Does Stokes' Theorem imply this? I'm learning Stokes' Theorem and this seems to be the case from what I can tell but the result seems unintuitive to me. If this is true is there an intuitive explanation for it or have I made a mistake?
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2$\begingroup$ If you are referring to the Kelvin-Stokes theorem, the vector field cannot be arbitrary: it must be the curl of another vector field, and that creates a dependency between the components. $\endgroup$– user65203Feb 13, 2019 at 0:06
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$\begingroup$ @YvesDaoust Thank you, this is what I intended to ask but accidentally missed that out. I have updated the question. $\endgroup$– FloseFeb 13, 2019 at 0:13
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2 Answers
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This is the model that made it click for me.
The curl of the vector field produces a divergence-free field.
Imagine something like a bubble wand, with a partially inflated bubble. This bubble is hit with electromagnetic radiation. Measure the flux across the surface of the bubble.
Some of the radiation passes through one wall of the bubble and out through the hole in the bubble wand. Some of it passes trough one wall of the bubble and out the opposite wall.
If the field is divergence-free, that radiation that passes though the bubble both on the way in and the way out has equal magnitude, and the cancels out. All that matters it what passes through the aperture in the wand.
The shape of the bubble become irrelevant. In fact, there doesn't even need to be a bubble at all. Just the skin of bubble film over the aperture will give the same flux.
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$\begingroup$ The OP is not talking about integrating divergence (flux) but about curl. $\endgroup$ Feb 13, 2019 at 0:27
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A counter example would be the half planes x>=0 and x<=0. They both have the same boundary but may have different integrals.
