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Invariant Galilean Transformations (FAQ) On All Laws (c) Eleaticus/Oren C. Webster Thnktank@concentric.net An obvious typo or two corrected. The Brittanica section revised to less 'pussy-footing' and to more directly anticipate the elementary measurement theory and basic analytic geometry that is applied to the transformation concept. ------------------------------ Subject: 1. Purpose The purpose of this document is to provide the student of Physics, especially Relativity and Electromagnetism, the most basic princ- iples and logic with which to evaluate the historic justification of Relativity Theory as a necessary alternative to the classical physics of Newton and Galileo. We will prove that all laws are invariant under the Galilean transformation, rather than some being non-invariant, after we show you what that means. We shall also show that another primal requirement that SR exist is nonsense: Michelson-Morley and Kennedy-Thorndike do indeed fit Galilean (c+v) physics. ------------------------------ Subject: 2. Table of Contents 1. Foreword and Intent 2. Table of Contents 3. The Principle of Relativity 4. The Encyclopedia Brittanica Incompetency. 5. Transformations on Generalized Coordinate Laws 6. The data scale degradation absurdity. 7. The Crackpots' Version of the Transforms. 8. What does sci.math have to say about x0'=x0-vt? 9. But Doesn't x.c'=x.c? 10. But Isn't (x'-x.c')=(x-x.c) Actually Two Transformations? 11. But Doesn't (x'-x.c+vt) Prove The Transformation Time Dependent? 12. But Isn't (x'-x.c')=(x-x.c) a Tautology? 13. But Isn't (x'-x.c')=(x-x.c) Almost the Definition of a Linear Transform? 14. But The Transform Won't Work On Time Dependent Equations? 15. But The Transform Won't Work On Wave Equations? 16. But Maxwell's Equations Aren't Galilean Invariant? 17. First and Second Derivative differential equations. ------------------------------ Subject: 3. The Principle of Relativity and Transformation If a law is different over there than it is here, it is not one law, but at least two, and leaves us in doubt about any third location. This is the Principle of Relativity: a natural law must be the same relative to any location at which a given event may be perceived or measured, and whether or not the observer is moving. The idea of location translates to a coordinate system, largely because any object in motion could be considered as having a coordinate system origin moving with it. If you perceive me moving relative to you - who have your own coordinate system - will your measurements of my position and velocity fit the same laws my own, different measurements fit? If a law has the same form in both cases it is called covariant. If it is identical in form, var- ables, and output values, it is called invariant. What we're asking is that if the x-coordinate, x, on one coordinate axis works in an equation, does the coordinate, x', on some other, parallel axis work? Speaking in terms of the axis on which x is the coordinate, x' is the 'transformed' coordinate. The situation is complicated because we're talking about coordinates - locations - but in most mean- ingful laws/equations, it is lengths/distances (and time intervals) the equations are about, and x coord- inates that represent good, ratio scale measures of distances are only interval scale measures on the x' axis. [See Table of Contents for discussion of scales.] So, if we have an x-coordinate in one system, then we can call the x' value that corresponds to the same point/location the transform of x. In particular, the Principle of Relativity is embodied in the form of the Galilean transformation, which relates the original x, y, z, t to x', y', z', t' by the transform equations x'=x-vt, y'=y, z'=z, t'=t in the simplified case where attention is focused only on transforming the x-axis, and not y and z. In the case of Special Relativity, the x' transform is the same except that x' is then divided by sqrt(1-(v/c)^2), and t'=(t-xv/cc)/sqrt(1-(v/c)^2). In either case, v is the relative velocity of the coordinate systems; if there is already a v in the equations being trans- formed use u or some other variable name. ------------------------------ Subject: 4. The Encyclopedia Brittanica Incompetency. One example of the traditional fallacious idea that an equation is not invariant under the galilean transformation comes from the Encyclopedia Brittanica: "Before Einstein's special theory of relativity was published in 1905, it was usually assumed that the time coordinates measured in all inertial frames were identical and equal to an 'absolute time'. Thus, t = t'. (97) "The position coordinates x and x' were then assumed to be related by x' = x - vt. (98) "The two formulas (97) and (98) are called a Galilean transformation. The laws of nonrelativ- istic mechanics take the same form in all frames related by Galilean transformations. This is the restricted, or Galilean, principle of relativity. "The position of a light wave front speeding from the origin at time zero should satisfy x^2 - (ct)^2 = 0 (99) in the frame (t,x) and (x')^2 - (ct')^2 = 0 (100) in the frame (t',x'). Formula (100) does not transform into formula (99) using the transform- ations (97) and (98), however." ................................................. Besides the trivially correct statement of what the Galilean 'transform' equations are, there is exactly one thing they got right. I. Eq-100 is indeed the correct basis for discussing the question of invariance, given that eq-99 is the correct 'stationary' (observer S) equation. [Let observer M be the 'moving'system observer.] In particular, eq-100 is of exactly the same form [the square