User:Rschwieb/Physics notes

The purpose of this page is to evolve how to think about various physical and mathematical physics concepts. Please keep comments in the comments section, as I would like to keep the main notes in my own words.

Notes

Quick links

• User:F=q(E+v^B)/Mathematical summary of physics: a handy cheat sheet for most of physics.

Mathematics

Use of manifolds

Manifolds are "warped up Euclidean space". Since the old Euclidean global coordinates no longer work, the manifold is covered by an atlas whose coordinates can be smoothly translated. In this way, the manifold is arbitrarily covered by "global coordinates". Notions of "straight lines" and "parallel lines" are modified for the warped space. (One of Macdonald's postulates for GR is "spacetime has a manifold structure.")

Tangent space

Because of the warped nature of manifolds, it becomes necessary to "flatten" the manifold at points (the tangent space at the point), and do regular calculus in the small on these flat pieces. The infinitesimal contributions are then gathered together to talk about lengths of curves in the manifold and their straightness. Under suitable conditions, given a full set of coordinate curves through a point, the partial derivatives furnish a corresponding collection of tangent vectors which are a basis for the tangent space attached at the point.

Tensor/tensor field

An individual abstract tensor can be thought of as a generalized vector. It is an object independent of any basis. A tensor field can be thought of as a generalized vector field. The field itself is an association of tensors to the points of a manifold, usually in a smooth way. Using vectors, for example, we can think of forces at each point as "vectors". When the quantity described is more complicated, as in the case of stress or energy–momentum density, then it may take more than a simple vector to describe the quantity. The metric tensor is a little different, as it is attaching a protractor to each point on the manifold, in order to measure angles.

Tensors "pack up" physical quantities

How do physicists use tensors? Although perhaps an oversimplification; properties which are separate using simpler maths (e.g. vector calculus) but inter-related are amalgamated into tensor components, i.e. physical tensors pack up a collection of physical quantities. Examples are already above - the 4-current J and the EM field tensor F. Instead of dealing with E, B, and J as separate vector fields each with their own components (as in vector calculus in the equations above), the entire EM field tensor has components which are the E and B fields, all into one object. A more extreme example is the stress-energy tensor (cf general relativity) which amalgamates energy and momentum densities and fluxes. So when doing manipulations/calculations, tensors effectively number-crunch the components of a collection of inter-related physical quantities all at once. I'm not sure how to decide on co/contra-variance from the start though, but once a 4-vector or tensor is decided to have co/contra-varaince its easy to change between co/contra-variant components using the metric. F = q(E+v×B) ⇄ ∑ici 13:58, 31 March 2012 (UTC)

Tensors really do form an algebra, and so tensor fields form an algebra. Superposition is the act of adding tensor fields pointwise. Where two tensors interact by forming a product, the pointwise product of two tensor fields produces a third tensor field.

Covariance/contravariance

I suspect these are just artifacts of our choice of basis when we do algebra. In some situations, index manipulation allows us to reexpress tensors so that they are "all covariant". This may just be a "trick" that works, and it still could be that contravariance is unavoidable and has physical significance.

Metric tensor

As described before, a "protractor" at every point in the manifold, allowing for the measurement of tiny angles and lengths. This is the correct connection between local coordinate frames and their straight rulers and global coordinates and geodesics in warped space. We work in the small and pretend everything is flat, but then we have to take into account that the large is bent. Macdonald describes that tiny measured distances dictate the metric at a point, which dictate the Christoffel symbols in the large, which dictate the geodesics in the large warped space.

Curvature

There are a few types of curvature which quantify how much a manifold is curved at a point. Some are intrinsic to the manifold (measurable by surface dwellers) and some are extrinsic, depending on a larger space containing the manifold. For example, on the surface of the Earth, one can measure the circumference of a circle laid out on the ground of radius r, and verify that it will be smaller than 2*pi*r, owing to the curvature of the Earth. As r tends to 0, the difference between the measured circumference and the actual circumference becomes a curvature scalar (I forget which).

Physics

Measurement

Measured quantities are always given with the understanding of some margin of error. We are limited by physical constraints in the process of measurement. For example, a quantity which is too small for our bert instruments to detect "is zero" for all intents and purposes. This is also exemplified in the statement of Noether's theorem, where "invariance" is not exact invariance, but rather a loosened invariance under some small epsilon.

Accellerometer

An accellerometer is a primitive notion which alerts us to action by inertial forces. A useful accellerometer may be imagined as a mass suspended in the center of a wireframe cube. Any inertial forces will decentralize the mass. Such an accellerometer can be rotated about an axis through the mass without moving the mass off the zero mark.

Motion

Motion in space is principally made up of displacement and rotation. Displacement of inertial objects occurs along geodesics. This seems to be analogous to "distance preserving transformations" being decomposable into rotations and translations.

Lorentz transformations

Lorentz transformations (boosts) are the coordinate transformations necessary to convert between the coordinates of two inertial frames. Algebraically they form the group of Minkowski metric preserving transformations of Minkowski 3+1 spacetime. Given two inertial frames, the second frame is translated so that the two origins coincide, and the Lorentz transformation carries one set of coordinates to the other. (I'd like to know why we may safely assume the origins coincide.)

Inertial frame

A frame of reference is said to be inertial if accellerometers affixed on the coordinate axes a small distance from the origin all read 0. This eliminates the possibility of any linear or angular accelleration. The accellerometers of an inertial frame will read 0 when viewed from any frame of reference. One can travel between every pair of inertial frames via a Lorentz transformation (and a translation?). Mathematically, there is an equivalence relation on the class of frames given by A~B iff there is a Lorentz transformation (and translation?) from A to B.

Postulates of general relativity

Roughly, from Macdonald's "GR in a nutshell":

1. Inertial frame postulate: if E is an event on the world line of an intertial object, an intertial frame can be constructed at E with the object at rest in the frame. (Macdonald also includes a postulate that spacetime is a manifold, and thus has global coordintes furnished by an atlas.)
2. Metric postulate: within an intertial frame, the interval between two events is given by the 3+1 metric. (This is a consequence of the constancy of the speed of light. The metric is increasingly accurate as the events are nearer to each other.)
3. Geodesic postulate: geodesic curves parameterized by their arclength satisfy a differential equation involving the Christoffel symbols.
4. Field equation postulate

Einstein's equations

Einstein determined that the presense of energy/matter at a location has a direct connection with the metric at the location. The equations, derived from the postulates, will tell you the metric field for the region, if you can tell it how the mass/energy is distributed. (Conversely dictating a metric tensor should determine the mass/energy distribution?) Determination of the metric determines geodesics and Christoffel symbols. The equations are differential equations, so that a solution consists of a collection of functions.

Conserved quantities/invariant quantities

It's a physical fact that there are some numbers attached to systems which do not change as they move within the system. This is to be kept distinct from invariant quantities. Invariant quantities are numbers attached to objects which are frame-independent. Some quantities may not be invariant as they move, but they can be conserved because the quantity flows to a neighboring frame within the system.

