mm-134 I have a funciton:F(u,v) = Integrate ( 0, 2*pi ) Exp(-i*a*(u*Cos(t)+v*Sin(t)) ) dta=constant>0 i=sqrt(-1)we note thatF(u,0) = 2*pi*J_0(u*a)F(0,v) = 2*pi*J_0(v*a)where J_0(x) = bessel function of 'rst kind; order zero.also if you assume that F(u,v) is continuous w.r.t u and v, anddifferentiate under the integral sign i *think* you can show thatu*dF/dv - v*dF/du = 0 where the above are taken as partial derivitives: not sure about thatidentity yet - might have slipped with my algebra. dont know if knowingthat will help resolve exactly what F is.i guess a good starting point would be knowing how one calculates theintegral of say F(u,0) to get the bessel function - however i cant recallthat at the moment...im assuming it might possibly be by series, and willwork on this.any ideas most appreciated.cheersmoth F(u,v) = Integrate ( 0, 2*pi ) Exp(-i*a*(u*Cos(t)+v*Sin(t)) ) dt>> a=constant>0 i=sqrt(-1)The OP has already solved the problem him(her)self, but I would like tosuggest an alternate solution that is much simpler:Rewrite u*Cos(t) + v*Sin(t) = r*Cos(t-phi), for some values r and phi.Then we have r = Sqrt(u^2 + v^2). The value of phi is irrelevant, since theintegral is over a complete period. The function F(u,v) is independent ofphi, and we may therefore just as well set phi equal to zero. The integralthen becomes:Integrate ( 0, 2*pi ) Exp(-i*a*(r*Cos(t)) ) dt,which just happens to be the original integral with u=r and v=0.-Michael. =Some more:>> have a funciton:>> F(u,v) = Integrate ( 0, 2*pi ) Exp(-i*a*(u*Cos(t)+v*Sin(t)) ) dt>> a=constant>0 i=sqrt(-1)>> we note that>> F(u,0) = 2*pi*J_0(u*a)> F(0,v) = 2*pi*J_0(v*a)>> where J_0(x) = bessel function of 'rst kind; order zero.>> also if you assume that F(u,v) is continuous w.r.t u and v, and> differentiate under the integral sign i *think* you can show that>> u*dF/dv - v*dF/du = 0>assuming the above is correct you can show via the method of characteristicsthat a function satisfying the above + the intiial condition of say F(u,0) =2*pi*J_0(u*a) isF(u,v) = 2*pi*J_0(a*Sqrt( u^2 + v^2 ))however is this actually a solution of the original integral as statedabove? (i think it may be...but would like to prove it)cheersmoth =some more again ;-)> Some more:> have a funciton:>> F(u,v) = Integrate ( 0, 2*pi ) Exp(-i*a*(u*Cos(t)+v*Sin(t)) ) dt>> a=constant>0 i=sqrt(-1)>> we note that>> F(u,0) = 2*pi*J_0(u*a)> F(0,v) = 2*pi*J_0(v*a)>> where J_0(x) = bessel function of 'rst kind; order zero.>> also if you assume that F(u,v) is continuous w.r.t u and v, and> differentiate under the integral sign i *think* you can show that>> u*dF/dv - v*dF/du = 0> assuming the above is correct you can show via the method ofcharacteristics> that a function satisfying the above + the intiial condition of say F(u,0) 2*pi*J_0(u*a) is>> F(u,v) = 2*pi*J_0(a*Sqrt( u^2 + v^2 ))>> however is this actually a solution of the original integral as stated> above? (i think it may be...but would like to prove it)>> cheers> mothi am sure this is correct now as i can verify that 2*Integrate [ 0, oo] r*Exp(-r^2)*J_0(2*Sqrt(t*(u^2+v^2)) )drequates toExp(-t*(u^2+v^2))which is what i started with. looks like i have answered my own problem ;-)sorry to waste board space.cheersmoth Yes, . means multiplication. I thought that that was a kind of universal>notation.No, because the . is too small and too low. A higher, bigger dot meansmultiplication.John Savardhttp://home.ecn.ab.ca/~jsavard/index.html Yes, . means multiplication. I thought that that was a kind of universal>notation.No, because the . is too small and too low. A higher, bigger dot means> multiplication.Not universally. English typographical preference used to be forrepresenting the decimal separator by such a raised dot.Phil-- Unpatched IE vulnerability: DNSError folder disclosureDescription: Gaining access to local security zonesReference: http://msgs.securepoint.com/cgi-bin/get/bugtraq0306/52.html== =Yes, . means multiplication. I thought that that was a kind of universal>notation.No, because the . is too small and too low. A higher, bigger dot means> multiplication.Yes, but AFAIK that cannot be obtained in ASCII. The closest thing there isis ..Jose Carlos Santos >Yes, . means multiplication. I thought that that was a kind of universalnotation.> No, because the . is too small and too low. A higher, bigger dot means> multiplication.Yes, but AFAIK that cannot be obtained in ASCII. The closest thing there is> is ..> Jose Carlos SantosSo many computer languages use *, that it has almost become the newsgroup defacto standard for multiplications. It is certainly less liable to be misinterpreted than .. >You seem to forget that the Canadians kicked our ass in two con§icts....>Really? Which two?One of them was vaguely contemporary with the War of 1812, as Irecall, but I donOt remember *another* one.However, I doubt very much that, given the *current* ratio of militarymight between our two countries, that Canadians are optimistic aboutbeing able to defend themselves against a determined attack from thesouth. Rather, they have faith in the good, peaceful, and democraticintentions of that nation.Perhaps, unfortunately, unlike some of its brothers to the south.But just today in the National Post, I found a *reason* for the UnitedStates to annex Canada.It seems that the sensible ordinary people in both countries, who wantthe same things, are being forced to put up with governments ofopposite kinds which are in many ways different from what they wouldwant because of bloc voting groups. In the States, they have the BibleBelt. In Canada, we have Quebec.So if we eachother out!That means, of course, it would be Bill Clinton and not George Bushwho would want to annex Canada.John Savardhttp://home.ecn.ab.ca/~jsavard/index.html >You seem to forget that the Canadians kicked our ass in two con§icts....>>Really? Which two?