From 80da24887ac760a9d18936634d8d46c0643521ee Mon Sep 17 00:00:00 2001 From: Preston Pan Date: Sun, 23 Jul 2023 09:12:03 -0700 Subject: add a lot of mindmap articles --- mindmap/Newtonian mechanics.org | 92 +++++++++++++++++++ mindmap/central force.org | 56 ++++++++++++ mindmap/conservative force.org | 37 ++++++++ mindmap/del operator.org | 103 ++++++++++++++++++++++ mindmap/derivative.org | 152 ++++++++++++++++++++++++++++++++ mindmap/dirac delta.org | 60 +++++++++++++ mindmap/electrostatics.org | 23 ++++- mindmap/everything.org | 1 + mindmap/function.org | 36 ++++++++ mindmap/group.org | 30 +++++++ mindmap/inheritance.org | 24 +++++ mindmap/inverse square.org | 190 ++++++++++++++++++++++++++++++++++++++++ mindmap/partial derivative.org | 14 +++ 13 files changed, 816 insertions(+), 2 deletions(-) create mode 100644 mindmap/Newtonian mechanics.org create mode 100644 mindmap/central force.org create mode 100644 mindmap/conservative force.org create mode 100644 mindmap/del operator.org create mode 100644 mindmap/derivative.org create mode 100644 mindmap/dirac delta.org create mode 100644 mindmap/function.org create mode 100644 mindmap/group.org create mode 100644 mindmap/inheritance.org create mode 100644 mindmap/inverse square.org create mode 100644 mindmap/partial derivative.org (limited to 'mindmap') diff --git a/mindmap/Newtonian mechanics.org b/mindmap/Newtonian mechanics.org new file mode 100644 index 0000000..c78b1e3 --- /dev/null +++ b/mindmap/Newtonian mechanics.org @@ -0,0 +1,92 @@ +:PROPERTIES: +:ID: 6e2a9d7b-7010-41da-bd41-f5b2dba576d3 +:END: +#+title: Newtonian mechanics +#+author: Preston Pan +#+html_head: +#+html_head: +#+html_head: +#+options: broken-links:t + +* Introduction +Newtonian mechanics is a mathematical model of the real world that is good +at approximating non-relatavistic macro phenomenon. In other words, it will +work for most engineers, and it has a few simple axioms: + +* Definition +First, we assume euclidean 3-dimensional space, and time as the parameterizing +variable, or in other words the variable that is constant for everyone and everything. +We assume objects can have a definite position, and have some sort of velocity +and mass. For these examples we will assume a system with only one particle in a vacuum +and then we will try to generalize. First we notice that this particle has a position +given by some vector \(\vec{r}\). Then, we say that if this position changes over time: +\begin{align*} +\vec{v} = \frac{d\vec{r}}{dt} +\end{align*} +where \(\vec{v}\) is velocity, the time-[[id:31d3944a-cddc-496c-89a3-67a56e821de3][derivative]] of position. We also define a mass \(m\) for this particle in question. Then, this particle will have a momentum: +\begin{align*} +\vec{p} = m\vec{v} +\end{align*} +And here we define /inertia/: +\begin{align*} +\vec{p}(t_{1}) = \vec{p}(t_{2}) +\end{align*} +Or the momentum of this particle will always stay the same throughout time, at least without other objects. Then we can define force: +\begin{align*} +\vec{F} := \frac{d\vec{p}}{dt} \\ += m\frac{d\vec{v}}{dt} \\ +\vec{a} := \frac{d\vec{v}}{dt} \\ +\vec{F} = m\vec{a} +\end{align*} +Now imagine we add another particle. This other particle will have its own position and momentum. These two particles together create +a new system. Because we want the laws of physics to work for all systems: +\begin{align*} +\vec{p}(t_{1}) = \vec{p}(t_{2}) +\end{align*} +Where \(\vec{p}\) becomes the momentum of the whole system. If \(\vec{p}_{1}\) is the momentum of the first particle and \(\vec{p}_{2}\) is the +momentum of the second particle, the momentum of the whole system is: +\begin{align*} +\vec{p} = \vec{p}_{1} + \vec{p}_{2} +\end{align*} +In general, the total momentum is defined to be: +\begin{align*} +\vec{p} = \sum_{i=0}^{n}\vec{p}_{i} +\end{align*} +And in real life, we observe that things can transfer momentum. That is: +\begin{align*} +\vec{p}_{1} = -\vec{p}_{2} +\end{align*} +When these two objects have the same position vector \( \vec{r}_{1} = \vec{r}_{2} \) (if they are point masses; don't have volume but have mass which is idealistic but works as an approximation). +Because this operation of momentum transfer is symmetrical: +\begin{align*} +\vec{p}_{2} = -\vec{p}_{1} +\end{align*} +Note that the fact that this operation is symmetrical must be the case to preserve the property: +\begin{align*} +\vec{p}(t_{1}) = \vec{p}(t_{2}) +\end{align*} +Of the entire system (otherwise you would be adding or subtracting momentum from the system). +In other words, the entire system must have inertia, and this statement itself is the conservation of momentum. Conservation of momentum along with transfer of momentum yields +a fully functional model of physics. + +* Textbook Formulation +The first law which we discussed is called /inertia/; the second law is the \( \vec{F} = m\vec{a} \) law as discussed; +we can get what is called Newton's Third Law as follows: +\begin{align*} +\frac{d\vec{p}_{2}}{dt} = -\frac{d\vec{p}_{1}}{dt} \\ +\vec{F}_{2} = -\vec{F}_{1} \\ +\vec{F}_{1} = -\vec{F}_{2} +\end{align*} +However, the third law follows from conservation of momentum (inertia) and transfer of momentum (not really the second or third law). I do not know why it exists, +and I think the formulation of Newtonian physics based off of less (and more descriptive) axioms is far more elegant, so I don't really know how this happened. +In any case, if you are using a classic textbook, you will be using this formulation of Newtonian mechanics. +* Newtonian Gravity +:PROPERTIES: +:ID: 158f53ba-5846-472b-ab39-336ed7f11251 +:END: +Gravity in Newtonian mechanics is defined via [[id:2a543b79-33a0-4bc8-bd1c-e4d693666aba][inverse square]] law. With the stipulation that mass can only be positive, +the gravitational force field has the same properties as all [[id:2a543b79-33a0-4bc8-bd1c-e4d693666aba][inverse square]] fields by [[id:4ed61028-811e-4425-b956-feca6ee92ba1][inheritance]]; therefore: +\begin{align*} +\vec{F}(\vec{r}) = \frac{Gm_{1}m_{2}}{r^{2}}\hat{r} +\end{align*} +Most of the inverse square results consequently carry over. diff --git a/mindmap/central force.org b/mindmap/central force.org new file mode 100644 index 0000000..99751fc --- /dev/null +++ b/mindmap/central force.org @@ -0,0 +1,56 @@ +:PROPERTIES: +:ID: c1e836e3-a80c-459d-8b68-396fa1687177 +:END: +#+title: central force +#+author: Preston Pan +#+html_head: +#+html_head: +#+html_head: +#+options: broken-links:t + +* Derivation +Central fields have the following properties: +1. The force happens at a distance between two particles. +2. The force gives both the particles an equal amount of force but in opposite directions (Newton's Third Law). +3. The force magnitude only depends on the distance between the two particles in question, and maybe some other property intrinsic to both the particles. +4. If you draw a straight line between the two points, the force vector has to be parallel to that line. +5. The force works in the same no matter where in the universe you are. + +Where these five properties are all common attributes of everyday forces. What these properties are basically saying is that we want +a function only dependent on the vector between two particles \(P_{1}\) and \(P_{2}\) that are experiencing the force, and it is +also parallel to this vector, as well as some symmetry constraints. Therefore, we can define a central force: + +* Definition +A central field is defined as follows: +\begin{align*} +\vec{f}(\vec{r}) = f(\vec{r})\hat{r}. +\end{align*} + +* Curl of Central Forces +Using the [[id:4bfd6585-1305-4cf2-afc0-c0ba7de71896][del operator]], we can find the curl of the central force. We will use the Cartesian del operator first +because it requires less math to understand, despite the calculation being longer. Therefore, we must switch coordinates: +\begin{align*} +f(\vec{r})\hat{r} = f(x, y, z)\frac{x\hat{i} + y\hat{j} + z\hat{k}}{(x^{2} + y^{2} + z^{2})^{\frac{1}{2}}} \\ +\vec{\nabla} \times \vec{f} = \begin{vmatrix} +\hat{i} & \hat{j} & \hat{k} \\ +\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ +f(x, y, z)\frac{x}{(x^{2} + y^{2} + z^{2})^{\frac{1}{2}}} & f(x, y, z)\frac{y}{(x^{2} + y^{2} + z^{2})^{\frac{1}{2}}} & f(x, y, z)\frac{z}{(x^{2} + y^{2} + z^{2})^{\frac{1}{2}}} +\end{vmatrix} +\end{align*} +If we can figure out one of these derivatives, then by symmetry we can figure out all these other derivatives. +\begin{align*} +\frac{\partial}{\partial y}f(x, y, z)\frac{z}{(x^{2} + y^{2} + z^{2})^{\frac{1}{2}}} = f'(x, y, z)\frac{z}{(x^{2} + y^{2} + z^{2})^{\frac{1}{2}}} + f(x, y, z)(\frac{z}{(x^{2} + y^{2} + z^{2})^{\frac{1}{2}}})' \\ +\frac{\partial}{\partial y}z(x^{2} + y^{2} + z^{2})^{-\frac{1}{2}} = -yz(x^{2} + y^{2} + z^{2})^{-\frac{3}{2}} \\ +\frac{\partial}{\partial y}f(x, y, z)\frac{z}{(x^{2} + y^{2} + z^{2})^{-\frac{1}{2}}} = f'(x, y, z)\frac{z}{(x^{2} + y^{2} + z^{2})^{\frac{1}{2}}} -yz f(x, y, z)(x^{2} + y^{2} + z^{2})^{-\frac{3}{2}} \\ +\frac{\partial}{\partial z}f(x, y, z)\frac{y}{(x^{2} + y^{2} + z^{2})^{-\frac{1}{2}}} = f'(x, y, z)\frac{y}{(x^{2} + y^{2} + z^{2})^{\frac{1}{2}}} -zy f(x, y, z)(x^{2} + y^{2} + z^{2})^{-\frac{3}{2}} \\ +(\frac{\partial}{\partial y}f(x, y, z)\frac{z}{(x^{2} + y^{2} + z^{2})^{-\frac{1}{2}}} - \frac{\partial}{\partial z}f(x, y, z)\frac{y}{(x^{2} + y^{2} + z^{2})^{-\frac{1}{2}}})\hat{i} = \vec{0} +\end{align*} +And finally, by symmetry, +\begin{align*} +\vec{\nabla} \times \vec{f} = \vec{0} +\end{align*} +Because the computation for \(\hat{j}\) and \(\hat{k}\) are the same. +Also, this implies that all central forces are conservative forces, so by [[id:4ed61028-811e-4425-b956-feca6ee92ba1][inheritance]]: +\begin{align*} +\vec{f} = \vec{\nabla}V +\end{align*} diff --git a/mindmap/conservative force.org b/mindmap/conservative force.org new file mode 100644 index 0000000..83d1c36 --- /dev/null +++ b/mindmap/conservative force.org @@ -0,0 +1,37 @@ +:PROPERTIES: +:ID: 6f2aba40-5c9f-406b-a1fa-13018de55648 +:END: +#+title: conservative force +#+author: Preston Pan +#+html_head: +#+html_head: +#+html_head: +#+options: broken-links:t + +* Definition +A conservative force has this property: +\begin{align*} +\oint\vec{f} \cdot d\vec{l} = 0 +\end{align*} +In other words, work done by \(\vec{f}\) is path independent, because in any closed loop integral, +you go from point \(\vec{a}\) to point \(\vec{b}\) and then back. If these forwards and backwards +paths end up canceling no matter what path you take, then it is clear that \(\vec{f}\) will do the +same amount of force no matter what path you take. Using Stokes' theorem: +\begin{align*} +\int_{S}(\vec{\nabla} \times \vec{f}) \cdot d\vec{a} = \oint\vec{f} \cdot d\vec{l} +\end{align*} +And therefore, if and only if \(\vec{\nabla} \times \vec{f} = \vec{0}\), this line integral is also \(\vec{0}\). Additionally, if you +integrate over \(\vec{f}\), we define \(V(\vec{r})\) such that: +\begin{align*} +\int_{\vec{a}}^{\vec{b}}\vec{f} \cdot d\vec{l} = V(\vec{a}) - V(\vec{b}) +\end{align*} +because it is path independent, we do not need to consider the infinite paths between \(\vec{a}\) and \(\vec{b}\), which +allows us to define this function \(V(\vec{r})\). Then by the fundamental theorem of calculus: +\begin{align*} +\vec{f} = -\vec{\nabla}V +\end{align*} +Therefore, conservative forces can be represented by a scalar field. Now taking the curl of both sides we get: +\begin{align*} +\vec{\nabla} \times \vec{f} = 0 +\end{align*} +Which is consistent with the result from above. diff --git a/mindmap/del operator.org b/mindmap/del operator.org new file mode 100644 index 0000000..7410e30 --- /dev/null +++ b/mindmap/del operator.org @@ -0,0 +1,103 @@ +:PROPERTIES: +:ID: 4bfd6585-1305-4cf2-afc0-c0ba7de71896 +:END: +#+title: del operator +#+author: Preston Pan +#+html_head: +#+html_head: +#+html_head: +#+options: broken-links:t + +* Definition +The operator /del/ in \( n \) dimensional euclidean space is defined as follows: +\begin{align*} +\vec{\nabla} := \sum_{i = 1}^{n} \hat{e_{i}}\frac{\partial}{\partial e_{i}} +\end{align*} +Where \( \frac{\partial}{\partial e_{k}}\) is the [[id:3993a45d-699b-4512-93f9-ba61f498f77f][partial derivative]] with respect to the \(k^{th}\) orthogonal axis, and \( \hat{e}_{k} \) is the +orthogonal basis vector pointing in that direction. In three dimensional euclidean +space using Cartesian coordinates, the del operator would look like: + +\begin{align*} +\vec{\nabla} = \begin{bmatrix} +\frac{\partial}{\partial x} \\ +\frac{\partial}{\partial y} \\ +\frac{\partial}{\partial z} +\end{bmatrix} += \hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} + \hat{k}\frac{\partial}{\partial z} +\end{align*} + +The del operator is what is called a /linear operator/ because it is consistent with operations +pertaining to linear algebra. +* Usage +The del operator is useful for representing the gradient, divergence, and curl of a given +scalar or vector field. + +** Gradient +:PROPERTIES: +:ID: 3587c3b4-c3d8-4ff1-b0ba-8eecb1ef0e4c +:END: +Multiplying the del operator by a scalar field yields a vector that is called the *gradient* +of a function: +\begin{align*} +\vec{\nabla}f = \begin{bmatrix} +\frac{\partial f}{\partial x} \\ +\frac{\partial f}{\partial y} \\ +\frac{\partial f}{\partial z} +\end{bmatrix} += \frac{\partial f}{\partial x}\hat{i} + \frac{\partial f}{\partial y}\hat{j} + \frac{\partial f}{\partial z}\hat{k} +\end{align*} +Where this vector points in the direction of the greatest rate of change, and has a magnitude corresponding +with the slope. The reason why is somewhat intuitive, if you think about it a little. + +** Divergence +:PROPERTIES: +:ID: 12a2d5b3-f98c-45e5-9107-5560288b5aa8 +:END: +Taking the dot product of the del operator with a vector field yields a scalar function, which is called the divergence: +\begin{align*} +\vec{\nabla} \cdot \vec{f} = \frac{\partial f_{x}}{\partial x} + \frac{\partial f_{y}}{\partial y} + \frac{\partial f_{z}}{\partial z} +\end{align*} +Where \( f_{n} \) is the \( n \) component of \( \vec{f} \). + +You can think of it as measuring the rate of change of the outwards or inwards direction of a vector field. +In order to think about this more clearly, we can think about the two dimensional case with just x and y. +Given a two-dimensional vector field, a two-dimensional divergence would look like this: +\begin{align*} +\vec{\nabla} \cdot \vec{f} = \frac{\partial f_{x}}{\partial x} + \frac{\partial f_{y}}{\partial y} +\end{align*} +and to explain this further, let's take a vector \( \vec{v} \), as well as two other vectors to compare it with, +\( \vec{v_{up}}\), and \( \vec{v_{right}} \). Then, we take \( \vec{r} = \vec{f}(\vec{v}) \) and compare it to +\( \vec{r_{up}} \) and \( \vec{r_{right}}\). We then compare the x component of the right vector with the original one, +and we compare the y component of the up vector with the original one, by taking the difference. We then sum these +differences, and what we are left with is a measurement of how spread apart the directions and magnitudes of these vectors +are in this local area. If these \( \vec{r} \) vectors are infinitely close to each other, we can consider this comparison to be analogous to the divergence at that point. +This argument naturally extends to three dimensions. + +** Curl +:PROPERTIES: +:ID: b25e0e44-c764-4f0a-a5ad-7f9d79c7660d +:END: +The curl of a vector field is defined as follows: +\begin{align*} +\vec{\nabla} \times \vec{f} = \hat{i}(\frac{\partial f_{z}}{\partial y} - \frac{\partial f_{y}}{\partial z}) - \hat{j}(\frac{\partial f_{z}}{\partial x} - \frac{\partial f_{x}}{\partial z}) + \hat{k}(\frac{\partial f_{y}}{\partial x} - \frac{\partial f_{x}}{\partial y}). +\end{align*} +Where the equation above is derived from the definition of the cross product. It represents the rate of change of a +vector field "perpendicular" to the divergence of the field. In fact, if you have any field \( \vec{f} \), +you can represent this field as an addition of a curl-less field and a divergence-less field. +Another way to think of it is that you are measuring the strength of rotational component of the vector field about a certain axis. + +** Laplacian +:PROPERTIES: +:ID: 65004429-a6b7-41f2-8489-07605841da3d +:END: +The Laplacian is defined as follows: +\begin{align*} +\nabla^{2}\vec{f} = \nabla \cdot \nabla\vec{f} +\end{align*} +It returns a scalar field and is the multivariable analogue to the second derivative. Because both the divergence +and gradient have been described, I feel it is trivial to understand the Laplacian. + +** Product Rules +The product rules pertaining to the del operator are consistent with that of linear algebra. +For example, \( \vec{\nabla} \times \vec{\nabla}f = 0\). You can show this yourself quite easily, so I find no need to go over it here. +When in doubt, just assume the del works the same way as any old vector, and you will usually be correct. diff --git a/mindmap/derivative.org b/mindmap/derivative.org new file mode 100644 index 0000000..be84116 --- /dev/null +++ b/mindmap/derivative.org @@ -0,0 +1,152 @@ +:PROPERTIES: +:ID: 31d3944a-cddc-496c-89a3-67a56e821de3 +:END: +#+title: derivative +#+author: Preston Pan +#+html_head: +#+html_head: +#+html_head: +#+options: broken-links:t +#+OPTIONS: tex:dvipng + +* Derivation +Let's say we want to know the rate of change of the [[id:b1f9aa55-5f1e-4865-8118-43e5e5dc7752][function]] \(f(x) = x^{2}\). Because this [[id:b1f9aa55-5f1e-4865-8118-43e5e5dc7752][function]] is not +linear (the [[id:b1f9aa55-5f1e-4865-8118-43e5e5dc7752][function]] is a parabola and is therefore curved), we can only take the rate of change by finding +the tangent line to a point on the curve. In other words, for any point \((x_{0}, y_{0})\), we want to find +the straight line that touches this point, and no other points on the parabola. + +The gameplan will look like this: first, we find the slope between two points \((x_{0}, y_{0})\) and \((x_{0} + dx, y_{0} + dy)\), +where \(dx\) and \(dy\) are the difference in x and y respectively, or in other words, just a small change in x and change in y. +Then, we make these points infinitely close to each other and see what the slope is. This resulting line will be infinitely close +to the tangent line. + +Now we want to find the equation for the slope between these two points: +\begin{align*} +m = \frac{y_{1} - y_{0}}{x_{1} - x_{0}} \\ +m = \frac{y_{0} + dy - y_{0}}{x_{0} + dx - x_{0}} \\ +m = \frac{dy}{dx} +\end{align*} + +Because both of these points need to satisfy the [[id:b1f9aa55-5f1e-4865-8118-43e5e5dc7752][function]] \(f(x) = x^{2}\): +\begin{align*} +y_{0} = x_{0}^{2} \\ +y_{0} + dy = (x_{0} + dx)^{2} \\ +dy = (x_{0} + dx)^{2} - y_{0} \\ +dy = (x_{0} + dx)^{2} - x_{0}^{2} \\ +m = \frac{(x_{0} + dx)^{2} - x_{0}^{2}}{dx} +\end{align*} + +Then we use a binomial expansion: +\begin{align*} +m = \frac{x_{0}^{2} + 2x_{0}dx + dx^{2} - x_{0}^{2}}{dx} \\ += \frac{2x_{0}dx + dx^{2}}{dx} \\ += 2x_{0} + dx +\end{align*} +Now we see that since \(dx\) is infinitely close to zero (but not zero because otherwise that would be dividing by zero), we can say that +the tangent line of \(x^{2}\) at this point is: +\begin{align*} +2x_{0} +\end{align*} +And since this works for all points over \(f(x)\), we can simply say: +\begin{align*} +\frac{dy}{dx} = 2x \\ +f'(x) = 2x +\end{align*} +These two notations are both valid. The first is called Leibniz notation, and the second is called Lagrange notation. + +* Definition +Note that you can easily show that the process we did for \(x^{2}\) works for most functions, and is defined as follows: +\begin{align*} +\frac{d}{dx}f(x) = \lim_{h\to0}\frac{f(x + h) - f(x)}{h} +\end{align*} +This \(lim_{h\to0}\) notation is a limit. It broadly dictates that \(h\) is going infinitely close to zero but is not exactly zero. You +will also see \(\frac{d}{dx}\) used as an operator on a [[id:b1f9aa55-5f1e-4865-8118-43e5e5dc7752][function]] much of the time, which also means you're taking a derivative with +whatever \(\frac{d}{dx}\) is multiplied with. + +** Higher Derivatives +The notation \( \frac{d^{n}}{dx^{n}}f(x) \) denotes taking \(n\) derivatives of \(f(x)\), one after the other. \(f''(x)\) works for second derivatives, and so on. +However, this gets annoying, so you can use \( f^{(n)}(x) \) as the \(n^{th}\) derivative of \( f(x) \) as well. + +* Derivative Rules +Usually, instead of using the definition in order to calculate derivatives, we use some simpler rules to do so. +We derive many of them here. +** Addition Rule +\begin{align*} +\frac{d}{dx}(f(x) + g(x)) = \lim_{h\to0}\frac{f(x + h) + g(x + h) - f(x) - g(x)}{h} = \lim_{h\to0}\frac{f(x + h) - f(x) + g(x + h) - g(x)}{h} = \lim_{h\to0}\frac{f(x + h) - f(x)}{h} + \frac{g(x + h) - g(x)}{h} \\ += \frac{d}{dx}f(x) + \frac{d}{dx}g(x) +\end{align*} +of course, subtraction works in the same way. +** Multiplication Rule +\begin{align*} +\frac{d}{dx}(f(x)g(x)) = \lim_{h\to0}\frac{f(x + h)g(x + h) - f(x)g(x)}{h} = \lim_{h\to0}\frac{f(x + h)g(x + h) - f(x)g(x + h) + f(x)g(x + h) - f(x)g(x)}{h} \\ += \lim_{h\to0}\frac{g(x + h)(f(x + h) - f(x)) + f(x)(g(x + h) - g(x))}{h} \\ += g(x)\lim_{h\to0}\frac{f(x + h) - f(x)}{h} + f(x)\frac{g(x + h) - g(x)}{h} = g(x)f'(x) + g'(x)f(x) +\end{align*} +And using the this rule as well as the chain rule and power rule which we will show later, the division rule is easily acquired. +** Chain Rule +:PROPERTIES: +:ID: ffd1bc3d-ab64-4916-9c09-0c89d2731b6d +:END: +The chain rule is a rule about nested functions in the form \( (f \circ g)(x) \). +Using Leibniz notation, it is easy to given an intuition on something called the chain rule: +\begin{align*} +\frac{dy}{dz}\frac{dz}{dx} = \frac{dy}{dx} +\end{align*} +Which, in other words, reads: if you have a [[id:b1f9aa55-5f1e-4865-8118-43e5e5dc7752][function]] \(y\) which has a [[id:b1f9aa55-5f1e-4865-8118-43e5e5dc7752][function]] \(z\) inside of it that is +dependent on \(x\), then \(y'(x) = y'(z(x))z'(x)\). We have manipulated things in the form \(dy\), \(dz\), \(dx\) +before all as regular variables, so although people say this is not rigorous, I would say that it in fact is. +You can treat these "differentials" as regular variables. +** Derivative Rules for Particular Functions +*** Power Rule +*** Sinusoidal Functions +*** Exponential Functions +By the definition of a derivative: +\begin{align*} +\lim_{h\to0}\frac{a^{x + h} - a^{x}}{h} = a^{x}\lim_{h\to0}\frac{a^{h} - 1}{h} +\end{align*} +The constant \(e\) is defined such that: +\begin{align*} +\lim_{h\to0}\frac{e^{h} - 1}{h} = 1; \\ +\frac{d}{dx}e^{x} = e^{x} +\end{align*} +Then by the chain rule: +\begin{align*} +\frac{d}{dx}a^{x} = \frac{d}{dx}(e^{\ln(a)})^{x} = \frac{d}{dx}e^{\ln(a)x}= \ln(a)e^{\ln(a)x} +\end{align*} +And therefore: +\begin{align*} +\lim_{h\to0}\frac{a^{h} - 1}{h} = \ln(a) +\end{align*} +* Implicit Differentiation +The equation of a circle centered at the origin is: +\begin{align*} +x^{2} + y^{2} = r^{2} +\end{align*} +This [[id:b1f9aa55-5f1e-4865-8118-43e5e5dc7752][function]] is clearly dependent on \(y\), and no, we don't need to do algebra to isolate the y (yet, we can do that later). instead, +we can simply take the derivative of both sides: +\begin{align*} +\frac{d(x^{2} + y^{2})}{dx} = \frac{d(r^{2})}{dx} +\end{align*} +the right hand side is obviously going to reduce to zero because it is a constant inside a derivative. +Because we consider \(y = y(x)\), taking the derivative of \(y\) in terms of \(x\) means we have to apply +the chain rule. +\begin{align*} +2x + 2y(x) * y'(x) = 0 +\end{align*} +Remember, the [[id:b1f9aa55-5f1e-4865-8118-43e5e5dc7752][function]] we are taking