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docs/physics/mechanics/1-classical-mechanics.md

@@ -4,13 +4,13 @@
 
 Each point $P \in \mathbb{R}^3$ can be written as $(x, y, z)$. However, we will utilize the convention below:
 
-$$\vb{r} = x \vu{x} + y \vu{y} + z \vu{z}$$
+$$\mathbf{r} = x \hat{\mathbf{x}} + y \hat{\mathbf{y}} + z \hat{\mathbf{z}}$$
 
 However, when working with ambiguous unit basis, it may be better to use the following:
 
-$$\vb{r} = r_1 \vu{e_1} + r_2 \vu{e_2} + r_3 \vu{e_3}$$
+$$\mathbf{r} = r_1 \hat{\mathbf{e_1}} + r_2 \hat{\mathbf{e_2}} + r_3 \hat{\mathbf{e_3}}$$
 
-Here, $r_1 = x, r_2 = y, r_3 = z$ and $\vu{e_1} = \vu{x_1}, \vdots$. That is, $\vb{r} = \sum r_i \vu{e_i}$.
+Here, $r_1 = x, r_2 = y, r_3 = z$ and $\hat{\mathbf{e_1}} = \hat{\mathbf{x_1}}, \ldots$. That is, $\mathbf{r} = \sum r_i \hat{\mathbf{e_i}}$.
 
 Vectors are an $\mathbb{R}$-module, with addition defined component-wise. Additionally, the scalar (dot) product and vector (cross) product are defined as usual.
 
@@ -24,27 +24,27 @@ Regarding time, classically, time is universal.
 
 **Definition**. The *mass* $m$ of an object characterizes its translational inertia, and is measured in kilograms (kg)
 
-**Definition**. The *force* $\vb{F}$ exerted on an object is a push or pull on said object and is measured in Newtons (N). Note that force is a vector.
+**Definition**. The *force* $\mathbf{F}$ exerted on an object is a push or pull on said object and is measured in Newtons (N). Note that force is a vector.
 
 ## Section 1.4 - Newton's First and Second Laws; Inertial Frames
 
 **Definition**. A *point mass* or *particle* is a convenient fiction, in which an object with mass has no size. It may move in space but has no internal degrees of freedom. Additionally, it may not have any rotational or vibrational kinetic energy.
 
-**Theorem**. Newton's First Law. In the absence of external forces, a particle moves with constant velocity $\vb{v}$.
+**Theorem**. Newton's First Law. In the absence of external forces, a particle moves with constant velocity $\mathbf{v}$.
 
-**Theorem**. Newton's Second Law. Given any particle with mass $m$, the net force $\vb{F}$ on the particle is always equal to the particle's mass times its acceleration. That is,
+**Theorem**. Newton's Second Law. Given any particle with mass $m$, the net force $\mathbf{F}$ on the particle is always equal to the particle's mass times its acceleration. That is,
 
-$$\vb{F} = m\vb{a} = m \ddot{\vb{r}}$$
+$$\mathbf{F} = m\mathbf{a} = m dot{\mathbf{r}}$$
 
-This can also be rewritten in terms of momentum. We know that momentum $\vb{p}$ can be written as $\vb{p} = m\vb{v} = m \dot{\vb{r}}$. Then,
+This can also be rewritten in terms of momentum. We know that momentum $\mathbf{p}$ can be written as $\mathbf{p} = m\mathbf{v} = m \dot{\mathbf{r}}$. Then,
 
-$$\vb{F} = m\vb{a} = m \dot{\vb{p}} = m \ddot{\vb{r}}$$
+$$\mathbf{F} = m\mathbf{a} = m \dot{\mathbf{p}} = m dot{\mathbf{r}}$$
 
-If we have a constant force $\vb{F} = F_0 \vu{x}$, we can write $\ddot{\vu{x}}(t) = \frac{F_0}{m}$. Then,
+If we have a constant force $\mathbf{F} = F_0 \hat{\mathbf{x}}$, we can write $dot{\hat{\mathbf{x}}}(t) = \frac{F_0}{m}$. Then,
 
-$$\dot{\vb{x}}(t) = \int \ddot{\vb{x}} dt = v_0 + \frac{F_0}{m}t$$
+$$\dot{\mathbf{x}}(t) = \int dot{\mathbf{x}} dt = v_0 + \frac{F_0}{m}t$$
 
-$$\vb{x}(t) = \int \dot{\vb{x}} dt = x_0 + v_0t + \frac{F_0}{2m}t^2$$
+$$\mathbf{x}(t) = \int \dot{\mathbf{x}} dt = x_0 + v_0t + \frac{F_0}{2m}t^2$$
 
 **Definition**. An *inertial frame* is a reference frame relative to some fixed frame if they are moving with constant velocity in regards to each other. Otherwise, the frames are *non-inertial*.
 