of argument one minus the square of argument two equals zero (argument three).] II. It is nonsense to say eq-99 should be derivable from eq-100; for one thing, the transforms are TO x' and t' from x and t, not the other way around, and the idea that either observer's equation should contain within itself the terms to simplify or rearrange to get to the other is ridiculous. As the transform equations say, the relationship of t', x' to t, x is based on the relative velocity between the two systems, but neither the original (eq-99) equation nor the M observer equation is about a relationship between coordinate systems or observers. One might as well expect the two equations to contain banana export/import data; there is no relevancy. The 'transform' equations are the relationships between x' and x, t' and t and have nothing to do with what one equation or the other ought to 'say'. The equations' content is the rate at which light emitted along the x-axes moves. III. Most remarkable, the True Believer SR crackpots who most despise the consequences of measurement theory (demonstrable fact) contained in this document are those who want to argue against our saying the Britt- anica got eq-100 right; They insist that the correct equation is derived directly from x'=x-vt and t'=t. Solve for x=x'+vt and replace t with t', then substitute the result in eq-99: (x'+vt')^2 - (ct')^2 = 0. Besides the fact that this results in an equation with arguments exactly equal to eq-99, they will insist the transform is not invariant. IV. A major justification they have for their idea of the correct M system equation on which to base the the discussion of invariance, is that the variables are M system variables, never mind the fact that the arguments are S system values. That argument of theirs is arrant nonsense. The velocity v that S sees for the M system relative to herself is the negative of what the M system sees for the S system relative to himself. In other words, x'+vt' is a mixed frame expression and it is x'+(-v)t' that would be strictly M frame notation, and that equation is far off base. [Work it out for yourself, but make sure you try out an S frame negative v so as not to mislead yourself.] V. In I. we said: "given that eq-99 is the correct 'stationary' equation. Let's look at it closely: x^2 - (ct)^2 = 0 (99) This whole matter is supposed to be about coordinate transforms. Is that what t is, just a coordinate? No. It isn't, in general. Suppose you and I are both modelling the same light event and you are using EST and I'm using PST. 'Just a time coordinate' is just a clock reading amd your t clock reading says the light has been moving three hours longer than my clock reading says. Well, that's what the idea that t is a coordinate means. Eq-99 works if and only if t is a time interval, and in particular the elapsed time since the light was emitted. Thus, that equation works only if we understand just what t is, an elapsed time, with emissioon at t=0. However, we don't have to 'understand' anything if we use a more intelligent and insightful form of the equation: (x)^2 - [ c(t-t.e) ]^2 = 0, where t.e is anyone's clock reading at the time of light emission, and t is any subsequent time on the same clock. Similarly, x is not just a coordinate, but a distance since emission. (x-x.e)^2 - [ c(t-t.e) ]^2 = 0 (99a) VI. In the spirit of 'there is exactly one thing they got right', the correct M system version of eq-99a is eq-100a: (x'-x.e')^2 - [ c(t'-t.e') ]^2 = 0 (100a) Every observer in the universe can derive their eq-100a from eq-99a and vice versa, not to mention to and from every other observer's eq-99a. Now, THAT's invariance. [You do realize that every eq-100a reduces to eq-99a, when you back substitute from the transforms, right? t.e'=t.e, x.e'=x.e-vt.] ------------------------------ Subject: 5. Transformations on Generalized Coordinate Laws The traditional Gallilean transform is correct: t' = t x' = x - vt. But remember this: a transform of x doesn't effect just some values of x, but all of them, whether they are in the formula or not. This is important if you want to do things right. The crackpot position is strongly against this sci.math verified position, and the apparently standard coordinate pseudo-transformation they suggest is perhaps the result. {See Table of Contents.] Let's use a simple equation: x^2 + y^2 = r^2, which is the formula for a circle with radius r, centered at a location where x=0. But what if the circle center isn't at x=0? Well, we'd want to use the form analytic geometry, vector algebra, and elementary measurement theory tells us to use, a form where we make explicit just where the circle center is, even if it is at x=x0=0: (x-x0)^2 + (y-y0)^2 = r^2. The circle center coordinate, x0, is an x-axis coordinate, just like all the x-values of points on the circle. So, in proper generalized cartesian coordinate forms of laws/equations we want to transform every occurence of x and x0 - by whatever name we call it: x.c, x_e, whatever. So, what is the transformed version of (x-x0)? Why, (x'-x0'); both x and x0 are x-coordinates, and every x-coordinate has a new value on the new axis. So, what is the value of (x'-x0') in terms of the original x data? >From the transform equations we see that x'=x-vt, which is also true for x0'=x0-vt: (x'-x0')=[ (x-vt)-(x0-vt) ]=(x-x0). In other words, when we use the generalized coordinate form specified by analytic geometry, we find that the value of (x'-x0') does not depend on either time or velocity in any way, shape, form, or fashion. Similarly for (y-y0). We can treat time the same way if necessary: (t-t0). The above is a proof that any equation in x,y,z,t is invariant under