Cap'n Refsmmat Posted August 3, 2011 Posted August 3, 2011 My argument that earth is in fact nearly spherical and not squished nearly flat (as the disputed frame of reference would have it) is based on a lot of earth science and common experience among earthlings since the "flat earth" era before science knew anything about earth's shape. And now we have the beautiful view of a spherical earth from space. Ah. So you have a mountain of evidence that suggests that Earth appears nearly spherical from this reference frame, and from similar ones. Do you have a similar mountain of observations from other reference frames? If not, how can you justify your conclusion? Relativity agrees that anyone sitting on Earth will agree it is very nearly spherical, so your evidence does not contradict it in the slightest. If travelers through our solar system want to know the distances between objects at any given moment, they will need their Lorentz transformation equations to translate what they see, zipping by at near 'C' (maybe seeing 1/8th of our AU, for instance) into the actual, objective, intrinsic, as it is, independent of various observational differences... distances. (The latter are of course best known from the end points, at rest with the distances measured.) Hm. But the Earth is zipping around the center of the galaxy at high speed. Surely they should use their Lorentz equations to translate what they see into the real rest frame, since Earth is moving. Right?
tar Posted August 4, 2011 Posted August 4, 2011 Dr. Rocket, I am quite confused about the difference between special and general relativity, and I am not sure about a theory that is a "special" case of another theory, with "gravity" left out. Where in reality do we find a situation where gravity is left out? And isn't it better to have a theory, that fits it back in? Regards, TAR2
swansont Posted August 4, 2011 Posted August 4, 2011 Dr. Rocket, I am quite confused about the difference between special and general relativity, and I am not sure about a theory that is a "special" case of another theory, with "gravity" left out. Where in reality do we find a situation where gravity is left out? And isn't it better to have a theory, that fits it back in? Regards, TAR2 As with many areas of the rest of physics, it's an idealization. But there are many cases where the effects are so small (or uniform) they don't matter, much like how you can use 1/2mv^2 for kinetic energy because the correct relativistic formula only differs out at the 15th digit. As a matter of scale, gravitational time dilation is a part in 10^16 per meter of height change, and effects from the sun are a thousand times smaller.
owl Posted August 4, 2011 Author Posted August 4, 2011 Ah. So you have a mountain of evidence that suggests that Earth appears nearly spherical from this reference frame, and from similar ones. Do you have a similar mountain of observations from other reference frames? If not, how can you justify your conclusion? Relativity agrees that anyone sitting on Earth will agree it is very nearly spherical, so your evidence does not contradict it in the slightest. Hm. But the Earth is zipping around the center of the galaxy at high speed. Surely they should use their Lorentz equations to translate what they see into the real rest frame, since Earth is moving. Right? Well, my theory is that, anywhere in the universe, the most accurate description/measurements of any object (or distance between objects) will be from the at rest frame relative to the object or distance described/measured. This seems like an unnecessary repeat, but I think the theory is verified by both kinds of epistemology mentioned in my "theory of common sense" post above. For one thing, length contraction on large scale has never been experimentally verified. (Also mentioned above.) Further, earth can not be both spheroid and radically flattened, so one shape must be true and the other false (or say a distorted image.) My money is on the spheroid for all the reasons recently argued. If we defined length contraction as the distorted optical image of objects due to the extreme frame of reference of flying by at near light speed and yet depending on light to convey the image... there would be no dispute here. But consider again the high speed voyagers flying by, seeing earth as squished at the poles... then (second pass) squished at the equator, and then, from orbit (at rest frame) as nearly spherical. Say they are on a mission to report the shapes of bodies in our solar system. What will they report? That earth changes with the three frames above? No. All agree that it does not radically change shape. So what will they report as good objective scientists? To the last challenge: I keeping with all of the above,... If dwellers near galactic center want to know earth's shape, they had best come close, preferably into earth orbit and find out without all those high speed (and long distance) image distortions. Say the voyagers above were on such a mission. I'm betting they reported a nearly spherical earth.
