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#1
April 7th 05
posted to sci.physics,sci.physics.relativity,alt.physics,sci.math
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(SR) Lorentz t', x' = Intervals
..answers groups snipped.
In sci.math, Eleaticus wrote on 06 Apr 2005 04:44:46 GMT : Disclaimer: approval for *.answers is based on form, not content. Opponents should first actually find out what the content is, then think, then request/submit-to arbitration by the appropriate neutral mathematics authorities. Flaming the hard- working, selfless, *.answers moderators evidences ignorance and atrocious netiquette. Version: 0.02.1 Archive-name: physics-faq/criticism/lorentz-intervals Posting-frequency: 15 days (SR) Lorentz t', x' = Intervals (c) Eleaticus/Oren C. Webster ------------------------------ Subject: 1. Introduction with the obvious debunking of the use of 'just coordinates' in any scientific formula. Defenders of the Special Relativity faith are especially fond of telling opponents of their space-time fairy tales that they do not know the difference between coordinates and magnitudes. There are a few quirks. That may often be so, but the fault lies ultimately with SR dogma. The Lorentz-Einstein transformations cannot possibly be 'just coordinates', which is the interpre- tation required to support the many sideshow carnival acts with which the SR faithful bedazzle the public, and establish their moral and intellectual superiority. If I get in my car and drive steadily for a few hours at 50 kilometers per hour, is 50t the distance I travel? A unit-conversion problem. I'm not sure how seriously anyone will take this proposal, but I propose a 'nil' to be the distance light travels in a nanosecond. (This would be exactly 0.299792458 m in more conventional measurements, or just shy of an Imperial foot.) Therefore, you're suggesting driving at velocity 74.5582 nils/s, or 268.410 kilonils per hour. (Lightspeed is 1 giganils/s.) At the end of an hour, of course, you've driven 268.410 microseconds -- so to speak. More precisely, you've driven a distance in an hour that would take light 268.410 microseconds to cover. Of course it gets more complicated if one adjusts one's clock properly; accelerating the car to 74.5582 nils/s will slow down the carborne clock by a factor of about 2.78 * 10^-15 for the duration of the trip; since the trip takes an hour one loses 10 picoseconds. This is probably a detail that can't be measured at these velocities. Of course not. The last time my hours-counting 'just coord- inates' clock was set to zero was when Zeno first reported one of his paradoxes to Parmenides. That was a long time ago, so my t is not useful for such purposes unless you also use my clock to established the starting time, perhaps t0, and use the formula 50(t-t0) to calculate the distance. In any case, my t is even then not 'just a coordinate' because it always represents particular elapsed times that can be used in the (t-t0) form to calculate perfectly good time intervals (elapsed times). Alternatively, I could (re)set my clock to zero at the start of some meaningful time interval, in which case my t shows a scientifically perfect current and/or end time. In which case, the Lorentz-Einstein t'=(t-vx/cc)/g is a function of an elapsed time interval (not 'just a coordinate') and a time interval (-vx/cc; the interval amount the t' clock is being screwed up at time t) and thus cannot be 'just a coordinate' since neither of the independent variables is such a 'just' thing. {Their meaning is shown below, step-by-step.] If it takes me 50 minutes to cross the Interstate highway, was x/50 my velocity crossing it? Your average velocity, in somethings per minute. (What units are *you* using?) Of course not. The origin of all my axes is at the very spot where Zeno first presented his first paradox to Parmenides. That makes my x equal a couple of thousands of miles, plus, and is not useful for such purposes unless you establish the starting x value, perhaps x0, and use the formula (x-x0)/50 to calculate my velocity. In any case, even then my x is not 'just a coordinate' because it always repesents particular distance intervals that can always be used in the (x-x0) form for any and every scientific purose. Alternatively, I could move my x-axis origin to the starting (zero) point of some meaningful distance, in which case my x shows a scientifically perfect current and/or end distance. In which case, the Lorentz-Einstein x'=(x-vt)/g is a function of a current/ending distance interval (not 'just a coordinate') and a distance interval (-vt; the interval amount the x' axis is being screwed up at time t) and thus cannot be 'just a coordinate' since neither of the independent variables is such a 'just' thing. {Their meaning is shown below, step-by-step.] ------------------------------ Subject: 2. Table of Contents 1. Introduction with the obvious debunking of the use of 'just coordinates' in any scientific formula. 