On Saturday, January 18, 2025 at 11:50:11 AM UTC-7 Jesse Mazer wrote:
On Sat, Jan 18, 2025 at 1:19 PM Alan Grayson <agrays...@gmail.com> wrote: On Saturday, January 18, 2025 at 10:24:01 AM UTC-7 Jesse Mazer wrote: On Sat, Jan 18, 2025 at 12:09 PM Alan Grayson <agrays...@gmail.com> wrote: On Saturday, January 18, 2025 at 9:15:17 AM UTC-7 Jesse Mazer wrote: On Sat, Jan 18, 2025 at 9:36 AM Alan Grayson <agrays...@gmail.com> wrote: On Saturday, January 18, 2025 at 7:19:41 AM UTC-7 Jesse Mazer wrote: On Sat, Jan 18, 2025 at 9:09 AM Alan Grayson <agrays...@gmail.com> 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 <agrays...@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-- *I mean, if one uses the LT, to transform from one frame to another frame, if the resultant parameters in the latter frame are not actually measured in the latter frame, I refer to those measurements as "apparent". What I'm stuggling with is what the LT actually results in. Does it tell us what is actually measured in the latter frame, or not? AG* Yes, it always tells you what is actually measured in the frame that you're transforming into, using that frame's own rulers and clocks. You haven't made it at all clear why you suspect otherwise. Jesse *I suspect otherwise because in the Andromeda problem, using the LT from the pov of the rest frame, at rest relative to the Earth, we get length contraction in the transformed frame (modeled as a rod moving toward the Earth)* I thought the problem was supposed to involve the assumption of a traveler going from Earth to Andromeda (who I imagined as riding in a rocket), with a rod that is at rest relative to Earth and Andromeda and whose length defines the distance between them in each frame? But here you seem to be talking about a rod moving relative to the Earth? So now I'm not even clear about what physical scenario you are imagining--please lay it out clearly by specifying all the physical objects you are imagining in the problem (like rocket, rod, Earth, Andromeda), which of them have nonzero velocity in the Earth's frame, and the rest length for any object you want to do length calculations for. Jesse *If you prefer, we can place the traveler to Andromeda in a spaceship moving with some velocity wrt the Earth, and now imagine a rod whose length is the distance from Earth to Andromeda.* So the rod is at rest relative to Earth and Andromeda, as I originally understood you to be saying? * Since motion is relative, we can imagine the spaceship is at rest, and the rod moving toward Earth (since the spaceship is ..imagined as moving toward Andromeda).* By "moving toward the Earth" do you mean moving *relative to the Earth* (i.e. distance between the Earth and either end of the rod is changing over time), or do you just mean that in the spaceship's frame, the rod is always moving in the direction of the Earth, but the Earth is moving at the same speed in the same direction so the end of the rod always coincides with the Earth (or remains at a fixed negligible distance from it)? If the latter, "the rod moving toward Earth" seems like confusing terminology for this, there would be much less room for confusion if you stuck to the language of talking about movement "relative to" a given object or observer as I suggested. * In the rest frame of the spaceship, the rod is contracted since it appears to be **moving. We can use the LT to calculate how much its length contracts due to the velocity of the rod. However, as I pointed out earlier, this measurement is NOT possible IN the frame of the moving rod**, since nothing is moving within this frame.* How much its length contracts in the frame of the spaceship, or in the frame of the rod (assuming the rod is at rest relative to Earth/Andromeda)? * This is an example of the LT NOT yielding a measurement in the target frame (in this case the frame containing the rod), which is an exception to the claim that the LT always predicts what a target frame will actually measure.* I don't know why you think that, but you'll have to clearly specify which frame you want to transform from, and which frame you want to use the LT to transform into (the 'target frame'). If you mean starting from the coordinates of the rod in the spaceship's rest frame, and using the LT to transform into the rest frame of the rod (so the rod's frame is the target frame), then if you actually do the algebra of the LT you will get the correct result that the rod is *longer* in its own rest frame than it was in the spaceship frame, not shorter. And if on the other hand you start with the coordinates in the rod's rest frame and use the LT to transform into the spaceship frame (so the spaceship frame is the target frame), you will get the correct result that the rod is shorter in the spaceship frame than in its own rest frame. Jesse *Let me try again. The spaceship is moving toward Andromeda from the Earth. Since motion is relative I can assume the spaceship is at rest, and a rod representing the distance from Earth to Andromeda is moving toward the Earth, since it is assumed the spaceship was originally moving in the opposite direction, toward Andromeda. * You didn't address my question of whether "moving toward the Earth" means moving *relative* to the Earth (i.e. rod is moving in Earth's rest frame, and other frames like the spaceship frame see the distance between the end of the rod and the Earth as changing over time), or if it just means that in the rod is moving in the direction of the Earth at the same speed that the Earth itself is moving in this frame, so the distance between any point on the rod and the Earth is unchanging. If you mean the latter, this is confusing terminology, definitely not the sort of thing you would find in any relativity textbook. Either way, can you *please* stick to defining movement and rest "relative to" specific objects or observers to prevent this kind of verbal ambiguity? *I assume, and you can as well, that the Earth is at rest, and the spaceship is moving toward Andromeda. Can you assume the Earth is at rest in this model and not allow us to get into a discussion of what "at rest" means? AG* *Now, from the pov of the rest frame, the spaceship, imagine a LT to determine the contraction of the rod.* Contraction of the rod in what frame? *I was explicit; from the rest frame of spaceship, apply the LT to determine the contraction of the rod, assuming you know its original length. Now ask yourself this question; in the rod frame, does the contraction information you've received from the LT correspond to any measurement result in the rod frame? (Answer; of course NOT!). This is what I've been trying to demonstrate; results given by the LT do NOT always give us what can be measured in the target frame of the LT, in this case the rod frame. AG (PS; I sent you a message on Facebook. You can reply here if you want.)* Are you saying that if we start with the coordinates of the ends of the rod in the spaceship frame (I think the rod is supposed to be moving in this frame, correct me if I'm wrong), and then use the LT to find the coordinates of the ends of the rod in the rod's own rest frame, you think we'd get the false prediction that the rod is contracted in the rod's rest frame? *The spaceship is in one frame, at relative rest WRT the rod; the rod is in another frame, in relative motion WRT to the spaceship, at the velocity which the spaceship was originally moving toward Andromeda. We know the rod's initial length and we calculate its contraction from the spaceship's frame, using the rod's relative velocity. AG* If so this is simply WRONG, that's not what you would find if you actually did the algebra of the Lorentz transformation equations. If you mean something different you need to be more clear about what frame we are starting with, what is the "target frame" we are transforming into using the LT, and which of these frames you want to know the "contraction of the rod" in. Jesse * Note that in the frame of the moving rod, its contraction, predicted by the LT, cannot be measured because although the rod is moving, nothing within the rod frame is moving. This, IMO, contradicts the claim that the LT always yields what is actually measured in the target frame of the LT, in this case the frame of the rod. OK? AG* * I think time dilation is an example of the opposite; a measurement which is realized in the target frame. AG * *but never length contraction as what's observed IN the rod frame. I realize that length contraction requires something moving, so it's not reasonable to expect this, although it does show there's an exception to what you allege the LT does. AG* 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/00103fd0-2c2f-4a61-ba6d-e1ac1d22e437n%40googlegroups.com.