the derivative of here is \((y(x))^{2}\), which is why the \(y'(x)\) term +appears; you're doing the chain rule on an inner [[id:b1f9aa55-5f1e-4865-8118-43e5e5dc7752][function]] that you don't know the value of but that you can represent +nonetheless. + +Now, we move everything to the other side in order to find \(y'(x)\): +\begin{align*} +y'(x) = -\frac{x}{y(x)} +\end{align*} + +and then we finally find \(y(x)\) and substitute it in: +\begin{align*} +y(x) = (r^{2} - x^{2})^{\frac{1}{2}} \\ +y'(x) = -\frac{x}{(r^{2} - x^{2})^{\frac{1}{2}}} +\end{align*} +The benefit of this strategy is that you can find the derivative of a circle (or as we will see later, many other curves) in terms of \(y\), +which is useful for converting coordinate systems. Implicit differentiation is also useful for some other things, like: +** Derivative of [[id:4f088813-cf40-4194-9251-b2392a50dc1c][Inverse]] Function diff --git a/mindmap/dirac delta.org b/mindmap/dirac delta.org new file mode 100644 index 0000000..c799c30 --- /dev/null +++ b/mindmap/dirac delta.org @@ -0,0 +1,60 @@ +:PROPERTIES: +:ID: 90574fea-88f4-4b80-9cda-32cff0bcb76d +:END: +#+title: dirac delta +#+author: Preston Pan +#+html_head: +#+html_head: +#+html_head: +#+options: broken-links:t + +* Definition +The dirac delta distribution can be defined by a series of curves. For example, take a rectangle with +length \(\frac{1}{x}\) and height \(x\), and center this rectangle symmetrically at the y axis, with the bottom of it +touching the x axis. Now if we take the area of this rectangle, it is always going to be \( \frac{x}{x} = 1\). +If we define this curve to be: +\begin{align*} +f_{n}(x) := \left\{ +\begin{array}{lr} +n, & \text{if } -\frac{1}{2n} \le x \le \frac{1}{2n} \\ +0, & \text{if }x < -\frac{1}{2n}, x > \frac{1}{2n} +\end{array} +\right\} +\end{align*} +then the dirac delta distribution is defined as \( \delta(x) = \lim_{n\to\infty}f_{n}(x)\). The resulting distribution exhibits the following properties: +\begin{align*} +\delta(x) = \left\{ +\begin{array}{lr} +\infty, & \text{if } x = 0 \\ +0, & \text{if } x \ne 0 +\end{array} +\right\} +\end{align*} +It is easy to extend this definition to two or more dimensions simply by altering the dimensions of this class of curves. +In physics, the most common distribution used is the three dimensional one, or \( \delta^{3}(x)\), but you can define a dirac delta distribution +for arbitrary dimensions: +\begin{align*} +\delta^{n}(\vec{r_{n}}) = \left\{ +\begin{array}{lr} +\infty, & \text{if } \vec{r_{n}} = \vec{0} \\ +0, & \text{if } \vec{r_{n}} \ne \vec{0} +\end{array} +\right\} +\end{align*} + +* Properties +Note that because the area under all the curves we used to define the delta function with is exactly one, here we expect no different: +\begin{align*} +\int_{a}^{b}\delta(x)dx = \left\{\begin{array}{lr} +1, & \text{if } a \le 0, b \ge 0 \\ +0 & \text{otherwise.} +\end{array} +\right\} +\end{align*} + +We will assume from now on that \(\delta(x)\) is within the integral range, because we know that it is zero when it isn't. When multiplied with another function, it has this property: +\begin{align*} +\int_{a}^{b}\delta(x)f(x)dx = f(0) \\ +\int_{a}^{b}\delta(x - a)f(x)dx = f(a) +\end{align*} +in which the dirac delta function "sifts" for a particular value of \(f(x)\). diff --git a/mindmap/electrostatics.org b/mindmap/electrostatics.org index e71a05e..55310de 100644 --- a/mindmap/electrostatics.org +++ b/mindmap/electrostatics.org @@ -26,7 +26,7 @@ on \( P_{1} \) is as follows: \end{align*} Where \( \hat{r} \) is the unit vector pointing in the direction of \( P_{2} \). Note that there are a -couple of interesting things about this force. First, it is an inverse square law, and the formula looks a lot like the one for gravitation, +couple of interesting things about this force. First, it is an [[id:2a543b79-33a0-4bc8-bd1c-e4d693666aba][inverse square]] law, and the formula looks a lot like the one for gravitation, only charge can be negative and mass cannot. Second, it is symmetrical, in the sense that the force felt by \( P_{2} \) is going to be the same, only \( \hat{r} \) is pointing in the other direction. Also, note that due to linearity, this force calculation follows the /superposition principle/. @@ -37,7 +37,7 @@ That is, if we have different electrostatic forces acting on one particle: \end{align*} Wait, where does the \( \frac{1}{4\pi\epsilon_{0}} \) term come from? Well, the surface area of a sphere -is \( 4\pi r^{2}\) , which explains both the inverse square law and this \( 4\pi \) term in the denominator, +is \( 4\pi r^{2}\) , which explains both the [[id:2a543b79-33a0-4bc8-bd1c-e4d693666aba][inverse square]] law and this \( 4\pi \) term in the denominator, but what about \( \epsilon_{0} \), what does it even mean? Well, it is simply a conversion of units from /speed of light/ terms to /SI unit terms/. If you @@ -95,3 +95,22 @@ Where \( \tau \) is the patch of volume we are integrating over, and \( \sigma \ which takes a position vector and returns the charge at that vector. Of course, surface and line integrals have their own analogues -- simply replace \( d\tau \) with \( da \) or \( dl \), and make sure your charge distribution is in the correct amount of dimensions. + +** [[id:12a2d5b3-f98c-45e5-9107-5560288b5aa8][Divergence]] and [[id:b25e0e44-c764-4f0a-a5ad-7f9d79c7660d][Curl]] of Electric Field +The divergence and curl of the electric field are essential to solving electrostatic configurations with more +ease, as well as proving some qualities about the electric field. Because the electric field is an [[id:2a543b79-33a0-4bc8-bd1c-e4d693666aba][inverse square]] +field: +\begin{align*} +\vec{\nabla} \cdot \vec{E} = \frac{\sigma(\vec{r_{1}})}{\epsilon_{0}} \\ +\oint\vec{E} \cdot d\vec{a} = \frac{q_{enc.