On Fri, Jan 17, 2025 at 8:40 PM Alan Grayson <[email protected]> wrote:
> > > On Friday, January 17, 2025 at 5:21:58 PM UTC-7 Alan Grayson wrote: > > On Friday, January 17, 2025 at 4:00:56 PM UTC-7 Jesse Mazer wrote: > > On Fri, Jan 17, 2025 at 5:18 PM Alan Grayson <[email protected]> wrote: > > > > On Friday, January 17, 2025 at 2:05:42 PM UTC-7 Jesse Mazer wrote: > > On Fri, Jan 17, 2025 at 1:37 PM Alan Grayson <[email protected]> wrote: > > > > On Friday, January 17, 2025 at 11:25:54 AM UTC-7 Jesse Mazer wrote: > > On Fri, Jan 17, 2025 at 12:31 PM Alan Grayson <[email protected]> wrote: > > > > On Friday, January 17, 2025 at 7:46:06 AM UTC-7 Jesse Mazer wrote: > > On Fri, Jan 17, 2025 at 9:38 AM Alan Grayson <[email protected]> wrote: > > > > On Friday, January 17, 2025 at 7:29:19 AM UTC-7 Jesse Mazer wrote: > > On Fri, Jan 17, 2025 at 7:51 AM Alan Grayson <[email protected]> wrote: > > > > On Thursday, January 16, 2025 at 5:52:52 PM UTC-7 Jesse Mazer wrote: > > On Thu, Jan 16, 2025 at 7:33 PM Alan Grayson <[email protected]> wrote: > > > > On Thursday, January 16, 2025 at 2:39:55 PM UTC-7 Jesse Mazer wrote: > > On Thu, Jan 16, 2025 at 2:43 PM Alan Grayson <[email protected]> wrote: > > > > On Thursday, January 16, 2025 at 11:36:48 AM UTC-7 Jesse Mazer wrote: > > On Tue, Jan 14, 2025 at 12:02 AM Alan Grayson <[email protected]> wrote: > > Using the LT, we have the following transformations of Length, Time, and > Mass, that is, > x --->x', t ---> t', m ---> m' > > > The length contraction equation is not part of the Lorentz transformation > equations, the x --> x' equation in the LT is just about the position > coordinate assigned to a *single* event in each frame. The length > contraction equation can be derived from the LT but only by considering > worldlines of the front and back of an object, and looking at *pairs* of > events (one on each of the two worldlines) which are simultaneous in each > frame--length in a given frame is just defined as the difference in > position coordinate between the front and back of an object at a single > time-coordinate in that frame, so it requires looking at a pair of events > that are simultaneous in that frame. The result is that for any inertial > object, it has its maximum length L in the frame where the object is at > rest (the object's own 'rest frame'), and a shorter length L*sqrt(1 - > v^2/c^2) in a different frame where the object has nonzero velocity v. > > The t ---> t' equation is likewise not the same as the time dilation > equation, it's just about the time coordinate assigned to a single event in > each frame, although it has a simpler relation to time dilation since you > can consider an event on the worldline that passes through the origin where > both t and t' are equal to 0, and then the time coordinates t and t' > assigned to some other event E on this worldline tell you the time elapsed > in each frame between the origin and E. And the LT don't include any mass > transformation equation. > > Jesse > > > You're right of course. TY. I see the LT as giving appearances because, > say for length contraction, the reduced length is not measured in the > primed frame, but that is the length measurement from the pov of the > unprimed or stationary frame. > > > In relativity one does not normally designate any particular frame to be > the "stationary frame", since all concepts of motion and rest are defined > in purely relative way; if one has two objects A and B in relative motion, > one could talk about the frame where A is stationary (A's 'rest frame') or > the frame where B is stationary (B's rest frame), but that's all. I'm not > sure what you mean by "the reduced length is not measured in the primed > frame"--which object's length are you talking about? If A's rest frame is > the unprimed frame and B's rest frame is the primed frame, then the length > of object A in the primed frame is reduced relative to its length in its > own rest frame, i.e. the unprimed frame. > > > *Let's consider a concrete example of a traveler moving at near light > speed to Andromeda. From the traveler's frame, the distance to Andromeda is > hugely reduced from its length of 2.5 MLY from the pov of a non-traveling > observer. This seems to imply that the reduced length is only measured from > the pov of the traveler, but not from the pov of the non-traveler, because > of which I describe the measurement from the pov of the traveler as > APPARENT. Do you agree that the traveler's measurement is apparent because > the non-traveler measures the distance to Andromeda as unchanged? TY, AG * > > > I don't know what you mean by "apparent", but there is no asymmetry in the > way Lorentz contraction works in each frame--if we assume there is a frame > A where Milky Way and Andromeda are