- #1
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
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
