How to express velocity gradient in cylindrical coordinates?

In summary: The electric and magnetic fields are functions of the spatial coordinates. The body force is the gravitational force which also is a function of the spatial coordinates. The body force is usually small compared to the electromagnetic force. The body force is sometimes ignored. The acceleration is a function of the spatial coordinates.
  • #1
Inquisitive Student
10
0

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]

537fc1493f5ed9.jpg

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?
 

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  • #2
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.
 
  • #3
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.
 
  • #4
Inquisitive Student said:
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}) ##.
 
Last edited:

Related to How to express velocity gradient in cylindrical coordinates?

1. What is velocity gradient in cylindrical coordinates?

The velocity gradient in cylindrical coordinates is a measure of the change in velocity with respect to a change in position in a cylindrical coordinate system. It describes the rate at which the velocity of a fluid changes in different directions at a specific point in space.

2. How is velocity gradient expressed in cylindrical coordinates?

In cylindrical coordinates, the velocity gradient is typically expressed using the components of the gradient matrix, which represents the change in velocity in the radial, circumferential, and axial directions. It can also be expressed using the partial derivatives of the velocity components with respect to the cylindrical coordinates.

3. What is the formula for calculating velocity gradient in cylindrical coordinates?

The formula for calculating velocity gradient in cylindrical coordinates is:

∇V = (∂V/∂r)î + (1/r)(∂V/∂θ)ĵ + (∂V/∂z)k̂
where ∇V is the velocity gradient vector, ∂V/∂r is the radial component of the gradient, (∂V/∂θ)/r is the circumferential component, and ∂V/∂z is the axial component.

4. How does velocity gradient affect fluid flow in cylindrical coordinates?

The velocity gradient in cylindrical coordinates plays a crucial role in determining the behavior of fluid flow. It is responsible for creating shear forces that can cause the fluid to deform, mix, or separate. These effects can have significant impacts on the performance of various engineering systems, such as pumps, turbines, and pipelines.

5. What are some real-world applications of velocity gradient in cylindrical coordinates?

The concept of velocity gradient in cylindrical coordinates is used in various fields, including fluid mechanics, aerodynamics, and chemical engineering. It is particularly relevant in the design and analysis of rotating machinery, such as wind turbines and jet engines, where the flow of fluids is highly influenced by the curvature of the system.

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