- #1

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- Homework Statement
- Transformation rules fot the fields in a lorentz boost in the z-direction.

- Relevant Equations
- Look at attempt at a solution.

Question:

Eq. 12.109:

My solution:

We’ll first use the configuration from figure 12.35 in the book Griffiths. Where the only difference is

that v_0 is in the z-direction. The electric field in the y-direction will be the same.

$$E_y = \frac{\sigma}{\epsilon _0}$$

Now we're going to derive the general rule, following just like how it's done in the book Griffiths. The surface current will look like:

$$\mathbf{K_{\pm}}=\mp \sigma v_0 \hat{\mathbf{z}}$$

By the right-hand rule the field would point in the positive x direction, with the magnitude as:

$$B_x = \mu_0 \sigma v_0.$$

The third system $\overline{S}$, would ofcourse travel in the z direction instead of the x direction.

The fields will look like:

$$ \overline{E_y} = \frac{\overline{\sigma}}{\epsilon _0}$$

$$\overline{B_x} = \mu_0 \overline{\sigma} \overline{v_0}$$

Now we just need to express $\overline{\mathbf{E}}$ and $\overline{\mathbf{B}}$ in terms of $\mathbf{E}$ and $\mathbf{B}$.

We will use the same equations 12.98, 12.99, 12.101 and 12.102 to do the algebra because they didn't change.

$$\overline{E_y} = \gamma \left( E_y + vB_x \right)$$

$$\overline{B_x} = \gamma \left( B_x + \frac{v}{c^2}E_y \right)$$

To find $E_x$ and $B_y$, the configuration will be identical to figure 12.36.

The fields in $S$ will be as follows:

$$E_x = \frac{\sigma}{\epsilon_0}$$

$$B_y = \sigma \mu_0 v_0$$

The fields in $\overline{S}$ will look like:

$$\overline{E_x} = \gamma \left( E_x + vB_y \right)$$

$$\overline{B_y} = \gamma \left( B_y + \frac{v}{c^2}E_x \right)$$

For the z component of the electric fields will be same and for the magnetic will also be the same, if we take a solenoid parallel to the z-axis.

But the solenoid being on the z-axis.

$$\overline{B_z} = B_z$$

$$\overline{E_z} = E_z$$

I'm not exactly sure if I'm right, because it all depends if your configuration is correct.

Eq. 12.109:

My solution:

We’ll first use the configuration from figure 12.35 in the book Griffiths. Where the only difference is

that v_0 is in the z-direction. The electric field in the y-direction will be the same.

$$E_y = \frac{\sigma}{\epsilon _0}$$

Now we're going to derive the general rule, following just like how it's done in the book Griffiths. The surface current will look like:

$$\mathbf{K_{\pm}}=\mp \sigma v_0 \hat{\mathbf{z}}$$

By the right-hand rule the field would point in the positive x direction, with the magnitude as:

$$B_x = \mu_0 \sigma v_0.$$

The third system $\overline{S}$, would ofcourse travel in the z direction instead of the x direction.

The fields will look like:

$$ \overline{E_y} = \frac{\overline{\sigma}}{\epsilon _0}$$

$$\overline{B_x} = \mu_0 \overline{\sigma} \overline{v_0}$$

Now we just need to express $\overline{\mathbf{E}}$ and $\overline{\mathbf{B}}$ in terms of $\mathbf{E}$ and $\mathbf{B}$.

We will use the same equations 12.98, 12.99, 12.101 and 12.102 to do the algebra because they didn't change.

$$\overline{E_y} = \gamma \left( E_y + vB_x \right)$$

$$\overline{B_x} = \gamma \left( B_x + \frac{v}{c^2}E_y \right)$$

To find $E_x$ and $B_y$, the configuration will be identical to figure 12.36.

The fields in $S$ will be as follows:

$$E_x = \frac{\sigma}{\epsilon_0}$$

$$B_y = \sigma \mu_0 v_0$$

The fields in $\overline{S}$ will look like:

$$\overline{E_x} = \gamma \left( E_x + vB_y \right)$$

$$\overline{B_y} = \gamma \left( B_y + \frac{v}{c^2}E_x \right)$$

For the z component of the electric fields will be same and for the magnetic will also be the same, if we take a solenoid parallel to the z-axis.

$$\overline{B_z} = B_z$$

$$\overline{E_z} = E_z$$

I'm not exactly sure if I'm right, because it all depends if your configuration is correct.