Comments

Inertial frames

Abridged discussion on inertial frames

It is a referance frame (i.e. arbitarily chosen coordinate system), defined to be "fixed to an observer", usually so that the observer is always at the orgin (but any fixed point would do). One way to look at this is Newton's 1st law modified: that inertial frames exist. Recalling basic classical mechanics - zero resultant force acting on an object implies zero resultant acceleration, and vice versa, and so a constant velocity. So anything at rest appears to have no force acting on the particle to accelerate it and change its velocity. What if the object moved at constant velocity, and the observer (somehow - never mind literally!) moved at exactly the same velocity as the object? Then from the observer's referance frame there is no relative velocity between the object and the observer, hence no resultant force acting on the object (zero relative velocity, zero acceleration). So in the observer's frame: any object which appears to be at rest, may not. The whole frame, including the observer (by definition) and the object (by coincidence) could be moving at any constant velocity (any non-infinite velocity, in principle!), where "zero velocity" is a special case (how can we actually tell?) This is an inertial frame - a frame where objects appear to be at rest, i.e. no relative motion. Consider another such observer, moving at another constant velocity. In that frame, the relative velocity between the object and 2nd observer will appear to be non-zero (we could calculate what the relative velcoity would be, but for now not necerssary, actually simple using vectors). Since the relative velocity is constant, the 2nd observer conlcudes there cannot be any force acting on the object. This 2nd frame is also an inertial frame - a frame where objects appear to be travelling at constant relative velocity. Inertial frames in general: relative velocties between frames are constant. Non-inertial frames are more complicated, but the idea is analagous (as far as I know): replace velocity by acceleration (including relative) in the above steps. Then acceleration is relative between accelerated observers, and this leads to the equivalence principle of GR: the relative acceleration beween "gravity" and an accelerating object is zero. Gravity is the same thing as curved space time (curvature changes velocity, implying acceleration). The equivalence principle is contained (in a complicated form) in the Einstein field equations. Hope this helps, F = q(E+v×B) ⇄ ∑ici 22:15, 2 April 2012 (UTC) Here is the diagram for the situation described above. The reference frames are inertial as there is constant relative velocity between the frames. The object appears to have no force acting on it. The reference frames are still inertial since they move relative to each other with constant velocity, so the motion of the object (which now does appear to have a force acting) may appear to be different in each frame because of the relative velocity between each frame (but still obeys physical laws). This difference in relative velocity is accounted for in the Lorentz transformations. Frames are now accelerated, since there is relative acceleration between the two frames. Again - motions may appear to be very different (but still obey physical laws). Inertial objects: Perhaps confusingly, "inertial object" in mechanics refers to almost the same thing stated in a couple of ways: the inertia (inertial mass) of an object is the resistance to the change in velocity (unless a force acts on it to accelerate it). Similarly an inertial object also means "not accelerated", without reference to mass, i.e. constant relative velocity to an observer. About frames of reference in general (you may already know, just making sure): Anything can have a reference frame, not just a "literal observer" (say a scientist/engineer making measurements or an alein on a spacecraft), we could transform to the frame of a nearby Lego brick, juggling ball, rocket, planet, electron, (examples of "the object") - all can have coordinate systems attached to them. From whichever reference frame you use, the laws of physics are the same. I think the circularity is because we usually talk about two observers or one observer in relation to some object, and the terminology between an object's inertia and that it happens to be in its own inertial reference frame, but may be accelerating in another frame, etc! F = q(E+v×B) ⇄ ∑ici 09:31, 3 April 2012 (UTC) I can see the difference between the mathematician's and the physicist's approaches. A mathematician wouldn't dream of starting from Maxwell's equations as a postulate, even though historically they are the origin of the discovery of special relativity. — Quondum☏✎ 10:53, 3 April 2012 (UTC) Not sure where else to start!? Relativity developed from the assumption that Maxwell's equations are correct but the notion of absolute space and time are wrong (sorry to keep paraphrasing). It would be rather absurd if Gauss' law and Ampère's circuital law (with Maxwell's correction) took different forms in different frames... F = q(E+v×B) ⇄ ∑ici 11:33, 3 April 2012 (UTC) We'd be going into a huge philosophical discussion here. Suffice to say that the mathematician's approach is to find a simple set of all-embracing postulates (or axioms) from which the whole of physics might be built. Starting from Maxwell's equations and the Lorentz equation (and let's say implicitly with flat space), we can only deduce special relativity as it applies to EM fields and electric charges. We could deduce energy–momentum conservation and special relativity for EM fields and guess that it applies to other phenomena, but we'd need similar postulates to cover those. On the other hand, if we assume that Poincaré symmetry (invariance of physics laws under displacement, rotations etc.) applied to all physical phenomena (again say in flat space), and postulate only that electric charge exists, is invariant (conserved) in any volume irrespective of vantage point, and exerts a force via a field that is linear in the charge, Maxwell's and the Lorentz equations result in toto. I find this development from the simpler but more powerful postulate of symmetry far more satisfying, and it gives one the insight that no other equations are consistent with the symmetry postulate. Thus, had we postulated any equations other than Maxwell's, we would have reach a theoretical inconsistency. From very similar postulates, one can show that the Einstein equation could not have had any other form (postulating only a gravitational source that is a conserved vector field), up to the value of two or three constants. As I suggested before, the historical discovery process does not dictate the shape of the theory; the observations merely constrain it. (Beware: some WP:OR claims) — Quondum☏✎ 14:12, 3 April 2012 (UTC) I had no implications to get into philosophy: even though an interesting and relaxing exercise for the mind and quite apart from calculations. Ok - Maxwell's equations as "where else to start?" is an understatement, since it's much better to just resort to the ultimate symmetry of nature: invariance of physical laws, as you say (in all frames - as I say above). Best since its obvious, simple, and does serve as the constraint in the theoretical formulation. Btw - should you find any mistakes in the images here, plase say so I can correct them asap (feel free to change the captions also). Thanks. F = q(E+v×B) ⇄ ∑ici 14:58, 3 April 2012 (UTC) I'll take some time to absorb the above. I'd like to add that mathematicians and physicists are not all that different in their science! In both cases, the devotees on both sides have built up systems with axioms, where the axioms are chosen so that things we usually take "for granted" are encoded into the system. Because it's so rare that we go back to those foundations, some students are fooled into thinking the axioms are inherently true. (That is the mystique of classical mathematics, anyway.) I like the diagrams, but I'm pretty sure I haven't grasped the whole idea. Can you talk about a single thing being inertial or is inertial always relative to another thing? I'm aware that (sufficiently smooth) constant velocity isn't detectable by a human observer, but that any sort of accelleration will cause apparently "inertial forces" to push around the human. Then I guess an object feeling such forces would be non-inertial. It would make sense then to try to detect inertial forces with an accellerometer like the one I described. Could we say: "An accellerometer with its mass in some equillibrium is inertial. Any motion of the mass indicates there are inertial forces acting on the accellerometer, and it isn't inertial during that period." Rschwieb (talk) 17:38, 3 April 2012 (UTC) Whether the concept of inertia makes sense when there is no object with respect to measure it is not exactly settled (see, for example, Mach principle). However, it is safest to assume that inertial frames make sense even in an empty universe. Anyway, even empty space is seething with virtual matter, so maybe the question is moot even in principle. Your accelerometer example is good: when isolated from other forces (or their effect is taken in to account), an accelerometer reading zero remains stationary with respect some inertial frame. So a bouncing ball while it is in the air follows an inertial frame according to the equivalence principle. — Quondum☏✎ 18:47, 3 April 2012 (UTC) I'm comfortable with an accellerometer reading 0 being inertial, but I still get worried when gravity is involved. Consider in a person in a rocket accellerating at 9.8m/s^2. Presumably they are not inertial, because they feel force pushing them against the back (bottom?) of the rocket. On Earth, gravity produces a similar sensation of being pushed: it is not a result of motion, but of curvature of space. While both experiments cause similar sensation, the person standing on Earth is inertial (?) and the person in the rocket is not (?). In both cases the accellerometer is at an equillibrium, but not at 0. Rschwieb (talk) 19:10, 3 April 2012 (UTC) It took so long I ran into another edit conflict again - sorry... You ask pretty deep questions... to the best of my understanding: An "inertial force" either means: a force which is accelerating the inertial mass of an object (not quite the same as gravitational mass, but equal by the equivalence principle), else is meaningless (I could be wrong). You statement about the accelerometer is else correct. Start from one reference frame again, and suppose an object moves with velocity. Adhering to the relativity postulates - how to measure the motion of this object? It "moves with velocity" but what velocity? For reasons stated above the observer in the frame can't observe it to be moving unless it moves relative to the observer, so using the coordinate frame the observer can detect and measure the apparent relative velocity of the object. The same reasoning applies when transforming to the object's reference frame, it will notice relative motion. The relative velocity v between the frames of the observer and object can be anything from 0 to less than but not equal to c (i.e. $0\leq v<c$ ), so when you ask: "can an inertial reference frame be considered alone, or is it always relative?" - the answer is yes and no at the same time. Since you can always transform from one frame and work from there, you can talk about events as an observer in one frame would measure them to be, or identically in any other frame, but the the term "relative" is automatic, since measurements are taken with respect to the frame. Another way to consider the difference between inertial and accelerated frames of reference, consider the Minkoski light-cone diagram (it suffices to consider just one space dimension x and one time dimension t). Complicated!? No: its literally a high-school "distance-against-time graph", but axes reversed to be a "time-against-distance graph"!! The curve plotted on the graph is the worldline of the observer. The gradient is the (inverse of) relative velocity of the frame in units of c (so gradient is dimensionless). So an observer moving at relative velocity v will have worldline as a straight line of constant gradient c/v (constant since v is constant). If the observer is accelerating, then the world line is a curve, since v is bounded the gradient is always bounded between $1<c/v\leq \infty$ (usualy the parameter $\beta =v/c$ is used. The diagrams at Lorentz transformation should help (produced by Maschen). "Absolute" and relative