Cap'n Refsmmat Posted August 4, 2011 Posted August 4, 2011 But optical distortions are a completely different effect from length contraction, and can be easily accounted for. Length contraction only becomes visible once you've removed all the strange optical effects that occur when you travel near the speed of light. Lorentz contraction does not come from relying on light to relay an image while traveling near the speed of light; only relativistic aberration and similar phenomena do.
swansont Posted August 4, 2011 Posted August 4, 2011 But optical distortions are a completely different effect from length contraction, and can be easily accounted for. Length contraction only becomes visible once you've removed all the strange optical effects that occur when you travel near the speed of light. Lorentz contraction does not come from relying on light to relay an image while traveling near the speed of light; only relativistic aberration and similar phenomena do. Put another way, if you measured length using the timing of events in each frame, you would conclude that the object is contracted in the moving frame. All sensors. No optical illusions. Optical illusions cannot account for the effects in collisions of relativistic nuclei. The nuclei really have to contract for the theory to work, and the theory works.
owl Posted August 4, 2011 Author Posted August 4, 2011 (edited) But optical distortions are a completely different effect from length contraction, and can be easily accounted for. Length contraction only becomes visible once you've removed all the strange optical effects that occur when you travel near the speed of light. Lorentz contraction does not come from relying on light to relay an image while traveling near the speed of light; only relativistic aberration and similar phenomena do. Thanks for the clarification. So, optical distortion is out. That leaves me with the following logic. (Epistemology includes logic as part of how we know what we think we know.) If a meter rod is immutable (in this context, like earth's shape doesn't change a lot) then it can not be both one ten millionth of earth's surface distance from equator to pole (through Paris), as derived, and also an eighth of that length, as we have discussed. It may "look" like the latter but it can not be both 100 centimeters and 12.5 centimeters. Tell me if you can, the fault in this logic. Edited August 4, 2011 by owl
Cap'n Refsmmat Posted August 5, 2011 Posted August 5, 2011 If we take the meter rod to exist in Minkowski space (a formulation of special relativity), with three spatial dimensions and one temporal dimension, then it may be immutable in the scenario you describe. If you look at the diagram on that page, different observers merely have a different "hypersurface of the present," which describes the slice of four-dimensional spacetime they observe from their reference frame. Someone better versed in SR will have to explain further.
Iggy Posted August 5, 2011 Posted August 5, 2011 Someone better versed in SR will have to explain further. Perhaps we don't need someone better versed in SR, but someone who can explain the concepts leading up to a variable hypersurface of the present in simple terms. Owl, please feel no need to respond. Tar, maybe this post will do something to answer... But be that as it may, there is something about reality that Owl and I have noticed, an "actual" nature to it, that four dimensional spacetime does not quite do justice. Perhaps if we understood exactly what it was saying, but I for one do not. A frame of reference... A person can always consider themselves the origin, at rest, at the center of their own coordinate system. Your frame of reference moves with you because you can never leave yourself behind or send yourself ahead. The following is a 2 dimensional coordinate system describing 'your' frame of reference. Each line marks 5 feet further from you. It shows you walking down the street past an intersection (perhaps not the best thought experiment, but it is the first that comes to mind), The positive numbers (to the right) mark the distance in front of you and negative numbers (to the left) mark distance behind you as you walk. At t=0 the street is 50 feet in front of you. At t=10 seconds you have reached the center of the intersection and at t = 20 the intersection is 50 feet behind you. If you carry a stopwatch with you, that is what it will read at those locations. The street moves a total of 100 feet in 20 seconds. Speed is change in position divided by change in time -- 100 ft / 20 sec = 5 feet/second -- so the speed of the street relative to you (i.e. its speed in your coordinate system or its speed in your frame of reference) is 5 feet per second or 3.4 miles per