On Fri, Oct 25, 2024 at 6:30 PM Jesse Mazer <[email protected]> wrote:
> > > On Fri, Oct 25, 2024 at 5:49 PM Alan Grayson <[email protected]> > wrote: > >> >> >> On Friday, October 25, 2024 at 11:34:13 AM UTC-6 Jesse Mazer wrote: >> >> On Fri, Oct 25, 2024 at 5:44 AM Alan Grayson <[email protected]> wrote: >> >> >> >> On Friday, October 25, 2024 at 2:44:06 AM UTC-6 Brent Meeker wrote: >> >> >> >> >> On 10/25/2024 1:36 AM, Alan Grayson wrote: >> >> >> >> On Thursday, October 24, 2024 at 11:07:18 PM UTC-6 Brent Meeker wrote: >> >> >> >> >> On 10/24/2024 5:46 PM, Alan Grayson wrote: >> >> >> >> On Thursday, October 24, 2024 at 1:30:32 PM UTC-6 Brent Meeker wrote: >> >> Here's how a light-clock ticks in when in motion. A light-clock is just >> two perfect mirrors a fixed distance apart with a photon bouncing back an >> forth between them. It's a hypothetical ideal clock for which the effect >> of motion is easily visualized. >> >> >> >> These are the spacetime diagrams of three identical light-clocks moving >> at *+*c relative to the blue one. >> >> >> *Three clocks? Black diagram? If only this was as clear as you claim. >> TY, AG* >> >> >> >> >> >> *You can't handle more than two? The left clock is black with a red >> photon. Is that hard to comprehend? Didn't they teach spacetime diagrams >> at your kindergarten? Brent * >> >> >> *What makes you think you can teach? * >> >> *That I have taught and my students came back for more.* >> >> *I can handle dozens of clocks. I know what a spacetime diagram. It was >> taught in pre-school. Why did you introduce a red photon? A joke perhaps? >> How can a clock move at light speed? * >> >> >> >> >> *None of the clocks in the diagram are moving at light speed. The black >> one and the red one are moving at 0.5c as the label says. What is it you >> don't understand about this diagram? Brent* >> >> >> *One thing among several that I don't understand is how the LT is >> applied. For example, if we transform from one frame to another, say in >> E&M, IIUC we get what the fields will actually be measured by an observer >> in the target or primed frame. (I assume we're transferring from frame S to >> frame S'). But when we use it to establish time dilation say, we don't get >> what's actually measured in the target frame, but rather how it >> appears from the pov of the source or unprimed frame. Presumably, that's >> why you say that after a LT, the internal situation in each transformed >> frame remains unchanged (or something to that effect). AG* >> >> >> Can you give a concrete example? If you some coordinate-based facts in >> frame S (source frame) and use the Lorentz transformation to get to frame >> S' (target frame), the result should be exactly what is measured in the >> target frame S' using their own system of rulers and clocks at rest >> relative to themselves (with their own clocks synchronized by the Einstein >> synchronization convention). >> >> Jesse >> >> >> *Glad you asked that question. Yes, this is what I expect when we use the >> LT. We measure some observable in S, use the LT to calculate its value in >> S', and this what an observer in S' will measure. But notice this, say for >> length contraction. Whereas from the pov of S, a moving rod shrinks as >> calculated and viewed from S, the observer in S' doesn't measure the rod as >> shortened! This is why I claim that the LT sometimes just tells how things >> appear in the source frame S, but not what an observer in S' actually >> measures. AG* >> > > But the length contraction equation isn't one of the Lorentz > transformation equations that map coordinates in S to coordinates in S'. It > is derivable from them, but the intended use of the length contraction > formula is for an observer at rest in S is if they want to predict how long > a rod will be in their own rest frame, given knowledge of its length in the > rod's rest frame S'. > > If you use the actual Lorentz coordinate transformation to derive it, one > way to do it would be to start in frame S' where the rod is at rest (so > position coordinates of its two ends are unchanging), then pick two events > on the worldline of either end which are not simultaneous in S' but are > simultaneous in S, and then use the inverse Lorentz transformation from S' > to S (equations given in the second box at > https://en.wikipedia.org/wiki/Lorentz_transformation#Coordinate_transformation) > to find the position coordinates of each end at a single moment in S (since > "length" in a given frame is understood as the distance between two ends of > an object at a single moment in time in that frame). > > For example suppose S' is moving at v=0.6c relative to S, so the gamma > factor in those equations is 1.25. And suppose in S' the rod's left end is > at rest at position coordinate x'=0 light years, and the right end is at > rest at position coordinate x'=12 light years. > Sorry, that should say that "the right end is at rest at position coordinate x'=10 light years", I changed the numbers as I was typing out this example but forgot to edit that one. > So for the event on the worldline of the left end, pick x'=0 light years, > t'=0 years; and for the event on the worldline of the right end, pick x'=10 > light years, t'=-6 seconds. In this case if you use the inverse Lorentz > transformation to find the coordinates of these events in frame S, the > event at the left end would be x=0 light years, t=0 years (the origins of > the two frames coincide according to the transformation), and the event on > the right end would work out to: > > t = gamma*(t' + v*x') = 1.25 * (-6 + 0.6*10) = 1.25 * (-6 + 6) = 0 > x = gamma*(x' + v*t') = 1.25 * (10 + 0.6*-6) = 1.25 * (6.4) = 8 > > So you get the conclusion that for a rod with rest length 10 light years > in its own frame S', in the S frame at time t=0 the left end is at position > x=0 light years and the right end is at position x=8 light years, so its > length is 8 light years in the S frame, which is exactly what's predicted > by the Lorentz contraction equation. > > > >> >> *On another point concerning time dilation; I demonstrated that given two >> inertial frames with relative velocity v < c, it's easy to synchronize >> clocks in both frames provided we know the distance of clocks from the >> location of juxtaposition, but I was mistaken in concluding this alone >> shows time dilation doesn't exist. * >> > > I don't think you've shown that, or at least you haven't clearly explained > what you mean--you didn't answer my question about your procedure that I > asked right before your back-and-forth exchange with Brent (the message > prior to the one where I asked for a concrete example). > > Jesse > >> -- >> You received this message because you are subscribed to the Google Groups >> "Everything List" group. >> To unsubscribe from this group and stop receiving emails from it, send an >> email to [email protected]. >> To view this discussion visit >> https://groups.google.com/d/msgid/everything-list/d2298d83-1930-4cb4-9a94-52a69cb3dd73n%40googlegroups.com >> <https://groups.google.com/d/msgid/everything-list/d2298d83-1930-4cb4-9a94-52a69cb3dd73n%40googlegroups.com?utm_medium=email&utm_source=footer> >> . >> > -- You received this message because you are subscribed to the Google Groups "Everything List" group. 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