velocity is sort of like gravitational or electric potential of a field, "absolute value" is arbitrary (i.e. gauge freedom) but the differences count and cause changes. In classical mechanics, since space and time are absolute, its possible to have an "absolute velocity" through the universe or relative to the "fixed background" (the "ether"). In relativity this is overturned by the relativity postulate above. F = q(E+v×B) ⇄ ∑ici 19:11, 3 April 2012 (UTC) @R: I see your question about rockets etc., has not been directly answered. The answer, according to the equivalence principle, is that the person in the rocket and the person standing on the earth are identical: the accelerometer reads 9.8 ms−2 in both cases and follows an accelerated (non-inertial) frame. A person on a free-falling lift is in an inertial frame. — Quondum☏✎ 20:11, 3 April 2012 (UTC) Sorry I didn't notice... I typed the above then pasted (including edit conflict). The explaination you give is nice. =) F = q(E+v×B) ⇄ ∑ici 20:28, 3 April 2012 (UTC) Really... so we may not consider ourselves inertial objects? (I cannot believe how often I'm typoing inertial. "Int" is almost automatic from my fingers, for some reason.)Rschwieb (talk) 20:32, 3 April 2012 (UTC) To a good approximation - yes we can (the Earth is locally inertial), but in actuality - no (the earth is a rotating frame of reference relative to outer space). To some level of precision in the "real" physical world, the inertial frame becomes an approximation since some amount of mass-energy will distort space-time. Maschen (talk) 20:52, 3 April 2012 (UTC) Just thought to add my opinion to the conversation, since F=q(E+v^B) has brought me over here, and you may notice the comment on the talk page with respect to this article here concerning the Basis.gif image. Thank you. Maschen (talk) 20:55, 3 April 2012 (UTC) I think you're right. =) F = q(E+v×B) ⇄ ∑ici 20:56, 3 April 2012 (UTC) Just to clarify terminology: an inertial object is any object that follows Newton's law of inertia, F=ma (or the relativistic equivalent). An inertial frame is a frame of reference that is based upon the paths of hypothetical inertial objects with zero force acting upon them. Thus, we are inertial objects, but since we have an accelerating force acting on us (via the ground), our frame of reference is not inertial. (Typos must be viral – I had to correct every instance of "inertial" I typed in this paragraph) — Quondum☏✎ 07:23, 4 April 2012 (UTC) Nice of you to clarify [it is basically what I have been saying above (and Maschen), though in long wordy paragraphs]. Importantly, in spite of its popularity, Newton's 2nd law in the form F = ma does superficially help, but is only true for a constant mass. The more general equation F = dp/dt (also an asserted definition) can derive it by the product rule (p = mv), and is true even in relativity, for both 3 and 4-momenta (though proper time is to be used for 4-momentum). I know that was obvious anyway though. I'm working away, albeit slowly. I need to start from Maschen's comment. There, it says "Earthdwellers are not technically inertial because of Earth's rotation" but I notice no comment was made on accelleration due to gravity. I have two comments/questions: To modify the Earth question again: for a massive object not rotating with respect to the universe, is a surfacedweller inertial or not? A while ago I thought about constant angular rotation and wondered if it was analogous to constant linear motion. A rotating observer might not notice it's rotating if it doesn't have a point of reference. But then I remembered the whole Mach principle anecdote, and realized that the observer would notice his/her arms pulled away from the torso. So it seems that constant angular velocity is different from constant linear velocity in this respect. Please let's resolve question 1 first, and maybe we can revisit this point later. Rschwieb (talk) 13:59, 4 April 2012 (UTC) As I use the terminology (and I think this is pretty standard), the surface-dweller would be called inertial (i.e. an inertial object), but the frame of reference stationary relative to him would not be an inertial frame. This presupposes the equivalence principle (i.e. the approach taken in general relativity in which gravity is not treated as a field). — Quondum☏ 14:14, 4 April 2012 (UTC) I've been hunting around a bit, and I cannot find support for my use of "inertial object". In fact, it does not seem to be a standard term at all. So let's just stick to the standard term inertial frame, so the question should be rephrased as "is a surface-dweller's frame of reference inertial or not?" — Quondum☏ 14:57, 4 April 2012 (UTC) We are coming back to the frame/object split again (which still feels chicken and the egg to me still). I was hoping to pin down one or the other, so it would be fine if the above question was answered with "frame" instead of "object". Here's another easy question: when do you say an object is "in a frame"? Maybe: "objects and frames which are motionless in a frame are "in" that frame". Rschwieb (talk) 15:02, 4 April 2012 (UTC) My answer to the above: First things first, rotation has a few attributes linear motion does not have. The mathematical definitions of angular quantities are basically identical to linear (position and angle, momentum and angular momentum etc., with minor exceptions of rotation needing unit vectors and cross products with the position vector). Yet tangential components of linear velocity, momentum, acceleration etc. of a point in the rotating system (even if just a coord frame), are accelerating because the velocity of the point is changing (even if the magnitude is constant the components are continuously changing). Rotation is necessarily acceleration but acceleration is not necessarily rotation (could be angular or linear). So about the question: "if a massive object does not rotate, is it inertial or not?", the answer depends on whether the object accelerates linearly relative to another inertial frame (in which case the frame is accelerated), or not (in which case it is inertial). On second thought it's misleading to say "relative to outer space/the universe" since that seems to imply space is absolute, which is certainly incorrect. A better answer would be "relative to nearby planets/moons", even just one planet or star would be enough. (The rotation of planets is complicated since it includes spin and orbital motions, to even further detail rotations are perturbed due to many-body interactions, in fact the motion is weakly chaotic). About gravitational acceleration while on Earth, all things locally are falling at the same rate (same acceleration), so there is no relative acceleration, and we can say we're in a local inertial frame. Does that clarify the first point? =) F = q(E+v×B) ⇄ ∑ici 15:13, 4 April 2012 (UTC) You can use "object" or "observer" interchangeably because as said above, we can always transform to any reference frame, and keep it fixed so we can analyze motion. I'm so sorry to confuse you over this, as said above let’s just talk about "frames with respect to whatever it is we fix the coord system to". When we say "in a coordinate system", we mean "use that coordinate system", i.e. "in a frame of referance" means to "use the coordinate system defined for that frame". F = q(E+v×B) ⇄ ∑ici 15:22, 4 April 2012 (UTC) OK, I accept "in a frame" means "with respect to that frame's coordinates" as a completely understandable definition. Now for the rotation thing. I see how "relative to the universe" is implying some sort of aboslute. But is being inertial not absolute? Can't we tell whether a frame is inertial or not by examining an accellerometer planted in it? Suppose we had a "angular accelleromter" which had rods on the sides that were flat against the side (0 angular velocity) or rising away from the sides to some degree (positive angular velocity). It would them seem possible to tell rotating frames apart from nonrotating ones, because we could simply examine that frame's angular accellerometer to see its behavior. Rschwieb (talk) 15:32, 4 April 2012 (UTC) "Is being inertial not absolute?" NO! - again, you can't tell what absolute velocity of an inertial frame is, only relative velocities, and can transform to any inertial frame. What is so "absolute" about inertial frames (in relativity)? "Can't we tell whether a frame is inertial or not by examining an accelerometer planted in it?" yes - if it reads zero its inertial! if not its non-inertial. "It would them seem possible to tell rotating frames apart from nonrotating ones" yes - because of relative acceleration, which is what I've been saying above. What's your point? Also you should stop "worrying about gravity" - treat it as an acceleration (which is, gravitational field strength g has units of acceleration, also force per unit mass).F = q(E+v×B) ⇄ ∑ici 15:46, 4 April 2012 (UTC) Dunno whether this'll help. Velocity is inherently relative, but acceleration is absolute. An inertial frame by definition has zero acceleration. — Quondum☏ 15:55, 4 April 2012 (UTC) How else to interpret gravity? How else can we understand free fall or weightlessness or for what matters the equivalence principle? While I don't undertstand GR in detial, curved space-time would lead acceleration, but isn't this also gravity? F = q(E+v×B) ⇄ ∑ici 16:02, 4 April 2012 (UTC) By the way - have we made a mistake somewhere? The article Inertial frame of reference says: "Physical laws take the same form in all inertial frames. By contrast, in a non-inertial reference frame the laws of physics vary depending on the acceleration of that frame with respect to an inertial frame, and the usual physical forces must be supplemented by fictitious forces." which actually makes more sense... We have said this is true for all observers without the inertial frame constrant, accelerations are quite complicated becuase they could be ficticous...F = q(E+v×B) ⇄ ∑ici 16:10, 4 April 2012 (UTC) A lot has taken place and I need to catch up: I think you answered: being inertial is not absolute This previous point seems to directly contradict your next answer: Can't we tell whether a frame is inertial or not by examining an accelerometer planted in it?" yes - if it reads zero its inertial! if not its non-inertial. Isn't being able to identify which objects are inertial from any frame whatsoever mean that property is absolute? That's what I had in mind anyway... And: Don't give up on me now! "Stop worrying about_____" is one of the last sounds I hear physics majors make before they get too impatient with me to continue. I am always astonished when some of my questions on fundamental-stumbling-blocks-to-my-entire-understanding-of-physics get dismissed as "things not to worry about" :P (The last time this happened, the guy gave up when he could not fathom how I saw a difference between the unit vector tangent to the unit circle at (1,0) and the unit vector tangent at (0,1)) Naturally we have different criteria for satisfaction with answers, so I hope you (and everybody else) continue to probe to find what I'm looking for with your explanations. Rschwieb (talk) 17:10, 4 April 2012 (UTC) There is no contradiction, my corresponding answers are: What do you mean by "being inertial is not absolute"? I interpret this as "whatever inertial frame of reference we choose, the velocity at which it moves is not absolute, it is relative to some other inertial frame" (again think of the electric potential analogy). How can the velocity of a frame be "absolute"? Which aspects lead to "absolution"? I hope you're not thinking space and time are absolute in the process... Again - what does the question mean to you? I interpret this as "given an accelerometer - how do we know the frame we're "in" is either inertial or not?". Put these together and what do we have? An inertial frame, whichever one we choose, is one where an accelerometer would read zero, because it moves at constant relative velocity to other inertial frames. There is zero contradiction! I'd be impressed if you could point out exactly why this is a contradiction!!! Frame of reference: select any object in the universe, and fix a coordinate system to it. The term "observer" is used for familiarity with human behaviour in making measurements (you might imagine standing at the origin of the chosen frame, looking around the place, labelling the positions and times in which events occur, hence the term "observer"), yet there is no need to take it so literally. Likewise for an object - do not be too literal. Just use the coordinate system