2. Table of Contents. 3. The Lorentz-Einstein transforms. 4. The 'just coordinates' argument. 5. Single-system, little-purpose ambiguity. 6. Relating two coordinate measures/systems. 7. Distances and moving coordinate axes. 8. Time intervals. 9. Einstein's (1905) derivations. 10. A word about intervals. 11. Intervals versus the Twins Paradox. 12. Summary ------------------------------ Subject: 3. The Lorentz-Einstein transforms Special Relativity's space-time circus is based on the 'transformation' equations by which it is believed one can relate a nominally 'stationary' system's space and time coordinates to those of an inertially (not accelerating) moving other observer. That moving observer's own physical body and coordinate system might have been identical in size to those of the stationary observer before the traveller began moving, but are 'seen' as very different by the stationary observer when the relative velocity of the two is great enough, a high percentage of the velocity of light. Concerning ourselves - as is customary - with just the spatial coordinate axis that lies parallel to the direction of motion, and with time, Einstein arrived at these formulas that relate the moving system measures or coordinates (x' and t') to the stationary system coordinates (x and t): x' = (x - vt)/sqrt(1-vv/cc) (Eq 1x) t' = (t - vx/cc)/sqrt(1-vv/cc) (Eq 1t) Be extremely careful here. Einstein in particular used chi and tau. I like to use (x_O, t_O) and (x_A, t_A), for clarity (tagging the coordinate systems with the name of the observers). The v is for the two systems' relative velocity as seen by the stationary observer, and is positive if the dir- ection is toward higher values of x. By concensus, the moving system x'-axis higher values also lie in that direction, and all axes parallel the other system's corresponding axis. Actually, it's a little more complicated in 4-space. However, the Lorentz reflects things more or less properly; if x_A = (x_O - v * t_O) * gamma(v), then it's clear that, as t_O increases, x_O must increase with positive v if x_A is to be constant (e.g., a certain point within A's car, snail shell, boat, or spacecraft). We used vv to mean the square of v but might use v^2 for that purpose below. Similarly for c. Poor editing. If you're going to post this tract once a week at least take the time to clean it up in a text editor *then* cut and paste it. Not that it's all that useful to post it anyway. Because it is believed that no physical object can reach or exceed c, the square-root term in both denominators is presumed always less than one, which means that the formulas say both x' and t' will tend to be greater than x and t, respectively. However, SRians call the x' result 'contraction' - which means shortening - and the t' result 'dilation' - which means increasing. Ah yes, that old chestnut. It's not really a contraction or an expansion, but a *hyperbolic twist*. The problem setup gets complicated, but here's one method. O and A are the customary observers. At time 0 they are coincident, for mathematical convenience. O has marked off two points, one at his local origin, and one some distance along the x-axis (call the distance d_O). How does A determine d_O, assuming the Lorentz? It turns out the simplest method by which A can measure d_O is by simply taking the time he crosses over the first mark (namely, 0), and the time he crosses over the second mark (call it t_A), subtracting the two, and doing some fiddling. How much fiddling? Well, the second mark is at x_A = (d_O - v * t_O) * gamma(v) t_A = (t_O - v * d_O / c^2) * gamma(v) where t_O is some convenient value (as the mark is expected to last indefinitely. Since A passes directly over the mark we have x_A = 0, or t_O = d_O / v. Therefore, t_A = (d_O / v - v * d_O / c^2) * gamma(v) = (d_O/v) * (1 - v^2/c^2) * gamma(v) = (d_O/v) / gamma(v) . Did the length contract or expand here? Another alternative is for A to shoot out a lightbeam at d_O the moment he crosses over the first mark, and then try to compute the length from a small reflective mirror that O has conveniently placed at the second mark. As O sees it, A emits a lightpulse at (0,0) and hits the mirror at (d_O, d_O/c), coming back to O at (0, 2*d_O/c). What does A see? Well, the mirror in A's coordinate system transforms into x_A = (d_O - v * d_O/c) * gamma(v) = d_O * (c - v)/c * gamma(v) t_A = (d_O/c - v * d_O/c^2) * gamma(v) = d_O * (c - v)/c^2 * gamma(v) Since A cannot see this directly (the light has to get back to him), the measured time -- call it t_m_A -- is t_A + x_A / c. This gives us t_m_A = 2 * d_O * (c - v)/c^2 * gamma(v) A might naively assume that the distance is c * t_m_A -- and he'd be wrong. Since c/(c-v) = 1/(1 - v/c) = 1/sqrt((1 - v/c)*(1 - v/c)) 1/sqrt((1 - v/c)*(1 + v/c)) = gamma(v) for positive v, t_m_A is again too short, although not by the same factor as the earlier measurement. ------------------------------ Subject: 4. The 'just coordinates' argument