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 * 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. 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