- #1
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In proving [itex]\nabla \times \vec{E} = \vec{0}[/itex] for electrostatic fields, Griffiths switches directly from the equation
[tex]\iint_{\mathcal{S}}(\nabla \times \vec{E})\cdot d\vec{a} =0[/tex]
To the conclusion [itex]\nabla \times \vec{E} = \vec{0}[/itex]. As for Gauss's theorem, I am wondering if there is a more precise mathematical proof of this. More precisely, the integral [itex]\iint_{\mathcal{S}}(\nabla \times \vec{E})\cdot d\vec{a}[/itex] can be 0 for three reasons
1) The vector [itex]\nabla \times \vec{E}[/itex] is perpendicular to [itex]d\vec{a}[/itex] everywhere.
2) The integral, as a sum, is worth 0 (i.e. some parts are positive, some negative, some nul such that the total is 0.)
3) [itex]\nabla \times \vec{E} = \vec{0}[/itex]
Is there a way to exclude the two first possibilities without referring to arguments such as "but this is evidently impossible for an electric field", but only by treating [itex]\vec{E}[/itex] as just another vector field? Many times I tought I had found the answer but later realized, I had not afterall.
My best attemps, I believe, gets rid of 1) by setting the surface integral as a sphere and supposing it is 0 because the field is everywhere perpendicular to it's surface. Evidently this is not possible, but I can't prove it. (Don't know enough vector field calculus to know where to start) Can this be done?
[tex]\iint_{\mathcal{S}}(\nabla \times \vec{E})\cdot d\vec{a} =0[/tex]
To the conclusion [itex]\nabla \times \vec{E} = \vec{0}[/itex]. As for Gauss's theorem, I am wondering if there is a more precise mathematical proof of this. More precisely, the integral [itex]\iint_{\mathcal{S}}(\nabla \times \vec{E})\cdot d\vec{a}[/itex] can be 0 for three reasons
1) The vector [itex]\nabla \times \vec{E}[/itex] is perpendicular to [itex]d\vec{a}[/itex] everywhere.
2) The integral, as a sum, is worth 0 (i.e. some parts are positive, some negative, some nul such that the total is 0.)
3) [itex]\nabla \times \vec{E} = \vec{0}[/itex]
Is there a way to exclude the two first possibilities without referring to arguments such as "but this is evidently impossible for an electric field", but only by treating [itex]\vec{E}[/itex] as just another vector field? Many times I tought I had found the answer but later realized, I had not afterall.
My best attemps, I believe, gets rid of 1) by setting the surface integral as a sphere and supposing it is 0 because the field is everywhere perpendicular to it's surface. Evidently this is not possible, but I can't prove it. (Don't know enough vector field calculus to know where to start) Can this be done?