as it has been fixed. From there, and only from there, can we describe motion and make sense of events. Inertial frame: Lets use the accelerometer you like! In an inertial frame - it reads zero. Simple as that. More specifically, Newton's 1st law is obeyed. Accelerated frame: Accelerometer gives a non-zero reading. Simple as that. More specifically, Newton's 1st law is not obeyed. However - the complication is that there may be forces observed in one frame which do not occur in another - ficticous forces (the coriolis force is a stereotypical example). Rotations are acceleration, but not necessarily vice versa, because the tangential components of linear velocity, momentum etc are changing. If you still have difficulty with gravity as acceleration, always resort to the equivalence principle: gravitational acceleration is the same as acceleration through spacetime. Step forward again. From now - I suggest you read this up further, and absorb the idea. These things are never understood overnight, you just have to think and think every now and then. =) F = q(E+v×B) ⇄ ∑ici 18:40, 4 April 2012 (UTC) Forgot to add: your statement from above "Isn't being able to identify which objects are inertial from any frame whatsoever mean that property is absolute?" is partially right, wrong, and obscure. Incorrect since from any inertial frame, all other inertial frames will appear to be inertial from the frame you choose, and again, because we can choose any frame and measure from there, the "inertiality of frames" is not absolute, it is relative. In a non-inertial frame, its awkward to answer, since the forces may be ficticious, rather than "absolute". In either case, just stick to the definitions of reference frames above. =)F = q(E+v×B) ⇄ ∑ici 19:52, 4 April 2012 (UTC) I will use "frame" rather than object from now on, judging from how much it bothers you. I have been imagining pointlike inertial objects as origins of inertial frames. I had thought there was little distinction (Maybe I'm missing something there?) We will eventually track down the source of my confusion. I think the major component is that I haven't definitions that make sense (to me) so far (that is what we're working on). I would appreciate less insinuation that I'm being careless or hasty about reading the above. Physics is simply a foreign language to me. I'd also like to add that I don't understand why you keep reexplaining some things to me, like the term "observer" and "object". I am imagining them exactly how you describe, and the concepts seem quite trivial. I have no idea what I said that makes you think reexplaining these is relevant. Rschwieb (talk) 20:17, 4 April 2012 (UTC) Rschwieb, please don't get me wrong. I'm really happy your asking these questions, and by no means faulting you for being careless or not reading the above or not paying attention - I know you are doing that and trying! =) If I sounded insinuating just then - it is only emphasis and reference to the previous parts of the discussion, not attacks, though apologies if you feel that way. F = q(E+v×B) ⇄ ∑ici 20:21, 4 April 2012 (UTC) Same about the repetition, its only to be sure. If you understand, thats fine. In response to the term "frame" instead of "object" just now (sorry if this paraphrases) was to becuase I want you to understand what inertial frames are, objects are what they are. Furthermore, you mention "imagining pointlike inertial objects as origins of inertial frames" - if that helps you, then use that interpretation. F = q(E+v×B) ⇄ ∑ici 20:26, 4 April 2012 (UTC) I apologize for the late response Rschwieb, by the look of this discussion, it has already been explained several times over by F=q(E+v^B), sort of (first at Wikipedia talk:WikiProject Physics, then Talk:Lorentz force, now here, never to the point)... About planet Earth: locally, since we cannot notice the rotating effects of the planet, we can say locally a patch of Earth is an inertial frame. But globally, because the Earth has mass (source of gravitational field) and is rotating (centripetal force), then in the same frame we chose before, to some level of precision, the effects of the slightly different directions of acceleration become apparent: the Earth is obviously a non-uniform mass distribution, so gravitational and centripetal forces (of what you refer to as "Earth-dwellers" - anything on the surface of the planet experiances a centripetial force due to the rotation of the planet, in addition to the gravitational acceleration) are not collinearly directed towards the centre of the Earth, also there are the perturbations of gravitational attraction from nearby planets, stars and the Moon. These accelerations pile up, so measurements with a spring balance or accelerometer would record a non-zero reading. Unfortunately I do not have such a clear understanding of GR either. The way I learned special relativity was by the Lorentz transformations and the Minkowski light cones more than anything, and that inertial frames are coordinate systems in which Newton's 1st law is obeyed. But everything I just said (I think) is already repeated by F=q(E+v^B) 10 times, so apologies for interrupting, then not being knowledgeable enough to help... Maschen (talk) 20:59, 4 April 2012 (UTC) I was not clear above - I meant the Earth rotating relative to the (say) solar system. If the Earth did not rotate... you may (or may not, depends) dislike this, but the answer I'd also give is the same as above. If the planet (somehow) accelerated relative to anything which happened to be an inertial frame, then the "Earthdweller" would be in an accelerated reference frame (since the Earthdwellers are at rest on the planet but the planet accelerates relative to an inertial frame, so the Earthdwellers also accelerate relative to the inertial reference frame...). If it travelled at constant relative velocity to another inertial frame, the Earthdwellers are in an inertial frame of reference. Does this make sense? Perhaps my explanation is genuinely puke... perhaps you should ask for more people at Wikiproject physics instead of me and F=q(E+v^B)? Maschen (talk) 21:24, 4 April 2012 (UTC) I think I've made a lot of progress (which may sound unbelievable until I summarize what I've discovered :) ) I'm taking an accellerometer as a primitive idea (that is, I accept it functions the way it does without further explanation.) An inertial frame is a frame for which an accellerometer affixed at the origin reads 0. A frame attached to a pointlike object in freefall is an inertial frame, as is a frame gliding through deep space at constant velocity. An frame for which the origin is accellerating, say driven by a force, is not inertial, and neither is a frame at a point on the surface of a planet (owing to the gravitational force, which will move the accellerometer off 0) I think we all agree that an inertial frame can be identified from any arbitrary frame, however it's probably wrong of me to describe that with the adjective "absolute". My question about detecting angular momentum of a frame with an angular accellerometer (the Mach Principle thing) appears to be an open question, according to absolute rotation. If it really is that hard I will just drop it indefinitely. On the last point, I've learned another nice angular accellerometer set up here, Newton's two spheres connected with a string. I have more to say but I have to go start dinner. F and Maschen, Thank you,I appreciate the comments. They will help me get into the physics mindset... however they will take me a long time to parse. Rschwieb (talk) 21:53, 4 April 2012 (UTC) One point on inertial frames: an accelerometer isn't necessarily affixed at the origin and reads zero there, it reads zero everywhere within that coordinate system, though its easier to use the origin, as you say. About measurements: using accelerometers, rulers and clocks etc - you have the correct idea (which you refer to as a primative idea) to assume they operate perfectly with zero uncertainty and 100% accuracy. For theoretical analysis there is no need to worry about the experimental errors, nor the actual mechanisms by which they operate (you say as "without further explaination"). If you already know that's fine, just checking. All points are correct (you'll be pleased to know). F = q(E+v×B) ⇄ ∑ici 07:30, 5 April 2012 (UTC) Don't worry about apologies, discussions with new people inevitably involve some toetrampling (which Q and I can attest to) but after we're past that it usually leads to lots of fruitful work (for example, the discussion page Q and I produced for GA. I'm quite happy with what was produced.) Still reading the above but posting a quick response now. I'm glad you've already confirmed something that I was about to ask. While driving home I realized inertial frames could not be spinning, and that rather than having an angular accellerometer and linear accellerometer at the origin, one could think of linear accellerometers on the coordinate axes a small distance from the origin. If all of those read 0, then I think there would be no accelleration of any sort. Barring something unforseen, I guess this would be equivalent to accellerometers at all points of the frame reading 0, as you mentioned. You also touched upon the reality that physicists necessarily have to work with error tolerances. I think I can make the recalibration required for this :) From now on, when I would like to say "exactly zero", I will try to think of it as "our best measurement is zero". I think that given sufficiently good measuring instruments, the accumulated errors would be tolerable. The business about accepting the way accellerometers work, or the absence of exact measurements might be a pitfall for undergraduate mathematicians, who have pretty romantic ideas about mathematics. Being more seasoned than that (I hope), I'll try to stay flexible. If abstract math has done anything for me it's given me mental flexibility :) Rschwieb (talk) 13:11, 5 April 2012 (UTC) Forgot to comment on what Maschen said above: while this was a better description of the Earth and acceleration etc than I gave, the centripetal acceleration is directed towards the Earth's spinning axis (approximately "cylindrically directed"), gravitational acceleration is radially towards the centre of mass ("spherically directed"), and the perturbations in planetary motion caused by the Moon, Sun, and other planets are weakly chaotic, so the directions of these components of acceleration are in very different directions, not slightly. Yet the non-gravitational accelerations are essentially negligible, so when these are vector added the only apparent acceleration we notice is gravitational acceleration.F = q(E+v×B) ⇄ ∑ici 13:34, 5 April 2012 (UTC) Probably a red herring as far as this discussion is concerned, though I'd like to add some comments for integrating into futire thoughts. #Physics §Measurements: the term measurement does not refer only to a physical experimental measurement, especially in QM. Hence, what is meant by a "measurement" could (and often does) mean something mathematically exact, especially within the framework of a theory. This includes the application of Noether's theorem. A second comment: apparatus consisting of two spheres connected by a string is not a suitable for detecting rotation unless gravitational effects are known to be zero, even in Newton's theory. Gravitational tidal effects will generate acceleration towards or away from each other, even in the absence of any other motion or rotation. — Quondum☏ 14:28, 5 April 2012 (UTC) On the first comment: OK, if any major problems arise with my usage of "measurement" I hope you let me know. On the second comment: I was only thinking of these accellerometers as imaginary ideal devices, but not as physically practical devices. In the presence of multiple accellerometers, I would not want their masses interfering with each other either (hopefully this is a reasonable exemption, since they are only imaginary.) This could be a mistake on my part though, because I might start having double standards about what physical forces apply to them. (Some rambling: If we were to consider the two objects gravitation attraction, the rotation would have to be rapid enough to produce centrifugal force overcoming the gravitational attraction to produce any tension. I suppose we'd be fooled into thinking the spinning system were inertial, if we used this method.) Hopefully my alteration to the inertial frame section above (using the regular accellerometers) does not suffer a similar defect? Rschwieb (talk) 14:49, 5 April 2012 (UTC)

