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Diffstat (limited to 'mindmap/inverse square.org')
| -rw-r--r-- | mindmap/inverse square.org | 6 |
1 files changed, 3 insertions, 3 deletions
diff --git a/mindmap/inverse square.org b/mindmap/inverse square.org index 52a3f61..132b322 100644 --- a/mindmap/inverse square.org +++ b/mindmap/inverse square.org @@ -63,7 +63,7 @@ And from 6 properties that seem like reasonable constraints, we derive the inver To formalize the sixth property by taking the flux through a sphere and showing it doesn't depend on \(\vec{r}\): \begin{align*} -kP_{1}P_{2}\oint\frac{\hat{r}}{r^{2}} \cdot d\vec{a} = kP_{1}P_{2}\oint\frac{1}{r^{2}}da = k\frac{P_{1}P_{2} 4\pi r^{2}}{r^{2}} = constant +kP_{1}P_{2}\oint\frac{\hat{r}}{r^{2}} \cdot d\vec{a} = kP_{1}P_{2}\frac{1}{r^{2}}\int da = k\frac{P_{1}P_{2} 4\pi r^{2}}{r^{2}} = constant \end{align*} Which is a result we will use later. @@ -127,7 +127,7 @@ k\int_{space}\sigma(\vec{r'})(3(x^{2} + y^{2} + z^{2})^{-\frac{3}{2}} - 3(x^{2} So is the divergence of this field zero? Well, not exactly. In order to understand why, we must ask: What happens to the divergence at \(\vec{0}\)? On first glance, it seems clearly undefined. After all, we're dividing by zero. However, after some amount of inspection, the assumption that our field's divergence is zero everywhere is incoherent. As we know from section one, he surface integral over a sphere of a point charge (flux) is a constant, -but if the divergence theorem is true, then the surface integral should also be zero, not constant. Because this flux is not \(\vec{r}\) dependent and therefore +but if the [[id:44e65b69-e5d5-464a-b1f3-8a914e1b7e9e][divergence theorem]] is true, then the surface integral should also be zero, not constant. Because this flux is not \(\vec{r}\) dependent and therefore is the same no matter how small the concentric ring is, the flux must come from the origin. We can model this behavior with the [[id:90574fea-88f4-4b80-9cda-32cff0bcb76d][dirac delta]] distribution. Our actual divergence is: @@ -147,7 +147,7 @@ includes the term \(\frac{1}{4\pi}\), we can just say: \vec{\nabla} \cdot \vec{f}(\vec{r''}) = k\sigma(\vec{r''}) \end{align*} -Where \(k\) in this case is a constant without the \(4\pi\) term. By the divergence theorem: +Where \(k\) in this case is a constant without the \(4\pi\) term. By the [[id:44e65b69-e5d5-464a-b1f3-8a914e1b7e9e][divergence theorem]]: \begin{align*} \int_{V}\vec{\nabla} \cdot \vec{f}d\tau = \int_{S}\vec{f} \cdot d\vec{a} = k\int_{V}\sigma(\vec{r''})d\tau = k\sigma_{enc.} \end{align*} |