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

1. Jan 11, 2004

### Arcon

E = -grad Phi - &A /&t

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

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

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 $$\displaystyle{\frac{\partial A}{\partial t}} = 0$$ so that

$$E = -\nabla \Phi$$

Does anyone find that confusing?

Note - Do to the nature of such a question please feel free to respond in PM.

Thanks

2. Jan 11, 2004

### Nereid

Staff Emeritus
Do they say why? If so, does their explanation make sense?

3. Jan 11, 2004

### DW

He didn't say what he here claims he said. He originally just felft off the vector potential term in a general statement for which bilge told him that his expression was not relativistically correct and waite told him just that he was neglecting the term. He then went and made statements concerning a choice of frame as if that was the intent all along.

4. Jan 12, 2004

### Arcon

They claim that I was saying that

$$\partial A/\partial t} = 0$$

is the relationship between the electric field and the Coulomb potentil and they ignore the fact that I was saying that it was an example in a given frame for a particular field.

Here is how it was described to them. I was explaining potential energy to someone and I wrote
They still didn't understand completely so I explained
The person to whom I was addressing understood at that point.

However for some reason someone else thought that there must have been a magnetic field in my example since they complained that what I described was "meaningless." In response to that claim it was apparent that they were not paying attention to the fact that this was the expression for a particular field in a particular frame of referance. So I explained further
Since they still didn't understand after this I won't quote more.

But I fail to see how anyone could be that confused. Since only two people responded and since both of them were confused I'm curious as to how they could be so confused. Since I assume that it's them and not the explanation I wanted an opinion from someone who is unbiased.

Do you find the three parts I've quoted above confusing?

Thanks

5. Jan 12, 2004

### DW

"THEY" did not say any such thing. You need to be specific when quoting someone. "They" were also not confused. Both of them are extremely well versed in general relativistic electrodynamics. You were just wrong for the reasons they explained which I mentioned.

6. Jan 19, 2004

### Phobos

Staff Emeritus
Arcon, DW - Enough with Waite & pmb. Discuss the issues at hand. Lose the baggage. Your discussions are welcome. The harassing is not. My patience is coming to an end.

7. Jan 19, 2004

Okay by me