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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 $\dot{\mathbf{P}} = \sum_\alpha \mathbf{F}_\alpha^{ext}$.
 
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