On Thu, Feb 6, 2025 at 4:48 AM Alan Grayson <[email protected]> wrote:
> > > On Monday, February 3, 2025 at 6:44:29 AM UTC-7 Jesse Mazer wrote: > > On Mon, Feb 3, 2025 at 3:16 AM Alan Grayson <[email protected]> wrote: > > > > On Sunday, February 2, 2025 at 11:37:48 PM UTC-7 Jesse Mazer wrote: > > On Mon, Feb 3, 2025 at 12:09 AM Alan Grayson <[email protected]> wrote: > > > > On Sunday, February 2, 2025 at 9:39:04 PM UTC-7 Jesse Mazer wrote: > > On Sun, Feb 2, 2025 at 11:04 PM Alan Grayson <[email protected]> wrote: > > *In mathematics, functions have domains and ranges, standard terminology. > A function maps domain sets to range sets. The image of a function is the > set containing its range. Again, standard mathematical terminology. The > contraction formula, derived for the LT, is a function. With me so far? AG* > > > In math the domain and range of a function have no physical > interpretation, they are just sets of mathematical objects with no comment > about what they "mean". For example, in purely mathematical terms, the > length contraction formula l = L*sqrt(1 - v^2/c^2) has as its domain values > of the variables L and v where L can be any member of the set of real > numbers > 0, and v is any member of the set of real numbers larger than or > equal to 0 but smaller than the value of c (in whatever units you're using > like 299792458 meter/second), and the range is the value of l which can > likewise be any member of the set of real numbers > 0. > > If you are speaking more metaphorically, basically just saying that any > equation like length contraction can be thought of as a sort of machine > that takes two values (a proper length L and a speed v) as input/"domain" > and spits out another value (a length l in whatever frame measured the > object to be moving at v) as output/"range", then I'm fine with that. > > > *Now about the substance. If coordinate frame O2 is the domain of the > formula function, the x values are elemments in its domain, and the x' > values are elements in its range.* > > > If you are talking about the LT function when applied to a problem where > we start with coordinates in O2 as input and get coordinates in O1 as > output, sure. If you're talking about the length contraction formula, no, > the only inputs to that are a proper length and a velocity in a single > frame, and the output is the length in that same frame. > > > *Here's where we disagree. ISTM, there's a convention, that the image > frame of the formula, is moving wrt the frame applying the contraction > formula,* > > > The length contraction formula is not exactly translating between > different frames at all, at least not in the sense that the input > exclusively consists of variables from one frame and the output a variable > from a different frame, the way the LT does. In the length contraction > formula, both the speed v which is used as input and the contracted length > L' which is given as output are measured in the SAME frame, the frame of > the observer who sees the object moving at speed v relative to their own > frame. The proper length L, which is also used as input, can be thought of > as length in a different frame, namely the object's own rest frame. > > Since there is a symmetry of motion where the speed of B relative to A is > the same as speed A relative to B, I suppose the v in the length > contraction formula *could* instead be defined as the speed of the observer > as measured in the object's rest frame, rather than defining it as the > speed of the object in the observer's frame as textbooks normally do. If > you think of it in that alternate way where v is the speed of the observer > in the object's frame, then you could consider both input variables L and v > to be measured in one frame (the object's rest frame) and the output > variable L' in another (the observer's frame). Even though v is not > normally described this way, it'd make no difference mathematically if you > did. But if we do think of it this way, I presume you'd then say the > object's rest frame is the frame "applying the contraction formula", and > the observer's frame is the "image frame"? > > > > * and L' is the contracted length in the frame in relative motion. If you > claim the contraction occurs in the same frame from which the formula is > applied, then won't we get no contraction? AG * > > > See above, the normal way of describing the length contraction formula > involves a mix of measurements in two different frames as input so there is > no single 'frame from which the formula is applied', but if you want, you > do have the option to think of the meaning of v in a different way such > that both L and v are defined in terms of measurements in a single input > frame (the rest frame of the object), in which case the contracted length > L' would be in a different output frame (the rest frame of the observer). > > > > > Also note that there is nothing in the LT formula that restricts which > frame you take as input and which you get as output. You can just as easily > start with the coordinates in O1 as input and use the formula to find the > coordinates in O2 as output. I just said this in the post above and even > offered to give you a numerical example of how this would work. > > > *Of course, but why do that? AG * > > > Because you asked how the LT could be used to predict a contracted > length--in order to do that, the output of the LT formula has to be defined > in a frame where the object has non-zero velocity. > > > > > * The contraction formula is a mapping or correspondence from coordinate > sets O2 to coordinate sets O1, the moving frame in relative motion wrt O2.