@@ -54,17 +54,17 @@ $$\vb{x}(t) = \int \dot{\vb{x}} dt = x_0 + v_0t + \frac{F_0}{2m}t^2$$
 
 **Definition**. Force pairs that operate in the same line as each other (eg. gravitational attraction) are called *central forces*.
 
-Recall that the change in momentum of any particle can be defined as $\dot{\vb{p}}_1 = \vb{F}_1 = \vb{F}_{12} + \vb{F}_{1}^{ext}$. Then, $\dot{\vb{p}}_2 = \vb{F}_{21} + \vb{F}_2^{ext}$. As $\vb{P} = \vb{p}_1 + \vb{p}_2$, we can see that
+Recall that the change in momentum of any particle can be defined as $\dot{\mathbf{p}}_1 = \mathbf{F}_1 = \mathbf{F}_{12} + \mathbf{F}_{1}^{ext}$. Then, $\dot{\mathbf{p}}_2 = \mathbf{F}_{21} + \mathbf{F}_2^{ext}$. As $\mathbf{P} = \mathbf{p}_1 + \mathbf{p}_2$, we can see that
 
-$$\dot{\vb{P}} = \vb{p}_1 + \vb{p}_2 = \vb{F}_{12} +\vb{F}_{21} + \vb{F}_{1}^{ext} + \vb{F}_{2}^{ext} = \vb{F}_{1}^{ext} + \vb{F}_{2}^{ext} = \vb{F}^{ext}$$
+$$\dot{\mathbf{P}} = \mathbf{p}_1 + \mathbf{p}_2 = \mathbf{F}_{12} +\mathbf{F}_{21} + \mathbf{F}_{1}^{ext} + \mathbf{F}_{2}^{ext} = \mathbf{F}_{1}^{ext} + \mathbf{F}_{2}^{ext} = \mathbf{F}^{ext}$$
 
-From this, we can see that if $\vb{F}^{ext} = 0$, then $\vb{P}$ is a constant.
+From this, we can see that if $\mathbf{F}^{ext} = 0$, then $\mathbf{P}$ is a constant.
 
-This argument can be generalized to multi-particle systems. Consider particle $\alpha$. Then, $\dot{\vb{p}}_\alpha = \vb{F}_\alpha = \sum_{\beta \neq \alpha} \vb{F}_{\alpha\beta} + \vb{F}_\alpha^{ext}$. We  can then see that
+This argument can be generalized to multi-particle systems. Consider particle $\alpha$. Then, $\dot{\mathbf{p}}_\alpha = \mathbf{F}_\alpha = \sum_{\beta \neq \alpha} \mathbf{F}_{\alpha\beta} + \mathbf{F}_\alpha^{ext}$. We  can then see that
 
-$$\dot{\vb{P}} = \sum_\alpha \vb{p}_\alpha = \sum_\alpha \sum_{\beta \neq \alpha} \vb{F}_{\alpha\beta} + \sum_\alpha \vb{F}_\alpha^{ext}$$
+$$\dot{\mathbf{P}} = \sum_\alpha \mathbf{p}_\alpha = \sum_\alpha \sum_{\beta \neq \alpha} \mathbf{F}_{\alpha\beta} + \sum_\alpha \mathbf{F}_\alpha^{ext}$$
 
-With $\sum_\alpha \sum_{\beta \neq \alpha} \vb{F}_{\alpha\beta} = \sum_\alpha \sum_{\beta > \alpha} \vb{F}_{\alpha\beta}( + \vb{F}_{\beta\alpha})$, we can see that $\dot{\vb{P}} = \sum_\alpha \vb{F}_\alpha^{ext}$.
+With $\sum_\alpha \sum_{\beta \neq \alpha} \mathbf{F}_{\alpha\beta} = \sum_\alpha \sum_{\beta > \alpha} \mathbf{F}_{\alpha\beta}( + \mathbf{F}_{\beta\alpha})$, we can see that $\dot{\mathbf{P}} = \sum_\alpha \mathbf{F}_\alpha^{ext}$.
 
