On Fri, Oct 25, 2024 at 11:35 PM Alan Grayson <[email protected]> wrote:
> > > On Friday, October 25, 2024 at 9:08:14 PM UTC-6 Jesse Mazer wrote: > > On Fri, Oct 25, 2024 at 10:00 PM Alan Grayson <[email protected]> wrote: > > > > On Friday, October 25, 2024 at 4:30:54 PM UTC-6 Jesse Mazer 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. 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 > > > *Well, one could set the two juxtaposed clocks to both read t=0, and then > set another clock, in either frame, at some known measured distance > removed, and send a signal, say sound waves, of known velocity, and then > set the time on this clock, advanced from the juxtaposed clock, depending > on how long it takes to signal to reach its destination. And one could do > this repeatedly for other clocks in either frame. AG* > > > OK, suppose you have two clocks in relative motion A1 and B1, and at the > moment B1 passes next to A1 (i.e. is spatially juxtaposed with it), they > both read t=0. Then you have another clock A2 at rest relative to A1, and > you used the Einstein synchronization convention in the A frame to set A2 > based on the reading on A1, assuming light moves at constant speed in both > directions in the A frame. You also have another clock B2 at rest relative > to B1, and you used the Einstein synchronization convention in the B frame > to set B2 based on the reading on B1, assuming light moves at constant > speed in both directions in the B frame. In this case, if the distances are > such that B2 is passing next to A2 when A2 reads t=0, then B2 will *not* > read t=0 at that moment, because using the Einstein synchronization > convention for different sets of clocks in different frames results in > different definitions of simultaneity for each frame. > > > *I don't follow. A2 never reads t=0. It starts at some t > 0, accounting > for the elapsed time on A1 when the signal reaches A2. And B2 is similarly > synchronized and never reads t=0. AG * > I was assuming A2 was synchronized with A1 at some moment before either read t=0, for example A could have transmitted the signal when it read t=-20. The details aren't important but my point was about what would happen assuming the synchronization for each pair of clocks (A1/A2 as well as B1/B2) happened prior to the moment when B1 passed A1, and prior to the moment the B2 passed A2. You could equally well assume the synchronization signal was sent at t=0 but the crossing of paths of the A's and the B's didn't happen until a later time, with the synchronization of each planned in such a way that A1 and B2 would read exactly the same time (say, t=10) at the moment they crossed paths. 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/CAPCWU3%2BRaeCgsps80466Ufh32pQLvZduyZyqHmUW%3DtX_d4XM6w%40mail.gmail.com.
