docs: lorentz transform

docs/SUMMARY.md

@@ -6,7 +6,7 @@
 
 - [Relativity](relativity/00-index.md)
   - [Special Relativity](relativity/special/00-index.md)
-    - [Light Speed & Reference Frames](relativity/special/01-reference-frames.md)
+    - [The Basics](relativity/special/01-reference-frames.md)
     - [Time Dilation](relativity/special/02-time-dilation.md)
     - [Length Contraction](relativity/special/03-length-contraction.md)
     - [Coordinate Transformations](relativity/special/04-transformations.md)

docs/relativity/general/00-index.md

@@ -10,24 +10,6 @@ further and leading to the emergence of gravity and the idea that space and time
 are not independent axes, but rather our reference frames are composed of a
 continuous _spacetime_.
 
-## New equations of motion
-
-The traditional equations of motion specify the position, velocity, and
-acceleration of an object with respect to time.
-
-$$
-\begin{align*}
-\text{position\quad} & \vec{s}(t) = \vec{r}_0 + \vec{v}_0t + \frac{1}{2}\vec{a}t^2 \\[0.5em]
-\text{velocity\quad} & \dot{\vec{s}}(t) = \vec{v}_0 + \vec{a}t \\[0.5em]
-\text{acceleration\quad} & \ddot{\vec{s}}(t) = \vec{a} \\[1em]
-\end{align*}
-\begin{gathered}
-\text{where\quad} & \vec{s} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \;
-\vec{v}_0 = \begin{bmatrix} v_{x0} \\ v_{y0} \\ v_{z0} \end{bmatrix}, \;
-\vec{a} = \begin{bmatrix} a_x \\ a_y \\ a_z \end{bmatrix}
-\end{gathered}
-$$
-
 ## Curved spacetime
 
 ```admonish warning

docs/relativity/special/01-reference-frames.md

@@ -50,8 +50,8 @@ This statement is
 seems, but that's a tale for another time.
 
 - $c$ is defined as the distance traveled by a photon in 1 second.
-- $c$ is _not a measurement_, it is a _definition_. In actuality, the definition
-  of $1 \text{ meter}$ is defined as the distance light travels in
+- $c$ is _not a measurement_, it is a _scientific constant_. In fact, the
+  definition of the meter is derived from the distance that light travels in
   $\frac{1}{299792458}$ seconds.
 - Science has accepted that this value, $299792458 \text{ m/s}$, is what we have
   decided to call the speed of light.

docs/relativity/special/02-time-dilation.md

@@ -43,7 +43,7 @@ takes for the photon to make one full round trip.
 
 **Problem**: Whose round-trip duration measurement is correct?
 
-![openstax example](https://openstax.org/books/university-physics-volume-3/pages/5-3-time-dilation)
+![openstax example](https://openstax.org/apps/archive/20241024.164013/resources/63398977f32f4f656b1c8973d4a7a68038c5155d)
 
 **Solution**: The astronaut and astronomer are both correct! It's not a trick
 question, this time---we can solve it with 8th-grade math and disciplined logic.
@@ -90,3 +90,10 @@ moving object, $\Delta \tau$, "dilates" compared to the elapsed time
 experienced by an observer at rest in the reference frame, $\Delta t$.
 
 ![time dilation](./assets/TimeDilationDemo.gif)
+
+Therefore, the time dilation equation is:
+
+$$
+t = \gamma t' \quad \text{where} \quad \gamma = 1/\sqrt{1
+-\Big(\frac{v}{c}\Big)^2} \\
+$$

docs/relativity/special/03-length-contraction.md

@@ -1,21 +1,35 @@
 # Special Relativity
 
-```admonish help
-In this chapter I will be using "prime" notation to distinguish elements in the
-two frames. The frame at rest has no primes, and the moving frame has primes next
-to the variables.
-```
-
 ## Length Contraction
 
-From [the previous chapter](./02-time-dilation.md) we know that the dilated
-time $t'$ is related to the proper time $t$ by the Lorentz factor $\gamma$.
+The relative speed between two inertial reference frames, $v$, is the same in
+both frames. Think about the astronaut and astronomer in the light clock example
+from [the previous chapter](./02-time-dilation.md).
+
+- The astronaut observes the astronomer moving at a constant speed, $v$.
+- The astronomer observes the astronaut moving at a constant speed, $v$.
+- The astronomer observes the elapsed time on the astronaut's clock, $t'$, to be
+  dilated by the Lorentz factor, $\gamma$.
+
+![length contraction](https://openstax.org/apps/archive/20241024.164013/resources/ca460689ad0a67fa34fbb0d7e07b49143a7800ea)
+
+Remember the definition of velocity, $v = x / t$. Since we know that the time
+interval $t$ is dilated by the Lorentz factor, $\gamma$, then the distance
