First, here's the link to my first post, talking about time vs speed of light. THE Horizon.

Next, here are some neat facts.

Here's the spreadsheet of cool facts from Brian Cox's lecture right before they found Higgs. That lecture had in it the following. First, everything's either the standard model or relativity and we can't tie them together yet. Then a divergence into the universe, introducing the Hubble constant, which is 1/13.4billion yrs, so that's roughly when everything was simultaneously here, i.e. the age of the universe. It's derived from known brightness of some supernovae. (distance) and red shifts (velocity) of everything we see. Assuming the redshift of the CMB is on that line, it's 13.4 billion years old. BTW, CMB is uwave frequency now, was literally red at first. After H0, Cox talked about particles. up,down, neutrino, electron being all you need. Two more columns of heavier versions, then photon & other force carriers. Nothing for gravity. Higgs field posited, the particle being very heavy & thus requiring high energy to make. After the switch to gravity & Einstein, he noted the light clock & Lorentz contraction as a consequence.

Well, that's just a lot of notes from the show. Then I started reading about relativity a bit, simultaneity & constancy of c. The first question is, what about a light-speed game of ping-pong, where the table is aligned with the velocity? The metaphor's imperfect (air hockey would be better) because there's no hypotenuse here. Instead just imagine the balls going straight back and forth. In the light clock meanwhile, photons go up and down between mirrors, perpendicular to v,"tick, tock." Onboard a speeding train, let's say the clock (tic,toc) and the game (ping pong) are perfectly synchronized, one second travel each way for both the pendulum and pingpong ball. That's my view aboard the train. What about to you, standing by the tracks?

I've got 3 things to discuss and quite difficult without pictures but here goes...

First, the standard light clock explanation. To you, the clock photons travel a diagonal, the hypotenuse of a triangle. (For ease of computation, say v =c/2 and a light-second's worth of distance is d, henceforward.) That lets you calculate, based on the root(5)* distance traveled after a tick and a tock, "the clocks on that train must be running slow if they think that's a second!" From the ratio of sqrt(5) to 2 you calculate the ratio of time slowing aboard the train.

{*This last is wrong, t' = t/sqrt(1-v^2/c^2), says many sources. I didn't immediately see my error, but it calculates to 15%, not 12% so I'll use that below. Update: wikipedia clarifies the answer, which is that the base leg is shortened because it's t', the quicker-ticking observer's clock from which frame we make this measurement, while the beat that determines the distance traveled is at the slower moving clock rate. Anyway, it complicates the algebra slightly.}

Second, the train has to shrink. That's because the pingpong and the clock are synchronized. The forward travelling pingpong ball also takes 1 sec (onboard time) and being synchronized, 1.15sec, observer time to cross the table one way, from "ping" to "pong" so to speak. Unlike the pendulum, those ping-pong photons will not travel the whole hypotenuse so to stay synchronized with the "tic-toc" of the clock the vehicle must stretch or shrink along the velocity direction so that the pingpong ball hits the paddle just at the "tock." With v = c/2 the numbers work out nicely. In a second's time, the ball travels the distance of the table, but the train has meanwhile moved half a second ahead. It will take two seconds for a photon to hit the second paddle, one to cross the table, and one more to catch up to the train. Only, it will actually be 2x 1.15sec since we're working in observer time where I've already noted the train clock is running ~15% slow. Whoops, a paradox! ...it

Now my third observation stumped me: it seems sure the return of the ping-pong ball back across the table will happen in an instant, since the train, and the "ping" paddle aboard it, is rushing forward to meet it. Remember the essence of this experiment is that the balls (being photons) travel at c w.r.t. all observers. Now the observer sees relative velocity between ball and paddle of 3c/2, and distance ~d/2, so elapsed t will be ~d/3c, or just 1/3 sec! (approximation 'cause I'm temporarily leaving out the 15% time dilation for easier calculating) How's

In the second it takes from ping to pong, the train moves half a light second (d/2) down the tracks and so it will take an extra half second for the "pong" to reach us. Along the way, the second "ping" is added because after all the report of pong is a photon and the ball is just as fast, so "ppionngg" will arrive all at once. Since I know the train's moving I expect each successive second's data to arrive an extra half second late. All together I hear ppionngg every 2.3 seconds, one sec for travel time of the ball, + one second's further train "entfernung" (distancing itself from me); so I guess it does all work out.

Next, here are some neat facts.

