1. The snowflake problem
   1. Animated Falling Snowflake 1
   2. Snow Storm
   3. Show the solution
2. The search light problem
   1. Search light 1
   2. Spot light
   3. Show the solution
3. What do we think actually happens?
   1. Is there are limiting speed?
   2. Does the spot grow or shrink or stay the same? Is there a limiting length?
4. Desmos analysis
   1. Spray of Photons
   2. Spray of Arcs
   3. Spray of Segments
   4. Calculate the point of first contact.
   5. Animated Falling Snowflake 2
   6. Animated Falling Snowflake 3
5. Conclusion

Is the speed of darkness faster than the speed of light?

Dr. William T Webber

Whatcom Community College

email: [email protected]

YouTube: Will@Whatcom

YouTube: PolyhedralPlayground

Materials from this talk are available at

https://www.wamap.org/course/public.php?cid=268

A video version of the talk is available on YouTube

https://youtu.be/ZEjiF-v3Apo

Animated Falling Snowflake 1

Snow Storm

A snowflake is falling straight down at a constant rate of 1 foot per second. It lands 3 feet away from the base of 15 foot tall streetlight. As the flake is falling, the streetlight casts a shadow of the snowflake on the ground. How fast is the shadow moving when the snowflake is a distance h above the ground?

Given: `(dh)/(dt) = -1` 

Question: `(ds)/(dt) = ?`  at height ` = h` 

`h/s=15/(3+s)` 

`h(3+s) = 15s` 

`3h+hs=15s` 

`3h=15s-hs` 

`s = (3h)/(15-h)` 

 `(ds)/(dt) = ((15-h)*3-3h(-1))/(15-h)^2(dh)/(dt)` 

 `(ds)/(dt) = (45-3h+3h)/(15-h)^2*(-1)` 

 `(ds)/(dt) = -45/(15-h)^2` 

Note: When `h = 0,`  we get `(ds)/(dt) = -45/(15^2) = -1/5.` 

Also note: If `h`  is chosen close enough to 15 then `(ds)/(dt)`  can be made arbitrarily large.

So, can the shadow move faster than the speed of light?

Is the speed of darkness faster than the speed of light?

Search light

Spot light

A search light is rotating at a rate of omega radians second. The light shines on a wall at a distance D from the light. Find the speed of the spot of light on the wall as the search light turns.

Given: `(d theta)/(dt) = omega` 

Question: `(ds)/(dt) = ?` 

`tan(theta) = s/D` 

 `sec^2(theta) (d theta)/(dt) = 1/D (ds)/(dt)` 

 `(ds)/(dt)=D sec^2(theta)(d theta)/(dt)` 

 `(ds)/(dt)=D omega sec^2(theta)` 

Note: `(ds)/(dt)`  will go to infinity when `theta`  goes to `pi/2.` 

What really happens?

1. Is there are limiting speed?
2. Does the spot grow or shrink or stay the same? Is there a limiting length?
3. Is there a difference between a cylindrical and conical beam of light?

Let's try this with slow light

1. Spray of Photons
2. Spray of Arcs
3. Spray of Segments

Consider a single photon shot at an angle of `theta.`   The time that it hits the wall is

 `T = theta/omega +(D*sec(theta))/c` 

We what to find the value of theta that minimizes `T.` 

`(dT)/(d theta) = 1/omega+D/c sec(theta)tan(theta)` 

`T`  is minimized when this derivative equals 0.

`0 = 1/omega+D/c sec(theta)tan(theta)` 

Let's use Newton's method.

`f(theta) = 1/omega+D/c sec(theta)tan(theta)` 

`f'(theta) = D/c(sec^3(theta)+sec(theta)tan^2(theta))` 

For this example we will use `c=3,`  `omega =1,`  `D=2.` 

`f(theta) = 1+2/3 sec(theta)tan(theta)` 

`f'(theta) = 2/3(sec^3(theta)+sec(theta)tan^2(theta))` 

Let `theta_0 = -1` 

 `theta_1 = -1-(1+2/3 sec(-1)tan(-1))/(2/3(sec^3(-1)+sec(-1)tan^2(-1))) = -.8942559844` 

`theta_2 = -.831468168` 

`theta_3 = -.8100103879` 

`theta_4 = -.8056896407` 

`theta_5 = -.8050137925` 

`theta_6 = -.8049140391` 

`theta_7 = -.8048994527` 

`theta_8 = -.8048973228` 

`theta_9 = -.8048970118` 

At this angle the point of contact is `2*tan(-.8048970118) =-2.079556741` 

Let's double check in a different way.

The time that is required for light to get to the point on the wall is

`T = theta/omega+sqrt(x^2+D^2)/c` 

`T = tan^(-1)(D/x)/omega+sqrt(x^2+D^2)/c` 

`(dT)/(dx)= 1/(omega(1+(D/x)^2))*(-D/x^2)+x/(c sqrt(x^2+D^2))` 

`(dT)/(dx)= -D/(omega(x^2+D^2))+x/(c sqrt(x^2+D^2))` 

This is minimum when

`0 = -D/(omega(x^2+D^2))+x/(c sqrt(x^2+D^2))` 

We note that the solution must be positive.

`D/(omega(x^2+D^2))=x/(c sqrt(x^2+D^2))` 

`(cD)/omega= x sqrt(x^2+D^2)` 

`(c^2D^2)/omega^2 = x^2(x^2+D^2)` 

`x^4+D^2x^2-(c^2D^2)/omega^2 = 0` 

 `x^2 = (-D^2+sqrt(D^4+(4c^2D^2)/omega^2))/(2)` 

 `x = sqrt(-D^2/2+D/2sqrt(D^2+(4c^2)/omega^2))` 

For our example: `D=2,`  `omega = 1,`  `c=3.` 

 `x = sqrt(-2^2/2+2/2sqrt(2^2+(4*3^2)/1^2)) = 2.07955652` 

1. Animated Falling Snowflake 2
2. Animated Falling Snowflake 3

Is the speed of darkness faster than the speed of light?

Dr. William T Webber

Whatcom Community College

email: [email protected]

YouTube: Will@Whatcom

YouTube: PolyhedralPlayground

Materials from this talk are available at

https://www.wamap.org/course/public.php?cid=268

A video version of the talk is available on YouTube

https://youtu.be/ZEjiF-v3Apo