hour. To add another frame of reference to the situation... as you walk along a girl named Sue passes you and walks ahead of you. She is trying to keep up with her dog we'll say. Your two coordinate systems now look like: According to Sue's frame the position of the intersection changes from +50 (50 feet in front of her) to -20 (20 feet behind her) while 10 seconds pass. The velocity of the street in Sue's coordinate system is... change in position (70ft) / change in time (10 s) = 7 ft/s. Sue also notices that her distance to you goes from zero to 40 in 20 seconds. Your velocity in her frame is 2 feet per second. Every second she gains 2 feet on you. Galilean Space-Time Diagram... Eliminate the vertical spatial axis that wasn't really being used and replacing it with a time axis makes a space-time diagram with one spatial dimension and one temporal dimension. This diagram doesn't involve special relativity -- just a Galilean space-time diagram, The objects that were dots are now lines. They are extended in space and time. Even though the diagram is not animated, it shows how the objects move and interact. At zero seconds (the very bottom of the diagram) Sue, the dog, and 'you' are all in the same spatial location. They are all at zero feet. The gray line is the street intersection. At zero seconds it has a starting position at x=50 feet. At t=0 it is 50 feet from the other two objects. As time moves along (moving up along the y-axis of the diagram) the intersection approaches the other three objects. The dog is the closest, followed by Sue, then 'you'. Two events are marked in this coordinate system. At P2 Sue crosses the center of the intersection. This happens in your coordinate system at x=14.29 feet, t=7.14 seconds -- 7.14 seconds into the thought experiment and 14.29 feet from you. At P1 (x=0, t=10) you cross the intersection. All of this will match the "your frame of reference" animation. All of the information in the animation is also in this diagram. For example, "how far is Sue from you at the end of the thought experiment?" is answered by how far the top of the red line is from the top of the black line. In a space-time diagram the tilt of a line tells you how fast that object is moving in that coordinate system. The gray line moves up by 2 units for every 10 units it moves across. For every 10 feet there are 2 seconds. Its speed is 10/2 or 5 feet per second. The black object (you) does not move horizontally on the diagram just like the black dot doesn't move on the "your frame of reference" animation because it is your reference frame. If you don't understand how the "your frame of reference" space-time diagram matches up with the "your frame of reference" animation then stop, go back and look at them and read again. Nothing else is going to make sense unless it is understood how these diagrams correspond. This is Sue's reference frame as a Galilean space-time diagram, Now in this frame with these coordinates, it is Sue who remains at x=0 ft. throughout the thought experiment. The events P1 and P2 are the same physical events, but in Sue's frame they are at P1 = -20 feet, 10 sec. P2 = 0 feet, 7.14 sec. To put that into words, Sue crosses the intersection at 7.14 seconds, and 'you' cross at 10 seconds when you are 20 feet behind Sue. To transform the events of one coordinate system to another, since this is Galilean (classical) space-time, the Galilean transforms are used, [math]x'=x-vt\,[/math] [math]t'=t \,[/math] where [math]v[/math] is the velocity between frames (2 ft/s in this case); [math]x[/math] and [math]t[/math] are the time and position in the first frame; and [math]x'[/math] and [math]t'[/math] in the second frame. In our first frame, 'your' frame, P1(x=0 , t=10), P2(x=14.29 , t=7.14). Solving Sue's frame gives, P1: [math]x'=0-2 \cdot 10 = -20[/math].... [math]t'=10[/math]... (-20,10) P2: [math]x'=14.29-2 \cdot 7.14 = 0[/math].... [math]t'=7.14[/math]... (0,7.14) Those numbers match our diagrams. The Galilean transformations reflect the classical rules of mechanics where space and time work the way Galileo and Newton expected them to. They tell us what should happen in one frame knowing what happens in another. They are the rules that tell us how to transform from one frame to another. The Galilean transforms assume a universal now -- if an event happens at [math]t[/math] in one frame then it happens at the same time ([math]t'=t[/math]) in any other frame. The transforms also assume absolute distance -- if you measure some distance between events, everyone else will measure that same distance in their frame. This method of kinematics is how people intuitively expect the universe to work so it is understandable that people have a hard time rejecting it, but it is