 ---
 

docs/physics/mechanics/2-projectiles-charged-particles.md

@@ -2,13 +2,13 @@
 
 ## Section 2.1 - Air Resistance
 
-**Definition**. The *drag*, or resistive force on an object due to the atmosphere, is denoted as $\vb{f}$. Note that this is **not** the force density, but the overall force. In most cases, this force directly opposes the direction of motion. If not, the other component is known as *lift*, however this is mostly negligible.
+**Definition**. The *drag*, or resistive force on an object due to the atmosphere, is denoted as $\mathbf{f}$. Note that this is **not** the force density, but the overall force. In most cases, this force directly opposes the direction of motion. If not, the other component is known as *lift*, however this is mostly negligible.
 
-We define air resistance as $\vb{f} = -f(v) \vu{v}$. We consider two types in this text: linear, where $f(v) = f_{lin} = bv$, and quadratic, where $f(v) = f_{quad} = cv^2$. Note that often times we consider both, and state that $f(v) = f_{lin} + f_{quad} = bv + cv^2$.
+We define air resistance as $\mathbf{f} = -f(v) \hat{\mathbf{v}}$. We consider two types in this text: linear, where $f(v) = f_{lin} = bv$, and quadratic, where $f(v) = f_{quad} = cv^2$. Note that often times we consider both, and state that $f(v) = f_{lin} + f_{quad} = bv + cv^2$.
 
 The linear term comes from viscous drag and is generally proportional to the viscosity of the medium, while quadratic drag tends to arise from the particle needing to accelerate the mass of air which it is constantly colliding against.
 
-In some cases, we can calculate these coefficients. With $D$ as the diameter of a spherical object, and $\beta$ and $\gamma$ as properties of the medium, we can state that $b = \beta D$ and $c = \gamma D^2$. In air at STP, $\beta = 1.6 \times 10^{-4} \text{N} \vdot \text{s}/\text{m}^2$, and $\gamma = 0.25 \text{N} \vdot \text{s}^2/\text{m}^4$.
+In some cases, we can calculate these coefficients. With $D$ as the diameter of a spherical object, and $\beta$ and $\gamma$ as properties of the medium, we can state that $b = \beta D$ and $c = \gamma D^2$. In air at STP, $\beta = 1.6 \times 10^{-4} \text{N} \cdot \text{s}/\text{m}^2$, and $\gamma = 0.25 \text{N} \cdot \text{s}^2/\text{m}^4$.
 
 Oftentimes, one factor is far more impactful than the other, meaning that the smaller of the two may be neglected. To do so, compute the following ratio:
 
@@ -18,9 +18,9 @@ Note that the result is expected to be of the same order of magnitude as the *Re
 
 ## Section 2.2 - Linear Air Resistance
 
-First, consider the case in which the drag force is negligible. Then, we see that with $\vb{F}_g = \vb{w} = mg$ and $\vb{F}_{drag} = \vb{f} = -b\vb{v}$, then Newton's second law tells us that
+First, consider the case in which the drag force is negligible. Then, we see that with $\mathbf{F}_g = \mathbf{w} = mg$ and $\mathbf{F}_{drag} = \mathbf{f} = -b\mathbf{v}$, then Newton's second law tells us that
 
-$$\vb{F} = m \dot{\vb{v}} = m\vb{g} - b\vb{b}$$
+$$\mathbf{F} = m \dot{\mathbf{v}} = m\mathbf{g} - b\mathbf{b}$$
 
 This separates into two equations:
 
@@ -58,7 +58,7 @@ We can find the horizontal range by setting $y(R) = 0$. This results in an equat
 
 ## Section 2.4 - Quadratic Air Resistance
 
-In the case of quadratic air resistance, we see that $\vb{f} = -c v^2 \vu{v}$. Then, we can see that in the horizontal case, $m\frac{dv}{v_x^2} = -c dt$. Thus,
+In the case of quadratic air resistance, we see that $\mathbf{f} = -c v^2 \hat{\mathbf{v}}$. Then, we can see that in the horizontal case, $m\frac{dv}{v_x^2} = -c dt$. Thus,
 
 $$v_x(t) = \frac{v_{x0}}{1+cv_{x0}t/m} = \frac{v_{x0}}{1+\tau/t}$$