> One of them was vaguely contemporary with the War of 1812, as I> recall, but I donOt remember *another* one.HeOs probably referring to the attempt to move into the land north of the 49th. As some sticklers here point out, there was no Canada then.However, I doubt very much that, given the *current* ratio of military> might between our two countries, that Canadians are optimistic about> being able to defend themselves against a determined attack from the> south. Rather, they have faith in the good, peaceful, and democratic> intentions of that nation.Probably not. IOm very lucky though, because if it ever came to that, I have a right to work in the UK, and IOd be gone, in a §ash.Perhaps, unfortunately, unlike some of its brothers to the south.But just today in the National Post, I found a *reason* for the United> States to annex Canada.It seems that the sensible ordinary people in both countries, who want> the same things, are being forced to put up with governments of> opposite kinds which are in many ways different from what they would> want because of bloc voting groups. In the States, they have the Bible> Belt. In Canada, we have Quebec.Interestingly though, Quebec in another one of those things that the US has taken interest in and theyOve made it clear, on more than one occasion, that they donOt want Quebec to leave, while some Canadians do. You had your civil war in the US, weOve just been dealing with the Quebec issue, it seems forever. Perhaps because we arenOt so anxious for civil war, never have been. Besides, the West has been making a lot of noises about being unhappy about things too.So if we could out!That means, of course, it would be Bill Clinton and not George Bush> who would want to annex Canada.Eiterh way, IOd leave, cross the pond, back to the land of my mother and my fatherOs ancestors.DJohn Savard> http://home.ecn.ab.ca/~jsavard/index.html IOm searching a conformal mapping that transform a rectangle in a ring>bounded by two circles with the same center, i.e., a disc with a>circle hole with the same center.>The horizontal edges of the rectangle must transform in the two circle>(with the same center), while the two vertical edges must coincide at>the end of the mapping and they must be transformed in a vertical>radius line going from the inner to the outer circles.>MAy anybody help me?Elliptic integrals.http://www.hypermaths.org/quadibloc/maps/mcf0703. htmOh, you donOt want to transform just a *plain* rectangle. I thinkyouOll have to go to the library and look this one up.John Savardhttp://home.ecn.ab.ca/~jsavard/index.html > Anyone know why Grad Schools will not accept your application unless>> you have a BachelorOs? Is it not possible to have the required>> knowledge and no degree? Is it so very hard to verify that oneOs claim>> of knowledge.> Occasionally someone _is_ admitted without a bachelorOs degree. But> these are rare exceptions. In mathematics, if an American applicant> had a top score on the Graduate Record Exam subject exam, then the math> department would probably be so eager to have him/her that they would> intervene with the admissions people to make an exception.Really? The GRE subject exam in math is kind of trivial -- tests> perhaps what a sophomore math major ought to know, and doesnOt> require any dif'cult reasoning, just techniques. IOd> think more weight would be given to the GRE Analytic score, which> I understand has now been or will shortly be eliminated(?). what exactly does the GRE analytic section (or for that matter theGMAT analytic section) really measure? I achieved SAT scores above1300 (in 1970 before the test was watered down) and GRE scores of720V/730M. The results on these tests and IQ tests always placed mein the 98th percentile or so, yet my GRE analytic scores were amediocre 60th percentile. What I remember of the test is that none ofthe items were particularly dif'cult, just that the items werelong-winded and highly detailed word problems; while I probablyanswered everything I saw correctly, I didnOt get to complete or evenlook at several of the other items on the test. Later on I read thatthis section (and the GMAT section I mention above) seem to be highlysensitive to test coaching techniques (Kaplan/Princeton Review). I.e.these sections have less to do with aptitude and more to do with testtaking knowledge: set up a matrix or table, eliminate wrong choices,and donOt spend too much time on any one of the items. Withaptitude sections like the math and verbal sections of the GRE andSAT you donOt have to keep these test taking techniques in mind, ifyou know the answer or how to compute it, you donOt have to waste alot of time getting to the point where you can write down your answer. IOve actually seen a trisection device with no moving parts at all.> ItOs basically a strip with one end having square and circular> attachments at the end. By placing the object in the right position> relative to a given angle, the trisectors of the angle are identi'ed.That rings a bell. Does it look like the image on the following page?http://www.math.cornell.edu/~dwh/books/eg00/prob14-4. htmlAnd hereOs a page with a bunch of stuff (mostly cranky), including onetrisection with a carpenterOs square:http://www.jimloy.com/geometry/trisect.htm-- Daniel W. Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W cf(e^(1/x) = [1, x-1, 1 1, 3x-1, 1, 1, 5x-1, 1, ...]This can be obtained by a transformation of LambertOs continued fraction for tan, or equivalently(e^(1/x)-1)/(e^(1/x)+1) = tanh(1/(2x)) = [0, 2x, 6x, 10x, ... ]If this is z, then e^(1/x) = (1+z)/(1-z)Now if y = (1+z)/(1-z), and z = 1/(2qx + z1), theny = 1 + 1/(qx-1 + 1/(1 + 1/y1)), then y1 = (1+z1)/(1-z1)and thuse^(1/x) = [1; 1768 or so.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 =Robert 5x-1, 1, ...]> This can be obtained by a transformation of LambertOs continued fraction > for tan, or equivalently(e^(1/x)-1)/(e^(1/x)+1) = tanh(1/(2x)) = [0, 2x, 6x, 10x, ... ]If this is z, then e^(1/x) = (1+z)/(1-z)Now if y = (1+z)/(1-z), and z = 1/(2qx + z1), then> y = 1 + 1/(qx-1 + 1/(1 + 1/y1)), then y1 = (1+z1)/(1-z1)> and thus> e^(1/x) = [1; x-1, 1, 1, 3x-1, 1, 1, 5x-1, 1, ... ]This all dates back to 1768 or so.