* > > > The contraction formula doesn't do that, the LT *can* be used to do that > if you start with coordinates in O2 as input and then get coordinates in O1 > as output, but as I say above it can just as easily map coordinates in O1 > taken as input to coordinates in O2 given as output. > > > * That is, the contraction formula maps O2, the frame with no rod, to O1, > the frame with the rod.* > > > I asked you several times in the last comment (and in a number before > that) to clarify if by "no rod" you just mean the rod is not *at rest* in > O2, or if you mean that O2 literally doesn't "see" the rod to assign it > coordinates, or something else. > > > *There is one rod is at rest in O1, and its frame in moving relatively wrt > O2. AG* > > > But you agree that the observer O2 can measure and assign coordinates to > the rod in O2's own rest frame? > > > > > * You say, and I now agree, that there's no contraction of the one and > only rod in O1.* > > > Yes. > > > * So what happened to contraction?* > > > The rod is contracted in other frames like O2, and as I said you're free > to use the LT to start with the coordinates in O1 used as input and then > use the LT formula to get the coordinates in O2 as output (once you know > the coordinates of the front and back of the rod in O2 it's easy to get the > length of the rod in O2 from that). And if you use the length contraction > formula rather than the LT it's an even simpler matter to derive the rod's > contraction in O2, you just need the rod's proper length L as well as its > velocity v in O2, you enter that L and v into the length contraction > formula as input and get the contracted length l in O2 as output. > > > *When we map from O2 to O1, you agreed, no contraction, so if we map from > O1 to O2, won't there also be no contraction?* > > > No, since the rod is at rest in O1 and moving in O2, if you map from O1 to > O2 using the LT you get a contracted length for the output. As I said I > could give you a numerical example showing this if you want. > > > *If rod is at rest in 01 and in relative motion wrt O2 (because O1 is the > moving frame) and we map from 02 to 01 we get no contraction, but if we map > in opposite direction, from O1 to O2, we get contraction? Could you explain > how you reach these conclusions? AG * > > > The quickest way to get this conclusion is just to know that the > predictions of the LT about length will always match those of the length > contraction equation which was derived from it, and if the rod has some > known proper length like L=10 and some nonzero velocity like v = 0.6c > relative to O2, then in O2 frame its length is 10*sqrt(1 - 0.6^2) = 8, but > since it has a velocity of v = 0 relative to O1, then in the O1 frame it > must have length 10*sqrt(1 - 0^2) = 10. The LT will agree with this so if > you use the equations with O2 as input and O1 as output, you'll have a > length of 8 as input and a length of 10 as output (no contraction in output > frame). But if you use the LT equations with O1 as input and O2 as output, > you'll have a length of 10 as input and a length of 8 as output > (contraction in the output frame). > > If you don't want to just trust the principle of "the LT agrees with the > length contraction equation", you can verify this by direct calculation > using the LT equations. Let's call O2 the unprimed frame and O1 the primed > frame. In the O2 frame we have the following equations for the position as > a function of time of each end of the rod (i.e. the worldline of each end): > > Back of the rod: x = 0.6c*t > Front of the rod: x = 8 + 0.6c*t > > These equations tell you that at t=0 the back of the rod is at x=0 and the > front of the rod is at x=8, so the rod has a length of 8 in the O2 frame, > and it's moving at 0.6c. > > Then in the primed O1 frame we have the following equations: > > Back of the rod: x' = 0 > Front of the rod: x' = 10 > > These equations tell you that at any choice of time coordinate t' the back > is always located at x'=0 and the front is always located at x'=10, i.e. > the rod is at rest in this frame and has a length of 10 (its proper length). > > One way to use the LT is to directly plug the full equations of motion in > one frame into the LT equations as input, and after a little algebra get > out the equations of motion in the other frame as output. That's what I did > in that post at > https://groups.google.com/g/everything-list/c/ykkIYDL3mTg/m/giZVF9PpDQAJ > going from the equations of motion in the frame where the rod was moving > (here the O2 frame) to the equations of motion in the frame where the rod > was at