both at rest (ignoring the fact that in > reality they have some motion relative to one another), and another frame B > where the rocket ship of the traveler is at rest, then in frame B the Milky > Way/Andromeda distance is shortened relative to the distance in their rest > frame, and the rocket has its maximum length; in frame A the the rocket's > length is shortened relative to its length in its rest frame, and the Milky > Way/Andromeda distance has its maximum value. The only asymmetry here is in > the choice of the two things to measure the length of (the distance between > the Milky Way and Andromeda in their rest frame is obviously huge compared > to the rest length of a rocket moving between them), the symmetry might be > easier to see if we consider two rockets traveling towards each other > (their noses facing each other), and each wants to know the distance it > must traverse to get from the nose of the other rocket to its tail. Then > for example if each rocket is 10 meters long in its rest frame, and the two > rockets have a relative velocity of 0.8c, each will measure only a 6 meter > distance between the nose and tail of the other rocket, and the time they > each measure to cross that distance is just (6 meters)/(0.8c). > > Jesse > > > *By apparent I just mean that the measurement the LT gives in this case, > is not what is actually measured in the target frame. Moreover, this is > differnt from the situation in the Twin Paradox as discussed in another > recent post on this thread. A*G > > > What do you mean by target frame? If the unprimed frame is the frame where > Milky Way/Andromeda are at rest and the primed frame is the frame where the > rocket is at rest, are you saying the primed frame does not actually > measure a shorter distance from Milky Way to Andromeda if we use the LT > starting from the coordinates of everything in the unprimed frame? Or are > you arguing something different? Are you using primed or unprimed as the > "target frame"? > > Jesse > > > *The target frame is the primed frame, the result of the LT. The unprimed > frame is the traveler's frame moving at some speed toward Andromeda. It's > often claimed that the result of applying the LT will yield the actual > measurement in the primed frame, but this isn't the case in this example. > AG* > > > OK, so you want the unprimed frame to be the frame where the rocket is at > rest and the Milky Way/Andromeda are moving? In that case the unprimed > frame will be the one where the distance between Milky Way/Andromeda is > contracted according to the length contraction equation, since they are > moving in that frame and at rest in the primed frame. And as I told you, > the LT is not the same as the length contraction equation, if you apply the > LT to the coordinates of the worldlines of Milky Way/Andromeda in the > unprimed frame, you will get the correct answer that in the primed frame > these worldlines have zero velocity (constant position as a function of > time) and a greater coordinate distance between them than they did in the > unprimed frame. > > Jesse > > > *Firstly, copied below is what I posted earlier today. Because I wanted to > contrast the Andromeda case with the TP, I used an SR solution for the > latter, and you will note that the line segment paths of the inscribed > polygon are inertial paths, and by infinitely refining the partition, I get > the circular motion for the return path of the traveling twin. You will > also note that the Earth-bound twin is at rest, and is analogous to the > rest bound observer in the Andromeda case. In the TP, the Earth bound twin > measures the traveling twin's clock running slower than his own clock, > using the LT, but this effect is real for the traveling twin; otherwise he > wouldn't see himself younger than his twin when they are juxtaposed upon > his return. In contrast, you have the Andromeda traveler also at rest, and > measuring the contracting distance in his frame, while the resting twin > measures time dilation in his frame. However, in the frame of the moving > rod representing the distance from Earth to Andromeda, according to your > analysis the observer in that frame does NOT measure his length > contracted;. only the rest frame measures the length contraction. * > > > First, there's a habit in your writings that I find ambiguous: your use of > the word "rest" and "moving" in an unqualified way, not clearly specifying > that you are just speaking in a relative way about the rest frame of a > particular object/observer in your thought-experiment. Sometimes this > actually leaves me unclear on which frame you are actually talking about, > at other times I can infer which one you're talking about but I wonder if > you're trying to implicitly suggest that your