> They had power and a superb elegance, those days! Wow.So IOve found a red wine of already more than 200 years...a nice present just at the birthday-weekend :-)Gottfried Helms =let I=k[x_1,x_2,..,x_n] is polinomial ring over 'eld of char=0 and> I_n - subspase gomogenius polinomial of power n. Let sl_2 - 3 -> dimesional simple lie algebra wich act at I_n in usual way. How 'nd a> irreducible components of decomposition of this representation? Need> 'nd something like as formulae of (Klebsh-Gordon)for tensor product.You use n in two different ways. Let I_k be the k-homogeneous part.What was the action of sl_2 again?. The way I see it I_k is naturallyacted upon by sl_n for all k. But these sl_n-modules are all simple(in characteristic zero) so there is no decomposition. Obviouslysomething is wrong.Jyrki Lahtonen, Turku, Finland =f and abs:R -> R, x -> |x|- =Frederick S. Bustamante> can someone help me with these practice problems?>> 1. Let f:X-->D for some connected metric space X and some discrete> space D. Show f is continuous iff f is constant.Any constant function is continuous, of course, and if f is not constant,consider the inverse images of two disjoint (perforce open) subsets of D.> 2. Let f:X-->R for some metric space X. If f is continuous, then |f|> is continuous.|f| is the composition of two continuous functions. (IOm trying to do thehinting thing:)LH =Is there a sequence S of sets such that lim card S_n < card lim S_n, where card denotes cardinality, both limits exist, and the inequality is strict? Here trans'nite limits are considered to exist as long as they are well-de'ned. Also, feel free to call on the axiom of choice.If so, does there exist such a sequence with lim S_n 'nite?-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu Is there a sequence S of sets such that lim card S_n < card lim S_n, > where card denotes cardinality, both limits exist, and the inequality > is strict? Here trans'nite limits are considered to exist as long > as they are well-de'ned. Also, feel free to call on the axiom of choice.If so, does there exist such a sequence with lim S_n 'nite?Yes, I think it is possible. LetOs try this:Let S_n = all rationals in [0,1) whose decimal expansion has onlyfindigits after the decimal point (not counting the in'nite number of0Os which follow). This may need to be tidied up a bit to removealternative formats, like .5 = .4999...Now card S_n = 10*n, thus lim card S_n = Aleph_0. However lim S_n isthe set of all reals on [0,1), thus card lim S_n is 2^Aleph_0.Thus lim card S_n < card lim S_n .Note: I am a little unsure about the argument that lim card S_n =Aleph_0, so maybe somebody can correct me if this is invalid. > Is there a sequence S of sets such that lim card S_n < card lim S_n, >> where card denotes cardinality, both limits exist, and the inequality >> is strict? Here trans'nite limits are considered to exist as long >> as they are well-de'ned. Also, feel free to call on the axiom of choice.>> >> If so, does there exist such a sequence with lim S_n 'nite?> Yes, I think it is possible. LetOs try this:> Let S_n = all rationals in [0,1) whose decimal expansion has only n> digits after the decimal point (not counting the in'nite number of> 0Os which follow). This may need to be tidied up a bit to remove> alternative formats, like .5 = .4999...> Now card S_n = 10*n, thus lim card S_n = Aleph_0. However lim S_n is> the set of all reals on [0,1), thus card lim S_n is 2^Aleph_0.No, because lim S_n is not the same thing as the closure of the limit.The only numbers that are eventually in S_n are the rationals with terminatingdecimals. You donOt even get 1/3, for example.> Thus lim card S_n < card lim S_n .No, card lim S_n = aleph_0 = lim card S_n < card (cl lim S_n) = c.> Note: I am a little unsure about the argument that lim card S_n Aleph_0, so maybe somebody can correct me if this is invalid.That part is ok.-- Dave SeamanJudge YohnOs mistakes revealed in Mumia Abu-Jamal ruling. Is there a sequence S of sets such that lim card S_n < card lim S_n, >where card denotes cardinality, both limits exist, and the inequality >is strict? Here trans'nite limits are considered to exist as long >as they are well-de'ned. Also, feel free to call on the axiom of choice.If by sequence you mean an ordinary sequence indexed by thepositive integers then no: This is clearly impossible if lim S_n is'nite (in that case there exists N so that S_n is contained in S_kfor all k > N), and if lim S_n is countably in'nite then card S_ncannot be bounded by a 'nite number - 'nally, if lim S_n isuncountable then the _union_ of countably many sets ofsmaller cardinality has smaller cardinality. (I believe thatlast bit needs AC.)If youOre considering more general sequences, for exampleindexed by ordinals, then the answer is yes - for examplethe countable ordinals constitute a sequence where thelimit of the cardinality is aleph_0 but the cardinality of thelimit is aleph_1.>If so, does there exist such a sequence with lim S_n 'nite?************************David C. Ullrich > >Is there a sequence S of sets such that lim card S_n < card lim S_n, >>where card denotes cardinality, both limits exist, and the inequality >>is strict? Here trans'nite limits are considered to exist as long >>as they are well-de'ned. Also, feel free to call on the axiom of choice.>> >>If by sequence you mean an ordinary sequence indexed by the>positive integers>I do.> then no: This is clearly impossible if lim S_n is>'nite (in that case there exists N so that S_n is contained in S_k>for all k > N),>It is way too early for me right now, but I believe this is false: E.g., let S_n = {1,n}.> and if lim S_n is countably in'nite then card S_n>cannot be bounded by a 'nite number - >How do we show that, exactly?>'nally, if lim S_n is>uncountable then the _union_ of countably many sets of>smaller cardinality has smaller cardinality. (I believe that>last bit needs AC.)>I see what you are getting at - that if card lim S_n > omega, then it must be sup card S_n - but I need more sleep to convince myself about the details of that one.