rest (O1). This does involve a little algebra though. A simpler way > of just checking that the LT map between those equations of motion is just > to pick the coordinates of some individual points that are along a given > end's worldline in the coordinates of one frame, and see that they always > map to coordinates of individual points that are along the same end's > worldline in the coordinates of the other frame. > > For example in the O2 frame, the points (x=0, t=0) and (x=3, t=5) and > (x=6, t=10) and (x=9, t=15) would all lie along the worldline of the BACK > of the rod given by x = 0.6c*t in this frame (I'm assuming we're using > units like light-seconds and seconds where c=1). If you take any of those > points as input and use the x-->x' LT to find the corresponding coordinates > of the point in the primed O1 frame as output, you will get (x'=0, t'=0) > and (x'=0, t'=4) and (x'=0, t'=8) and (x'=0, t'=12) [note that the x-->x' > equation in this case is x' = 1.25*(x - 0.6c*t) and the t-->t' equation is > t' = 1.25*(t - 0.6*x)]. You can see that all of these points do lie along > the line x'=0, the equation for the BACK of the rod in the O1 frame. > > Similarly, in the O2 frame, the points (x=8, t=0) and (x=11, t=5) and > (x=14, t=10) and (x=17, t=15) all like along the worldline of the FRONT of > the rod given by x = 8 + 0.6c*t in this frame. If you take any of those > points as input and use the x-->x' LT to find the corresponding coordinates > in the primed O1 frame as output, you will get (x'=10, t'=-6) and (x'=10, > t'=-2) and (x'=10, t'=2) and (x'=10, t'=6). You can see all these points > lie along the line x'=10, the equation for the FRONT of the rod in the O1 > frame. > > Then if you want to go in reverse, starting with O1 coordinates as input > and getting O2 coordinates as output, you can use the LT equations for > x'-->x which in this example is x = 1.25*(x' + 0.6c*t'), and the one for > t'-->t which is t = 1.25*(t' + 0.6*x'). You can verify that the reverse > mapping here works too, for example if you take (x'=10, t'=2) as input (a > point I listed above as being on the worldline of the front of the rod) you > get back (x=14, t=10) as output. So you can verify this way that if you are > given the equations x' = 0 and x' = 10 for the rod in the O1 frame > (corresponding to a rod at rest with length 10 in the O1 frame), if you > pick any point along those lines and map to the O2 frame with the LT, as > output you always get points along x = 0.6c*t and x = 8 + 0.6c*t > (corresponding to a rod of length of length 8 and velocity 0.6c in the O2 > frame). So, here the input coordinates represent a rod with its proper > length of 10, and the output coordinates represent a rod with a contracted > length of 8. > > Jesse > > > Let's start from the beginning to make sure we're on the same page. Using > the contraction formula L' = L * sqrt ( 1 - (v/c)^2), where L is the rest > length of rod in frame f1, moving at velocity v wrt frame f2, and L' is the > contracted length of the rod as calculated from the pov of f2. Do you agree > with my interpretations of these variables? Do you agree that the measured > length of rod in f1 is never L', but always its rest length L? These are > Yes or No questions. TY, AG > As long as we are considering a case where f2 is different from f1 (which is normally the only situation where anyone would bother to use the formula), I agree with everything you just wrote. But the one caveat I'd add is that one is technically free to consider the special case where f2 = f1, i.e. you are imagining that the frame f2 of the observer who is using the equation is the same as f1, the rest frame of the rod, so in this case v=0 and L' = L. In my first paragraph above (the one starting with 'The quickest way to get this conclusion') I had L=10, and I considered both the case of an observer O2 moving at 0.6c relative to the rod who used L=10 and v=0.6c as input to the formula to output a length L' = 8 in his frame, and also the case of observer O1 who was at rest relative to the rod and used L=10 and v=0 as input to the formula to output a length of L' = 10 in his frame. Of course the latter is a trivial case and it's easier to just remember that rest length/proper length is what will be measured in the frame where the rod is at rest, but I included it to illustrate the point that either observer can apply the formula to get the correct length in their frame as the output. Jesse > > > > > > * And if we map from O2 to O2, using the rod's coordinates in O2 as input, > won't there also be no contraction? AG * > > > In this case there is no change in the length from input to output, but > that length is still contracted relative to the object's proper length > (this calculation going from O2 to O2 wouldn't have the proper length > appearing on either side of the equation though, you'd need to do a > different calculation to find it). > > Jesse > > > > * So, AFAICT, contrary to what relativity claims, contraction doesn't > exist! Note also what happens to the Parking Pardox. No contraction of any > object in any frame. Paradox solved! Are we