Consistency of Maxwell's laws with the Lorentz Transformation

[mjordan2nd]

Homework Statement

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.

Homework Equations

In this case, I know that

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

The field transformations were given by the book.

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

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

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.

 

[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 $\vec{E'} = 5\hat{x'} + 6\hat{x'}+7\hat{z'}$.

 

[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?

 

[TSny]

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

 

[mjordan2nd]

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

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

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.

 

[TSny]

Did you leave out a term on the right?

 

[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.

 

[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.

 

[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.

 

[TSny]

Good work!