the galilean transforms. Just use the generalized coordinate form, with (x-x0)/etc, in the transformation process, not the incompetently selected privileged form, with just x/etc. [The form is "privileged" because it assumes the circle center, point of emission, whatever, is at the origin of the axes instead at some less convenient point. After transform the coordinate(s) of the circle center/origin are also changed but the privileged form doesn't make this explicit and screws up the calculations, which should be based on (x'-x0') but are calculated as (x'-0).] The value of (x'-x0') is the same as (x-x0). That makes sense. Draw a circle on a piece of paper, maybe to the right side of the paper. On a transparent sheet, draw x and y coordinate axes, plus x to the right, plus y at the top. Place this axis sheet so the y-axis is at the left side of the circle sheet. Now answer two questions after noting the x-coordinate of the circle center and then moving the axis sheet to the right: (a) did the circle change in any way because you moved the axis sheet (ie because you transformed the coordin- nate axis)? (b) did the coordinate of the circle center change? The circle didn't change [although SR will say it did]; that means that (x'-x0') does indeed equal (x-x0). The coordinate of the circle center did change, and it changed at the same rate (-vt) as did every point on the circle. That means that x0'<>x0, and the fact the circle center didn't change wrt the circle, means that the relationship of x0' with x0 is the same as that of any x' on the circle with the corresponding x: x'=x-vt; x0'=x0-vt. This is to prepare you for the True Believer crackpots that say 'constant' coordinates can't be transformed; some even say they aren't coordinates. These crackpots include some that brag about how they were childhood geniuses, btw. QED: The galilean transformation for any law on generalized Cartesian coordinates is invariant under the Galilean transform. The use of the privileged form explains HOW the transformed equation can be messed up, the next Subject explains what the screwed up effect of the transform is, and how use of the generalized form corrects the screwup. ------------------------------ Subject: 6. The data scale degradation absurdity. The SR transforms and the Galilean transforms both convert good, ratio scale data to inferior interval scale data. The effect is corrected, allowed for, when the transforms are conducted on the generalized coordinate forms specified by analytic geometry and vector algebra. Both sets of transforms are 'translations' - lateral movements of an axis, increasing over time in these cases - but with the SR transform also involving a rescaling. It is the translation term, -vt in the x transform to x', and -xv/cc in the t transform to t', that degrades the ratio scale data to interval scale data. In general, rescaling does not effect scale quality in the size-of-units sense we have here. SR likes to consider its transforms just rotations, however - in spite of the fact Einstein correctly said they were 'translations' (movements) - and in the case of 'good' rotations, ratio scale data quality is indeed preserved, but SR violates the conditions of good ro- tations; they are not rigid rotations and they don't appropriately rescale all the axes that must be rescaled to preserve compatibility. The proof is in the pudding, and the pudding is the combination of simple tests of the transformations. We can tell if the transformed data are ratio scale or interval. Ratio scale data are like absolute Kelvin. A measure- ment of zero means there is zero quantity of the stuff being measured. Ratio scale data support add- ition, subtraction, multiplication, and division. The test of a ratio scale is that if one measure looks like twice as much as another, the stuff being measured is actually twice as much. With absolute Kelvin, 100 degrees really is twice the heat as 50 degrees. 200 degrees really is twice as much as 100. Interval scale data are like relative Celsius, which is why your science teacher wouldn't let you use it in gas law problems. There is only one mathematical operation interval scales support, and that has to be between two measures on the same scale: subtraction. 100 degrees relative (household) Celsius is not twice as much as 50; we have to convert the data to absolute Kelvin to tell us what the real ratio of temperatures is. However, whether we use absolute Kelvin or relative Celsius, the difference in the two temperature readings is the same: 50 degrees. Thus, if we know the real quantities of the 'stuff' being measured, we can tell if two measures are on a ratio scale by seeing if the ratio of the two measures is the same as the ratio of the known quant- ities. If a scale passes the ratio test, the interval scale test is automatically a pass. If the scale fails the ratio test, the interval scale test becomes the next in line. It isn't just the bare differences on an interval scale that provides the test, however. Differences in two interval scale measures are ratio scale, so it is ratios of two differences that tell the tale. Let's do some testing, and remember as we do that our concern is for whether or not the data are messed up, not with 'reasons', excuses, or avoidance. ------------------------------------------------------ Are we going to take a transformed length (difference) and see whether that length fits ratio or interval scale definitions? Of course, not. Interval scale data are ratio after one measure is subtracted from another. That is the major reason the SR transforms can be used in science. Let there be three rods, A, B, C, of length 10, 20, 40, respectively. These lengths are