How to express velocity gradient in cylindrical coordinates?

[Inquisitive Student]

Homework Statement

The vlasov equation is (from !Introduction to Plasma Physics and Controlled Fusion! by Francis Chen):

$$\frac{d}{dt}f + \vec{v} \cdot \nabla f + \vec{a} \cdot \nabla_v f = 0$$

Where $$\nabla_v$$ is the del operator in velocity space. I've read that $$\nabla_v = \frac{\partial}{\partial v_r} \hat{v_r} + \frac{1}{v_r}\frac{\partial}{\partial \theta_v} \hat{v_\theta} + \frac{\partial}{\partial v_z} \hat{z}$$ which I think is the operator expressed in cylindrical coordinates in velocity space. I would like to express this operator in cylindrical coordinates in regular space (i.e. $$\hat{\rho},\hat{\theta},\hat{z}$$).

Homework Equations

(diagram)[/B]

The Attempt at a Solution

$$v_r = \sqrt{(v_\rho)^2 + (v_\phi)^2} = v$$

$$\theta_v = \phi + \tan^{-1}(\frac{v_\phi}{v_\rho})$$

$$\frac{\partial}{\partial v_r} = \frac{\partial}{\partial v_\rho}\frac{\partial v_\rho}{\partial v_r} + \frac{\partial}{\partial v_\phi}\frac{\partial v_\phi}{\partial v_r} = \frac{v_r}{v_\rho}\frac{\partial}{\partial v_\rho} + \frac{v_r}{v_\phi}\frac{\partial}{\partial v_\phi} = \frac{v}{v_\rho}\frac{\partial}{\partial v_\rho} + \frac{v}{v_\phi}\frac{\partial}{\partial v_\phi}$$

$$\frac{\partial}{\partial \theta_v} = \frac{\partial}{\partial v_\rho}\frac{\partial v_\rho}{\partial \theta_v} + \frac{\partial}{\partial v_\phi}\frac{\partial v_\phi}{\partial \theta_v} = -\frac{v^2}{v_\phi}\frac{\partial}{\partial v_\rho} + \frac{v^2}{v_\rho}\frac{\partial}{\partial v_\phi}$$

$$\vec{v} = v*\hat{v_r} = v (cos(\theta_v - \phi) \hat{\rho} + sin(\theta_v - \phi) \hat{\phi}) \rightarrow \hat{v_r} = cos(\theta_v - \phi) \hat{\rho} + sin(\theta_v - \phi) \hat{\phi} = \frac{v_\rho}{v}\hat{\rho} + \frac{v_\phi}{v} \hat{\phi}$$

$$\hat{v_\theta}$$ is 90 degrees rotated from $$\hat{v_r}$$ thus:

$$\hat{v_\theta} = -sin(\theta_v - \phi) \hat{\rho} + cos(\theta_v - \phi) \hat{\phi} = -\frac{v_\phi}{v} \hat{\rho} + \frac{v_\rho}{v} \hat{\phi}$$

So:

$$\frac{\partial}{\partial v_r} \hat{v_r} = (\frac{v_\rho}{v}\hat{\rho} + \frac{v_\phi}{v} \hat{\phi}) * (\frac{v}{v_\rho}\frac{\partial}{\partial v_\rho} + \frac{v}{v_\phi}\frac{\partial}{\partial v_\phi}) = (\frac{\partial}{\partial v_\rho} + \frac{v_\rho}{v_\phi}\frac{\partial}{\partial v_\phi})\hat{\rho} + (\frac{v_\phi}{v_\rho}\frac{\partial}{\partial v_\rho} + \frac{\partial}{\partial v_\phi})\hat{\phi}$$

$$\frac{1}{v_r}\frac{\partial}{\partial \theta_v} \hat{v_\theta} = \frac{1}{v}(-\frac{v_\phi}{v} \hat{\rho} + \frac{v_\rho}{v} \hat{\phi}) * (-\frac{v^2}{v_\phi}\frac{\partial}{\partial v_\rho} + \frac{v^2}{v_\rho}\frac{\partial}{\partial v_\phi}) = (\frac{\partial}{\partial v_\rho} - \frac{v_\phi}{v_\rho}\frac{\partial}{\partial v_\phi})\hat{\rho} + (-\frac{v_\rho}{v_\phi}\frac{\partial}{\partial v_\rho} + \frac{\partial}{\partial v_\phi})\hat{\phi}$$

Putting everything together:
$$\frac{\partial}{\partial v_r} \hat{v_r} + \frac{1}{v_r}\frac{\partial}{\partial \theta_v} \hat{v_\theta} + \frac{\partial}{\partial v_z} \hat{z} = (2\frac{\partial}{\partial v_\rho} + (\frac{v_\rho}{v_\phi} - \frac{v_\phi}{v_\rho})\frac{\partial}{\partial v_\phi})\hat{\rho} + (2\frac{\partial}{\partial v_\phi} + (\frac{v_\phi}{v_\rho} - \frac{v_\rho}{v_\phi})\frac{\partial}{\partial v_\rho})\hat{\phi} + \frac{\partial}{\partial v_z} \hat{z}$$

However this does not look right.
Shouldn't $$\nabla_v = \frac{\partial}{\partial v_\rho} \hat{\rho} + \frac{\partial}{\partial v_\phi} \hat{\phi} + \frac{\partial}{\partial v_z} \hat{z}$$ ?

Have I made a mistake somewhere?

 

[Charles Link]

I don't think you can do what you are trying to do because $f=f(\vec{r},\vec{v}, t)$. The distribution function $f$ is basically in a 6 dimensional space of $\vec{r}$ and $\vec{v}$. $\\$ At each location $\vec{r}$ there is a distribution of velocities. The distribution function $f$ supplies information about how the particle velocities are distributed at this point. The velocity vectors are in a completely different space from the coordinate vectors.

 

[Inquisitive Student]

Ok, but then how does one evaluate $$\vec{a} \cdot \nabla_v$$ in the vlasov equation? The acceleration vector is in spatial coordinates I believe.

 

[Charles Link]

Ok, but then how does one evaluate $$\vec{a} \cdot \nabla_v$$ in the vlasov equation? The acceleration vector is in spatial coordinates I believe.

Acceleration $\vec{a}=\vec{F}/m$ , and the vector $\vec{F }$ is usually replaced by the electromagnetic force $q(\vec{E}+\vec{v} \times \vec{B})$.