The 'just coordinates' argument is so patently ridiculous that even opponents have a hard time accepting just how simple and obvious its debunking can be, as shown in this section. However, further sections take a more arithmet- ical approach that you'll maybe find more professorial. The 'just coordinates' argument is that t is mot a duration, not a time interval; it's just an arbitrary clock reading. But what if the moving system observer comes speeding by while you make your annual 'spring forward' or 'fall back' change? The formula says that the moving system clock's 'just coordinate' reading can be calculated from yours: t' = (t - vx/cc)/sqrt(1-vv/cc) (Eq 1t) Imagine the moving system oberver's confusion if his clock changes its reading while he's looking at it! But clocks *always* change their reading while one's looking at it. If one takes one second to read a clock, the clock advances by one second as one is reading it. As it is, most scientists ignore this issue anyway, using GMT or just using a high-precision interval timer. If one wants to call that a duration as opposed to a reading, fine -- SR has the implicit assumption that space is isotropic and that one can move and rotate the coordinate system using any rigid transform (combination of translations and rotations), prior to the start of the actual measurement -- and after the measurement if one is careful about one is doing (one has to avoid, for example, Sagnac's effect). So this particular objection verges on the silly. [snip for brevity] ------------------------------ Subject: 5. Single-system, little-purpose ambiguity. Since we're going to be comparing measurements on two coordinate systems in the next section, Are we? Be extremely careful here. let's go to our supply cabinet and get our yard-stick (which we use to measure things in inches) and our meter-stick (which we use to measure things in centimeters). Here, I'm getting mine. Oh! Oh! I'm going to get my nilstick. It's just under a foot long and works very well -- it's also made purely of light rays. There's an ant on mine, and he ... she ... sure is hanging on, right at the 3.5 inch mark of the yard- stick. And how long, wide, and thick is this ant? Let's see if I can wave the stick around enough that she'll let go. Nope. Sagnac's effect. However, before I gave up I waved the stick and the ant 'all over the place". That's about as precise as me pointing vaguely eastward and stating "London is thataway". Always, however, the ant was at the 3.5" mark on the yard-stick, and always 3.5" away from the end of the stick, however far and wide I have transported her. In the stick's coordinate system, yes. To properly characterize the motion in the world's coordinate system I'd need some more details. Neither of those 3.5" facts means very much. Of the two, the distance aspect meant almost nothing. So the distance was 3.5" from the end. So what? That length, distance, was not in use. And only maybe the ant might have been concerned with just what location, 'just coordinate', on the stick she was at. Just so with x and t. You do have a point. Any (O,A) problem has to reflect the fact that light travels at lightspeed back to the observer. In the computations above I took pains to ensure exactly that. Otherwise, one has what I might call a "phantom observation". Any x != 0 for either coordinate system might fall into this category. Even OWLS measurements might fall into this category, as an OWLS measurement requires the difference between two measurements of two clocks -- and these two clocks are by their very nature *in two different coordinate systems*. A TWLS measurement merely requires a mirror, a good clock, a good marker, and a light source. So, is the 3.5" reading just a coordinate? Or a distance/length? It's ambiguous in and of itself, and really makes no difference what you say until you try to make use of the number. Hey, my address is 5047 Newton Street. If you are looking for me and you're at 4120 Newton, it is helpful information, because it tells you which direction to go. Is that 'just coordinate'? There are a large number of Newton Streets. My Newton Street might be somewhere in San Francisco. Yours might be somewhere in Dallas. His might be in Concord -- and it gets *really* squirrely as there are a fair number of cities named Concord! Small wonder the USPS created long ago the notion of 5-digit (or 9-digit) numbers to zip the mail to where it's intended to go. Where it really becomes useful, perhaps, is in telling you how far away I am. That's not just a coordinate value, that's a distance, length, interval. Not really. In some streets, houses are numbered every 2; others might decide to use 10 as a delta. Still other locales might use 100 -- usually for large commercial buildings. Conversion into yard, meters, nils, or miles is at best approximate, and at worst downright impossible. However, it is subtracting 4120 from 5047 that tells you which direction and how far. Not