Velocity Dependent Potential

[CNX]

Homework Statement

Consider a particle of mass m and charge q that moves in an E-field $\vec{E}=\frac{E_0}{r}\hat{r}$ and a uniform magnetic field $\vec{B}=B_0\hat{k}$. Find the scalar potential and show the vector potential is given by $\vec{A}=\frac{1}{2}B_0 r \hat{\theta}$. Then obtain the Lagrange equations of motion and identify the conserved quantities

Homework Equations

Lagrange equations

The Attempt at a Solution

Using cylindrical coords,

$$L=T-V=\frac{1}{2}m \left ( \dot{r}^2+\dot{\theta}^2+\dot{z}^2 \right) - e\phi + e\vec{v} \cdot \vec{A}$$

$$L = \frac{1}{2}m \left ( \dot{r}^2+\dot{\theta}^2+\dot{z}^2 \right) - eE_0\ln(r) + \frac{1}{2}e\dot{\theta}B_0r$$

Using the Lagrange equation,

$$0 = m\ddot{r} + eE_0\frac{1}{r} - \frac{1}{2}e\dot{\theta}B_0$$

$$0=m\ddot{\theta}$$

[tex]0=m\ddot{z}[/itex]

Correct?

 

[gabbagabbahey]

Homework Statement

Consider a particle of mass m and charge q that moves in an E-field $\vec{E}=\frac{E_0}{r}\hat{r}$ and a uniform magnetic field $\vec{B}=B_0\hat{k}$. Find the scalar potential and show the vector potential is given by $\vec{A}=\frac{1}{2}B_0 r \hat{\theta}$. Then obtain the Lagrange equations of motion and identify the conserved quantities

It took me a minute to realize that you are using $(r,\theta,z)$ as your cylindrical coordinates...ewww :yuck:... that makes things confusing since one usually uses $\vec{r}$ to represent the position of the particle, and hence it is more natural for $r$ to be used as the spherical polar coordinate corresponding to the distance from the origin so that $\vec{r}=r\hat{r}$. In this case however, $\vec{r}=r\hat{r}+z\hat{z}$ which is not very aesthetic...but, I digress...

The Attempt at a Solution

Using cylindrical coords,

$$L=T-V=\frac{1}{2}m \left ( \dot{r}^2+\dot{\theta}^2+\dot{z}^2 \right) - e\phi + e\vec{v} \cdot \vec{A}$$

It is always a good idea to check your units...does the angular term in $\vec{v}=\dot{r}\hat{r}+\dot{\theta}\hat{\theta}+\dot{z}\hat{z}$ have the correct units?

$$L = \ldots - eE_0\ln(r) + \ldots$$

Looks like you have a sign error here...remember, $\vec{E}=-\vec{\nabla}}\phi$

Using the Lagrange equation,

$$0=m\ddot{\theta}$$

Correct?

Be careful with this equation, remember that it is a full time derivative (not a partial derivative) in the Euler-Lagrange equation:
$$\frac{d}{dt}\frac{\partial L}{\partial \dot{\theta}}\neq m\ddot{\theta}$$

 

[CNX]

My attempt at corrections (Note the change in coordinate labels):

Using cylindrical coords $(\rho, \theta, z)$,

$$L=T-V=\frac{1}{2}m \left ( \dot{\rho}^2+\rho\dot{\theta}^2+\dot{z}^2 \right) - e\phi + e\vec{v} \cdot \vec{A}$$

$$L = \frac{1}{2}m \left ( \dot{\rho}^2+\rho\dot{\theta}^2+\dot{z}^2 \right) + eE_0\ln(\rho) + \frac{1}{2}e\dot{\theta}B_0 \rho^2$$

Using the Lagrange equation,

$$0 = m\ddot{\rho} - m\rho\ddot{\theta}^2 - eE_0\frac{1}{\rho} - e\dot{\theta}B_0\rho$$

$$0=m\rho^2 \ddot{\theta} + 2m\rho\dot{\rho}\dot{\theta} + e B_0\rho\dot{\rho}$$

[tex]0=m\ddot{z}[/itex]

 

[gabbagabbahey]

My attempt at corrections (Note the change in coordinate labels):

Using cylindrical coords $(\rho, \theta, z)$,

$$L=T-V=\frac{1}{2}m \left ( \dot{\rho}^2+\rho\dot{\theta}^2+\dot{z}^2 \right) - e\phi + e\vec{v} \cdot \vec{A}$$

$$L = \frac{1}{2}m \left ( \dot{\rho}^2+\rho\dot{\theta}^2+\dot{z}^2 \right) + eE_0\ln(\rho) + \frac{1}{2}e\dot{\theta}B_0 \rho^2$$

Surely you meant to put some brackets in there:

$$L=T-V=\frac{1}{2}m \left ( \dot{\rho}^2+(\rho\dot{\theta})^2+\dot{z}^2 \right) - e\phi + e\vec{v} \cdot \vec{A}$$

$$L = \frac{1}{2}m \left ( \dot{\rho}^2+(\rho\dot{\theta})^2+\dot{z}^2 \right) + eE_0\ln(\rho) + \frac{1}{2}e\dot{\theta}B_0 \rho^2$$

Using the Lagrange equation,

$$0 = m\ddot{\rho} - m\rho\ddot{\theta}^2 - eE_0\frac{1}{\rho} - e\dot{\theta}B_0\rho$$

$$0=m\rho^2 \ddot{\theta} + 2m\rho\dot{\rho}\dot{\theta} + e B_0\rho\dot{\rho}$$

[tex]0=m\ddot{z}[/itex]

The first one should be:

$$0 = m\ddot{\rho} - m\rho\dot{\theta}^2 - eE_0\frac{1}{\rho} - e\dot{\theta}B_0\rho$$

Other than that everything looks good ...You can double check your answers by comparing them to what you get from the Lorentz Force law; you should get the same equations of motion.

How about the conserved quantities...what are you getting for those?

 

[CNX]

Thanks. Wouldn't $\dot{z}$ be the conserved quantity?

 

[gabbagabbahey]

Sure, the axial component of the velocity ( $\dot{z}$ ) or the axial momentum ( $p_z=m\dot{z}$ ) is conserved, but is that the only conserved quantity?...How does one typically go about finding the conserved quantities in Lagrangian or Hamiltonian dynamics?

 

[CNX]

Thanks I got them