on a known ratio scale, our original x-axis, with one end of each rod at the origin, where x=0, and the other end at the coordinate that tells us the correct lengths. Note that these x-values are ratio scale only because one end of each rod is at x=0. That may remind you of the correct way to use a ruler or yard/meter-stick: put the zero end at one end of the thing you are measuring. Put the 1.00 mark there instead of the zero, and you have interval scale measures. Let A,B,C, be 10, 20, 40. Let a,b,c be x' at v=.5, t=10. x'=x-vt. A B C a b c ---------------- -------------------- 10 20 40 5 15 35 ---------------- -------------------- B/A = 2 b/a = 3 C/A = 4 c/a = 7 C/B = 2 c/b = 2.333 Obviously, the transformed values are no longer ratio scale. The effect is less on the greater values. C-A = 10 b-a = 10 C-A = 30 c-a = 30 C-B = 20 c-b = 20 Obviously, the transformed values are now interval scale. This will hold true for any value of time or velocity. (C-A)/(B-A) = 3 (c-a)/(b-a) = 3 (C-B)/(B-A) = 2 (c-b)/(b-a) = 2 Obviously, the ratios of the differences are ratio scale, being identical to the ratios of the corresponding original - ratio scale - differences. The main difference between these results and the SR results is that the differences do not correspond so neatly to the original, ratio scale, differences. This is due only to the rescaling by 1/sqrt(1-(v/c)^2). The ratios of the differences on the transformed values do correspond neatly and exactly to the ratio scale results. Using the generalized coordinate form, such as (x-x0), the transform produces an interval scale x' and an interval scale x0'. That gives us a ratio scale (x'-x0'), just like - and equal to - (x-x0). ------------------------------ Subject: 7. The Crackpots' Version of the Transforms. It has become apparent - whether misleading or not - that the crackpot responses to the obvious derive from a common source, whether it be bandwagoning or their SR instructors. Below, in the sci.math subject, we see that all sci.math respondents agree with the basic "controversial" position of this faq: every coordinate is transformed, whether a supposed "constant" or not. Think about it, the generalized coordinate of a circle center, x0, applies to infinities upon infinities of circle locations (given y and z, too); it is a constant only for a given circle, and even then only on a given coordinate axis. And even "variables" are often held 'constant' during either integration or differentiation. The utility of a "variable" is that you can discuss all possible particular values without having to single out just one. That utility does not make particular - singled out - values on the variable's axis not values of the variable just because they have become named values. In any case, all that is preamble to the incompetent idea they have proposed for a transform of coordinates. It is based on the idea that the circle center, point of emission, whatever, has coordinates that cannot be transformed. Let there be an equation, say (x)^2 - (ict)^2 = 0. What is the transformed version of that equation? Answer: (x')^2 - (ict')^2 = 0. That's the one thing the Brittanica got right. Note that the leading crackpot just criticized this faq for presuming to correct the Britt- anica, but it then and before poses the incompetent pseudo- transform we analyze here in this section. x to x' and t to t' are obviously coordinate transforms; the x and t coordinates have been replaced by the coord- inates in the primed system. A tranform of an equation from one coordinate system to another is NOT a substitution of the/a definition of x for itself; that is not a coordinate transformation. The most that can said for such a substitution is that it is a change of variable. But the crackpots are calling this a coordinate trans- form of the original equation: (x'+vt)^2 - (ict')^2 = 0. It is not a coordinate transform, of course, except accidentally. (x'+vt) is not the primed system coordinate, it is another form/expression of x. They get that substitution by solving x'=x-vt for x; x=x'+vt. So, by incompetent misnomer, they accomplish what they have been railing against all along. It has been the generalized coordinate form in question all this time: (x-x0)^2 - (ict)^2 = 0. Here they substitute for x instead of transforming to the primed frame: (x'+vt-x0)^2 - (ict')^2. ----- ^ | ^ | It is still x ^ but see what they have accomplished by their mis/malfeasance: [x'+vt-x0]=[x'+(vt-x0)]=[x'-(x0-vt)]. =[x'-x0'] The crackpots have been bragging about how you don't have to transform the circle center's coordinate to transform the circle center's coordinate. Bragging that what they were doing was not what they said they were doing. This does give us insight as to some of the crackpot variations on their x0'<>x0-vt theme, which in all the variations will be discussed in later sections.. They are used to seeing the mixed coordinate form, (x'+vt-x0) without realizing what it respresented, so - accompanied with a lack of understanding of the term 'dependent' - they are used to seeing just the one vt term, and not the one hidden in the defi- nition of x' and are used to imagining it makes the whole expression time dependent and thus not invariant. About which, let x=10, let, x0=20, v=10, and t variously 10 and 23: (x-x0)=-10. Using their (x'+vt-x0): For t=10, we have (x'+vt-x0) = [ (10-10*10) + (10*10) - (20) ] = -90 + 100 - 20 = -10 = (x-x0) For t=23, we have (x'+vt-x0) = [ (10-10*23) + (10*23) - (20) ] = -220 + 230 - 20 = -10 = (x-x0) The result depends in no way on the value of time; we showed the obvious for a couple of instances of t just so you can see that the crackpots not only do not understand the obvious