-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu >> >Is there a sequence S of sets such that lim card S_n < card lim S_n, >where card denotes cardinality, both limits exist, and the inequality >is strict? Here trans'nite limits are considered to exist as long >as they are well-de'ned. Also, feel free to call on the axiom of choice. >>If by sequence you mean an ordinary sequence indexed by the>>positive integers> I do.> >> then no: This is clearly impossible if lim S_n is>>'nite (in that case there exists N so that S_n is contained in S_k>>for all k > N),> It is way too early for me right now, but I believe this is false: > E.g., let S_n = {1,n}.> But what is the limit of this in your opinion? In mine it would just beOtheO two point set {*,&}, lim has a Ouniversal propertyO categorical interpretation which IOmcertainly using, and I would think David is too, but possibly withoutwanting to say so.here it means ifS_1 --> S_2 --> S_3 ...has a limit S, it is OtheO object (set in this case) that has mapsto it from every S_i, with the property that the map S_j to S is the sameas mapping to S_k (k>i) and then mapping to S.in your example, the maps between S_i and S_(i+1) I assume takes 1 to 1but what does it do to i? if it sends it to i+1, then youOre identifyingthem, and the limit would be the two point set.ItOs not just the sets you need to consider but the map between sets.>> and if lim S_n is countably in'nite then card S_n>>cannot be bounded by a 'nite number - > How do we show that, exactly?> Other wise after a certain point in the sequence you are just identifyingelements of the set, thatOs what it means if there is a uniform bound onthe size of the sets. The way the limit gets bigger is to have bigger andbigger nested subsets.>>'nally, if lim S_n is>>uncountable then the _union_ of countably many sets of>>smaller cardinality has smaller cardinality. (I believe that>>last bit needs AC.)> I see what you are getting at - that if card lim S_n > omega, then it > must be sup card S_n - but I need more sleep to convince myself about > the details of that one. >> >> >Is there a sequence S of sets such that lim card S_n < card lim S_n, >>where card denotes cardinality, both limits exist, and the inequality >>is strict? Here trans'nite limits are considered to exist as long >>as they are well-de'ned. Also, feel free to call on the axiom of choice.> >If by sequence you mean an ordinary sequence indexed by the>positive integers> I do.>> > then no: This is clearly impossible if lim S_n is>'nite (in that case there exists N so that S_n is contained in S_k>for all k > N),> It is way too early for me right now, but I believe this is false: >> E.g., let S_n = {1,n}.>> >>But what is the limit of this in your opinion? In mine it would just be>OtheO two point set {*,&}, ??? What he said _is_ a counterexample to what IOd written (becausebut the limit of the sets {1,n} is the one-point set {1}.>lim has a Ouniversal propertyO categorical interpretation which IOm>certainly using, and I would think David is too, but possibly without>wanting to say so.>>here it means if>>S_1 --> S_2 --> S_3 ...>>has a limit S, it is OtheO object (set in this case) that has maps>to it from every S_i, with the property that the map S_j to S is the same>as mapping to S_k (k>i) and then mapping to S.>>in your example, the maps between S_i and S_(i+1) I assume takes 1 to 1>but what does it do to i? if it sends it to i+1, then youOre identifying>them, and the limit would be the two point set.>>ItOs not just the sets you need to consider but the map between sets.This may well be a reasonable and standard notion, but itOs certainlynot what I meant by lim S_n, and not what I think he meant either.The notion of the limit of a sequence of sets has at least one otherperfectly standard meaning that has nothing to do with what youwrite here: S = lim S_n means that every element of S is in S_nfor large n and every non-element of S is a non-element of S_nfor large n. (Ie, the indicator function of S_n converges pointwiseto the indicator function of S.)> and if lim S_n is countably in'nite then card S_n>cannot be bounded by a 'nite number - > How do we show that, exactly?>> >>Other wise after a certain point in the sequence you are just identifying>elements of the set, thatOs what it means if there is a uniform bound on>the size of the sets. The way the limit gets bigger is to have bigger and>bigger nested subsets.>'nally, if lim S_n is>uncountable then the _union_ of countably many sets of>smaller cardinality has smaller cardinality. (I believe that>last bit needs AC.)> I see what you are getting at - that if card lim S_n > omega, then it >> must be sup card S_n - but I need more sleep to convince myself about >> the details of that one.************************David C. Ullrich >>But what is the limit of this in your opinion? In mine it would just be>>OtheO two point set {*,&}, ??? What he said _is_ a counterexample to what IOd written (because> but the limit of the sets {1,n} is the one-point set {1}.> Not if you were to de'ne a map between {1,n} and {1,n+1} that, purely asa map of sets, identi'ed 1 with 1 and n with n+1. As sets, the limit isgoing to be a set with two elements, whih=ch i just denoted by * and & forthe sheer hell of it.The problem is that my notion of lim requires maps between the sets.If you were to send both 1 and n to 1 in S_{n+1} then the limit would bethe one point set.>>lim has a Ouniversal propertyO categorical interpretation which IOm>>certainly using, and I would think David is too, but possibly without>>wanting to say so.>>here it means if>>S_1 --> S_2 --> S_3 ...>>has a limit S, it is OtheO object (set in this case) that has maps>>to it from every S_i, with the property that the map S_j to S is the same>>as mapping to S_k (k>i) and then mapping to S.>>in your example, the maps between S_i and S_(i+1) I assume takes 1 to 1>>but what does it do to i? if it sends it to i+1, then youOre identifying>>them, and the limit would be the two point set.>>ItOs not just the sets you need to consider but the map between sets.This may well be a reasonable and standard notion, but itOs certainly> not what I meant by lim S_n, and not what I think he meant either.