Clarifying Electrodynamics: Relativity and the Faraday Tensor

[Arcon]

I would like your opinion regarding an explanation I gave elsewhere. I hold that the explanation below is straight forward. However it appears as if some were confused by it.

In a certain frame of referance, for a particular electromagnetic field, the relation $$\partial A/ \partial t = 0$$ holds true. Such a condition will hold in the case of a time independant magnetic field. The equation

$$E = - \nabla \Phi - \dfrac{\partial A}{\partial t}$$

in this example and in this frame reduces to

$$E = - \nabla \Phi$$

Does anyone think that this is relativistically incorrect?

I know this seems like a dumb question but some people claim that this is relativistically incorrect. Such a claim is obviously wrong. However I can't understand why they're having such a difficult time understanding this. Is it what I explained above confusing?

The 4-potential, $$A^{\alpha}$$, is defined in terms of the Coulomb potential, $$\Phi$$, and the magnetic vector potential, A as

$$A^{\alpha} = (\Phi/c, A) = (\Phi/c, A_x, A_y, A_z)$$

The Faraday tensor, $$F^{\alpha \beta}$$, is defined as

$$F^{\alpha \beta} = \partial^{\alpha} A^{\beta} - \partial^{\beta} A^{\alpha}$$

[See "Classical Electrodynamics - 2nd Ed.," J. D. Jackson, page 551, Eq. (11.136). I'm using different units]

The $$F^{0k}$$ components of this relationship for k = 1,2,3 are, respectively

$$\displaystyle{\frac{E_{x}}{c}} = \partial^{0} A^{1} - \partial^{1} A^{0} = - \displaystyle{\frac{1}{c}} \displaystyle{\frac{\partial A_{x}}{\partial t}} - \displaystyle{\frac{1}{c}} \displaystyle{\frac{\partial \Phi}{\partial x}}$$

$$\displaystyle{\frac{E_{y}}{c}} = \partial^{0} A^{2} - \partial^{2} A^{0} = - \displaystyle{\frac{1}{c}} \displaystyle{\frac{\partial A_{y}}{\partial t}} - \displaystyle{\frac{1}{c}} \displaystyle{\frac{\partial \Phi}{\partial y}}$$

$$\displaystyle{\frac{E_{z}}{c}} = \partial^{0} A^{3} - \partial^{3} A^{0} = - \displaystyle{\frac{1}{c}} \displaystyle{\frac{\partial A_{z}}{\partial t}} - \displaystyle{\frac{1}{c}} \displaystyle{\frac{\partial \Phi}{\partial z}}$$

These can be expressed as the single equation

$$E = -\nabla \Phi - \displaystyle{\frac{\partial A}{\partial t}}$$

This equation and the equation B = curl A are equation (11.134) in Jackson on page 551. In fact Jackson uses these two equations to define $$F^{\alpha \beta} = \partial^{\alpha} A^{\beta} - \partial^{\beta} A^{\alpha}$$

In the example stated above $$\partial A/\partial t = 0$$ so that

$$E = -\nabla \Phi$$

Does anyone find that confusing?

 

[mfb]

That's trivially correct.