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Photon clock docs (#40)
* checkpoint * checkpoint * docs: photon clock * docs: photon clock demo
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docs/devlog.md

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@@ -71,3 +71,8 @@ visualizations described
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## 2024-12-30
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More relativity docs.
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## 2024-01-01
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More time making an animation to show time dilation with the light clock
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example.

docs/relativity/01-special.md

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@@ -128,6 +128,9 @@ takes for the photon to make one full round trip.
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- The astronomer observes the photon moving at exactly the speed of light, $c$.
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- The astronaut and astronomer disagree on the time it takes for the photon to
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make one full round trip in the light clock.
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- The astronaut observes the photon completing one round trip in
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$\Delta \tau$.
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- The astronomer observes the photon completing one round trip in $\Delta t$.
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**Problem**: Whose round-trip duration measurement is correct?
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- The light clock is at rest so the photon travels $2D$ in the time it takes to
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complete one round trip.
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- The time it takes for the photon to make one full round trip is
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$\Delta t = \frac{2D}{c}$.
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$\Delta \tau = \frac{2D}{c}$.
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In the astronomer's inertial frame:
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- The photon moves at speed $c$.
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- The light clock is moving so the photon travels a longer distance in the time
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it takes to make one full round trip.
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- The light clock is moving relative to the astronomer so the photon travels a
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longer distance in the time it takes to make one full round trip in the clock.
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- The photon travels sideways a distance $v \Delta t$ in addition to the
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distance $2D$ it travels up and down between the mirrors.
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- The total distance the photon travels can be found by pythagorean theorem.
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$$
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s = \sqrt{(2D)^2 + (v \Delta t)^2}
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$$
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- The time it takes for the photon to make one full round trip is
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$\Delta t = \frac{2 \sqrt{D^2 + (v \Delta t)^2}}{c}$.
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```
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We see here that both of Einstein's postulates are satisfied, and we have not
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```
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### Exploring time dilation
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In the astronomer's reference frame, the photon moves across more space over the
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course of one "tick" of the clock. The only way the speed of the photon, $c$,
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can be the same in all reference frames and yet travel across more distance in
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one reference frame than in another is if the elapsed time experienced by a
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moving object, $\Delta \tau$, "dilates" compared to the elapsed time
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experienced by an observer at rest in the reference frame, $\Delta t$.
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![time dilation](./assets/TimeDilationDemo.gif)
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docs/relativity/image.png

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