really. For starters, the direction is not directly along the street -- in most locales even numbers and odd numbers occupy opposite sides. It gets even wackier when Newton Street sits exactly on a city boundary, which means City A gets to number the houses on its side of the street as it likes, whereas City B numbers its houses according to its desires. Much hilarity -- if one can call it that -- may ensue. It is only because both 5047 and 4120 are distances from the same point - ANY same point - that the result means anything. They don't mean that much, even if all of the preceding are discounted. Most housing developments have varying-size lots. I've also neglected to mention corners and bends. My x - my yardstick reading - is always a distance or length; it is impossible to be otherwise with an honest, competently designed yardstick. At least on a yardstick one has a fighting chance of uniform lengths -- unless you're waving it around, which induces various forces on your ant because of centripetal considerations. Whether or not its reading is of good use in some particular scientific formula depends on whether I put the zero end of the yardstick at some useful place. As in the introduction, we should either put it at the starting location/end, or use two readings from it: (x-x0). And there's also the issue of the size of the tick -- or the ant. ------------------------------ Subject: 6. Relating two coordinate measures/systems. Taking care to not damage our brave little ant, I place my yard-stick onto the table, zero end to the left, 36" end to the right. Now I place the 'just coordinate' meter-stick on the table in the same orientation, in a random location, and find that the ant's coordinate on the meter-stick is 51. And I place my nil stick on top of both of yours (it is transparent and has a channel; the ant is safe) and get a reading of 19.105. Your point is valid but very long-winded. [snip for brevity] But, you say, the Lorentz transform contain a -vt term. That it does, out of necessity. Even the conventional Galilean transform: x_A = (x_O - v * t_O) t_A = (t_O) has such a term. To omit that term freezes the problem. Or has it ocurred to you that the ant, after you've made it yardstick-sick, might want to move along its length to somewhere else to avoid your armwaving? So after you've put the stick down it's at the 3.5 mark. However, 1 second later, the ant, crawling at a uniform velocity of 0.5 in/sec, will end up at the 4.0 mark. 2 seconds later, it will be at 4.5 mark, and it will eventually crawl off the stick, and presumably to someplace a little less whirly, like her queen's anthill... :-) Just for giggles, the ant is wearing a wris****ch, and has a tiny little ruler of her own, anchored permanently on her thorax. Not much use, perhaps (it's always 0 at her thorax), but it's clear that that v * t_O term, assuming her watch remains synchronized (or nearly so) with the table-clock, is necessary. ------------------------------ Subject: 7. Distances and moving coordinate axes. We discovered x'=x.z' + x/g as the correct formula for relating one coordinate to another system's. But the Lorentz transform contains another term, -vt/sqrt(1-vv/cc). What is it? Let's start with our x'=51 cm, x=3.5", x.z'=42.11 example. Every minute, let's move the meter-stick one inch to our right. At minute 0, the cm reading was 51 cm. At minute 1, the cm reading is now 50 cm. At minute 2, the cm reading is now 49 cm. In this instance, v=1 inch/minute. And t was 0, 1, 2. What has happened is that we have made our x.z' a lie, and increasingly so. -vt/.3937 is the change in x.z'. x' = (x.z - vt/.3937) + x/.3937. Obviously, vt/.3937 is not a coordinate; even most SRians wouldn't imagine it was. It is an interval, the distance over which the moving system has moved since t=0. And, of course, x/.3937 is the distance of our brave little ant from the point where x=0 and the centimeter reading is x.z'-vt/.3937. Yes, every minute the meter- stick moves to the right and the meter-stick coordinate of the spot where x=0 gets less and less - and eventually negative. Make sure you understand that every minute the x' coordinate, because of -vt/g, becomes a better measure of, say, the 3.5" paper we might be measuring with the yard-stick, given that 51 was too big a number and -vt is negative. That is, until the two origins coincide at x'=x=0, and then it gets worse and worse. With -vt positive (because v0) the situation is different. With 51 and -vt positive, x' just gets worse and worse over time. Quite obviously, the fact that we now have the correct formula for relating an x interval to an arbitrary x' coordinate even when the x' axis is moving, does not mean that x'is anything more than nonsense for use in any scientific formula. Unless we were smart enough to put the x zero point in a useful location, and use (x'-x.z'+vt/.3937) in the scientific formula. (x'-x.z'+vt/.3937) equals the useful, Ratio Scale value x/.3937. Actually, it