Absoluteness

Abridged discussion on absoluteness

I want to wrap up the point in the archived discussion above where I thought two of F's answers seemed contradictory. I wanted to say "being inertial is absolute", but I probably just misused the word. I meant that "all observers in all frames agree on which objects are inertial." So, to me, the two statements "being inertial is not absolute" and "everyone agrees upon inertial frames" are patently contradictory statements. Later, F, you provided your interpretation which explains why you disagreed: "What do you mean by "being inertial is not absolute"? I interpret this as "whatever inertial frame of reference we choose, the velocity at which it moves is not absolute, it is relative to some other inertial frame" (again think of the electric potential analogy). How can the velocity of a frame be "absolute"? Which aspects lead to "absolution"? I hope you're not thinking space and time are absolute in the process... Suggesting that some velocity is absolute is so absurd I did not even think of it, so I'm glad you indicated that above. I hope I made the "inertial frames are recognizable from observers in all frames" interpretation clear at some point (maybe after that paragraph). It seems that "all observers agree on which frames are inertial" is a true statement. So is "absolute" really the wrong word? After I read a little more I've begun to think that the proper use of "absolute" implies some sort of distinguished thing against which all other things are compared. Maybe the right word for "agreed upon by all observers in all frames" would be "invariant"? I don't like invariant at all really... still looking for a better word. Rschwieb (talk) 20:01, 5 April 2012 (UTC) The statements "being inertial is absolute" and "all observers in all frames agree on which objects are inertial" simply do not equate to each other, because of constant relative velocity. For any number of inertial frames, all observers mutually agree that their accelerometers read zero. It really is misuse of the term "absolute" with inertial frames in relativity (but not classical mechanics). Yes - "absolute" genuinely is the wrong word for inertial frames, and certainly not to be misused for invariant - which is correctly applicable to inertial frames: the laws of physics are, without fail, always the same (invariant). Invariant only means something does not change. Absolute means "only one entity" (say, the value of a quantity). There is an absolute temperature in thermodynamics, but how do you compare this with velocity (in relativity)? Suppose we are in an inertial frame, with no others locally around (so we are "alone in the universe"), in which Newton's 1st law applies. Since there is zero resultant acceleration of the frame, what velocity does the frame travel? Why? Now suppose another inertial frame comes along. What velocity does our frame travel at now (forget the other frame)? Since can use our coordinate system to measure the motion of the other inertial frame (moving) we have something to refer to when we ask "how fast are we going?" The observer in the other inertial frame will also follow the same reasoning. The first question is rhetorical - can it be answered!? The second can.F = q(E+v×B) ⇄ ∑ici 22:36, 5 April 2012 (UTC) Before you ask about absoluteness of non-inertial frames, when Quondum said acceleration is acceleration is absolute (while velocity is not), I'm now on the boundary between "yes I agree" and "no I disagree", have to research and think about this myself some more. A non-inertial frame does have an absolute acceleration relative to an inertial frame, but what about two non-inertial frames? Acceleration must be relative if the equivalence principle holds (but then relative acceleration is zero for that case). You mentioned Absolute means "only one entity" (say, the value of a quantity). I guess I was just tempted to call the 0 measurement on all accellerometers of inertial frames an "absolute property" that all inertial frames share. Your above comment seems to jive with what I suggested about abosluteness referring to a distinguished object (like absolute space), against which others would be referenced. I could understand that, and I would be sure to be careful before trying to use "absolute" again. Rschwieb (talk) 02:33, 6 April 2012 (UTC) You've convinced me that "absolute" does not mean "frame independent". I had the feeling that "relative" might be the antonym of "absolute". I was tempted to say that "relative" was close to "frame dependent" and "absolute" close to "frame independent". I was given feedback that my attempt was a failure . My second attempt to pinpoint absoluteness is: (*)"If blah is absolute, that implies there is a distinguished thing X which is the baseline for blah". Example: Newtonian physics - "the universe" is a distinguished frame which we can compare all frames to. Example: thinking that time is absolute - there is a single clock somewhere that is "the correct" time, and the way to determine if a clock in the universe is "correct" is to compare it to the universal clock. Can you comment if I'm heading in the right direction with statement (*)? Rschwieb (talk) 12:59, 6 April 2012 (UTC) Rschwieb, there was no friction, nor criticism, or any need to apologize. =) As to *, yes - the answer is correct, if you're interpreting X as some inherent feature of "blah"(!) as the base measurement (i.e. zero value) of "blah". However - there is the subtlety of this X set to a non-zero value becuase of the units used. 1st example: Yes, absolutley true. "blah" is space, X is the "background universe". The subtlety is: where is the origin? Its arbitrarily chosen, but once chosen it can be fixed, and define all positions in the universe. =) 2nd example: About the clocks, well, what do you mean by "correct time"? This is just units of absolute time, if all clocks are synchronized to start at simaltaneously, and each continue to tick at the same time interval (whatever the unit of time is chosen to be), then yes all clocks agree that each tick is simultaneous. So absolute time means that the unit interval of time is the same for all clocks independent of their motion. Hence "blah" = absolute time , X = the start and finish times (not a contradiction - clock/modulo arithmetic!). The subtlety in action: Two clocks ticking, each in different units, mean the ticks will not be simultaneous, but this is just units which clearly does not affect the physics (does affect calculations obviously!). =) F = q(E+v×B) ⇄ ∑ici 15:36, 6 April 2012 (UTC) Another example: This is exactly the concept of absolute temperature I gave above (I know this detracts from relativity to thermodynamics, but its a good example). Details aside, here is the idea: Molecular motion is easily found experimentally to be temperature-dependent. Again - whatever units we use, if we continue to decrease the temperature (of whichever system we analyze) we find, at some point, there is a temperature when molecular motion "ceases" (not quite because of QM zero-point energies in molecules), put another way the entropy of the whole universe reaches a minimum. What does this mean? That molecular motion is bounded by temperature, that there is a minimum temperature - called absolute zero temperature. So here X = absolute zero temperature, "blah" = temperature. Now for the subtlety: because of the way the Kelvin is defined, the relation between Kelvin and Celsius temperatures is conveniently linear: $T({}^{\circ }C)=T(K)-273.15$ so temperature change in kelvin = temp change in celculuis, and absolute zero corresponds to zero Kelvin but -273.15 Celsius. Obviously just a matter of units though, independent of the physical phenomenon. =) F = q(E+v×B) ⇄ ∑ici 15:37, 6 April 2012 (UTC) OK, I'll be sure to keep the temperature example in my pocket too. Being able to change the units on the axes of the frame makes sense to me too. In Euclidean space, it just amounts to multiplying the coordinates by a diagonal matrix, "stretching/compressing" the axes to get between such systems. Since the transformation does not change direction or angles, it's relatively easy to accept as "harmless". I think I'm OK with absoluteness for now. Starting another question in the new section. Rschwieb (talk) 18:03, 6 April 2012 (UTC)