> The notion of the limit of a sequence of sets has at least one other> perfectly standard meaning that has nothing to do with what you> write here: S = lim S_n means that every element of S is in S_n> for large n and every non-element of S is a non-element of S_n> for large n. (Ie, the indicator function of S_n converges pointwise> to the indicator function of S.)> The category theory style one says that an element of the limit is in oneof the sets, and then every one after that too, and its images areidenti'ed. The natural Idea of lim youOre using (and it is much more natural, andeasier to think about on the surface) is better than mine because it candeal with suggestive things like the S_i in the counter example, where wekeep special meanings for the letters in the set. I would have to tell youhow to OannihilateO elements which in general I canOt, because IOm hidingsomething about what I really mean by lim: you need an indexing categoryand some functors lying around, but I suspect you either know this alreadyor donOt care.I can force mine to be equivalent by saying Osuppose I want to deal withsituations where S_i OmapsO to S_{i+1} but i donOt want to explicilty map ito anything (for the 'rst time) (again keeping with the above counter example). Then I simply remove i from the set, and all its preimages from the previous sets.OAs it is, if one is allowed to just kill off elements at will, then it isalways possible to create these examples of lim card not being related tocard lim, necessarily.Take a vector space X of abitrary dimension over a 'eld of abitrarycardinality, de'ne a chainX --> X --> X > ...with the zero map at every arrow, then the lim is the zero vectorspace in my way of thinking.If you want it to be X then use the identity map between them.Oh, and IOve not apologized for presuming to tell you what you werethinking, sorry.It wasnOt very clearly explained, was it? sorry.>> and if lim S_n is countably in'nite then card S_n>>cannot be bounded by a 'nite number - > How do we show that, exactly?> >>Other wise after a certain point in the sequence you are just identifying>>elements of the set, thatOs what it means if there is a uniform bound on>>the size of the sets. The way the limit gets bigger is to have bigger and>>bigger nested subsets.>>'nally, if lim S_n is>>uncountable then the _union_ of countably many sets of>>smaller cardinality has smaller cardinality. (I believe that>>last bit needs AC.)> I see what you are getting at - that if card lim S_n > omega, then it > must be sup card S_n - but I need more sleep to convince myself about > the details of that one.************************David C. Ullrich >>But what is the limit of this in your opinion? In mine it would just be>OtheO two point set {*,&}, >> >> ??? What he said _is_ a counterexample to what IOd written (because>> but the limit of the sets {1,n} is the one-point set {1}.>> >>Not if you were to de'ne a map between {1,n} and {1,n+1} that, purely as>a map of sets, identi'ed 1 with 1 and n with n+1. As sets, the limit is>going to be a set with two elements, whih=ch i just denoted by * and & for>the sheer hell of it.>>The problem is that my notion of lim requires maps between the sets.ThatOs a problem, all right.LetOs think about this. Evidently there are at least two things thenotion of the limit of a sequence of sets might mean. One of themis in fact the limit of a sequence of sets - the other one is _not_actually a notion of the limit of a sequence of sets, itOs thelimit of a structure involving a bunch of sets together withmaps between them.Now we have a problem about the limit of a sequence of sets.With no maps given. DoesnOt it seem plausible to assume thatthis was about the notion that only involves a sequence ofsets?>If you were to send both 1 and n to 1 in S_{n+1} then the limit would be>the one point set.>lim has a Ouniversal propertyO categorical interpretation which IOm>certainly using, and I would think David is too, but possibly without>wanting to say so.>>here it means if>>S_1 --> S_2 --> S_3 ...>>has a limit S, it is OtheO object (set in this case) that has maps>to it from every S_i, with the property that the map S_j to S is the same>as mapping to S_k (k>i) and then mapping to S.>>in your example, the maps between S_i and S_(i+1) I assume takes 1 to 1>but what does it do to i? if it sends it to i+1, then youOre identifying>them, and the limit would be the two point set.>>ItOs not just the sets you need to consider but the map between sets.>> >> This may well be a reasonable and standard notion, but itOs certainly>> not what I meant by lim S_n, and not what I think he meant either.>> The notion of the limit of a sequence of sets has at least one other>> perfectly standard meaning that has nothing to do with what you>> write here: S = lim S_n means that every element of S is in S_n>> for large n and every non-element of S is a non-element of S_n>> for large n. (Ie, the indicator function of S_n converges pointwise>> to the indicator function of S.)>> >>The category theory style one says that an element of the limit is in one>of the sets, and then every one after that too, and its images are>identi'ed. >>The natural Idea of lim youOre using (and it is much more natural, and>easier to think about on the surface) is better than mine because it can>deal with suggestive things like the S_i in the counter example, where we>keep special meanings for the letters in the set. I would have to tell you>how to OannihilateO elements which in general I canOt, because IOm hiding>something about what I really mean by lim: you need an indexing category>and some functors lying around, but I suspect you either know this already>or donOt care.HereOs what IOm curious about. Say we de'ne a sequence of realsby letting x_n be sqrt(2) to n decimals: x_1 = 1.4, x_2 = 1.41, etc.Can we say that lim x_n = sqrt(2), or would that be keeping specialmeanings for the digits in the decimal expansions?>I can force mine to be equivalent by saying Osuppose I want to deal with>situations where S_i OmapsO to S_{i+1} but i donOt want to explicilty map i>to anything (for the 'rst time) (again keeping with the above counter >example). Then I simply remove i from the set, and all its preimages from >the previous sets.O>>As it is, if one is allowed to just kill off elements at will, then it is>always possible to create these examples of lim card not being related to>card lim, necessarily.