empirically wrong. The actual thing that should be the same between every frame is not distance and not duration, it is a particular velocity. The speed of light, 300 thousand km/s, is the same in every inertial reference frame. This is not something that one can easily get their mind around, so let's look at it in terms of space-time transformations. In a Galilean transform, just like we saw before, everything is skewed over from one frame to another. Here we have two frames. You are the black line and Sue is again the red line. You shoot a laser to the right and Sue chases the laser at 0.6 times the speed of light. This is your frame and hers according to classical mechanics (i.e. galileo transforms), The three lines and the coordinate system have been skewed over like a deck of cards. This might be more apparent if we put both coordinate systems on one image: It seems to make sense, if light is moving away from you at 1c and Sue is chasing that light at 0.6c then Sue is going to think the light is moving 0.4c... just like the diagram shows. This means that both reference frames share the same present. To make this clear, the green events that I marked all happen at the same instant for both observers. They both happen when all of the clocks in both reference frames read 0.5 seconds. But, this also means that light does not have the same velocity in both frames. For light to have the same velocity in both frames we us Minkowski space-time with the Lorentz transformations. Minkowski Space-Time diagram... here is an empirically correct diagram of the same thought experiment depicted above with both coordinate systems: It is messy, so first keep in mind that this shows the same situation where Sue is chasing the laser from 'your' perspective. Find the event marked with a green dot. To see where this event is in your coordinate system (the black coordinate system again) follow from the green dot down the black line to the black spatial axis. It is at one light-second. Also follow the green dot left along the black line to the black time axis. It is at one second. The event is at one second and one light-second in the black coordinate system. The green event is along the ray of light, so we have found that light moves one light-second per second or 1c according to 'you'. In the red coordinate system follow the red line down from the green event to the red spatial axis. It is 0.5 light-seconds. Again follow the red line left to the red time axis. It is 0.5 seconds. Light moves 0.5 light-seconds per 0.5 seconds or 1c according to Sue. The same thing can be done with any event that can be marked on the diagram. You would find that Sue moves 0.6c to the right in the black coordinate system and you move 0.6c to the left according to Sue's coordinate system, and both of you find that light moves 1c. This corresponds to the empirically correct set of transformations -- the Lorentz transformations. The horizontal black lines are a single moment in time in the black coordinate system. All of the events along those lines happen at the same time according to 'your' clock. The horizontal-ish red lines mark a single moment in time in the red coordinate system. There is no absolute now across different reference frames. To make this clear I'll mark a present instant at t=1 in both reference frames: All of the events along the green line happen at the same time for 'you' -- at t=1 on your clock. All of the events along the blue line happen at the same time for Sue -- at t=1 on her clock. They do not share the same present in Minkowski space-time and Minkowski space-time is the construct that keeps the speed of light invariant. That diagram got very complicated, so let me simplify it... Here is a corollary description of absolute time and Galileo space-time... http://www.phy.syr.edu/courses/modules/LIGHTCONE/galilean.html http://www.phy.syr.edu/courses/modules/LIGHTCONE/maxwell.html And of special relativity and Minkowski Spacetime... http://www.phy.syr.edu/courses/modules/LIGHTCONE/minkowski.html If you can't make a Minkowski diagram then you would have no hope of discussing the ontology of space-time. 3