logic of the algebra { (x'-x0')=[ (-vt)-(x0-vt) ]=(x-x0) } - which shows that the transform has no possible time term effect - but they don't understand even a simple arithmetic demonstration of the facts. Oh. Their (x'+vt-x0) or (x'+vt'-x0) reduces the same way since t'=t: (x-vt+vt-x0)=(x-x0). Their process, which says (x'+vt') is the transform of x, says that (x'+vt') is the moving system location of x, but it can't be because x is moving further in the negative direction from the moving viewpoint. That formula will only work out with v<0 which is indeed the velocity the primed system sees the other moving at. However, that formula cannot be derived from x'=x-vt, the formula for transformation of the coordinates from the unprimed to the primed, ------------------------------ Subject: 8. What does sci.math have to say about x0'=x0-vt? The crackpots' positions/arguments were put to sci.math in such a way that at least two or three who posted re- sponses thought it was your faq-er who was on the idiot's side of the questions. Their responses: ---------------------------------------------------------- I. x0' = x0. In other words: x0' <> x0-vt, or "constant values on the x-axis are not subject to the transform". AA: ==================================================================== No. x0' = x0 - vt. Well, if you want, you could define "constant values on the x-axis", but in the context of the question that is not relevant. The relevant fact is that if the unprimed observer holds an object at point x0, then the primed observer assigns to that object a coordinate x0' which is numerically related to x0 by x0'= x0 -vt. AA: ==================================================================== EE: ==================================================================== What does this mean? The line x=x0 will give x'=x-v*t=x0-vt', so if x0' is to give the coordinate in the (x',t',)-system, it will be given by x0'=x0-v*t': ie., it is not given by a constant. Thus, being at rest (constant x-coordinate) is a coordinate-dependent concept. EE: ==================================================================== GG: ==================================================================== Sounds very false. We can say that the representation of the point X0 is the number x0 in the unprimed system, and x0' in the primed system. Clearly x0 and x0' are different, if vt is not zero. However one may say that (though it sounds/is stupid) the point X0 itself "is the same throughout the transformation". However that expression sounds meaningless, since a transform (ok, maybe we should call it a change of basis) is only a function that takes the point's representation in one system into the same point's representation in another system. It is preferrable to use three notations: X0 for the point itself and x0 and x0' for the points' representations in some coordinate systems. GG: ==================================================================== ------------------------------ Subject: 9. But Doesn't x.c'=x.c? That idea is one of the most idiotic to come up, and it does so frequently. And in a number of guises. The idea being that x.c' <> x.c-vt, with x.c being what we have called x0 above; the notation makes no difference. Some crackpots have managed to maintain that position even after graphs have illustrated that such an idea means that after a while a circle center represented by x.c' could be outside the circle. The leading crackpot just make that explicit, as far as one can tell from his befuddled post in response to a line about "active" transforms, which are actually moving body situations, not coordinate transformations: -------------------------------------------------------------------- e>An active transform is not a coordinate transform, ... Right, it is a transform of the center (in the opposite direction) done to effect the change of coordinates without a coordinate transform. ... E: Transform of the center? Center of a circle? He really is saying a circle center moves in the opposite direction of the circle! Right? -------------------------------------------------------------------- If r=10 and x.c was at x.c=0, then the points on the circle (10,0), (-10,0), (0,10) and (0,-10) could at some time become (-10,0), (-30,0), (-20,10), and (-20,-10), but with x.c'=x.c, the circle center would be at (0,0) still! The circle is here but its center is way, way over there! Indeed, although a change of coordinate systems is not movement of any object described in the coordinates, the x.c'=x.c crackpottery is tantamount to the circle staying put but the center moving away. Or vice versa. ------------------------------ Subject: 10. But Isn't (x'-x.c')=(x-x.c) Actually Two Transformations? One crackpot puts the (x'-x.c')=(x-vt - x.c+vt) relationship like this: (x-vt+vt - x.c). See, he says, that is transforming x (with x-vt - x.c) and then reversing the transform (x-vt+vt - x.c). That's just another crackpot form of the idiocy that x.c' <> x.c-vt. You'll have noticed the implication is that there is no transform vt term relating to x.c. ------------------------------ Subject: 11. But Doesn't (x'-x.c+vt) Prove The Transformation Time Dependent? That particular crackpottery is perhaps more corrupt than moronic, since it includes deliberately hiding a vt term from view, and pretending it isn't there. [However, we have seen above that it is a familiar incompetency, and not likely an original.] "Look," the crackpots say, "there is a time term in the transformed (x' - x.c+vt). The transform isn't invariant! It's time dependent!" Just put x' in its original axis form, also, which reveals the other time term, the one they hide: (x'-x.c+vt) = (x-vt - x.c+vt) = (x-x.c). So, at any and all times, the transform reduces to the original expression, with no time term on which to be dependent. Then there is the fact that if you leave the equation in any of the various notation forms - with or without reducing them algebraicly - the arithmetic always comes down to the same as (x-x.c). That means nothing to crack- pots, but may mean something to you. ------------------------------ Subject: 12. But Isn't (x'-x.c')=(x-x.c) a Tautology? My dictionary relates 'tautology' to needless repetition. That's another form of the x.c' <> x.c-vt idiocy. The repetition involved is the vt transformation term. Apply the -vt term to the x term, and it is needless repetition to apply it anywhere again? The 'again' is to the x.c term. The x.c' = x.c crackpot idiocy. The repetition of the vt terms is required by the presence of two x values to be transformed. Be sure to note the next section. ------------------------------ Subject: 13. But Isn't (x'-x.c')=(x-x.c) Almost the Definition of a Linear Transform? Now, how on earth can we relate a tautology to a basic definition in math? >From the top, bottom, middle, and other books in the stack we get this definition: -------------------------------------------------------------- A linear transformation, A, on the space is a method of corr- esponding to each vector of the space another vector of the space such that for any vectors U and V, and any scalars a and b, A(aU+bV) = aAU + bAV. ------------------------------------------------------------- Let points on the sphere satisfy the vector X={x,y,z,1}, and the circle center satisfy C={x.c,y.c,z.c,1}. Let a=1, and b=-1. Let A= ( 1 0 0 -ut ) ( 0 1 0 -vt ) ( 0 0 1 -wt ) ( 0 0 0 1 ) A(aX+bC) = aAX + bAC. aX+bC = (x-x.c, y-y.c, z-z.c, 0 ). The left hand side: A( x - x.c , y - y.c, z - z.c, 0 ) = ( x-x.c , y-y.c, z-z.c, 0 ). The right hand side: aAX= ( x-ut, y-vt, z-wt, 1 ). bAC= (-x.c+ut, -y.c+vt, -z.c+wt, -1 ). and aAX+bAC = ( x-x.c, y-y.c, z-z.c, 0 ). Need it be said? Sure: QED. On the galilean transform the definition of a linear transform, A(aU+bV)=aAU + bAV, is completely satisfied. The generalized form transforms exactly and non-redundantly - with ONE TRANSFORM, not a transform and reverse transform - and non- tautologically, just as the very definition of a linear transform says it should. And does so with absolute invariance, with this galilean transformation. ------------------------------ Subject: 14. But The Transform Won't Work On Time Dependent Equations? The main crackpot that has asserted such a thing was referring to equations such as in Subject 4, above. The Light Sphere equation; for which we have shown repeatedly elsewhere that the numerical calculations are identical for any primed values as for the unprimed values. The presence - before transformation - of a velocity term seems to confuse the crackpots. It turns out there is ex- treme historical reason for this, as you will see in the subject on Maxwell's equations. ------------------------------ Subject: 15. But The Transform Won't Work On Wave Equations? See Subject 17, below, for a discussion of Second Derivative forms and the galilean transforms. ------------------------------ Subject: 16. But Maxwell's Equations Aren't Galilean Invariant? Oh? Just what is the magical term in them that prevents (x'-x.c')=(x-vt - x.c+vt)=(x-x.c) from holding true? It turns out not to be magic, but reality, that interferes with the application of the galilean transforms to the gen- eralized coordinate form(s) of Maxwell: there are no coordi- nates to transform! When True Believer crackpots are shown the simple demonstration that the galilean transform on generalized cartesian coordinates is invariant, their first defense is usually an incredibly stupid "x0'=x0, because the coordinate of a circle center, or point of emission, etc, is a constant and can't be transformed." The last defense is "but Maxwell's equations are not invariant under that coordinate transform." When asked just what magic occurs in Maxwell that would prevent the simple algebra (x'-x0')=[ (x-vt)-(x0-vt) ]=(x-x0) from working, and when asked them for a demonstration, they will never do so, however many hundreds of times their defense is asserted. The reason may help you understand part of Einstein's 1905 paper in which he gave us his absurd Special Relativity derivation: THERE ARE NO COORDINATES IN THE EQUATIONS TO BE TRANSFORMED. Einstein gave the electric force vector as E=(X,Y,Z) and the magnetic force vector as B=(L,M,N), where the force components in the direction of the x axis are X and L, Y and M are in the y direction, Z and N in the z direction. Those values are not, however, coordinates, but values very much like acceleration values. BTW, the current fad is that E and B are 'fields', having been 'force fields' for a while, after being 'forces'. So, when Einstein says he is applying his coordinate transforms to the Maxwell form he presented, he is either delusive or lying. (a) there are no coordinates in the transform equations he gives us for the Maxwell transforms, where B=beta=1/sqrt(1-(v/c)^2): X'=X. L'=L. Y'=B(Y-(v/c)N). M'=B(M+(v/c)Z). Z'=B(Z+(v/c)M). N'=B(N-(v/c)Y). X is in the same direction as x, but is not a coordinate. Ditto for L. They are not locations, coordinates on the x-axis, but force magnitudes in that direction. Similarly for Y and M and y, Z and N and z. (b) the v of the "coordinate transforms" is in Maxwell before any transform is imposed; Einstein's transform v is the velocity of a coordinate axis, not the velocity of a particle, which is what was in the equation before he touched it. (c) if they were honest Einsteinian transforms, they'd be incompetent. The