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## 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

[tex]

\frac{\partial}{\partial x'} = \gamma \frac{\partial}{\partial x} + \frac{\gamma v}{c^2} \frac{\partial}{\partial t}

[/tex]

[tex]

\frac{\partial}{\partial t'} = \gamma \frac{\partial}{\partial t} + \gamma v \frac{\partial}{\partial x}

[/tex]

[tex]

\frac{\partial}{\partial y'} = \frac{\partial}{\partial y}

[/tex]

[tex]

\frac{\partial}{\partial z'} = \frac{\partial}{\partial z}

[/tex]

[tex]

E_1 '= E_1

[/tex]

[tex]

E_2'=\gamma (E_2 - \beta B_3)

[/tex]

[tex]

E_3'=\gamma (E_3 + \beta B_2)

[/tex]

[tex]

B_1' = B_1

[/tex]

[tex]

B_2'=\gamma (B_2 + \beta E_3)

[/tex]

[tex]

B_3'=\gamma (B_3 - \beta E_2)

[/tex]

The field transformations were given by the book.

## The Attempt at a Solution

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

[tex]

(\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'}.

[/tex]

Is this equal to

[tex]

\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}

[/tex]

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

Thanks.

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