On Thu, Jan 16, 2025 at 7:33 PM Alan Grayson <agrayson2...@gmail.com> 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 <agrays...@gmail.com> 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 <agrays...@gmail.com> 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 > > > 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 everything-list+unsubscr...@googlegroups.com. > To view this discussion visit > https://groups.google.com/d/msgid/everything-list/2dcb7689-d812-444c-a901-30e160e9edc1n%40googlegroups.com > <https://groups.google.com/d/msgid/everything-list/2dcb7689-d812-444c-a901-30e160e9edc1n%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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