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 > > > > > > > > > > > > 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/4b7933f9-ed18-4c0f-8f0e-0537fa087cacn%40googlegroups.com > <https://groups.google.com/d/msgid/everything-list/4b7933f9-ed18-4c0f-8f0e-0537fa087cacn%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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