Relativistic rotation thought experiment

Collapsed discussion

Here is a thought experiment I've had in mind for a long time, but I don't know how to understand what happens. Imagine a large disc with a radius drawn from center to edge, with the two ends and midpoint of the radius clearly marked. Select a convenient inertial frame of reference with two spatial axes in the plane of the disc, and the last axis perpendicular. Suppose the disc is then sent spinning so that the instantaneous speed of the edge point is near the speed of light, as seen from our selected frame. In our everyday nonrelativistic experience, we're aware that the instantaneous speed of the edgepoint is greater (twice as great?) than that of the midpoint, but of course, something stranger is going to happen as the midpoint approaches the speed of light (the edgepoint is certainly not going to go twice as fast). I find this especially hard to think about, and I imagine it involves explaining some lack of simultaneity. Will the radius appear to be curved? Will Lorentz contraction do anything weird? (Naively one might think the circumference of the disc shrinks.) Rschwieb (talk) 18:03, 6 April 2012 (UTC) I think this one is trickier (more controvertial) than you might realize. See Ehrenfest paradox. One thing to realize is that the concept of a rigid body is not compatible with special relativity. If you insist that the disc rotates "rigidly" in some sense (and also take the low-mass limit to avoid gravitational issues), I doubt the midpoint will ever exceed half the speed of light; you'd just keep pouring energy into getting the periphery moving closer to the speed of light. I think it would be more instructive (and a good deal simpler) to examine a "rigid" rod being accelerated lengthwise, and see how rigid it seems when examined from different inertial frames of reference. — Quondum☏ 19:24, 6 April 2012 (UTC) It would not surprise me at all if it were really hard. I intentionally didn't suggest anything about the physical characteristics of the disc other than its shape, because it's purely imaginary. Certainly if you consider it to be massive and require energy to accellerate that's a different discussion. I had a rigid ideal object in mind, I would not be surprised if you were right about the concept being inconsistent with relativity. Q about what you said abou the midpoint: Suppose you believe the edge can be brought up to .6*c. Then somebody throws aside a curtain and reveals what you thought was the edge is actually the midpoint of a larger disc. Wouldn't this imaginary situation mean that midpoints can exceed half the speed of light? Maybe if we actually write down the relativistic equations of motion we would actually get somewhere. Question #1: "how does contraction come into the rotational picture?". Rschwieb (talk) 00:33, 7 April 2012 (UTC) Mathematically, there is no problem with what you propose. Suppose that the disc is defined as a set of points moving in time, along trajectories that an observer who is stationary with respect to the centre of the disc considers to stay in radial lines rotating with a uniform angular velocity and each point at a constant radius. This picture can be extended out to infinity: there is no limit to the radius of the disc. At some radius the rotating points will be moving at the speed of light, and beyond that they will be moving faster in proportion to the radius. The only difference is that inside this radius, the world lines are timelike, at the radius they are lightlike, and outside it they are spacelike. It is only when you start to impose impossible conditions such as that they should be physical objects or that the disc should be "rigid" that problems occur. As far as contraction comes into the picture, if you have an observer on one of the timelike trajectories, as the disk (suddenly for purposes of illustration) speeds up, he will see the neighbouring points move away due to e.g. non-simultaneity in the time of acceleration of each point from his perspective. Mathematically, if you wanted to analyze this, it would probably be simplest to consider an inertial frame of reference in which some point on the disc is momentarily stationary. — Quondum☏ 06:56, 7 April 2012 (UTC) Lets try it with relativistic mass (I'll use a possible frame you propose, one with say origin of coord system at the centre of the disk, the z axis through the axis of rotation, the xy plane in the disk's plane, and measure the tangential velocity v of a point on the disk, parallel to the y axis (or equivalently the x axis, doesn't matter), observed and measured from the origin, i.e. from (0,0,0) look at a point (0,r,0) [or (r,0,0), or even (a,b,0) where a2+b2=r2 but that would be stupid...] where r is the radius of the point where |v| is nearly c.) Because of mass-energy equivalence, the mass of an object increases when relative velocities approach the speed of light which is where the energy goes, given by $m=\gamma (\mathbf {v} )m_{0},\,\gamma (\mathbf {v} )={\dfrac {1}{\sqrt {1-\mathbf {v} \cdot \mathbf {v} /c^{2}}}}$ (m0 = rest mass) since we're considering an instantaneous component of (relative) velocity v at a point (in the inertial frame just stated), the v remains the same magnitude and direction (when the disk is not angularly accelerating), the disk turns around continuously the component is observed there continuously. though most likley there are subtleties I'm not including (such as accounting for gravitation due to a spinning mass). Anyway... → At some radius r where |v| ≲ c, the mass of the disk from the inertial frame would appear to be increasing to infinity, which is why it can't exceed c, and the energy has to come from somewhere (this formula arises from applying conservation of mass and energy). Even in principle, to asymptotically increase the speed of the disk around r to c you need an infinite amount of energy as the mass approaches infinity. → The parts of this infinite disk beyond r which move at |v| ≥ c can't be observed, put another way their worldlines are space-like (in other words, inaccessible parts of spacetime) described so nicely above. F = q(E+v×B) ⇄ ∑ici 10:29, 7 April 2012 (UTC) The principle of increase in mass is not controvertial, but the interpretation of "inaccessible" I don't think we should delve into yet (after all, QED necessarily includes spacelike propagators in its formalism with observable consequences). What might be helpful as a first step in grasping what it looks like is to consider observers sitting on the disk up to lightspeed radius, to develop a coordinate system for each, and to work out how far away he neighbours end up (stretch factors), and how much their clocks are shifted according to their position. The reference frame I suggested will correspond to that of one oberver on the disc at some instant. The transform from the central observer to the moving frame will be $t'=\gamma (t-\omega rx/c^{2})$ $x'=x$ $y'=\gamma (y-\omega rt)$ $z'=z$ $\gamma =1/{\sqrt {1-(\omega r)^{2}/c^{2}}}$. The neighbour has similar equations, but with a rotation. After acceleration of the disc, a neighbour at the same radius is at a constant separation both in time and space from the orbiting observer's perspective; it'll take a little algebra to parameterise these separations by location and time (I'm feeling a tad lazy to do that at the moment). But I hope you see what I'm getting at: each neighbour's clock will have gained an offset, the disc will have distorted (stretched circumferentially locally and maybe shrunk on the far side, and quite possibly the circles may become non-circles. Clocks further out will have slowed, those further in will have sped up. All within the framework of special relativity, and in principle without mathematical difficulties but with definite predictions. — Quondum☏ 12:13, 7 April 2012 (UTC) OK, I am adding this to my paper outline of things to remember when thinking about this problem again. I'm reminded of my lack of practice even with basic linear problems. Since a lot of the explanation is "over the head" of my current ability, I'll be postponing any more questions along this line for a while. Thanks for the above input and explanation. Rschwieb (talk) 14:25, 7 April 2012 (UTC) If I could add... I just wanted to provide Rschwieb some background on what you said earlier about energy, and how it can be understood using an inertial frame. The Lorentz transformations you give above do look correct since we only consider an instantaneous relative velocity, and $|\mathbf {v} |=\omega r$, but I didn't write them since I wasn't 100% sure we could transform the points which move on the disk in this way (and have no experience with transformations incorporating rotations either...). Just to formalize Rschwieb's original question mathematically, we want to see what happens to coordinates in the limit $|\mathbf {v} |=\omega r\rightarrow c$. For fixed $\omega$ we can change r. For $r\approx 0\,\Rightarrow \,\gamma \approx 1\Rightarrow \,t'\approx t$ (clocks near the centre almost tick simultaneously), for r when the edge of the disk rotates at tangential speed c, given by the limit $r\rightarrow c/\omega \,\Rightarrow \,\gamma \rightarrow \infty \,\Rightarrow \,t'\gg t$ (i.e. clocks further away slows down). Same for y, or better yet an infinitesimal length dy, when measured in the inertial frame at time t $y'=\gamma (y-\omega rt),\,y'+dy=\gamma (y+dy-\omega rt)\,\Rightarrow \,dy'=\gamma dy\,\Rightarrow \,dy=dy'/\gamma$ (just length contraction), so for $r\approx 0\,\Rightarrow \,\gamma \approx 1\Rightarrow \,dt\approx dy'$ (infinitesimal lengths circumferentially almost the same in each frame) and $r\rightarrow c/\omega ,\,\gamma \rightarrow \infty \,\Rightarrow dy\rightarrow 0$ (shrinks further away)... Nice thinking Quondum!... =) F = q(E+v×B) ⇄ ∑ici 15:17, 7 April 2012 (UTC) Suppose we took the disc we have been discussing and just slice it along two radii meeting at an acute angle to produce a sector. I imagine all the above analysis holds? Is the sector's shape modified as the system rotates at reletavistic speeds? Rschwieb (talk) 16:12, 7 April 2012 (UTC) That’s an interesting question =). I think it would apply considering an instantaneous moment while the sector passes in front the observer, and can measure the instantaneous relative velocity then. The shape would still be contracted circumferentially and time dilated further away as it rotates all the way around, since the observer in the inertial frame could turn his view to (say the x direction after the y) we could pick an instant when the sector passes in front (of the x axis), and repeat the analysis. Notice the coordinate frame is not rotating with the view of the observer. F = q(E+v×B) ⇄ ∑ici 16:42, 7 April 2012 (UTC) This was why I wanted to treat the disc as a bunch of individual atoms, so that rigidity was not an issue. If we consider narrow sectors with each part held in place by whatever centripetal force is necessary to keep it on the same circle, yes, the sector will get narrower, more the further out you are. Gaps will open between the sectors, producing a flower shape (as I understand it). It may make sense to think of things like the ladder paradox first, this dealing with linear motion and hence being simpler. — Quondum☏ 17:52, 7 April 2012 (UTC) The "flower from a sector" visualization illustrates nicely how the contraction occurs on a rotating disk from an inertial frame. Even though it was a very interesting question, it's certainly better (though probably less gripping) to understand Lorentz transformations for linear motion before anything, since the Lorentz transformations for boosts (in streight lines) are just matrices and 4-vectors. By the way, velocity addition in relativity is intrinsically non-linear (and that's just for relative velocities in streight lines, nevermind acceleration or rotation...that was a random but relevant aside...) F = q(E+v×B) ⇄ ∑ici 22:24, 7 April 2012 (UTC) As you go out radially along the sector, the proper arclength becomes larger but the contraction also becomes