Schrödinger's hat Posted August 5, 2011 Posted August 5, 2011 Thanks for the clarification. So, optical distortion is out. That leaves me with the following logic. (Epistemology includes logic as part of how we know what we think we know.) If a meter rod is immutable (in this context, like earth's shape doesn't change a lot) then it can not be both one ten millionth of earth's surface distance from equator to pole (through Paris), as derived, and also an eighth of that length, as we have discussed. It may "look" like the latter but it can not be both 100 centimeters and 12.5 centimeters. Tell me if you can, the fault in this logic. (Note that the below uses the geometric interpretation of relativity, ie. space is four dimensional. To the best of my knowledge this is the only mainstream interpretation of the mathematics. There are others but I do not think they are relevant right now.) I think your logic is sound but you are acting on a faulty premise. Namely, that space and time are separate things. So you have two events on earth (an event is a point in four dimensions), in one frame they are at the same time, but the distance between them is 100cm. If you look at them from a different reference frame they might be 12.5cm and 1 microsecond apart. Neglecting the microsecond is where the apparent paradox comes in. This is highly counter-intuitive (like most of the universe once you get away from our nice, medium sized scale), but there are very good reasons for the adoption of this model. A two/three dimensional (Euclidean) analogue can be useful (ignore time being a dimension for now). Bear in mind this analogy won't stretch very far, but it illustrates a similar concept. Imagine you are constrained to a flat table on which you have been living your life and you can (somehow) see and interact with a circle which is floating above the table. If someone were to rotate the circle so it became (from your point of view) an ellipse, it would seem quite outrageous and counter to your normal expectations. The (2D) distance from one side to the other would be smaller in every 2D experiment you could perform. This is because you can't measure the difference in height between the two sides.
Cap'n Refsmmat Posted August 5, 2011 Posted August 5, 2011 (Note that the below uses the geometric interpretation of relativity, ie. space is four dimensional. To the best of my knowledge this is the only mainstream interpretation of the mathematics. There are others but I do not think they are relevant right now.) I think your logic is sound but you are acting on a faulty premise. Namely, that space and time are separate things. So you have two events on earth (an event is a point in four dimensions), in one frame they are at the same time, but the distance between them is 100cm. If you look at them from a different reference frame they might be 12.5cm and 1 microsecond apart. Neglecting the microsecond is where the apparent paradox comes in. It's worth noting, then, that if an object is "immutable," it is the spacetime intervals which measure it which are invariant from one reference frame to the next. A spacetime interval s is defined by: [math]s^2 = \Delta x^2 - c^2 \Delta t^2[/math] If you measure the distance and time between two events in any reference frame, then calculate the spacetime interval, you will get the same answer from any reference frame. If you were to only measure distance and neglect that time is another dimension, you would see discrepancies.
swansont Posted August 5, 2011 Posted August 5, 2011 Thanks for the clarification. So, optical distortion is out. That leaves me with the following logic. (Epistemology includes logic as part of how we know what we think we know.) If a meter rod is immutable (in this context, like earth's shape doesn't change a lot) then it can not be both one ten millionth of earth's surface distance from equator to pole (through Paris), as derived, and also an eighth of that length, as we have discussed. It may "look" like the latter but it can not be both 100 centimeters and 12.5 centimeters. Tell me if you can, the fault in this logic. The principle of relativity indicates that this notion is misguided. "Immutable" has no meaning when you go to a new frame of reference. It's not unlike the concept of conservation of energy: true within a frame, but not between frames. In a frame where I am at rest, a moving train has a large amount of kinetic energy. In a frame where the train is at rest, I have a small amount of kinetic energy. They are not equal. One might ask where the energy went, but the question is nonsensical. The energy didn't "go" anywhere. There is no reason to think that the energy measured in two different frames will be the same; that's not what conservation of energy (immutable) means. Similarly, it turns out there is no valid reason to think that an object's length will be the same in different frames. "If a meter rod is immutable (in this context, like earth's shape doesn't change a lot) then it can not be both one ten millionth of earth's surface distance from equator to pole (through Paris), as derived, and also an eighth of that length" is a true and valid statement, and since there is a contradiction, by reductio ad absurdum, we must conclude that the premise is false. The length of the rod is not immutable. Length is not an absolute.
