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. Of course, if you preferred you could just use the Einstein synchronization convention only for the two A clocks, and set both B clocks to read 0 at the moment they passed an A clock. But this would be an arbitrary choice to favor the A frame's definition of simultaneity--you could just as easily have done the opposite and used the Einstein synchronization convention only for the two B clocks, and set both A clocks to read 0 at the moment they passed a B clock. So you haven't given a procedure that gives a unique definition of simultaneity. 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/c56ac48c-cf6b-4d95-b485-e6968e68b3dbn%40googlegroups.com > <https://groups.google.com/d/msgid/everything-list/c56ac48c-cf6b-4d95-b485-e6968e68b3dbn%40googlegroups.com?utm_medium=email&utm_source=footer> > . > -- 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/CAPCWU3J7YEpWo9xMPvRsnjpZXK3DNKvozPMBXg3VHSa02g6%3DEg%40mail.gmail.com.