>>Take a vector space X of abitrary dimension over a 'eld of abitrary>cardinality, de'ne a chain>X --> X --> X > ...>>with the zero map at every arrow, then the lim is the zero vector>space in my way of thinking.>If you want it to be X then use the identity map between them.>>Oh, and IOve not apologized for presuming to tell you what you were>thinking, sorry.>>It wasnOt very clearly explained, was it? sorry.> and if lim S_n is countably in'nite then card S_n>cannot be bounded by a 'nite number - > How do we show that, exactly?>>Other wise after a certain point in the sequence you are just identifying>elements of the set, thatOs what it means if there is a uniform bound on>the size of the sets. The way the limit gets bigger is to have bigger and>bigger nested subsets.>'nally, if lim S_n is>uncountable then the _union_ of countably many sets of>smaller cardinality has smaller cardinality. (I believe that>last bit needs AC.)> I see what you are getting at - that if card lim S_n > omega, then it >> must be sup card S_n - but I need more sleep to convince myself about >> the details of that one.>> >> ************************>> >> David C. Ullrich************************David C. Ullrich >Is there a sequence S of sets such that lim card S_n < card lim S_n, >where card denotes cardinality, both limits exist, and the inequality >is strict? Here trans'nite limits are considered to exist as long >as they are well-de'ned. Also, feel free to call on the axiom of choice. >>If by sequence you mean an ordinary sequence indexed by the>>positive integers>I do.> then no: This is clearly impossible if lim S_n is>>'nite (in that case there exists N so that S_n is contained in S_k>>for all k > N),>It is way too early for me right now, but I believe this is false: >E.g., let S_n = {1,n}.There was a typo there - I meant to say that if lim S_n is 'nitethen there exists N such that lim S_n is contained in S_k forall k > N.>> and if lim S_n is countably in'nite then card S_n>>cannot be bounded by a 'nite number - >How do we show that, exactly?This almost seems like either you havenOt thought aboutit or weOre using different de'nitions of lim S_N. Sojust for the record, the de'nition I have in mind is this:lim S_n = S if (i) for every x in S there exists N such thatx is in S_n for all n > N and (ii) for every x not in S thereexists N such that x is in S_n for no n > N.It follows that if S = lim S_n and A is a 'nite subset ofS then there exists N such that A is contained in S_nfor all n > N (this follows from the de'nition and thefact that given any 'nite set of positive integers hasa largest element.) Hence if S = lim S_n and S isin'nite then for every positive integer k there existsN such that S_n has at least k elements for all n > N.Hence if S_n has at most k elements for all n andS = lim S_n then S has at most k elements.>>'nally, if lim S_n is>>uncountable then the _union_ of countably many sets of>>smaller cardinality has smaller cardinality. This sentence is not true. ItOs also not what I really meant,and not what we need. WhatOs true, and what we need here,is that if lim S_n is uncountable then the union of countably many sets of cardinality bounded below card lim S_nhas cardinality smaller than card lim S_n.>(I believe that>>last bit needs AC.)>I see what you are getting at - that if card lim S_n > omega, then it >must be sup card S_n - but I need more sleep to convince myself about >the details of that one.LetOs say S = lim S_n, and A = card S. Suppose that lim card S_n = B < A. Then card S_n <= B for all but'nitely many n; since lim S_n is unchanged by altering'nitely many S_n, wlog card S_n <= B for all n. Socard S <= card union S_n <= B.************************David C. Ullrich Is there a sequence S of sets such that lim card S_n < card lim S_n, > where card denotes cardinality, both limits exist, and the inequality > is strict? Here trans'nite limits are considered to exist as long > as they are well-de'ned. Also, feel free to call on the axiom of choice.If so, does there exist such a sequence with lim S_n 'nite?I would imagine that this depends on what youOre taking your limit over.YouOve suggestively written n implying over the naturals. Instead howabout taking the ('ltered direct) limit indexed by some suitable category(or net, as I;ve been reminded recently of for analysis). ItOs too nastyfor me to consider before coffee. Any one any ideas on that? =I am looking for the volume of the solid of revolution obtained byrotating the region bounded by y=x and y=x^2 around the verify this? I am looking for the volume of the solid of revolution obtained by> rotating the region bounded by y=x and y=x^2 around the x-axis?>> The answer I have come up with is 2pi/15.>> Can anyone verify this?Show us your work... if you can. =Heres the work I did:x = x^2; x= 1, 0 these are the points which bound the solid. V = | [pi(x)^2 - pi(x^2)^2] dxV = pi | [x^2 - x^4] dxV = pi (x^3/3) - (x^5/5)evaluated at 1 and 0 = 2pi/15 Heres the work I did:>> x = x^2; x= 1, 0 these are the points which bound the solid.Yup.> V = | [pi(x)^2 - pi(x^2)^2] dx>> V = pi | [x^2 - x^4] dx>> V = pi (x^3/3) - (x^5/5)>> evaluated at 1 and 0 = 2pi/15Very good.There is also another method to use. You used the §at disks method:V = pi | (f[x]^2 - g[x]^2) dx (b,a)You can also use the thin shell method:V = 2pi |(f[x]-g[x])dx(a,b) but rotated about the y axis. Heres the work I did:> x = x^2; x= 1, 0 these are the points which bound the solid.> V = | [pi(x)^2 - pi(x^2)^2] dx> V = pi | [x^2 - x^4] dx> V = pi (x^3/3) - (x^5/5)> evaluated at 1 and 0 = 2pi/15Also check out the theorems of Pappus-Guldinus, one of which states that thevolume of a body of revolution is equal to the product of the generatingarea and the distance travelled by the centroid of the area while the bodyis being generated. Heres the work I did:>> x = x^2; x= 1, 0 these are the points which bound the solid.