How Does Velocity Affect Potential in Electromagnetic Fields?

[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}

0=m\ddot{z}[/itex]<br /> <br /> 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 ... 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}

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}

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

<br /> <br /> The first one should be:<br /> <br /> 0 = m\ddot{\rho} - m\rho\dot{\theta}^2 - eE_0\frac{1}{\rho} - e\dot{\theta}B_0\rho<br /> <br /> Other than that everything looks good <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f642.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":smile:" title="Smile :smile:" data-smilie="1"data-shortname=":smile:" />...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.<br /> <br /> 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