owl Posted August 5, 2011 Author Posted August 5, 2011 swansont: ...we must conclude that the premise is false. The length of the rod is not immutable. Length is not an absolute. Cap ‘n R; post 108: If we take the meter rod to exist in Minkowski space (a formulation of special relativity), with three spatial dimensions and one temporal dimension, then it may be immutable in the scenario you describe. (my bold) In the earth- of- different- shapes debate (as seen from different frames of reference), he also said that it’s not about earth changing shape but rather that it IS different shapes in different frames of reference. Still, it can't be both spherical and severely oblate, as an object with intrinsic properties. It must be one or the other. Maybe you guys should talk it over. Btw; Just to check my understanding of Minkowski"s four dimensions (forgetting spacetime as a malleable medium for the moment), is it not simply three spatial dimensions (line, plane, and volume) and time, which adds the dynamic of objects moving through 3-d space over time i.e., not a static 3-D universe?
Cap'n Refsmmat Posted August 5, 2011 Posted August 5, 2011 swansont and I did not contradict each other. The physical length of the rod is not immutable; however, in four-dimensional Minkowski space, the rod itself is immutable. Its physical length as measured in different frames will change, but only because different reference frames see a different three-dimensional hypersurface out of the same four-dimensional object. Minkowski space is somewhat more mathematically complicated than I am qualified to describe, but I think a reasonable abbreviated version would be "three spatial dimensions plus one time dimension", yes. The key is that time is a dimension.
swansont Posted August 5, 2011 Posted August 5, 2011 If we take the meter rod to exist in Minkowski space (a formulation of special relativity), with three spatial dimensions and one temporal dimension, then it may be immutable in the scenario you describe. My bold. That's what you missed. Length vs spacetime interval. length is not immutable/invariant. In the earth- of- different- shapes debate (as seen from different frames of reference), he also said that it’s not about earth changing shape but rather that it IS different shapes in different frames of reference. Precisely. It is a different length in a new frame. It does not change, because that concept is misapplied. Still, it can't be both spherical and severely oblate, as an object with intrinsic properties. It must be one or the other. Length is not an intrinsic property.
Iggy Posted August 6, 2011 Posted August 6, 2011 There is no absolute "at rest", to answer your question. There is only relatively at rest... with whatever is the object of investigation/measurement. Velocity is relative... that is interesting... are you saying that an object's velocity changes from one frame of reference to another? The earth's velocity changes? If that is the case, why do people not fly off the surface into deep space when the earth's velocity changes from zero to near the speed of light?
owl Posted August 6, 2011 Author Posted August 6, 2011 Velocity is relative... that is interesting... are you saying that an object's velocity changes from one frame of reference to another? The earth's velocity changes? If that is the case, why do people not fly off the surface into deep space when the earth's velocity changes from zero to near the speed of light? This challenge makes no sense to me. Par for the course in our history of almost perfect lack of communication. I was replying to swansont's challenge to substantiate what he thought (I think) was a claim that an at rest frame of reference was in some way an absolute reference point. I explained that I made no such claim. I gave examples of how a frame at rest with what is measured is preferable to extreme frames of reference for measuring earth's size and shape, and examining micro-organisms in a lab, etc... like zipping by at near 'C' vs under a microscope. I also said that GPS clocks have different velocities (and tick at different rates) than earth-surface clocks, which require adjustments for positioning accuracy. Your continuing attempts to distort what I say will continue to fail, but I will not continue to reply.
swansont Posted August 6, 2011 Posted August 6, 2011 If two observers are moving with respect to each other, how do we decide which measurement is the correct one? Relativity predicts that each will see the other's frame as being length contracted, and their time dilated. Their own frame will not. Who is right?