>> V = | [pi(x)^2 - pi(x^2)^2] dx>> V = pi | [x^2 - x^4] dx>> V = pi (x^3/3) - (x^5/5)>> evaluated at 1 and 0 = 2pi/15Yup.Jon Miller The problem really only has one answer, despite the fact that it> appears as a paradox. If we merely redescribe the same problem using> only mathematical and set terms, we get the answer right away.> Consider the following redescription:>> Let the set S0 = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 } {1}. Now> recursively de'ne the set Sn as follows:>> Sn = Sn+1 U { 10n+1, 10n+2, ..., 10n+10 } {10n+1}.What the heck is that?Sorry for the confusion, as it should actually read:S0 = Empty setSn+1 = Sn U { 10n+1, 10n+2, ..., 10n+10 } {n+1} for n>0Sn represents the state of the container at each step n.> So you say no balls are left? Certainly the balls are removed one at a> time.Yes.> What was the last ball removed if there are no balls left?There is no last ball. The 'nal set S = lim Sn.> I would like toknow what natural number (thatOs the way they are labeled)> is on that ball. Certainly you can tell me that number right away.As stated above, there is no last ball.Jonathan HoyleGene Codes Corporation > >> Probability is countably additive. When you add together a countable>> number of 0Os, as in the coin problem, the result is 0. The real numbers>> in [0,1] are uncountable, and therefore countable additivity does not>> apply.>> > On second thought, weOre not adding an in'nite number of probability> 0Os at all -- weOre multiplying an in'nite number of probability 1Os.> Any idea if probability is countable multiplicative as well?>> > An in'nite product can be converted to an in'nite sum by taking logarithms.>> >> Yes, but then youOre not dealing with a countable sum of probabilities>> any more -- youOre dealing with a countable sum of log-probabilities. >> Are log-probabilities countably additive?>> For each n, we can compute the probability p_n that a given ball is still>> in the bucket after the n-th step. Since p_n -> 0 as n -> oo, we can>> conclude that the probability of that particular ball being in the bucket>> at noon is 0.> Right.>> After that, we use countable additivity over all the balls to show that the>> bucket is empty at noon with probability 1.> Here I donOt follow. If p_k is the probability that ball number k is> in the basket at noon, the probability that the basket is empty at> noon is prod(1-p_k). I donOt immediately see how the fact that> probabilities are countably additive helps evaluate that product.The probability that any ball is in the bucket at noon is bounded aboveby the sum of all the p_k, which is zero. For example, in the case oftwo probabilities, the probability that at least one is left at noon is1 - (1-p_1)*(1-p_2) = p_1 + p_2 - p_1 * p_2 <= p_1 + p_2. Use inductionto show that a similar inequality holds for any 'nite number of balls,and then take the limit.-- Dave SeamanJudge YohnOs mistakes revealed in Mumia Abu-Jamal ruling. Here I donOt follow. If p_k is the probability that ball number k is> in the basket at noon, the probability that the basket is empty at> noon is prod(1-p_k). I donOt immediately see how the fact that> probabilities are countably additive helps evaluate that product.The probability that any ball is in the bucket at noon is bounded above> by the sum of all the p_k, which is zero. For example, in the case of> two probabilities, the probability that at least one is left at noon is> 1 - (1-p_1)*(1-p_2) = p_1 + p_2 - p_1 * p_2 <= p_1 + p_2. Use induction> to show that a similar inequality holds for any 'nite number of balls,> and then take the limit.> The expected number of balls remaining is sum p_k.-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ > Here I donOt follow. If p_k is the probability that ball number k is>> in the basket at noon, the probability that the basket is empty at>> noon is prod(1-p_k). I donOt immediately see how the fact that>> probabilities are countably additive helps evaluate that product.>> >> The probability that any ball is in the bucket at noon is bounded above>> by the sum of all the p_k, which is zero. For example, in the case of>> two probabilities, the probability that at least one is left at noon is>> 1 - (1-p_1)*(1-p_2) = p_1 + p_2 - p_1 * p_2 <= p_1 + p_2. Use induction>> to show that a similar inequality holds for any 'nite number of balls,>> and then take the limit.>The expected number of balls remaining is sum p_k.Right, thatOs much easier.-- Dave SeamanJudge YohnOs mistakes revealed in Mumia Abu-Jamal ruling. =Here is the problem:What is the area between the curve x=sqrt(16 - y^2) and the y-axis?I realize this is a semi-circle, IOve solved for y (+ and -), yet IdonOt see the area between the curve and the axis this is looking for. Is there a solution to this problem? Here is the problem:What is the area between the curve x=sqrt(16 - y^2) and the y-axis?I realize this is a semi-circle, IOve solved for y (+ and -), yet I> donOt see the area between the curve and the axis this is looking for.> Is there a solution to this problem?The region is a semicircular disk of radius 4 with bounding diameter having endpoints at (0,4) and (0,-4) on the y axis and lying to the right of the y axis (x >= 0).Take half the area of the appropriate circle. Here is the problem:>>What is the area between the curve x=sqrt(16 - y^2) and the y-axis?>>I realize this is a semi-circle, IOve solved for y (+ and -), yet I>donOt see the area between the curve and the axis this is looking for.Why not?> Is there a solution to this problem?Yes. The intersections are at y=-4 and y=4, so the area is equal tothe integral from y=-4 to y=4 of sqrt(16-y^2)dy.-- about what I accept as reality. Calvin (Calvin and Hobbes) Here is the problem:>> What is the area between the curve x=sqrt(16 - y^2) and the y-axis?>> I realize this is a semi-circle, IOve solved for y (+ and -), yet I> donOt see the area between the curve and the axis this is looking for.> Is there a solution to this problem?Yes. > Here is the problem:>> What is the area between the curve x=sqrt(16 - y^2) and the y-axis?