having fun yet? AG* > > > On Sunday, February 2, 2025 at 7:52:55 PM UTC-7 Jesse Mazer wrote: > > On Sun, Feb 2, 2025 at 8:17 PM Alan Grayson <[email protected]> wrote: > > On Sunday, February 2, 2025 at 4:31:48 PM UTC-7 Jesse Mazer wrote: > > On Sun, Feb 2, 2025 at 5:08 PM Alan Grayson <[email protected]> wrote: > > On Sunday, February 2, 2025 at 2:43:09 PM UTC-7 Jesse Mazer wrote: > > On Sun, Feb 2, 2025 at 12:44 PM Alan Grayson <[email protected]> wrote: > > I will study your post and respond later. For now, let me say that the GPS > situation is irrelevant. It just shows that time dilation is real. Nothing > to do with length contraction. Also, after reading some of your earlier > comments, I agree that in the frame containing the rod, its length is not > contracted. This is the rest frame with the rod at the origin. The frame > from which the LT is applied has an observer at the origin, but no rod, and > is in relative motion compared to the frame with the rod. I hope you have > no objections to this comment. If you have any objections, please let me > know. AG > > > Mostly sounds fine but the only thing I'd want to double check is that > when you say "The frame from which the LT is applied has an observer at the > origin, but no rod", do you just mean that the rod is not at rest in this > observer's frame? The rod is still measurable and can be assigned > coordinates with changing position as a function of time in this observer's > frame (the observer you called O2 in your earlier post), agreed? > > > I want two frames with the rod in one, which I thought was your initial > model. The rod is situated and fixed at the origin, and there is no rod in > the frame using the LT; > > > But as I asked you repeatedly, when you say no rod "in" the O2 frame do > you just mean there is no rod that's *at rest relative to* the O2 frame, or > are you somehow denying that any given physical object like a rod is > assigned coordinates by *all* frames including the O2 frame in which the > rod is moving? > > > You still haven't answered this question, and it seems like it might be > important given some of your other phrases below... > > > > > or if you prefer we can model the situation with a rod in each frame, at > rest, both at origin, and their rest lengths are unimportant. > > > No need for two rods, provided you agree above that the O2 frame still > assigns coordinates to the rod even though the rod is not at rest in that > frame. > > > *OK. One rod, and frame with rod is given coordinates in both frames. For > me, x ---> x' means a LT from frame with no rod, * > > > Does "frame with no rod" just mean "frame with no rod at rest in it", or > are you somehow claiming there's a frame that doesn't "see" the rod at all > in terms of being able to measure it and assign coordinates to it? > > > > *to frame with rod, and from this there's the implication of contraction > in x' frame, with rod. Do you agree or not? AG* > > > > Drawing on the GPS situation, from any rod/frame applying the LT, the > formula IMO predicts the measured length in the frame it is observing, > > > GPS is distinct because the clocks don't just tick at their natural rate > > > *They tick naturally and are then reset to presumably synchronize them > with orbiting clock. AG* > > but are artificially adjusted, as I said. If the rod is at rest in O1 and > moving relative to O2, can we assume we are initially given the coordinates > of the rod as measured in O2, then then O2 frame is the one "applying" the > LT to predict the coordinates in the frame O1, so that O1 would be "the > frame it is observing" in your statement above? > > > *Yes, except we don't have to assume the moving rod has coordinates in O2. > AG * > > > Do you just mean it doesn't have *fixed* coordinates in O2, or do you mean > it isn't assigned coordinates at all in O2? If the latter, are you > imagining it's somehow invisible to the O2 observer? If so that's not how > things work in relativity, the rod is just an ordinary physical object, of > course the O2 observer is going to be able to measure it as it passes by > his own system of rulers and clocks, and say things like "when the clock > attached to the 3-light-second mark on my ruler showed a time of 5 seconds, > the back of the rod was passing right next to it (as seen in a photo taken > at that location at that moment, for example), therefore the worldline of > the back of the rod passes through the coordinates x=3 light seconds, t=5 > seconds in my coordinate system" > > > > > > similar to the Earth bound clocks in GPS which predict the time delays in > the orbiting clocks. For this reason, in the contraction case, the > frame/observer applying the LT, doesn't predict the contraction in > observer's own frame (which doesn't exist if there's one rod in the model), > but in the frame with the rod. > > > No, as my numerical example shows, if we start with the coordinates of the > rod in O2 and use the LT to predict its coordinates in O1, we get a > prediction of NO contraction of the rod in O1; the prediction will be that > the rod has its "proper length" in the O1 frame. > > > However, and