thought-experiment shows that > we must accept some notion of an objective truth about which observer is > "really at rest", as opposed to the standard physicist's understanding that > rest and motion can only be defined in relative terms. > > So, can I ask that if you are just using these terms in a relative way, > would you please always phrase it in a way that specifies whose rest frame > you're talking about? (eg "the rocket's rest frame", "the Earth's rest > frame" etc.) And second, if you *do* mean to make some argument that one of > your thought-experiments suggests a concept of objective/absolute rest, in > that case could you be explicit that you are making such an argument by > talking about "absolute rest" or some similar term? Please let me know if > you are willing to make this change to your way of writing before > addressing my more specific questions below. > > Now, when you say "the frame of the moving rod representing the distance > from Earth to Andromeda" and "only the rest frame measures the length > contraction", do you mean to introduce a rod at rest relative to > Earth/Andromeda into the thought-experiment, and when you call this a > "moving rod" you are talking about the perspective of the rest frame of the > rocket which is in motion relative to Earth/Andromeda, and the rocket's > rest frame is what you mean by "the rest frame"? If so, it is of course > true that the rod would be contracted in the rocket's rest frame but not in > the rod's own rest frame, and similarly true that the rocket would be > contracted in the rod's rest frame but not in the rocket's own rest frame. > This symmetry is similar to time dilation in that if you have two clocks A > and B in inertial relative motion, in the rest frame of clock A it will be > clock B that's running slow while clock A is running normally, and in the > rest frame of clock B it will be clock A that's running slow while clock B > is running normally. > > Also when you say "You will also note that the Earth-bound twin is at > rest" are you suggesting that the conclusion of this thought experiment is > that the frame where Earth is at rest is more "correct" in some absolute > sense, or just saying that this is what's true by convention if we analyze > the problem from the perspective of the Earth rest frame? Would you agree > or disagree that we could analyze the whole problem from the perspective of > any other specific inertial frame, like a frame where the Earth is moving > at 0.99c the whole time and the traveling twin is sometimes moving faster > and sometimes moving slower during different sections of its non-inertial > path, and we would get exactly the same answer to the question of how much > each twin has aged at the moment they reunite at the same location (a > 'local physical fact' in the sense I discussed before)? > > Jesse > > > *When I refer to the non-traveling twin in the TP, do you find this > ambiguous? Do you find the traveling twin's frame ambiguous? If you do, I > don't how I can be more specific. * > > > It's usually understood that "non-traveling twin" just means the twin that > moves inertially between the two meetings, and "traveling twin" means the > one that changed velocities at least once on its path between the meetings. > > > *The non-traveling twin is at rest on the Earth throughout. I never heard > of any other concept of the non-traveling twin. The traveling twin is > moving with respect to the Earth. I never heard of any other concept of the > traveling twin. AG* > > > All that's important to the twin paradox is that one twin is inertial and > the other is not. For example if twin A is moving away from Earth > inertially the whole time, and twin B is at rest relative to Earth for a > while and then accelerates to catch up with twin A, then twin A is the > inertial twin and twin B is the non-inertial one during the time between > their meetings, and so it's guaranteed that twin B will have aged less than > twin A when they reunite. > > > *On and off over many years I've read about the TP. Never, not once, have > I read it described as you do. Moreover, for the moving observer to leave > and return, it's impossible for that observer to be totally inertial. If > you model any observer leaving Earth, that observer cannot be inertial. AG* > > > > > If you mean something different, like the idea that there is some > objective/absolute sense that the "non-traveling twin" is at rest rather > than moving, then I would object to that. But as stated, those phrases > don't involve the word "moving" or "at rest" without qualification, which > is what I was objecting to in my comments above. Will you agree in future > to specify what object/observer "rest" and "moving" are relative to if you > mean them in a relative way (which can easily be specified with a phrase > like 'moving relative to [some object]' or 'at rest relative to [some > object]'), or else to explicitly use some phrase like "absolute rest" and > "absolute movement" if you mean them in a non-relative way? > > > *No absolute anything. All motion is relative to something. AG * > > > OK, then are you willing to alter your way of writing about these things > to prevent ambiguity, to always use phrases like "moving relative to > [object]" or "at rest relative to [object]" to specify the relative > motion/rest you are thinking of? > > > > > *Concerning the Andromeda frames, there's a frame with a moving rod, > representing the distance between the Earth and Andromeda,* > > > "Moving" in an absolute or relative sense? If in a relative sense, moving > relative to what? Are you talking about a rod which is at rest relative to > Earth and Andromeda, and moving relative to the rocket? > > > *The observer in Andromeda case is traveling, moving with respect to the > Earth. Then we can assume this observer is at rest, relative to a moving > rod which represents the distance from Earth to Andromeda. AG * > > > > > * and the frame of an observer using this frame to determine the length > contraction.* > > > Are you talking about an observer on the rocket which is moving relative > to Earth/Andromeda, > > > *Yes. AG* > > > using his own rest frame to determine the length contraction of the rod > which is at rest relative to Earth/Andromeda? > > > *The rod is moving, the observer is stationary in his rest frame, from > which he calculates the length contraction. AG * > > > The rod is moving relative to this observer A, and is thus contracted in > A's frame. > > > *But not, at that time, also contracted in B frame. This is different from > the TP where time is dilated in frame B, the frame of traveling twin. IOW, > using the LT in the TP, what is measured from the Earth or stationary > frame, is what's actually measured in the moving frame. Not so in Andromeda > problem. AG* > > > *Let me clarify the problem I'm trying to resolve; notice the T in LT. It > stands for Transformation, presumably from one frame to another. It's > claimed that the LT will produce the measured result in the target frame, > based on parameters of the source frame for the transformation. And this > seems to be the case in the TP; from the source frame, the frame at rest on > the Earth, the LT tells us what will be measured in the traveling frame. > And it seems to do just that, since the clock in the traveling frame > actually ticks slower as predicted.* > See my point above about the need to specify which of the two segments of the non-inertial twin's path you mean when you refer to "the traveling frame"--the LT only deals with inertial frames, there is no inertial frame where the non-inertial twin is at rest during the entire journey! And if for example by "traveling frame" you mean the frame where the non-inertial clock is at rest during the outbound leg, you seem to have gotten things completely backwards here--the LT would *not* predict the non-inertial clock is ticking slower in this frame during the period where it's at rest in this frame, instead it would predict the non-inertial clock shows no time dilation during this leg of the trip, while the clock of the inertial twin is ticking slower. Jesse > * If it didn't, the traveling observer would not age slower than the > stationary observer. But when we consider the Andromeda problem, the LT > seems NOT to predict what the frame of the moving rod will measure. Maybe > it does, as you indicated, but only when a measurement is taken, unlike the > case of the TP, where the result seems inherent, and not requiring a > measurement. AG * > > > > And likewise if we assume the observer A is on board a rocket as I > suggested, the rocket is moving relative to an observer B who is at rest > relative to Andromeda, and so the rocket is contracted in the frame of > observer B. So length contraction is completely symmetric between inertial > frames, as is time dilation--are you saying otherwise? > > Jesse > > > > > > > > > *IOW, the cases are similar except for the fact that one involves time > dilation and other involves length contraction, but what is measured in the > target frames is hugely different. This is the puzzle I am struggling with; > namely, why is time dilation a measurable result for the traveling twin, > but length contraction is NOT a measurable result for the frame of the > moving rod in the Andromeda case, even though the frames doing the > measuring in both cases have measurable results in their frames, but not in > their respective moving frames? AG* > > *PS for Clark; I am halfway through the video you posted. IMO, there can > be several