}}{\epsilon_{0}} \\ +\vec{\nabla} \times \vec{E} = \vec{0} \\ +\oint\vec{E} \cdot d\vec{l} = \vec{0} +\end{align*} + +* Electrostatic Potentials +Because \(\vec{E}\) is a [[id:6f2aba40-5c9f-406b-a1fa-13018de55648][conservative field]]: +\begin{align*} +\vec{E} = -\vec{\nabla}V \\ +\nabla^{2}V = -\frac{\sigma(\vec{r_{1}})}{\epsilon_{0}} \\ +V(\vec{r}) = \frac{1}{4\pi \epsilon_{0}}\int\frac{\sigma(\vec{r_{2}})}{r}dr +\end{align*} diff --git a/mindmap/everything.org b/mindmap/everything.org index 8e93fe7..8c92bd9 100644 --- a/mindmap/everything.org +++ b/mindmap/everything.org @@ -6,6 +6,7 @@ #+html_head: #+date: <2023-06-10 Sat> +#+OPTIONS: tex:dvipng * Everything? This article is about everything. No, really, it is. Every possible thought you could have is encapsulated diff --git a/mindmap/function.org b/mindmap/function.org new file mode 100644 index 0000000..07b86cb --- /dev/null +++ b/mindmap/function.org @@ -0,0 +1,36 @@ +:PROPERTIES: +:ID: b1f9aa55-5f1e-4865-8118-43e5e5dc7752 +:END: +#+title: function +#+author: Preston Pan +#+html_head: +#+html_head: +#+html_head: +#+options: broken-links:t + + + +* Definition +A function \( f(x) \) is a set \( S \) of ordered pairs that map the first value of the ordered pair to the second value +in the ordered pair, where the first value may not have duplicates in the set \(S\). The map from the first value to the +second value has the notation \(f(x) = y \) for some x and y, where \(f\) is the mapping. Note that we can also define rules +for \(f\) and do not therefore have to explicitly define all the mappings: +\begin{align*} +S = \{(x, y): x^{2} = y, x, y \in \mathbb{R} \} +\end{align*} +Which is an example of a parabolic function. \(x\) and \(y\) can both conceptually be any object, but usually they are +mathematical objects. Some examples of such objects include tensors and scalars. +* ordered pair +:PROPERTIES: +:ID: 1b1b522e-d4de-4832-9ca4-c6d1cfee27e6 +:END: +However, we must find a way to define what an ordered pair is. Sets have no order by default, so we need to +add order by doing the following: +\begin{align*} +(x_{0}, y_{0}) := \{x_{0}, \{x_{0}, y_{0}\}\} +\end{align*} +Where the element that is not explicitly a set gives us the definition of the first element. +* Function Group +Let \((S, \circ)\) define a [[id:ba7b95b0-0ce6-4b33-9a79-5e5fddaea710][group]] where \(S\) is the set of all functions, and \(\circ\) is the composition +binary operator. Then \(f(x) = x\) is the identity element, and an inverse of a function is defined +as \( (f \circ f^{-1})(x) = (f^{-1} \circ f)(x) = x \). diff --git a/mindmap/group.org b/mindmap/group.org new file mode 100644 index 0000000..5fb0498 --- /dev/null +++ b/mindmap/group.org @@ -0,0 +1,30 @@ +:PROPERTIES: +:ID: ba7b95b0-0ce6-4b33-9a79-5e5fddaea710 +:END: +#+title: group +#+author: Preston Pan +#+html_head: +#+html_head: +#+html_head: +#+options: broken-links:t + +* Definition +A group is an [[id:1b1b522e-d4de-4832-9ca4-c6d1cfee27e6][ordered pair]] \((G, *)\) where \(G\) is a set and \(*\) is a binary operation (operation defined between two members of set G) defined such that: +\begin{align*} +a * b \in G \\ +\exists e : a * e = a +\end{align*} +where the operation \(*\) is said to be closed under \(G\), and \(e\) is called the /identity/ of group \((G, *)\). +** Associativity +This is the property such that: +\begin{align*} +(a * b) * c = a * (b * c) +\end{align*} +** inverse +:PROPERTIES: +:ID: 4f088813-cf40-4194-9251-b2392a50dc1c +:END: +An inverse is defined as follows: +\begin{align*} +\forall a \exists a^{-1} : a * a^{-1} = e +\end{align*} diff --git a/mindmap/inheritance.org b/mindmap/inheritance.org new file mode 100644 index 0000000..dc3f21d --- /dev/null +++ b/mindmap/inheritance.org @@ -0,0 +1,24 @@ +:PROPERTIES: +:ID: 4ed61028-811e-4425-b956-feca6ee92ba1 +:END: +#+title: inheritance +#+author: Preston Pan +#+html_head: +#+html_head: +#+html_head: +#+options: broken-links:t + +* Definition +In programming, inheritance is the adoption of attributes of a child class by that of a parent class. +For instance, in this example: +#+begin_src python :results output both +class Animal: + def __init__(self, birthday): + self.weight = weight + self.birthday = birthday + +class Dog(Animal): + pass +#+end_src +The class ~~Dog~~ will have the same fields as that of animal. In general, you can think of inheritance as +taking on attributes from a node higher in the abstraction hierarchy. diff --git a/mindmap/inverse square.org b/mindmap/inverse square.org new file mode 100644 index 0000000..52a3f61 --- /dev/null +++ b/mindmap/inverse square.org @@ -0,0 +1,190 @@ +:PROPERTIES: +:ID: 2a543b79-33a0-4bc8-bd1c-e4d693666aba +:END: +#+title: inverse square +#+author: Preston Pan +#+html_head: +#+html_head: +#+html_head: +#+options: broken-links:t + +* Derivation +Say you're doing some physics modeling, and you want to describe a force that has the following properties (and one more that I will introduce later, +and many of which inherit from the definition of a [[id:c1e836e3-a80c-459d-8b68-396fa1687177][central force]]): +1. The force happens at a distance between two particles. +2. The force gives both the particles an equal amount of force but in opposite directions (Newton's Third Law). +3. The force magnitude only depends on the distance between the two particles in question, and maybe some other property intrinsic to both the particles. +4. If you draw a straight line between the two points, the force vector has to be parallel to that line. +5. The force works in the same way no matter where in the universe you are. + +We will call our mysterious force field \(\vec{f(\vec{r})}\). We also want to consider two particles, and for simplicity we can say these +particles have no volume, so they are just points. These point particles we will call \(P_{1)}\) and \(P_{2}\). For simplicity, we will also +consider two dimensions instead of three, realizing that generalizing to arbitrary dimensions in euclidean space is trivial in this