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 Apr 5 answered Why don't we hear sound reflecting from buildings, mirrors, etcetera? Jan 10 comment Do people wearing glasses have different field of view than those who don't? Glasses: yes; contacts: no. Sep 28 revised When I boil a kettle, what stops all the water from turning (exploding!) in to steam in one go once it reaches 100°C? added 1320 characters in body Sep 18 revised When I boil a kettle, what stops all the water from turning (exploding!) in to steam in one go once it reaches 100°C? added 996 characters in body Sep 18 answered When I boil a kettle, what stops all the water from turning (exploding!) in to steam in one go once it reaches 100°C? Jul 9 revised Special relativity: where does this naive calculation go wrong? added 503 characters in body Jul 9 accepted Special relativity: where does this naive calculation go wrong? Jul 9 revised Special relativity: where does this naive calculation go wrong? added 1138 characters in body Jul 9 comment Special relativity: where does this naive calculation go wrong? And, brilliant, the distance 5.196 is precisely two (contracted) ship lengths: $2\times 2.6\ \text m$ which is because the light pulse is half a $c$ faster than the ship. Just like a point moving at 100 km/h needs two car lengths of road to pass 50 km/h car. Jul 9 comment Special relativity: where does this naive calculation go wrong? Okay, this makes sense now. The time of exit of the pulse is at a different location (front of ship versus back), and so it appears further delayed, and the time blows up to 17.3 nanoseconds, not 11.5. Jul 9 comment Special relativity: where does this naive calculation go wrong? Ah, I think this must be due to signs being treated in other than the expected way in the coordinate transform. This result matches what we would expect if the light pulse were entering from the front and leaving from the back, traversing less than 2.6 meters. Jul 9 answered Why is displacement negative during free fall? Jul 9 comment Special relativity: where does this naive calculation go wrong? The thing is, how can the distance traveled by the light be seen as shorter than the contracted length of the ship's cabin of 2.60 meters. That's one of the counter-intuitive facts here. A pulse of light enters the back of the ship, which is 2.6 meters long and moving forward at $c/2$; how can that light leave the ship's front only 1.73 meters down the path? Jul 9 comment Special relativity: where does this naive calculation go wrong? I don't understand the results intuitively. $\Delta t'$ works out to be only 5.77 nanoseconds. And $\Delta x'$ works out to 1.73 m, the familiar number from earlier calculations which now has a different interpretation. So in fact, the time between the beam entering/leaving events is faster in the standing reference frame and it so happens that the total distance traveled by the beam of light is 1.73 m. Jul 9 comment Special relativity: where does this naive calculation go wrong? ... so, I think it's the $-v\Delta x/c^2$ term in the time dilation that is at the root of this: we must account for the change in position when judging the time dilation. If that is the case, I find my intuition weakening here. But wait, $\Delta x$ is in the $S$ frame, I see? So in other words, aha, because the two events are at different positions in the $S$ frame (rear of ship versus front), we have to take that into account. The simple form of the time dilation only applies to one place, like a single clock in the spaceship. Jul 9 comment Special relativity: where does this naive calculation go wrong? Thanks. I, by the way, had tried the $v\Delta t$ term in the length calculation, suspecting that the error may be that the two events are sampled at different times. In my example, all that did was produce the 2.6 meter condensed length: $\Delta x$ is the 4.33 m, and $v\Delta t$ is the 1.73 m movement. I.e. the more general form with the $v\Delta t$ term simply lets us confirm that the ship is 2.6 meters long even though we sample the position of its front and rear at different times: the beam crossings separated by 11.54 ns. Jul 9 revised How do we know that $F = ma$, not $F = k \cdot ma$ Grammar. Jul 9 asked Special relativity: where does this naive calculation go wrong? May 14 awarded Yearling Mar 18 awarded Good Answer