direction of the particle's movement is x, which means it is X and L that are supposed to be transformed, not Y and M, and Z and N. And when SR does transform more than one axis, each axis has its own velocity term; using the v along the x-axis as the v for a y-axis and z-axis transform is thus trebly absurd: the axes perpendicular to the motion are not changed according to SR, the v used is not their v, and the v is not a transform velocity anyway. (d) as everyone knows, the effect of E and B are on the particle's velocity, which is a speed in a particular direction. Both the speed and direction are changed by E and B, but v - the speed - is a constant in SR. As absurd as are the previously demonstrated Einsteinian blunders, this one transcends error and is an incredible example of True Believer delusion propagating over decades. The components of E and B do differ from point to point, and in the variations that are not coordinate free, they are subject to the usual invariant galilean trans- formation when put in the generalized coordinate form. ------------------------------------------------------------- The SR crackpots don't know what coordinates are. The various things they call coordinates include coordin- nates, but also include a variety of other quantities. ------------------------------------------------------ 1. One may express coordinates in a one-axis-at-a-time manner [like x^2+y^2=r^2] but it is the use of vector notation that shows us what is going on. In vector notation the triplet x,y,z [or x1,x2,x3, whatever] represents the three spatial coordinates, but there are so-called basis vectors that underlie them. Those may be called i,j,k. Thus, what we normally treat as x,y,z is a set of three numbers TIMES a basis vector each. 2. These e*i, f*j, g*k products can have a lot of meanings. If e, f, j are distances from the origin of i,j,k then e*i, f*j, g*k are coordinates: distances in the directions of i,j,k respectively, from their origin. That makes the triplet a coordinate vector that we describe as being an x,y,z triplet; perhaps X=(x,y,z). The e*i, f*j, g*k products could be directions; take any of the other vectors described above or below and divide the e,f,g numbers by the length of the vector [sqrt(e^2+f^2+g^2)]. That gives us a vector of length=1.0, the e,f,g values of which show us the direction of the original vector. That makes the triplet a direction vector that we describe as being an x,y,z triplet; perhaps D=(x,y,z). The e*i, f*j, g*k products could be velocities; take any of the unit direction vectors described above and multiply by a given speed, perhaps v. That gives a vector of length v in the direction specified. That makes the triplet a velocity vector that we describe as being an x,y,z triplet; perhaps V=(x,y,z). Each of the three values, e,f,g, is the velocity in the direction of i,j,k respectively. The e*i, f*j, g*k products could be accelerations; take any of the unit direction vectors described above and multiply by a given acceleration, perhaps a. That gives a vector of length a in the direction specified. That makes the triplet an acceleration vector that we describe as being an x,y,z triplet; perhaps A=(x,y,z). Each of the three values, e,f,g, is the acceleration in the direction of i,j,k respectively. The e*i, f*j, g*k products could be forces (much like accel- erations); take any of the unit direction vectors described above and multiply by a given force, perhaps E or B. That gives a vector of length E or B in the direction specified. That makes the triplet a force vector that we describe as being an x,y,z triplet; perhaps E=(x,y,z) or B=(x,y,z). Each of the three values, e,f,g, is the force in the direction of i,j,k respectively. Einstein's - and Maxwell's - E and B are not coordinate vectors. ============================================================ There is another variety of intellectual befuddlement that misinforms the idea that Maxwell isn't invariant under the galilean transform: confusions about velocities. Velocities With Respect to Coordinate Systems. ----------------------------------------------- Aaron Bergman supplied the background in a post to a sci.physics.* newsgroup: =============================================================== Imagine two wires next to each other with a current I in each. Now, according to simple E&M, each current generates a magnetic field and this causes either a repulsion or attraction between the wires due to the interaction of the magnetic field and the current. Let's just use the case where the currents are parallel. Now, suppose you are running at the speed of the current between the wires. If you simply use a galilean transform, each wire, having an equal number of protons and electrons is neutral. So, in this frame, there is no force between the wires. But this is a contradiction. ================================================================ First of all, the invariance of the galilean transform (x'-x.c') =(x-x.c), insures that it is an error to imagine there is any difference between the data and law in one frame and in another; the usual, convenient rest frame is the best frame and only frame required for universal analysis. [Well, (x'<>x, x,c'<>x.c, but (x'-x.c')=(x-x.c).] Second, given that you decide unnecessarily to adapt a law to a moving frame, don't confuse coordinate systems with meaningful physical objects, like the velocity relative to a coordinate system instead of relative to a physical body or field. In other words, what does current velocity with respect to a coordinate system have to do with physics? Nothing. Certainly not anything in the example Bergman gave. What is relevant is not current velocity with