case. +Now, the direction of force must be solely dependent on the orientation of the other particle, which follows from property 4. +We call this direction \(\hat{r}\) and the distance between them \(r\); the vector that represents the direction and the distance together +we are going to define to be \(\vec{r}\). + +If we imagine drawing \(P_{1}\) and \(P_{2}\) randomly on a piece of paper, we can imagine using a compass to draw a circle by centering at \(P_{1}\) +and putting the end of the compass at \(P_{2}\). If we imagine putting \(P_{2}\) at any other point on the perimeter of this circle, the force magnitude remains the +same by property 3. This is what is called radial symmetry. Note that this is true in three dimensions as well, only in three dimensions +we will have to use a sphere. Now, let's say there's another particle further away than \(P_{2}\) called \(P_{3}\). That particle will also have a circle, but this +circle will have a total circumference larger than that of \(P_{2}\). Here we introduce the last property: let's say that these two circles stop becoming hypothetical +and start becoming an infinite amount of particles. If we ignore property 2 for a second, the total force felt by the ring at \(P_{2}\) should be the same as the +total force felt by the ring at \(P_{3}\). In other words, the field gives each ring around it an equal amount of force, or in more technical terms, the flux though +all the possible concentric rings is the same, or the flux through a given concentric ring is independent of \(\vec{r}\). + +Think about the implications of this last rule. The circumference of a circle is \(2\pi r\), which means if both rings are getting the same amount of force, in the +case of the \(P_{3}\) ring, we are distributing the same amount of force to a greater amount of particles (it is farther away and the amount of particles is going +to be proportional to the circumference). In other words, we are dividing some amount of force \(k\) by the circumference of a circle \(2\pi r\), or in other words, +the /field intensity/ drops off proportionally to \(\frac{1}{r}\). +Therefore, our force has to look something like: +\begin{align*} +\frac{k}{2\pi r}\hat{r} +\end{align*} +but in three dimensions, that isn't quite right. + +Spheres have a surface area of \(4\pi r^{2}\). In three dimensions, instead of dividing among the circumference of a circle, we are dividing among the surface area +of a sphere. Therefore, in three dimensions we get: +\begin{align*} +\frac{k}{4\pi r^{2}}\hat{r} +\end{align*} + +by property 3, there might be some property intrinsic to each particle in question that makes the force vector have a higher magnitude (like mass for gravity), so we have to scale +this force by our measurement of that property. By property 2, the force must be the same magnitude but opposite direction on the two particles. In order to +preserve that property, this force must scale by both particles' properties: +\begin{align*} +\vec{f}(\vec{r}) = \frac{kP_{1}P_{2}}{4\pi r^{2}}\hat{r} +\end{align*} +in this article I am going to consider this \(4\pi\) constant as a part of \(k\), so the final form will be: +\begin{align*} +\frac{kP_{1}P_{2}}{r^{2}}\hat{r} +\end{align*} +And from 6 properties that seem like reasonable constraints, we derive the inverse square law. + +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 +\end{align*} +Which is a result we will use later. + +Also, from here on out, we will be dealing with a field that is not dependent on \(P_{2}\); we will be dealing with systems with many particles and therefore +simply not considering \(P_{2}\) and multiplying it in at the end will be more convenient. Additionally, usually we simply don't know the value of \(P_{2}\); +we usually in the real world are given a system that can interact with arbitrary values of \(P_{2}\), or in other words, in the real world systems often +interact with a lot of different objects, so we want to consider the field of the system independent of the unknown other object that might interact with it, +which leads us to: + +* Definition +Inverse square fields for point particles are fields which have the form: +\begin{align*} +\frac{k\sigma}{r^{2}}\hat{r}. +\end{align*} +where \(\vec{r}\) is the distance between the source particle (the particle that gives the force) and test particle (the particle that receives the force), +and sigma defines the strength of the source point. This would be mass in gravity or charge in electrodynamics. + +Inverse square laws follow the superposition principle, and therefore: +\begin{align*} +\vec{f}_{total} = \sum_{i = 0}^{n}\vec{f}_{i} +\end{align*} +Which means statements made about fields of single point particles is most likely true of systems of more than one particle. +Inverse square fields also have a continuous distribution form: +\begin{align*} +\vec{f}(\vec{r''}) = \int_{space}\frac{k\sigma(\vec{r'})}{r^{2}}\hat{r}d\tau +\end{align*} +Where sigma becomes a smooth distribution. The integral is for a continuous distribution of points that follow the inverse square law, where +the field generated by each individual point contributes in a weighted sum. + +\(\vec{r'}\) is displacement vector of the field source, whereas \(\vec{r''}\) is the displacement of the test particle. \(\vec{r} := \vec{r''} - \vec{r'}\). +* Divergence of Inverse Square Fields +Using the [[id:4bfd6585-1305-4cf2-afc0-c0ba7de71896][del operator]]: +\begin{align*} +\vec{\nabla} \cdot \vec{f(\vec{r''})} = \int_{space}k\sigma(\vec{r'}) * \vec{\nabla} \cdot \frac{\hat{r}}{r^{2}}d\tau +\end{align*} +In this example, I will only be using the Cartesian del operator because although the calculation is harder +the math needed in order to understand Cartesian coordinates is lesser. +If we are using the Cartesian del operator, we must also use Cartesian coordinates. Therefore, we do unit +conversions to replace \(\vec{r}\): +\begin{align*} +\vec{\nabla} \cdot \vec{f} = k * \int_{space} \sigma(\vec{r'}) * \vec{\nabla} \cdot \frac{1}{x^{2} + y^{2} + z^{2}}\frac{x\hat{i} + y\hat{j} + z\hat{k}}{(x^{2} + y^{2} + z^{2})^{\frac{1}{2}}}d\tau +\end{align*} +If we can solve for one of these partial derivative terms, we can solve for the other two by symmetry: +\begin{align*} += k\int_{space} \sigma(\vec{r'}) * \vec{\nabla} \cdot (\frac{x\hat{i}}{(x^{2} + y^{2} + z^{2})^{\frac{3}{2}}} + \frac{y\hat{j}}{(x^{2} + y^{2} + z^{2})^{\frac{3}{2}}} + \frac{z\hat{k}}{(x^{2} + y^{2} + z^{2})^{\frac{3}{2}}})d\tau \\ += k\int_{space}\sigma(\vec{r'})(\frac{\partial}{\partial x}\frac{x}{(x^{2} + y^{2} + z^{2})^{\frac{3}{2}}} + \frac{\partial}{\partial y}\frac{y}{(x^{2} + y^{2} + z^{2})^{\frac{3}{2}}} + \frac{\partial}{\partial z}\frac{z}{(x^{2} + y^{2} + z^{2})^{\frac{3}{2}}})d\tau +\end{align*} +We use the chain rule to solve for this: +\begin{align*} +\frac{\partial}{\partial x}x(x^{2} + y^{2} + z^{2})^{-\frac{3}{2}} = (x^{2} + y^{2} + z^{2})^{-3/2} - x * 2x * \frac{3}{2} (x^{2} + y^{2} + z^{2})^{-\frac{5}{2}} \\ += (x^{2} + y^{2} + z^{2})^{-3/2} - 3x^{2}(x^{2} + y^{2} + z^{2})^{-\frac{5}{2}} +\end{align*} +Then it will look like this inside the integral: +\begin{align*} +k\int_{space}\sigma(\vec{r'})((x^{2} + y^{2} + z^{2})^{-3/2} - 3x^{2}(x^{2} + y^{2} + z^{2})^{-\frac{5}{2}} + (x^{2} + y^{2} + z^{2})^{-3/2} - 3y^{2}(x^{2} + y^{2} + z^{2})^{-\frac{5}{2}} + (x^{2} + y^{2} + z^{2})^{-3/2} - 3z^{2}(x^{2} + y^{2} + z^{2})^{-\frac{5}{2}})d\tau +\end{align*} +If we factor out the \((x^{2} + y^{2} + z^{2})^{-\frac{5}{2}}\) term and collecting the like terms: +\begin{align*} +k\int_{space}\sigma(\vec{r'})(3(x^{2} + y^{2} + z^{2})^{-\frac{3}{2}} - 3(x^{2} + y^{2} + z^{2})(x^{2} + y^{2} + z^{2})^{-\frac{5}{2}})d\tau = k\int_{space}(3(x^{2} + y^{2} + z^{2})^{-\frac{3}{2}} - 3(x^{2} + y^{2} + z^{2})^{-\frac{3}{2}})d\tau = 0. +\end{align*} +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 +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: +\begin{align*} +k\sigma(\vec{r'})4\pi\delta^{3}(\vec{r}) +\end{align*} +So now we can take the volume integral of this quantity: +\begin{align*} +\vec{\nabla} \cdot \vec{f}(\vec{r''}) = 4\pi k\int_{space}\delta^{3}(\vec{r''} - \vec{r'})\sigma(\vec{r'})d\tau \\ += 4\pi k\sigma(\vec{r''}) +\end{align*} +As an analogy, if we have a point mass (an object with finite mass in a single point), the density of any volume containing the point will be zero, but +the gravitational field generated by that point mass will not be zero. Obviously just taking the density blindly will not accurately account for the +actual distribution of mass within space. You need to use something like the [[id:90574fea-88f4-4b80-9cda-32cff0bcb76d][dirac delta]] distribution to model this behavior. Also, since the \(k\) constant +includes the term \(\frac{1}{4\pi}\), we can just say: +\begin{align*} +\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: +\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*} + +* Curl of Inverse Square Fields +Because an inverse square field is a special case of a [[id:c1e836e3-a80c-459d-8b68-396fa1687177][central force]], this result is also the same as for all general central fields. That is: +\begin{align*} +\vec{\nabla} \times \vec{f} = \vec{0} +\end{align*} +Which implies the field in question is a [[id:6f2aba40-5c9f-406b-a1fa-13018de55648][conservative force]]: +\begin{align*} +\oint \vec{f} \cdot d\vec{l} = 0 +\end{align*} +It also implies: +\begin{align*} +\vec{f} = -\vec{\nabla}V +\end{align*} +Where \(V\) is known as the potential of this field. + +* Potential of Inverse Square fields +Now we want to find the specific potential scalar field for inverse square fields. In an inverse square field, +given the identity: +\begin{align*} +\int_{\vec{a}}^{\vec{b}}\vec{f} \cdot d\vec{l} = V(\vec{a}) - V(\vec{b}) +\end{align*} +We can set a reference point as \(\vec{a}\) -- something analogous to setting sea level or setting your coordinate system. Here, we set it infinitely far away and integrate +in a straight line such that the path is parallel to \(\vec{r}\), with respect to a point particle at the origin: +\begin{align*} +k\sigma\int_{\vec{\infty}}^{\vec{b}}\frac{1}{r^{2}}\hat{r} \cdot d\vec{l} = V(\vec{\infty}) - V(\vec{b}) \\ +k\sigma\int_{\infty}^{b}\frac{1}{r^{2}}dr = V(\vec{\infty}) - V(\vec{b}) \\ +-\frac{\sigma}{4\pi \epsilon_{0}r}\Big|^{b}_{\infty} = - \frac{k\sigma}{\infty} + \frac{k\sigma}{b} = \frac{k\sigma}{b} \\ +V(\vec{r}) := \frac{k\sigma}{r} +\end{align*} +It is easy to prove that \(V\) follows the superposition principle. Thus, we can define a continuous distribution for the potential field as well: +\begin{align*} +V(\vec{r}) := k\int_{space}\frac{\sigma(r')}{r}d\tau +\end{align*} +Note that because this field does not require keeping track of vector orientation, it is significantly easier to solve for \(V\) then convert to \(\vec{f}\). Additionally, +setting a reference point to something that is not infinity would be valid as well -- we just choose infinity because it cancels off the constant term. However, +the /difference/ in potentials is absolute and does not require any constant adjustment. diff --git a/mindmap/partial derivative.org b/mindmap/partial derivative.org new file mode 100644 index 0000000..ac1c5d2 --- /dev/null +++ b/mindmap/partial derivative.org @@ -0,0 +1,14 @@ +:PROPERTIES: +:ID: 3993a45d-699b-4512-93f9-ba61f498f77f +:END: +#+title: partial derivative +#+author: Preston Pan +#+html_head: +#+html_head: +#+html_head: +#+options: broken-links:t + +* Definition +The operator \(\frac{\partial}{\partial x_{n}}\) takes the [[id:31d3944a-cddc-496c-89a3-67a56e821de3][derivative]] of a function \(f(\vec{x})\) with respect to \(x_{n}\) in +the direction of the \(x_{n}\) axis. In other words, we take the derivative \(f(\vec{x})\) as if all variables +except \(x_{n}\) is constant. -- cgit