doesn't. Of course your v = 1.4121 * 10^-12 c will yield gammas on the order of (1 - 10^-24), which would basically be an error of about 15 atomic nuclei per astronomical unit (if that) -- but it's not exactly zero. ------------------------------ Subject: 8. Time intervals. [snip for brevity, as it's redundant] ------------------------------ Subject: 9. Einstein's (1905) derivations. When we return to Einstein's derivations of the transform formulas with a well-focused eye, we find he was a wee bit confused - or at least self-contradictory. When he set up his (at first unknown) tau=moving system time formulas, he created three particular instances of tau. Tau.0 is the time at which light is emitted at the moving origin toward a mirror to the right that is moving at rest wrt that moving origin and at a constant distance from that origin. He lets the stationary time slot have the value t, a constant, the stationary system starting time. Tau.1 is the time at which the light is reflected. Tau.1 is a phantom time; it cannot be directly observed. He lets the stationary time be t+x'/(c-v); t is still a constant and x'/(c-v) is the time interval since t. Tau.2 is the time at which the light gets (back) to the moving origin. The stationary time value is put as t + x'/(c-v) + x'/(c+v); t is still a constant and x'/(c-v) + x'/(c+v) is the time interval since t. On the thesis that the moving observer sees the time to the mirror as the same as the time back to the origin, he sets .5[ tau.0 + tau.2 ] = tau.1. Tau.0 completely drops out of the analysis and leaves no trace, and has no effect. That's what YOU think. tau.2 = tau.0 + delta.tau; this means tau.1 = tau.0 + 1/2 delta.tau, whatever delta.tau might be. Granted, this is easily fixed by translation of the coordinate systems. It's not dropped out at all, and a careful analysis will show this. Further, the t you see in tau.0, tau.1, and tau.2 also completely drops out with no trace and no effect, leaving us with exactly what you'd get if you had explicilty said t' is an interval and so is t. Same issue, same fix. What doesn't drop out in the stationary time values is x'/(c-v) and x'/(c+v), the time interval it takes for light to get to the fleeing mirror, and the time interval it takes for light to get back to the approaching origin. Thus, his resultant t' formula is strictly based on time intervals in the stationary system. Time intervals since some starting time, yes, but time intervals. There is absolutely nothing in the derived formulas that depends on arbitrary coordinates like the constant t in the stationary time arguments. Let's look at the x dimension; it is x'=x-vt [as x increases by vt, the effect over time is x'=(x+vt)-vt)], which Einstein explicitly sets up as a constant stationary distance. He uses that x' not just in the time interval parts of the stationary time arguments, but also in the x (distance) stationary system argument for the tau at the time light is reflected. x' can't be the stationary system coordinate of the mirror at that time. That value is x'+vt. x' is explicitly an interval, distance. Thus, the whole tau derivation of the t' formula is fully and explicitly based on x' - a spatial length/distance/interval - and the two time interals x'/(c-v) and x'/(c+v). While we're at it, if the starting t is not zero, his x'=x-vt formula is complete nonsense also. Given that there was some L that was the mirror x-location and length when the light is emitted, if t was already, say, 500, then x'=L-vt could have been a very negative length. The issues here are fixable. The conventional SR coordinate system has both observers O and A coincident at time t_O = t_A = 0. (Or, if you prefer, t = tau = 0.) The same is true for x_O = x_A = 0 at time 0. If one prefers, one can use tensors. I'd have to look up the details but from a functorial standpoint (if that's the word I want), one can define x_A = f(x_O, t_O) t_A = g(x_O, t_O) and work from there. The pair (f,g) constitute some sort of tensor. Einstein goes on to prove that the tensor must ultimately resolve to the Lorentz. ------------------------------ Subject: 10. A word about intervals. There are intervals, and there are intervals. If we put our yard stick zero point at one end of a piece of paper and read off the coordinate at the other end of the paper, we have a good measure of the paper's length, a Ratio Scale measure. [Absolute temperature scales are ratio scale.] Actually, we do not. How did you orient that paper? Did you measure along an edge? Did you just drop the stick in any old orientation on a sheet of paper that just happened to be on the table first? Is there a Rubik's Cube under the paper, distorting it? Did your dog (assuming you have one) chew the paper first? What if your 3-year-old son (ditto) decided to play "crumple the tax form, Daddy"? What, precisely, are you measuring? Even if one assumes an edge measurement of an idealistically flat 8 1/2" x 11" 20-pound bond sheet with