$

If you're not familiar with the $\nabla$ operator you can look it up; applied to a real-valued function like this it's called a gradient. In the case where $f$ is in cartesian coordinates it's pretty trivial:

$\nabla f = \left(\dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y},\dfrac{\partial f}{\partial z}\right)$

The gradient in cylindrical and spherical coordinates is somewhat more complicated. There's a useful table here.

The components of $\vec{u}$ are just the cartesian coordinates in this case, and the $x_i$'s are the cylindrical coordinates. So for instance for the first cylindrical coordinate ($r$) you would get:

$\dfrac{\partial f}{\partial r} = \left(\dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y},\dfrac{\partial f}{\partial z}\right)\cdot\left(\dfrac{\partial x}{\partial r},\dfrac{\partial y}{\partial r},\dfrac{\partial z}{\partial r}\right) = \dfrac{\partial f}{\partial x}\dfrac{\partial x}{\partial r} + \dfrac{\partial f}{\partial y}\dfrac{\partial y}{\partial r} + \dfrac{\partial f}{\partial z}\dfrac{\partial z}{\partial r}$

Or, if you wanted to just write the operator:

$\dfrac{\partial}{\partial r} = \left(\dfrac{\partial x}{\partial r}\right)\dfrac{\partial }{\partial x} + \left(\dfrac{\partial y}{\partial r}\right)\dfrac{\partial }{\partial y} + \left(\dfrac{\partial z}{\partial r}\right)\dfrac{\partial }{\partial z}$

Using this method you can derive the derivatives $\dfrac{\partial}{\partial x}$, $\dfrac{\partial}{\partial z}$ and $\dfrac{\partial}{\partial z}$ in terms of the cylindrical coordinates. You can also look up the answer in just about any reference on the topic (good way to check your answer), but it's probably worth going through the derivation at least once to get a deeper understanding.

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  • $\begingroup$ Oh and by the way, it's likely that as you work through this and apply it to your NS equation, it will get messy with lots of terms... then they should start to cancel, or many may be zero, and you should eventually get where you want to be. Just might be a bit of a bumpy road. $\endgroup$ – Kyle Jul 10 '13 at 18:30
  • $\begingroup$ I got to the point where I have d/dy(Vx) + d/dx(Vy). I have found what d/dx and d/dy are using your method. Now how do I know what Vx and Vy are? I am lost at this point $\endgroup$ – Ben Jul 10 '13 at 22:57
  • $\begingroup$ As Nick points out in his answer, $v_x$ is just $\frac{\partial x}{\partial t}$, so if you replace x with it's representation in terms of cylindrical coordinates, you can just do the math from there. Remember $v_r=\frac{\partial \theta}{\partial t}$ and similarly for $v_\theta$. $\endgroup$ – Kyle Jul 10 '13 at 23:03

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