- #1
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 [tex]\partial A/ \partial t = 0[/tex] holds true. Such a condition will hold in the case of a time independant magnetic field. The equation
[tex]E = - \nabla \Phi - \dfrac{\partial A}{\partial t} [/tex]
in this example and in this frame reduces to
[tex]E = - \nabla \Phi[/tex]
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, [tex]A^{\alpha}[/tex], is defined in terms of the Coulomb potential, [tex]\Phi[/tex], and the magnetic vector potential, A as
[tex]A^{\alpha} = (\Phi/c, A) = (\Phi/c, A_x, A_y, A_z)[/tex]
The Faraday tensor, [tex]F^{\alpha \beta}[/tex], is defined as
[tex]F^{\alpha \beta} = \partial^{\alpha} A^{\beta} - \partial^{\beta} A^{\alpha} [/tex]
[See "Classical Electrodynamics - 2nd Ed.," J. D. Jackson, page 551, Eq. (11.136). I'm using different units]
The [tex]F^{0k}[/tex] components of this relationship for k = 1,2,3 are, respectively
[tex]\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}}[/tex]
[tex]\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}}[/tex]
[tex]\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}}[/tex]
These can be expressed as the single equation
[tex]E = -\nabla \Phi - \displaystyle{\frac{\partial A}{\partial t}}[/tex]
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 [tex]F^{\alpha \beta} = \partial^{\alpha} A^{\beta} - \partial^{\beta} A^{\alpha} [/tex]
In the example stated above [tex]\partial A/\partial t = 0[/tex] so that
[tex]E = -\nabla \Phi [/tex]
Does anyone find that confusing?
In a certain frame of referance, for a particular electromagnetic field, the relation [tex]\partial A/ \partial t = 0[/tex] holds true. Such a condition will hold in the case of a time independant magnetic field. The equation
[tex]E = - \nabla \Phi - \dfrac{\partial A}{\partial t} [/tex]
in this example and in this frame reduces to
[tex]E = - \nabla \Phi[/tex]
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, [tex]A^{\alpha}[/tex], is defined in terms of the Coulomb potential, [tex]\Phi[/tex], and the magnetic vector potential, A as
[tex]A^{\alpha} = (\Phi/c, A) = (\Phi/c, A_x, A_y, A_z)[/tex]
The Faraday tensor, [tex]F^{\alpha \beta}[/tex], is defined as
[tex]F^{\alpha \beta} = \partial^{\alpha} A^{\beta} - \partial^{\beta} A^{\alpha} [/tex]
[See "Classical Electrodynamics - 2nd Ed.," J. D. Jackson, page 551, Eq. (11.136). I'm using different units]
The [tex]F^{0k}[/tex] components of this relationship for k = 1,2,3 are, respectively
[tex]\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}}[/tex]
[tex]\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}}[/tex]
[tex]\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}}[/tex]
These can be expressed as the single equation
[tex]E = -\nabla \Phi - \displaystyle{\frac{\partial A}{\partial t}}[/tex]
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 [tex]F^{\alpha \beta} = \partial^{\alpha} A^{\beta} - \partial^{\beta} A^{\alpha} [/tex]
In the example stated above [tex]\partial A/\partial t = 0[/tex] so that
[tex]E = -\nabla \Phi [/tex]
Does anyone find that confusing?
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