perfect 90 degree corners, there's still the issue of which edge. If instead we put the one end of the paper at the one inch mark (or the zero end of the stick one inch 'into' the length of the paper) we get measures that are one inch off the true, ratio scale length. The two messed up measures are still intervals, but they are Interval Scale measures. [Household temperature scales are interval scale, which is why your physics and chemistry professors won't let you use them without first converting to the ratio scale absolute temperatures.) t'=t/g and x'=x/g represent ratio scale measures, given that t and x were ratio scalae to start with. t'=t.z'+t/g and t'=t/g-vx/gcc are both interval scale measures, even given a good ratio scale t and a good ratio scale x. x'=x.z'+x/g and x'=x/g-vt/g are both interval scale measures, even given a good ratio scale x and a good ratio scale t. Look for the "(SR) Lorentz t', x' = degraded measures" document soon at a newsgroup near you. So convert them to a convenient coordinate system first. Oh, look. There's a *second* sheet of paper underneath the first, offset to (but in the same orientation as) it. *Now* what are you measuring? Same issue as your "offset mark" -- and there's no real good coordinate system, just convenient ones. ------------------------------ Subject: 11. Intervals versus the Twins Paradox. t'=(t-vx/cc)/g shows t' being greater than t. Be extremely careful here. As you may have noticed in my computations above, the time interval measured by the moving observer depends on precisely how he measures it. In fact, t' = 2 * d_O * (c + v)/c^2 * gamma(v) might apply, if one assumes marks (in O's space) at x_O=0 and x_O = -d_O, and A fires a beam of light at the mirror attached to the mark at the instant his origin passes over O's. (This is merely t_m_A with -v substituted for v.) The reason Special Relativity will not allow the use of its basic time equation in determining what SR has to say about the twins' ages, is that t' and x' are supposedly just coordinates, and they say you have to take the coordinate pairs (t',x') and (x,t) into consideration in both the time and place the twins' separation started and the time and place the twins reunited. Actually, SR isn't all that useful here anyway. The twin traveling in the rocket is experiencing acceleration, which is explicitly forbidden in SR problems. With the usual formulations of the twin paradox, admittedly, the rocket twin will be converted into meat jam (if the rocket doesn't simply disintegrate because of the very heavy forces during acceleration!). Assuming Newtonian ideals the acceleration of the rocket twin to near lightspeed will take the better part of a year (g = 9.805 m/s/s, c = 299792458 m/s, t = c/g = 30575467.4 s, 1 Gregorian year = 31556952 seconds). (It will also take a ridiculously large amount of rocket fuel.) [rest snipped] Here are a few thoughts for you. [1] Your comments aren't too bad -- but extremely long-winded. My requirements for any thought experiment are these: [a] Define your coordinate systems properly. [b] Make sure you know what you're measuring, and how you're measuring it, and in which coordinate system you're measuring. [c] Explicitly state your assumptions. [d] Remember that x_O != 0 and x_A != 0 are "phantom measurements"; the observer is forever anchored at his origin and cannot move therefrom. Since lightspeed is constant the adjustment is relatively simple, of course -- just add (or subtract) x_A / c, and you've got your time interval. [2] There's nothing wrong with shifting coordinate systems, as long as one follows [1][a] and uses rigid transformers (translations or rotations). Note that moving something at a uniform velocity might be construed as a shear -- although in SR's case it's more like a hyperbolic rotation. [3] SR and GR have been validated by a bevy of scientific experiments, from the MMX (which admittedly is a weak validation as it also can be used to justify BaT/emissive theory; the only thing MMX disproves is the notion, apparently common at the time, that lightspeed is c relative to a single unique fixed source) to the extremely sophisticated measurements being currently conducted by Gravity Probe B and the computations of Mercury's orbital perturbations and various orbiting binary stars -- some of which contain pulsars. Derivatives of SR and GR must also be used in design of such things as particle accelerators, which routinely accelerate particles faster than light -- if one believes in Newtonian theory, which doesn't work at such power levels. It turns out the accelerators see mass gains, which must be compensated for. [4] There is no absolute velocity. Even in the case of light, there's only an absolute speed. [5] Nothing you've stated here disproves SR, from a mathematical standpoint. It is merely a warning that one has to be careful during problem setup. -- #191, It's still legal to go .sigless. 
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