# Field transformations in the z-direction

• milkism
In summary, the conversation discusses the derivation of a general rule for electric and magnetic fields in different configurations. The summary includes equations and conclusions from different configurations and mentions a possible mistake in the first solution. The conversation also mentions the use of Purcell's book for further understanding and addresses the issue of "bumping" a thread on a forum.
milkism
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.

Trying reading Purcell's as a complement. It uses SR to introduce E&M.

Last edited:
ChiralSuperfields said:
Trying reading Purcell's as a complement. It uses SR to introduce E&M.
Can't find anything about field transformations in his book, can you point out what I did wrong?

Okay I have tried to do it with different configurations.
New solution:

At configuration 1 (top configuration), I concluded the following:
$$E_y = \frac{\sigma}{\epsilon _0}$$
$$\mathbf{K_{\pm}} = \pm \sigma v_0 \mathbf{\hat{z}}$$
$$B_x = -\mu _0 \sigma v_0$$
$$\overline{E_y} = \gamma \left( E_y - vB_x \right)$$
$$\overline{B_x} = \gamma \left( B_x - \frac{v}{c^2} E_y \right)$$
At configuration 2 (middle left), I concluded the following:
$$\overline{E_z}=E_z$$
At configuration 3 (bottom left), I concluded the following:
$$\overline{B_z} = B_z$$
At configuration 4 (bottom right), I concluded the following:
$$E_x = \frac{\sigma}{\epsilon_0}$$
$$B_y = \mu _0 \sigma v_0$$
$$\overline{E_x} = \gamma \left( E_x + vB_y \right)$$
$$\overline{B_y} = \gamma \left( B_y + \frac{v}{c^2} E_x \right)$$

Personal questions:
Since in griffiths they have boosted at a negative x directio, should I also have boosted in a negative z direction?
I think my second solution is more correct than my first one.

milkism said:
Can't find anything about field transformations in his book, can you point out what I did wrong?
Sorry I have only read a bit of purcell's so I am not up to that level yet.

milkism
The transformations in the second solution are correct, but indeed you should have moved the capacitor in the negative z-direction because this corresponds to you moving in the positive z-direction relative to the rest of space and thus a positive boost in the z-direction.

## 1. What is a field transformation in the z-direction?

A field transformation in the z-direction refers to the change in a physical quantity, such as electric or magnetic field, as it moves in the z-direction, or vertical direction, in a given space.

## 2. Why are field transformations in the z-direction important?

Field transformations in the z-direction are important because they allow us to understand how a physical quantity changes as it moves in a specific direction, which can have significant implications in various fields such as electromagnetism, fluid dynamics, and optics.

## 3. How are field transformations in the z-direction calculated?

The calculation of field transformations in the z-direction depends on the specific physical quantity and the equations governing its behavior. In general, it involves using mathematical tools such as vector calculus, differential equations, and boundary conditions.

## 4. What factors can affect field transformations in the z-direction?

Several factors can affect field transformations in the z-direction, including the properties of the medium through which the field is propagating, the shape and size of the objects in the field, and the presence of other fields or sources in the space.

## 5. How can field transformations in the z-direction be applied in real-world situations?

Field transformations in the z-direction have many practical applications, such as in designing antennas for wireless communication, predicting the behavior of fluids in pipes or channels, and understanding the behavior of light in optical devices. They are also crucial in the development of technologies such as MRI machines and particle accelerators.

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