Schrödinger's hat Posted August 6, 2011 Posted August 6, 2011 (edited) I gave examples of how a frame at rest with what is measured is preferable to extreme frames of reference for measuring earth's size and shape, and examining micro-organisms in a lab, etc... like zipping by at near 'C' vs under a microscope. That's fine for individual objects, and this is where we use the term 'proper'. Ie. The proper time it takes me to go to the shops, is the time taken in a frame where I am at rest. The proper shape of earth is a (very very slightly oblate and bumpy) spheroid, not a highly oblate spheroid. The proper distance between the poles is however many thousands of km (~14000?). However, that doesn't mean the distance is always 14000. In some frames the poles might be 1km and a few seconds apart. Arguing over whose frame is right is like having two people on opposite sides of a table. They have two drinks on the table and are both arguing that the one on the left is theirs. Edit: Deleted response to Iggy written when my sarcasm detector failed. Edited August 6, 2011 by Schrödinger's hat 1
Iggy Posted August 7, 2011 Posted August 7, 2011 (edited) This challenge makes no sense to me. I believe you. I was replying to swansont's challenge to substantiate what he thought (I think) was a claim that an at rest frame of reference was in some way an absolute reference point. No, you are either obfuscating, mistaken, or lying. You were replying to something really rather simple: how can you tell who is moving and who is at rest Your answer was also rather simple: There is no absolute "at rest", to answer your question. There is only relatively at rest. "at rest" means "zero velocity". Are we on the same page now? Earth, according to you, doesn't have a single velocity. It has zero velocity according to one frame of reference and a great deal of velocity according to another. By your reasoning: earth changes its velocity, does it not? How is this not, again by your reasoning, subjective idealism? Do you think a person's perspective changes the velocity of an object? If you don't believe perspective changes velocity then you must believe earth has a single, unchanging, immutable, intrinsic velocity. Either earth has a single absolute velocity, or you are (by your own understanding of the term) a subjective idealist. So, which is it? 1. Earth has a single velocity that does not depend on various points of view. 2. Earth's velocity changes with perspective. <edit>I should add... it is, of course, entirely agreeable that velocity is frame dependent. The implications attributed to frame dependent quantities by Owl are disagreeable (eg they are subjective idealism)</edit> Btw; Just to check my understanding of Minkowski"s four dimensions (forgetting spacetime as a malleable medium for the moment), is it not simply three spatial dimensions (line, plane, and volume) and time, which adds the dynamic of objects moving through 3-d space over time i.e., not a static 3-D universe? No, Minkowski space-time is not "simply three spatial dimensions (line, plane, and volume) and time". You could describe space-time that way, but Minkowski space-time is a specific instance of space-time which is different from other space-times. For example, Minkowski space-time has a metric signature (+---) (or -+++ depending on sign convention) and Galilean -- a Euclidean space-time -- has a metric signature (++++). Minkowski's metric signature preserves time-like, null, and space-like vectors under Lorentz transforms where the other does not. This means a time-like vector is always time-like in Minkowski space, and the same with null and space-like vectors. This is important because preserving a null vector is the same as saying "the speed of light is constant". This was the point of my previous very lengthy and laborious post. The speed of light is constant in Minkowski space-time while it is not in classical space-time. Edited August 7, 2011 by Iggy
tar Posted August 9, 2011 Posted August 9, 2011 Iggy, I am still reading #109. Thank you for doing that. I have had problems with my computer that I usually use, so I am not keeping up. Will "respond" after catching up with the thread. Regards, TAR2
DrRocket Posted August 9, 2011 Posted August 9, 2011 Dr. Rocket, I am quite confused about the difference between special and general relativity, and I am not sure about a theory that is a "special" case of another theory, with "gravity" left out. Where in reality do we find a situation where gravity is left out? And isn't it better to have a theory, that fits it back in? Regards, TAR2 Special relativity was developed before general relativity. It was developed to explain effects that are primarily electrodynamic and specifically ignores effects due to gravity. SR was announced in 1905. Einstein spent the next 10 years trying to extend the special theory of relativity to include gravity. The result was general relativity. General relativity explains gravity in terms of the curvature of spacetime viewed as a Lorentzian manifold. A manifold is an object that "locally looks like" ordinary euclidean space. A simple example is the surface of a sphere, like the Earth, which can be described in small patches as a flat plane. Similarly the curved Lorentzian manifold of general relativity is described in small patches Minkowski space. The geometry of Minkowski space is mathematically just special relativity. You can think of it as general relativity on flat space, which since gravity is a manifestation of curvature is just general relativity without gravity. Alternately you can think of SR as the local version of GR, and just realize that all manifolds are "almost flat" in a sufficiently small patch. So SR provides the local, linearized picture, and GR puts the small patches together and "fits it (gravity) back in" just as you suggest. Where you find a situation where gravity is left out is in freefall -- as with astronauts in orbit experiencing "weightlessness". A local reference frame in freefall is Lorentzian and is a convenient local frame/patch for use in general relativity. In this frame the equations of special relativity hold locally and this is the physical reflection of the statement that "SR is the localization of GR".