Here's the spreadsheet of cool facts from Brian Cox's lecture right before they found Higgs. That lecture had in it the following. First, everything's either the standard model or relativity and we can't tie them together yet. Then a divergence into the universe, introducing the Hubble constant, which is 1/13.4billion yrs, so that's roughly when everything was simultaneously here, i.e. the age of the universe. It's derived from known brightness of some supernovae. (distance) and red shifts (velocity) of everything we see. Assuming the redshift of the CMB is on that line, it's 13.4 billion years old. BTW, CMB is uwave frequency now, was literally red at first. After H0, Cox talked about particles. up,down, neutrino, electron being all you need. Two more columns of heavier versions, then photon & other force carriers. Nothing for gravity. Higgs field posited, the particle being very heavy & thus requiring high energy to make. After the switch to gravity & Einstein, he noted the light clock & Lorentz contraction as a consequence.

Well, that's just a lot of notes from the show. Then I started reading about relativity a bit, simultaneity & constancy of c. The first question is, what about a light-speed game of ping-pong, where the table is aligned with the velocity? The metaphor's imperfect (air hockey would be better) because there's no hypotenuse here. Instead just imagine the balls going straight back and forth. In the light clock meanwhile, photons go up and down between mirrors, perpendicular to v,"tick, tock." Onboard a speeding train, let's say the clock (tic,toc) and the game (ping pong) are perfectly synchronized, one second travel each way for both the pendulum and pingpong ball. That's my view aboard the train. What about to you, standing by the tracks?

I've got 3 things to discuss and quite difficult without pictures but here goes...

First, the standard light clock explanation. To you, the clock photons travel a diagonal, the hypotenuse of a triangle. (For ease of computation, say v =c/2 and a light-second's worth of distance is d, henceforward.) That lets you calculate, based on the root(5)* distance traveled after a tick and a tock, "the clocks on that train must be running slow if they think that's a second!" From the ratio of sqrt(5) to 2 you calculate the ratio of time slowing aboard the train.

{*This last is wrong, t' = t/sqrt(1-v^2/c^2), says many sources. I didn't immediately see my error, but it calculates to 15%, not 12% so I'll use that below. Update: wikipedia clarifies the answer, which is that the base leg is shortened because it's t', the quicker-ticking observer's clock from which frame we make this measurement, while the beat that determines the distance traveled is at the slower moving clock rate. Anyway, it complicates the algebra slightly.}

Second, the train has to shrink. That's because the pingpong and the clock are synchronized. The forward travelling pingpong ball also takes 1 sec (onboard time) and being synchronized, 1.15sec, observer time to cross the table one way, from "ping" to "pong" so to speak. Unlike the pendulum, those ping-pong photons will not travel the whole hypotenuse so to stay synchronized with the "tic-toc" of the clock the vehicle must stretch or shrink along the velocity direction so that the pingpong ball hits the paddle just at the "tock." With v = c/2 the numbers work out nicely. In a second's time, the ball travels the distance of the table, but the train has meanwhile moved half a second ahead. It will take two seconds for a photon to hit the second paddle, one to cross the table, and one more to catch up to the train. Only, it will actually be 2x 1.15sec since we're working in observer time where I've already noted the train clock is running ~15% slow. Whoops, a paradox! ...it

**take that long because it's got to stay synchronized with the clock! It's got to get there in just one (dilated) second. This is why Einstein (or Lorentz) said the train has to shrink. From the observer's perspective, we calculate the unknown distance from known time and speeds, yielding that the length of the squished table d(squished) = (c-v)*1.118 which tells us the train has to shrink to just 18% more than***can't***it's "real" (at-rest) length. Ok, fine. Mind bending, but I get it.***half*Now my third observation stumped me: it seems sure the return of the ping-pong ball back across the table will happen in an instant, since the train, and the "ping" paddle aboard it, is rushing forward to meet it. Remember the essence of this experiment is that the balls (being photons) travel at c w.r.t. all observers. Now the observer sees relative velocity between ball and paddle of 3c/2, and distance ~d/2, so elapsed t will be ~d/3c, or just 1/3 sec! (approximation 'cause I'm temporarily leaving out the 15% time dilation for easier calculating) How's

*that*gonna synchronize with the tick-tock? Last night talking with Miles I convinced myself it was a consequence of the time when I make the observations, which time is itself obviously subject to lightspeed delays. Now, I'm just confused again. However, check this out:In the second it takes from ping to pong, the train moves half a light second (d/2) down the tracks and so it will take an extra half second for the "pong" to reach us. Along the way, the second "ping" is added because after all the report of pong is a photon and the ball is just as fast, so "ppionngg" will arrive all at once. Since I know the train's moving I expect each successive second's data to arrive an extra half second late. All together I hear ppionngg every 2.3 seconds, one sec for travel time of the ball, + one second's further train "entfernung" (distancing itself from me); so I guess it does all work out.

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