+traveled by the astronaut, $x'$, must also be stretched by $\gamma$ to keep the
+velocity constant.
+
 
 $$
-t' = \gamma t \quad \text{where} \quad \gamma = 1/\sqrt{1
--\Big(\frac{v}{c}\Big)^2} \\
+v = x/t, \quad v = x'/t', \quad t = \gamma t' \\
+v = \frac{x}{t} = \frac{x'}{t'}, \quad \gamma = t / t' \\
+$$
+$$
+x' = \frac{xt'}{t} = \frac{x}{\gamma}
+$$
+$$
+x' = x \sqrt{1 - \Big(\frac{v}{c}\Big)^2}
 $$
 
-Notice how the relationship between $t$ and $t'$ depends on velocity, $v$. The
-definition of velocity is the displacement divided by the time, $v = x / t$, so
-what if we wanted to solve for the displacement $x$ in terms of $x'$?
+For any velocity $v$ less than the speed of light, $c$, the length contraction
+factor is less than 1, meaning the length of the object in the moving frame is
+shorter than the length of the object in the stationary frame.

docs/relativity/special/04-transformations.md

@@ -1,10 +1,5 @@
 # Special relativity
 
-In this section we'll see how relativity affects equations of motion for events
-measured between inertial reference frames. The majority of this work is based
-on the Relativity chapter in
-[OpenStax University Physics Volume 3](https://openstax.org/books/university-physics-volume-3/pages/5-5-the-lorentz-transformation).
-
 ## Coordinate transformations between inertial reference frames
 
 An _event_, $s$, is a location and time coordinate relative to an inertial
@@ -25,8 +20,6 @@ $$
 But what if we want to define the same event relative to a moving reference
 frame, $S'$?
 
-![moving reference frame](https://openstax.org/apps/archive/20241024.164013/resources/a2b3997dff7a717555902ca3279bc1723837f76d)
-
 ### Galilean Transformations
 
 Suppose $S'$ is moving with respect to $S$ at velocity $\vec{v}$.
@@ -91,20 +84,49 @@ set of equations that describe how to transform coordinates between two
 inertial reference frames that are in relative motion and are consistent with
 the speed of light postulate.
 
-![moving reference frame](https://openstax.org/apps/archive/20241024.164013/resources/a2b3997dff7a717555902ca3279bc1723837f76d)
-
+```admonish example
 Recall our [previous example](02-time-dilation.md) with the astronaut and the
 astronomer. Let's say the astronomer's frame of reference is $S$ and the
 astronaut's frame of reference is $S'$, moving at velocity $v$ relative to $S$
 and the $x$ axis is the direction of motion.
 
-The astronomer observes the origin of $S'$ at time $t$. The astronomer also
-observes the displacement of a photon in the astronaut's light clock from the
-origin of $S'$ to be $x'\sqrt{1 - \frac{v^2}{c^2}}$.
+The astronomer observes the origin of $S'$ at time $t$ to have a displacement
+$x$. The astronomer also observes the displacement of a photon in the
+astronaut's light clock from the origin of $S'$ as $x'$.
+
+![moving reference frame](https://openstax.org/apps/archive/20241024.164013/resources/a2b3997dff7a717555902ca3279bc1723837f76d)
+
+The origin of $S'$ is moving at velocity $v$ relative to $S$. An event occurs at
+coordinate $(x', 0, 0, t')$ in $S'$ and at coordinate $(x, 0, 0, t)$ in $S$.
+
+- The displacement of the origin of $S'$ is $vt$.
+- The displacement of the event in $S'$ is $x'$.
+- THe displacement of the event in $S$ is the displacement of $S'$ plus the
+  displacement $x'$ after accounting for relativity.
+
+**Problem:** What is the displacement of the event in $S$?
+
+**Solution:**
+In previous chapters we derived the [time dilation](./02-time-dilation.md) and
+[length contraction](./03-length-contraction.md) equations that relate the time
+and distance between two inertial reference frames. Let's apply them here to
+transform between $S$ and $S'$.
 
 $$
-\begin{align*}
-x' &= \gamma (x - vt)\\
-t' &= \gamma (t - \frac{vx}{c^2})
-\end{align*}
+t = \gamma t', \quad x = \frac{x'}{\gamma} \quad \text{where} \quad \gamma = 1/\sqrt{1
+-\Big(\frac{v}{c}\Big)^2} \\
+$$
+therefore
 $$
+x = vt + x' \sqrt{1 - \Big(\frac{v}{c}\Big)^2}, \quad \text{and} \quad x' =
+\frac{x - vt}{\sqrt{1 - \Big(\frac{v}{c}\Big)^2}}
+$$
+
+That was a lot of work, and it was only for the $x$ coordinate!
+```
+
+Shorthand for this operation is to call it a _Lorentz boost_. The inverse
+Lorentz boost is the same thing but with the velocity reversed.
+([source](https://en.wikipedia.org/wiki/Lorentz_transformation)).
+
+## Space-time interval