greater. At one point I wondered if the two radii would not merely just look increasingly parallel the further out you went, but the description of a petal makes it sound like you're sure the contraction outstrips the increase in proper arclength. Is there an easy way to see that? Rschwieb (talk) 01:23, 8 April 2012 (UTC) Yes - look at the calculations above: "$dy=dy'/\gamma ,\,r\rightarrow c/\omega \,\Rightarrow \,\gamma \rightarrow \infty \,\Rightarrow \,dy\rightarrow 0$ [shrinks further away (to zero length - so contraction dominates)]...". The gamma factor is the behind the numbers. F = q(E+v×B) ⇄ ∑ici 01:59, 8 April 2012 (UTC) I spent time reexamining them and I see why. Briefly, the contracted length of the arclength is approximately (using a wedge of 1 radian for convenience) $r{\sqrt {1-{\frac {\omega ^{2}r^{2}}{c^{2}}}}}$ which I see does tend to zero. (As an aside, always bear in mind my unfamiliarity with the equations. Referring me to a stack of equations is OK if you really can't think of a clearer snapshot though :) In this case being forced to work with them helped out quite a bit. A handful of encounters with engineers making mistaken arguments with similar stacks of equations and limits has taught me to exercise caution using such arguments.) Analytically that equation reaches a max at $r={\frac {c}{\omega {\sqrt {2}}}}$. I'm not fully sure the figure works exactly to find the widest point of the petal (because we are making some arguments about some curved edges being approximately straight), but it's probably close. Rschwieb (talk) 14:34, 8 April 2012 (UTC) This sounds right. If you model each sector as a series of chains in arcs, unconnected to each other each but each constrained to remain at its own radius and not to fold, with the centre of every chain on the same (rotating) radius, this approach will work (exactly) without concern about curvature. — Quondum☏ 15:29, 8 April 2012 (UTC) Starting from the arc length for 1 rad: $s=r{\sqrt {1-{\frac {\omega ^{2}r^{2}}{c^{2}}}}}$, while I calculated the maximum ds/dr = 0 and agree on the value you give, I'm not sure why you've used this equation, or interpreted it correctly. For 1 rad (quite a large angle ≈57°) you are considering an arc length s corresponding to that angle (and radius r) which will also quite big. Yet s has to be very small to approximate a streight part of a curve, which would mean r would then have to be very small or tend to c/ω. The above analysis for dy and dy' is correct as these are infinitesimals, but you seem to be assuming a substantially greater (curved) arc length (?) can be treated approximately straight (when it isn't for 1 rad)? This doesn't seem consistent, please correct where I'm wrong. =( F = q(E+v×B) ⇄ ∑ici 16:29, 8 April 2012 (UTC) On the contrary, using $r=c/{\sqrt {2}}\omega$ (which is ≈0.707 the radius of wherever the disk rotates at speed c ), the contracted maximum arc length from the inertial frame is $s=c/2\omega$ (=0.5 the radius of wherever the disk rotates at speed c )... $r\rightarrow c/\omega$. It seems quite reasonable, somehow... because s beyond that would continue to increase slightly but the contraction overrides it completley for the short remainder ≈0.29 of the luminal velocity radius... what do you two think? F = q(E+v×B) ⇄ ∑ici 17:00, 8 April 2012 (UTC) (ec, I have not read the new stuff you posted yet) I'm sure you are comfortable plunging directly to the differential, but in my case, I feel more comfortable working from the approximations up. It's easy to overstep bounds if one is too free with limits and differentials. I'm hoping I'm erring on the side of caution. Surely for any nonzero theta, the arc will be curved. Of course, the larger r is and the smaller theta is, the straighter the arc will appear. You seem to be implying there is a threshold where I may or may not call a bent line "approximately straight"? They're all bent! :) Rschwieb (talk) 17:24, 8 April 2012 (UTC) Yes spacetime is really curved, but the point previously made is that the arc corresponding to 1 rad is certainly noticeably more curved than dy and dy'. I hoped you would find differentials more helpful; the equation $dy=dy'/\gamma$ is far simpler to understand using differential lengths than worrying about arc length approximations. I'm simply saying not completley comfortable with your version of the analysis above... F = q(E+v×B) ⇄ ∑ici 18:18, 8 April 2012 (UTC) I think the only differentials I'm at home with are those in calculus, like the slope of a tangent line, and "relative change". Thinking of the differential form of the metric as "measuring difference beween neighboring points" hurts my brain, because I think they should be distance 0 apart. It requires a little unlearning to understand differentials, I think. Wait a sec, I think I caught something you're saying that I missed. Of course, the Lorentz contraction comes into play when something is moving in a straight line, but are you saying that it's a general principle that it's valid to apply Lorentz contraction to differentials in the direction of motion? If I am measuring something moving along the x axis, then I would multiply it with the Lorentz factor to contract it... but I feel like the differential would be perpendicular to the motion so... I'm lost! Rschwieb (talk) 19:40, 8 April 2012 (UTC) Don't consider a varying metric at this stage: think in flat (Minkowski) space. Space only becomes curved around mass, and we're dealing with the limit of light masses. Yes, the Lorentz factor applies in the direction of motion (x) and in time (t); the factor in the perpendicular directions (y and z) is 1. It's generally best to draw the object and coordinates in each frame on an x–t graph. And it is valid to apply Lorentz contraction to infinitesimal bits of bodies and integrate the result. Velocity is the only important thing in this; acceleration does not affect the infinitesimal picture. — Quondum☏ 20:23, 8 April 2012 (UTC) Clarifying Lorentz contraction: Its perfectly possible to Lorentz transform coordinates in an arbitary direction (obviously more complicated), but still whatever the direction of motion, the Lorentz contraction is always along that direction. There is no perpendicular warping. For this case remember the tangential velocity v for the rotaton was measured parallel to the y axis because of how the system was set up (clearly it would not matter to use dx and dx' instead, correspondingly y = y', when measuring along the y axis, i.e. an instant when v is parallel to the x axis and perp to y). F = q(E+v×B) ⇄ ∑ici 21:49, 8 April 2012 (UTC) I had known contraction only takes place in the direction of motion, but learning that it's valid to use on differentials is new to me. Doing this with relativistic vector algebra is also awkward for me. From polar coordinates, y=r*sin(t), whence dy=r*sin(t)dt. Applying Lorentz contraction to the differential dy, the contracted length $dy'={\sqrt {1-{\frac {\omega ^{2}r^{2}}{c^{2}}}}}r\cos(t)dt$. In this expression, as r approaches c/omega, dy' approaches 0. Is this basically the argument that was given above? I suppose eventually I might want to look at something that helps explain why we can treat differentials like long straight rods. At the moment, I'm only familiar with the fact that "the length of a long straight rod is contracted by this factor", and that seems pretty different from this infinitesimal object. Rschwieb (talk) 00:21, 9 April 2012 (UTC) Every object can be modelled as a framework built out lots of little molecule-sized rods. Applying Lorentz contraction works at any scale. So applying it to differentials is a straightforward process of taking the limit as the length of the rods diminishes. Take care with your expression, however; is not correct when you move off the x-axis, because the direction of motion is no longer in the y direction. Once cos ωt reaches zero (the particle having moved π/2 radians), the velocity and hence contraction is in the −x direction, not in the y direction. So you'd have to rework it for a full expression involving x and y once you get off the x-axis, as suggested by F (it involves a matrix). Alternately, do the analysis constrained to the x-axis for simplicity, and rotate the result using the symmetry of the problem. — Quondum☏ 05:17, 9 April 2012 (UTC) Hm, that's a great deal more complex than I was intending at this point... I was only looking at a small sector with one side on the x-axis because looking at any other small sector would be exactly the same (which is what you're talking about, I think, Q, constraining the analysis then relying on symmetry.) But I see what you're saying about a full expression, and it'd probably be good practice for me. I realize that the contraction in the y direction will gradually transfer to the x direction until pi/2, then begin to transfer back to the y direction approaching pi etc. It makes sense to me to parameterize the path of a particle along a circular orbit around the center. Then you can form a differential (not sure with respect to what variable) and do arclength. I think I'm trapped in a Newtonian frame of mind right here, because I don't know officially what happens next. (Intellectually I gather that we will be transforming each tangent vector with a Lorentz matrix boosting in the direction of motion, but read the next sentence to see what I'm really asking.) If we are applying a Lorentz transformation to a tangent vector, it's because we are moving between two frames? What two frames are those? Rschwieb (talk) 13:16, 9 April 2012 (UTC) Yes, the full treatment does become a little tedious, and would be inadvisable for getting the hang of the idea. Your description seems good. Just regard dy as the length of a tiny but otherwise arbitrary-length vertical rod embedded in the rotating sector as it crosses the x-axis as seen by a "stationary" observer, and dy′ as its length measured by an observer in the frame of the moving rod; we are only interested in ratios of differentials. Yes, the Lorentz transformation corresponds to a change of frames, or equivalently, a change of coordinates between the two observers as given by special relativity, which I gave higher up. Put a "d" in front of each side and you've got your Lorentz transform for this example. — Quondum☏ 13:50, 9 April 2012 (UTC) Can we say exactly which two frames am I travelling between when I apply the transformation? Rschwieb (talk) 13:55, 9 April 2012 (UTC) When you lorentz transform coordinates from one frame to another, you choose which frame you're already in and which one to transform to, so yes you can. F = q(E+v×B) ⇄ ∑ici 14:22, 9 April 2012 (UTC) (ec) The first is the (t,x,y,z) frame, which corresponds to the observer sitting placidly in the centre of the spinning disc. The second is the (t′,x′,y′,z′), which corresponds to the observer travelling tangential to the disc (briefly travelling alongside –tangential with– the rod dy under consideration), with his 4D origin chosen to coincide with that of the stationary observer. I would call it "switching between" rather than "travelling between": there is no "travel" involved, only a (passive) change of coordinates in 4D space. — Quondum☏ 14:26, 9 April 2012 (UTC) Q's on the mark, I'm literally asking what the two frames are (although I may not have been clear enough.) OK, I will say "switching" rather than "travelling" to avoid the idea of actual movement. I think we might have scrambled coordinate notations from before, so let me try to say the whole thing here. So, from the frame (t,x,y,z) the stationary observer staring down the length of the x-axis sees the tangential observer's frame (t',x',y',z') zip across the x-axis. An infinitesimal tangential rod with proper length dy' appears to be contracted to dy=dy'/γ where γ is the appropriate contraction factor. I think that this amalgamation of the Newtonian work with the relativistic work is sitting better with me after this talk. Thanks! Rschwieb (talk) 15:11, 9 April 2012 (UTC) Spot on. I cannot improve on your synopsis. — Quondum☏ 15:30, 9 April 2012 (UTC)