respect to a coordinate system, but current velocity with respect to wires and/or a medium. The velocity of an imaginary coordinate sys- tem has absolutely nothing to do with meaningful physical vel- ocity. You can - if you are insightful enough and don't violate item (e) - identify a coordinate system and a relevant physical object, but where some v term in the pre-transformed law is in use, don't confuse it with the velocity of the coordinate transform. Velocities With Respect to ... What? ----------------------------------------------- Albert Einstein opened his 1905 paper on Special Relativity with this ancient incompetency: =============================================================== The equations of the day had a velocity term that was taken as meaning that moving a magnet near a conductor would create a current in the conductor, but moving a conductor near a wire would not. This was belied by fact, of course. The important velocity quantity is the velocity of the magnet and conductor with respect to each other, not to some absolute coordinate frame (as far as we know) and not to an arbitrary coordinate system. One possible cause was the idea: "but the equation says the magnet must be moving wrt the coordinate system" or "... the absolute rest frame". There not being anything in the equation(s) to say either of those, it is amazing that folk will still insist the velocity term has nothing to do with velocity of the two bodies wrt each other. ----------------------------------------------------------- ------------------------------ Subject: 17. First and Second Derivative differential equations. One of the intellectually corrupt ways of denying the very simple demonstration of galilean invariance of all laws expressed in the generalized coordinate form demanded by analytic geometry, vector analysis, and measurement theory [ (x'-x.c')=[ (x-vt)-(x.c-vt) ]=(x-x.c) ] is the assertion that those equations 'over there' (usually Maxwell or wave) are somehow immune to the elementary laws of algebra used to demon- strate the invariance. [Unfortunately, the assertions are never accompanied by reference to the magical math that makes elementary al- gebra invalid. Wonder why that is?] Part of the time it is based on the old lore based on the incompetent transformation of the privileged form of an equation instead of the correct form. [Evidence of this is any reference to an effect due to the velocity of the transform; it falls out algebraicly - as you see above - and cancels out arith- metically - as you can see above.] But usually it is just whistling in the dark, waving the cross (zwastika, I'd say) at the mean old vampire. The most general equation that could be conjured up is a differential with either First or Second Derivatives. Let's examine the plausibility of such magical magical, non-invariance assertions. (a) to get a Second Derivative you must have a First Derivative. (b) to get a First Derivative you must have a function to differentiate. (c) to get a Second Derivative you must have a function in the second degree. So, let us examine the question as to whether any such common Maxwell/wave equation will differ for (a) the common, privileged form, represented as ax^2, with a being an unknown constant function. (b) the generalized cartesian form, represented as a(x-x.c)^2 = ax^2 -2ax(x.c) + ax.c^2, with a being an unknown constant function. (c) the transformed generalized cartesian form, represented as a(x-vt -x.c+vt)^2, same as for (b), = ax^2 -2ax(x.c) + ax.c^2, of course, with a being an unknown constant function. I. for (a), remembering that x.c is a constant, and that this version is only correct because x.c=0, otherwise (b) is the correct form: d/dx ax^2 = 2ax (d/dx)^2 ax^2 = 2a II. for (b), remembering that x.c is a constant. d/dx (ax^2 -2ax(x.c) + ax.c^2) = 2ax - 2ax.c (d/dx)^2 (ax^2 -2ax(x.c) + ax.c^2) = 2a III. for (c); same as for (b). So, what we have seen so far is (1) differential equations in the second degree - the wave equations - must clearly be the same for all forms: the privileged form in x, the generalized cartesian form in x and the centroid, x.c, or the transformed generalized cartesian form. That is, anyone who imagines that correct usage gives different results for galilean transformed frames is at first showing his ignorance, and in the end showing his intellectual corruption. (2) As far as the First Derivatives are concerned, the only cases in which there really is a difference between the two forms is where x.c <> 0, and in that case, the use of the privileged form is obviously incompetent. So, how do you correctly use the differential equations? If you are using rest frame data with the centroid at x=0, etc, you can't go wrong without trying to go wrong. If you are using rest frame data with the centroid not at x=0, you must use (x-x.c) anyplace x appears in the equation. If you are using moving frame data, you must use the moving frame centroid as well as the light front (or whatever) moving frame data itself, perhaps first calculating (x'-x.c'), which equals (x-x.c) which is obviously correct, and which is obviously the plain old correct x of the privileged form. Unless, of course, there really is some magical term or expression that invalidates the obvious and elemen- tary algebra of the invariance demonstration. Or maybe you just whistle when you don't want basic algebra to hold true. Eleaticus !---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---? ! Eleaticus Oren C. Webster ThnkTank@concentric.net ? ! "Anything and everything that requires or encourages systematic ? ! examination of premises, logic, and conclusions" ? !---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?