Cartesian velocity to polar velocity (Velocity Field Context)

Question

I'm trying to derive the polar functions for the Velocity Potential Function $\Phi$ from its cartesian definitions of:

\begin{align} \frac{dΦ}{dx} &= u \\\\\frac{dΦ}{dy} &= v \end{align}

where I believe $u$ and $v$ to be velocities in the $x$ and $y$ direction respectively.

I am trying to get to the following equation for $V_r$: \begin{align} \frac{dΦ}{dr} &= V_r \end{align} where $V_r$ is the velocity in the $r$ direction from the origin in polar coordinates. The only way I can currently complete this derrivation is by using \begin{align} V_r = u\cos(θ) \end{align} with θ being the angle from the x-axis in radians. If I accept this then I can follow through with the chain rule using $r = x\cos(θ)$ and differentiating to get $\frac{dr}{dx} = \cos(θ)$. Then it becomes \begin{align} \frac{dΦ}{dx} = \frac{dΦ}{dr} \times \frac{dr}{dx} = \frac{dΦ}{dr}\times \frac{1}{\cos(θ)} = \frac{V_r}{\cosθ} \end{align}

However, I have a feeling this is incorrect, and I cannot for the life of me work out how $V_r = u\cos(θ)$. Could anyone explain whether this is right and if so why?

Any help is greatly appreciated, thanks.

Answer 1

I'm going to assume you mean to use partial derivatives everywhere, otherwise there's something I'm missing.

In that case, your equation $$V_r=u\cos(\theta)$$ is, in fact, not correct. How could it be? Then your velocity would be $0$ for points in the $yy$-axis, even if $v$ were not zero. You can come up with similar arguments but it makes no sense for $V_r$ to depend only on $u$ and not $v$. These are independent velocities and $V_r$ should be some combination of both.

I think one of the main problems here lies in misuse of the chain rule, as $\Phi$ is a function of two variables, and not just one. For example, it is also not generally true that $$\frac{\partial\Phi}{\partial x}=\frac{\partial\Phi}{\partial r}\frac{\partial r}{\partial x}$$ What is correct is considering, for example, both polar variables $r$ and $\theta$ and writting $$\frac{\partial\Phi}{\partial x}=\frac{\partial\Phi}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial\Phi}{\partial \theta}\frac{\partial \theta}{\partial x}$$ as you can see here. And you can in fact use the chain rule the same way as above to determine how to differentiate $\Phi$ with respect to $r$: $$\frac{\partial\Phi}{\partial r}=\frac{\partial\Phi}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial\Phi}{\partial y}\frac{\partial y}{\partial r}$$ Since $x=r\cos(\theta)$ and $y=r\sin(\theta)$, the above becomes $$\frac{\partial\Phi}{\partial r}=u\cos(\theta)+v\sin(\theta)$$

Answer 2

In cartesian coordinates, $\delta \phi = u\delta x + v \delta y$ where $u, v$ are local velocity components.

But $ \delta \phi = \displaystyle \frac{\partial \Phi}{\partial x} \delta x + \frac{\partial \Phi}{\partial y} \delta y$ which gives us

$u = \displaystyle \frac{\partial \Phi}{\partial x}, \, v = \frac{\partial \Phi}{\partial y}$.

In polar coordinates,

$\delta \phi = v_r \delta r + v_{\theta} r \delta \theta$

But $\phi$ is a function of $(r, \theta$),

$\delta \phi = \displaystyle \frac{\partial \Phi}{\partial r} \delta r + \frac{\partial \Phi}{\partial \theta} \delta \theta$

So, $v_r = \displaystyle \frac{\partial \Phi}{\partial r}, v_{\theta} = \displaystyle \frac{1}{r} \frac{\partial \Phi}{\partial \theta}$

Conversion -

$x = r \cos \theta, y = r \sin \theta$
$u = v_r \cos \theta, v = v_r \sin \theta$

$v_r = \displaystyle \frac{\partial \Phi}{\partial r} = v_r (\cos^2 \theta + \sin^2 \theta) = u \cos \theta + v \sin \theta$

$v_{\theta} = \displaystyle \frac{1}{r} \frac{\partial \Phi}{\partial \theta} = \frac{1}{r} (\frac{\partial \Phi}{\partial x} \frac{\partial x}{\partial \theta} + \frac{\partial \Phi}{\partial y} \frac{\partial y}{\partial \theta}) = v \cos \theta - u \sin \theta$