# Consistency of Maxwell's laws with the Lorentz Transformation

1. Mar 12, 2013

### mjordan2nd

1. The problem statement, all variables and given/known data

I am trying to prove that Maxwell's laws are consistent with special relativity if one frame is moving in the x direction with another.

2. Relevant equations

In this case, I know that

$$\frac{\partial}{\partial x'} = \gamma \frac{\partial}{\partial x} + \frac{\gamma v}{c^2} \frac{\partial}{\partial t}$$
$$\frac{\partial}{\partial t'} = \gamma \frac{\partial}{\partial t} + \gamma v \frac{\partial}{\partial x}$$
$$\frac{\partial}{\partial y'} = \frac{\partial}{\partial y}$$
$$\frac{\partial}{\partial z'} = \frac{\partial}{\partial z}$$
$$E_1 '= E_1$$
$$E_2'=\gamma (E_2 - \beta B_3)$$
$$E_3'=\gamma (E_3 + \beta B_2)$$
$$B_1' = B_1$$
$$B_2'=\gamma (B_2 + \beta E_3)$$
$$B_3'=\gamma (B_3 - \beta E_2)$$

The field transformations were given by the book.

3. The attempt at a solution

Faraday's law relates two vector quantities. If I have $\nabla' \times E'$ I can plug in the appropriate derivatives and field transformations, but do I need to transform the basis vectors as well? For instance

$$(\nabla' \times E')_x \hat{x'} = \left( \frac{\partial E_3'}{\partial y'} - \frac{\partial E_2'}{\partial z'} \right) \hat{x'} = \left( \frac{\partial}{\partial y} \left[ \gamma (E_3 + \beta B_2) \right] - \frac{\partial}{\partial z} \left[ \gamma (E_2-\beta B_3) \right] \right) \hat{x'}.$$

Is this equal to

$$\left( \frac{\partial}{\partial y} \left[ \gamma (E_3 + \beta B_2) \right] - \frac{\partial}{\partial z} \left[ \gamma (E_2-\beta B_3) \right] \right) \hat{x}$$

or does $\hat{x} \neq \hat{x'}$, and so on for y and z? I'm not sure I really understand how the basis vectors transform.

Thanks.

Last edited: Mar 12, 2013
2. Mar 12, 2013

### TSny

You don't transform the unit vectors, just the components of the vectors. For example, if the components of the electric field in the unprimed frame are $E_x = 2, E_y = 3, E_z = 4$ and if the components of the field in the primed frame are $E'_x = 5, E'_y = 6, E'_z = 7$, then $\vec{E} = 2\hat{x} + 3\hat{y}+4\hat{z}$ and $\vec{E'} = 5\hat{x'} + 6\hat{x'}+7\hat{z'}$.

3. Mar 13, 2013

### mjordan2nd

I'm a bit stuck with Faraday's law.

Based on what I established above and the definition of the curl I have written

$$\nabla' \times E' = \left( \frac{\partial}{\partial y} \left[ \gamma (E_3 + \beta B_2) \right] - \frac{\partial}{\partial z} \left[ \gamma (E_2-\beta B_3) \right] \right) \hat{x} - \left[ \left( \gamma \frac{\partial}{\partial x} + \frac{\gamma v}{c^2} \frac{\partial}{\partial t} \right) \gamma (E_3 + \beta B_2) - \frac{\partial E_1}{\partial z} \right] \hat{y} + \left[ \left( \gamma \frac{\partial}{\partial x} + \frac{\gamma v}{c^2} \frac{\partial}{\partial t} \right) \gamma (E_2 - \beta B_3) - \frac{\partial E_1}{\partial y} \right] \hat{z}$$

$$= -\left( \gamma \frac{\partial}{\partial t} + \gamma v \frac{\partial}{\partial x} \right) \left( B_1 \hat{x} + \gamma (B_2 + \beta E_3) \hat{y} + \gamma (B_3 -\beta E_2) \hat{z} \right) = -\frac{\partial B'}{\partial t'}$$

I'm not sure where to go from here. Nothing seems obvious to reduce this in the unprimed frame into the same, or similar format as in the primed frame. Any suggestions?

4. Mar 13, 2013

### TSny

Work with just one component, say the $\hat{x}$ component, on each side. Expand the derivative expressions and see if you can simplify.

5. Mar 13, 2013

### mjordan2nd

If I work with just the x component I get something like

$$\gamma \left[ \left( \nabla \times E \right)_x + \beta \left( \frac{\partial B_2}{\partial y} + \frac{ \partial B_3}{\partial z} \right) \right ] = - \gamma \frac{\partial B_1}{\partial t}.$$

That said, I don't know what to do with the second term on the LHS of the equation. Furthermore, since I've boosted it in the x direction, this creates an assymetry in the way my terms look in the different dimensions. In the y and z dimensions I can't figure out how to get anything that looks like the curl of E. The Lorentz factor doesn't factor out in those dimensions quite as nicely.

6. Mar 13, 2013

### TSny

Did you leave out a term on the right?

7. Mar 13, 2013

### mjordan2nd

Genius! Now I should get a divergence of B which goes away. Okay, so now I'm at a point where the x-component of the transformation does look like what I expect. Have I made a mistake in my algebra with the gamma terms for the y and z components, though? Those are driving me nuts.

Edit: Hmm... actually I'm close for the x-component. The beta and v term don't really cancel -- I must have made a mistake somewhere. That said, what's been driving me nuts on and off for the last 36ish hours is the damned y and z components.

8. Mar 13, 2013

### TSny

Don't worry, peace is close at hand. If you are not using units where c = 1, then Faraday's law will have a factor of c in the denominator of the right hand side: $\frac{\partial }{c\partial t}$

You'll see how wonderfully those pesky factors of $\gamma$ and $\beta$ come together.

Last edited: Mar 13, 2013
9. Mar 13, 2013

### mjordan2nd

Ha, I did it!! It was all an issue of units. Setting c=1 and realizing

$$v^2=1-\gamma^{-2}$$

made it all pop out. Thanks a lot for your help, it is very much appreciated. Now on to the Maxwell-Ampere law! With these realizations I think that should be a lot easier.

10. Mar 13, 2013

Good work!