owl Posted August 9, 2011 Author Posted August 9, 2011 (edited) Schrodinger’s hat: (Note that the below uses the geometric interpretation of relativity, ie. space is four dimensional. I think your logic is sound but you are acting on a faulty premise. Namely, that space and time are separate things. "Space is four dimensional" requires the following distinction: 3-D space plus time or 4-D space? I think you mean the former, so... My premise is that space is empty volume (except, of course, where occupied by objects) with three dimensions, and that time is event duration as objects move through space. Neither is a thing, and combining them into ‘spacetime’ still does not make spacetime a malleable medium which is curved by mass/energy. That s what the ontological debate is about. If you want to call my above premises faulty, it behooves you to show how so. If you say (or anyone says) that reality is dependent on frames of reference, you throw philosophical realism out and subscribe to a form of idealism in which things ARE as they are seen, having no intrinsic properties of their own, independent of frames of reference. The rest is window dressing, including the claim that a severely oblate earth is just as accurate/real as the “proper” nearly spherical shape that is so well documented that we can confidently teach our children that it is not, in fact, squished nearly flat. Edited August 9, 2011 by owl
Iggy Posted August 9, 2011 Posted August 9, 2011 If you say (or anyone says) that reality is dependent on frames of reference, you throw philosophical realism out Because you say that velocity depends on frame of reference, you have thrown out philosophical realism? Can't answer? If you say (or anyone says) that reality is dependent on frames of reference, you throw philosophical realism out and subscribe to a form of idealism in which things ARE as they are seen, having no intrinsic properties of their own, independent of frames of reference. The speed of light, the space-time interval, and a host of other properties do not depend on frame. "no intrinsic properties... independent of frames of reference" is a strawman, and you've already been told this. 1
swansont Posted August 9, 2011 Posted August 9, 2011 My premise is that space is empty volume (except, of course, where occupied by objects) with three dimensions, and that time is event duration as objects move through space. Neither is a thing, and combining them into ‘spacetime’ still does not make spacetime a malleable medium which is curved by mass/energy. That s what the ontological debate is about. If you want to call my above premises faulty, it behooves you to show how so. Does light go straight in space where no masses are nearby and bend when it goes around a massive object? I'll spare you the effort: the answer to this is yes. Can you model this and other gravitational effects geometrically? Again, yes. There is plenty of evidence which shows GR to be an accurate description of how nature behaves. So clearly, the spacetime that GR describes is indeed curved. (Your claim that it isn't a malleable medium continues to be a strawman and a red herring, because nobody here is claiming that it is, and in fact relativity claims that there is no medium) If you want to argue that GR is just a model and isn't "reality," come up with a way to TEST for this. It is incumbent on YOU to find a way to test your proposal. You can't pass the buck and assume it's true until proven false. That's yet another logical fallacy. So: How does one show a valid scientific model to be "truth" or not?
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