this is where I get my prediction which you object to; in > this frame, the frame with the rod, the only prediction possible is zero > contraction. > > > If you are talking about the type of calculation I describe above, I > *agree* the prediction would be zero contraction in the O1 frame, which > matches the fact that no contraction is MEASURED in the O1 frame. > > > *Yes, this is what I've been saying. AG* > > > It was you who claimed that there was some prediction using the LT that > would conflict with the fact we both agree on that no contraction is > measured in the frame where the rod is at rest. > > > *I changed my pov when reading one of your previous posts. But since > there's no contraction measured in frame where the rod exists* > > > Are you saying the rod literally does not "exist" in other frames in the > sense of not being measured at all, or are you just saying the rod is not > at rest in other frames? If you're somehow saying the rod is not assigned > coordinates at all in the O2 frame, that doesn't make sense, see above. > > > > * and is at rest (even though the frame is in relative motion), the LT has > no other possible predictions, so it seems that length contraction never > occurs!* > > > Sure length contraction occurs, in the example it occurs for the O2 > observer who sees the rod in motion. If the rod has a proper length of 10 > light-seconds, and the O2 observer says the BACK of the rod passed by x=3 > and t=5 in his coordinate system, and the FRONT of the rod passed by x=11 > and t=5 in his system, then the distance between the front and back at the > single moment t=5 in this frame must be 11 - 3 = 8, so that's what the O2 > judges the rod's length to be, a length which is shorter than the rod's > proper length measured by the O1 observer. > > If you want to start from coordinates in one frame and then use the LT to > predict coordinates in another, and you're asking about how this could ever > lead to a prediction of contraction, one option would be to *start* from > the coordinates in O1's frame where the rod is at rest (unlike in my > numerical example where I started with the rod's coordinates in the frame > where the rod was moving), then apply the LT to *predict* the coordinates > in O2's frame where the rod is moving. If we did that instead of the > opposite, in that case we *would* get a prediction of a contracted length > in the predicted frame O2. I could give a different numerical example > illustrating this, if you would actually be interested in reading through > it. > > > > * This is where the rubber hits the road in our disagreement about what > the LT predicts, and what is measured. If contraction is never measured > because it never occurs, the "predictions" of the LT are worthless to the > point of not existing. I hope you're not going to tell me now, that x' in x > --> x' refers to the frame applying the LT. AG* > > > > Hence, the LT doesn't do as you claim, and it doesn't function like the > GPS situation. Moreover, I recall you used spacetime diagrams to show > length *expansion *in your frame at relative speed, but never before have > I heard or read of such a claim, which raises the proverbial red flag. AG > > > As I stated repeatedly, by "expansion" I just meant the length would be > predicted as larger in the second frame (O1 above) compared to the first > frame where the rod was moving (O2 above), > > > *How could that be if the result of the LT is x', not x?* > > > In my numerical example I treated the primed frame as the one where the > rod is at rest, i.e. O1 in your terminology. So, going from x-->x' takes > you from O2 to O1. > > Also note however that the complete set of LT equations include both > x-->x' and x'-->x, so you're free to go in either direction. > > > > * The rod is not contracted in O1 even though its moving relative to O2, > and the length of the rod is not in the image of anything in O2?* > > > If you want to do the LT from O2 to O1 you have to start with the > coordinates of the rod in O2, I don't know if that's what you mean by "the > image of anything" or if you just mean that we don't obtain the O2 > coordinates as a result of a LT. > > > > * Remember; the LT prediction in O1 is no contraction. I think this is > exactly where we disagree about what the formula predicts.* > > > What do you think we are disagreeing about in terms of what the formula > predicts? I've said that if you start with the coordinates in O2 and use > the LT to predict O1, the prediction is no contraction in O1. > > Jesse > > > > * Maybe in your reply we can finally resolve this issue. AG* > > I wasn't talking about expansion relative to the rod's proper length. The > prediction will be that the rod has its proper length in frame O1, i.e. "no > contraction" relative to the proper length. > > 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/b362c4c1-4de6-45d7-804e-34f21529c823n%40googlegroups.com > <https://groups.google.com/d/msgid/everything-list/b362c4c1-4de6-45d7-804e-34f21529c823n%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]. 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