Lorentz and Poincaré groups

(I'd like to know why we may safely assume the origins coincide.)

The symmetry group of Minkowski space is the Poincaré group, just as the Euclidean group is the symmetry group of Euclidean space. That is to say, it is the group of all orthogonal transformations of one rectilinear coordinate frame to another, and includes translations, rotations, Lorentz boosts and compositions thereof, exactly the symmetry one would expect of an affine space. The Lorentz group is a subgroup thereof restricted so that a selected point (which we could call the origin) is unaffected by the group action. So to answer the query, it is a matter of definition, primarily to simplify algebraic formulae when translation is of little interest. — Quondum^☏ 16:24, 9 April 2012 (UTC)

Umm. Now that I think of it, the Poincaré group is the symmetry group of the affine Minkowski space, and the Lorentz group (aka Lorentz transformation when passive) is the symmetry group of the Minkowski vector space. The latter is applicable to deltas (displacements and differentials) – in this case it is not merely a convenience. — Quondum^☏ 16:45, 9 April 2012 (UTC)

I'd like to know more along these lines. When studying isometries of the plane, I learned the elementary fact that isometries decompose into translations and rotations(=isometries that leave 0 fixed). This does not induce a direct product of groups, but rather a semidirect product, owing to the different behavior of the translations and rotations. So, it makes complete sense to me to view the Poincare group as the (3+1 metric) isometries of Minkowski spacetime, and the Lorentz group as the subgroup of rotations in this Poincare group. The feeling I'm geting is that translations are "invisible" to general relativity... is that about right? Rschwieb (talk) 17:35, 9 April 2012 (UTC)

It depends what you mean by "invisible". If you mean it is an exact symmetry of the theory, yes, in the same way as a rotation is. An experimental setup cannot determine where/when it is, cannot determine how fast it is moving and cannot determine which way it is facing, except in relation to local features. However, "local features" might be rather extensive, such as the rest of the universe, which in our case does at least give us a "when" and "how fast". The dropping of translations when going from affine to vector spaces is inherent mathematically in the definition of these types of spaces. — Quondum^☏ 17:49, 9 April 2012 (UTC)

I'm trying to say "translations are invisible" because it looks like we don't have to distinguish between frames that differ by translations. That kind of makes sense since a frame and a translated frame are moving with each other, and they both look motionless to each other. Maybe I need to develop a feel for "affine space", because currently I have none. Rschwieb (talk) 18:09, 9 April 2012 (UTC)

Translations are fairly innocuous in that they are particularly simple to add or remove from the full Euclidean/Poincaré groups; when your coordinates are linear, it is a simple matter of subtracting a constant from all coordinates. This is so obvious that one tends to do so automatically, so in this case "invisible"≈"simple" (or "uninteresting"). But it is not invisible mathematically: you'll see that a change of velocity (of the observer) can move a distant event from the past into the future and vice versa (provided it is outside the light cone). Acceleration swings your coordinates (including time) around as dramatically as the steering does in a car. I definitely recommend you look into affine space – it is actually a very simple concept, amounting to taking a vector space an "forgetting" where the origin is (and thus introducing greater symmetry). Real (Euclidean) geometric space is the prototypical example: there is no special point somewhere in space that is "the origin". You can subtract points, obtaining a displacement vector (that lives in a separate space), but you cannot add points except when you weight them with a total weight of 1 (and in so doing get a "centre of mass"). — Quondum^☏ 18:55, 9 April 2012 (UTC)

I think we're now getting close to the borders of my knowledge again (a good thing). I think we've discussed the "forgetting the origin" idea before, and intellectually I can understand it as a vector space acting on a set A. I'll have to reread the affine space article a few times... Rschwieb (talk) 19:34, 9 April 2012 (UTC)

I can't remember what I read, but I think I read that a Lorentz boost can be decomposed into a spatial rotation and a "pure boost"? First determine if that is crazy talk. Then help me with this: Suppose there are two inertial frames, one which appears motionless to me, and another which appears to be moving away with uniform velocity. Can I think of a "pure boost" between these frames as a sudden impulse which makes the stationary frame travel with the moving one (with zero velocity between them)? Rschwieb (talk) 16:52, 12 April 2012 (UTC)

Yes to all (except to it being crazy talk, and we call the combination a Lorentz transformation, not a Lorentz boost). The decomposition does not commute. But: the frame in which each action is defined as "pure" almost certainly affects the decomposition, making it non-unique. Nevertheless, once the two frames for defining "pure" are specified, in (3+1)D such a decomposition is always possible and unique. I imagine you could define the two frames in terms of the object being boosted/rotated, in which case the decomposition would be invariant. — Quondum^☏ 17:26, 12 April 2012 (UTC)