ways to solve a problem and acceleration is one legitimate way > because there IS accelation for the traveling twin, and acceleration IS > equivalent to gravity, and gravity DOES slow clocks. I will finish the > video out of curiosity for the author's supposed solution, but it seems > obvious that acceleration is one possible solution. AG* > > *"Houston, we have a problem!" Now let's consider time dilation using SR > in the Twin Paradox. Imagine the traveling twin moving in a circle and > returning to Earth, and imagine the circle contains a polygon consisting of > straight paths, which will later be infinitely partitioned, whose limit > will be that circle. As measured by the stationary twin, the traveling > twin's clock is dilated along each segment, so when the twins are > juxtaposed, the traveling twin's elapsed time is LESS than clock readings > for the stationary twin. If this is correct, it demostrates that what the > stationary twin measures, is actually what the traveling twin's clock > reads. IOW, what happens to time dilation in this case is OPPOSITE to what > happens to the frames for the trip to Andromeda! Do you understand what I > am alleging -- that length contraction acts in an opposite manner compared > to time dilation, when I would expect them to behave similarly? AG* > > > > > About mass, since the measured mass grows exponentially to infinity as v > --> c, isn't this derivable from the LT, but in which frame? AG > > > The notion of a variable relativistic mass is just an alternate way of > talking about relativistic momentum, often modern textbooks talk solely > about the latter and the only mass concept they use is the rest mass. For > example the page at > https://courses.lumenlearning.com/suny-physics/chapter/28-5-relativistic-momentum > has a box titled "Misconception alert: relativistic mass and momentum" > which says the following (note that they are using u to denote velocity): > > "The relativistically correct definition of momentum as p = γmu is > sometimes taken to imply that mass varies with velocity: m_var = γm, > particularly in older textbooks. However, note that m is the mass of the > object as measured by a person at rest relative to the object. Thus, m is > defined to be the rest mass, which could be measured at rest, perhaps using > gravity. When a mass is moving relative to an observer, the only way that > its mass can be determined is through collisions or other means in which > momentum is involved. Since the mass of a moving object cannot be > determined independently of momentum, the only meaningful mass is rest > mass. Thus, when we use the term mass, assume it to be identical to rest > mass." > > I'd say there's nothing strictly incorrect about defining a variable > relativistic mass, it's just a cosmetically different formalism, but it may > be that part of the reason it was mostly abandoned is because for people > learning relativity it can lead to misconceptions that there is more to the > concept than just a difference in how momentum is calculated, whereas in > fact there is no application of relativistic mass that does not involve > relativistic momentum. Momentum is needed for situations like collisions or > particle creation/annihilation where there's a change in which objects have > which individual momenta, but total momentum must be conserved. It's also > used in the more general form of the relation of energy to rest mass m and > relativistic momentum p, given by the equation E^2 = (mc^2)^2 + (pc)^2, > which reduces to the more well-known E=mc^2 in the special case where p=0. > > By the way, since relativistic momentum is given by p=mv/sqrt(1 - > v^2/c^2), you can substitute this into the above equation to get E^2 = > (m^2)(c^4) + (m^2)(v^2)(c^2)/(1 - v^2/c^2), and then if you take the first > term on the right hand side, (m^2)(c^4), and multiply it by (1 - > v^2/c^2)/(1 - v^2/c^2) and gather terms, you get E^2 = [(m^2)(c^4) - > (m^2)(v^2)(c^2) + (m^2)(v^2)(c^2)]/(1 - v^2/c^2), and two terms cancel each > other out so this simplifies to E^2 = (m^2)(c^4)/(1 - v^2/c^2), and then if > you take the square root of both sides you get E = γmc^2. So the original > equation for energy as a function fo rest mass m and relativistic momentum > p can be rewritten as E=Mc^2 where M is the relativistic mass defined as M > = γm, again showing that relativistic mass is only useful for rewriting > equations involving relativistic momentum. > > 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/9eba4b44-e843-41e1-9a40-d1b761d9699bn%40googlegroups.com > <https://groups.google.com/d/msgid/everything-list/9eba4b44-e843-41e1-9a40-d1b761d9699bn%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/CAPCWU3KAgih9-kLbE%2Bqiw77GTC41mKdSKE%3D9wrdt2zh-KNr_SQ%40mail.gmail.com.