>> I realize this is a semi-circle, IOve solved for y (+ and -), yet I> donOt see the area between the curve and the axis this is looking for.> Is there a solution to this problem?>> Yes.>>I would start by 'nding where the maximum is.xO=2ysqrt(16-y^2)0=2y(sqrt(16-y^2))y=0, y=4Integral(0,4) sqrt(16-y^2) dyy=4 sin xdy=4 cos x dx16integral(0,pi/2) (sin x)^2 dx16integral(0,pi/2) 1/2-1/2 cos 2x dx1/2 x - 1/4 sin 2xpi/4-1/4-- David MoranChief MeteorologistOklahoma Storm Team How to solve this equation (by elementary methods!):>n^p+m^p=(2n-m)^p>n>0;m>0 et p>2WLOG gcd(m,n) = 1.Butm^p = (2n-m)^p - n^p is divisible by (2n-m)-n = n-mand n^p = (2n-m)^p - m^p is divisible by (2n-m)-m = 2(n-m).So the only way to have gcd(m,n)=1 is if |n-m| = 1 (and n even).If m=n+1, 2n-m = m-2, so that wonOt work. So weOre left with the case m=n-1, and the equationn^p + (n-1)^p - (n+1)^p = 0Mod n, we have (-1)^p == 1, so p is even (after eliminating n=2).Mod n^3, since (1-n)^p == 1 - p n + (p choose 2) n^2 and(1+n)^p == 1 + p n + (p choose 2) n^2, we have -2 p n == 0, so 2 p == 0 mod n^2. In particular, p > n. Now write the equation as 1 + (1-1/n)^p = (1+1/n)^p.Left side < 2, right side > (1+1/n)^n > 2 for n > 2[ for this, note that n ln(1+1/n) = 1 - 1/(2n) + 1/(3n^2) - ...is an alternating series, so n ln(1+1/n) > 1-1/(2n) ].Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 =blang> How to solve this equation (by elementary methods!):> n^p+m^p=(2n-m)^p> n>0;m>0 et p>2I think I see a way forward. We havem^p = (2n-m)^p - n^pand since (2n-m) - n divides the right side, it divides the left. So, eithern - m = 1 or -1orn-m has a prime factor in common with m.In the latter case, m and n are not relatively prime. By induction, we comedown to the casen - m = +-1.LH > > Induction hypothesis - If the k graphs intersect at exactly (n+1)> > vertices, there is a k-coloring of G.>> We note that : m = k(k+1)/2>> Induction step - To prove that if the k graphs intersect at n vertices> (n < m), there is a k-coloring of G.>> Consider an intersection point (i,j,...,l) [There is atleast one> intersecting point (i,j,...,l) because n < m]. We construct GO from G> such that the intersection point (i,j,...,l) in G is split as> (i,j,...) and (i,l) in GO.DoesnOt that mean K_i will have k+1 points in GO?I think not. In G, K_i can intersect with the other (k-1) completegraphs (at?) atmost (k-2) points [since atleast one intersecting pointoccurs between more than 2 complete graphs]. Hence, in GO, K_i cancontain atmost k-1 intersecting points, which does not violate theconstruction rules. For instance, if we have an intersecting point(1,2,3,4) in G, splitting it as (1,2,3) and (1,4) in GO cannot makeK_1 intersect at more than k-1 points.> By induction hypothesis, GO has a> k-coloring. A k-coloring of GO is also a k-coloring of G by assigning> the color of (i,j,...) in GO to (i,j,...,l) in G.How do you know that will be a *proper* k-coloring of G?Hmm... I have no answer :(. Back to the drawing board for me. --Pradip. details, I do recall that theAlgebra FX 2.0(+) is more clever at handling complex numbers than anyof its predecessors. Ordinarily this wouldnOt help you, but Christmasis coming... Cordially,Richard KanarekP.S. Try before you buy! Arrrghh.... this calculator is really annoying me.>>IOve put my fx-991MS calculator into CMPLX mode, and I can get it to>calculate and show me, for example, the real and imaginary parts of>1 ->>3i>(simple I know but it de'nitely works). However I canOt get it to>show>>me>the real and imaginary parts of, for example, 1 - sqrt3i. It says>Math>ERROR. This also happens with *any* other complex numbers calculation>involving roots or powers.>> If you mean (sqrt(3))*i, then you can almost certainly overcome the> problem by putting in some parentheses.>>I still canOt raise a complex expression to a power though e.g. (1 +>3i)^3.>>The only exception is squaring it - the calulator has a separate button>>speci'cally for squaring i.e. I donOt have to put in the ^ then the>3>>(or any power), I simple press the square button. Any idea on the correct>>parenthesis for this?>>IOve tried (1 + 3i)^3, (1 + 3i)(^3), ((1 + 3i)(^3)) - none seem to work.> If you mean sqrt(3i) then you are up against a restriction on the> operations allowed on your calculator. (Note that sqrt is> multi-valued.)>>Perhaps this is the reason I canOt square the expression?>> Well, probably for a similar reason. One approach to programming the>> ^ operation on a calculator is to take a logarithm, then do a>> multiplication, then do an antilogarithm. I vaguely remember one>> calculator I have owned (or it may have been an early implementation>> of BASIC) refuse to do things like (-2)^3, and I assumed that the>> reason is something to do with the non-existence (in reals) of>> log(-2). In complex numbers the log function is multi-valued. In both>> cases, negative reals and complex numbers, it is possible to work>> around the problem; my current calculator is perfectly happy with>> (-2)^3.>> I donOt know the exact algorithms used on calculators. But they will>> have been optimised for real arithmetic. When programming the complex>> number operations, the designers will have had to decide when to adapt>> the routines to allow operations with complex numbers, and when not to>> bother. It seems that Casio decided that general exponentiation of>> complex numbers was not worth the bother.>>IOve got a Casio CFX-9850G, and the manual speci'caly states that complex>operation is limited to simple addition, subtraction, multiplication and>division, along with the square, square root and reciprocal functions.>Complex-speci'c functions for the modulus, argument and conjugate are>provided, so all other functions (powers, logs, trig) can be implemented>using laws/equations are Galilean invariant when expressed in the generalized cartesian coordinates demanded by basic analytic geometry, vector algebra, and measurement theory. = Opponents of the content should 'rst actually 'nd out what it is, then think, then request/submit-to arbitration by the appropriate neutral mathematics authorities. Flaming the hard- working, sel§ess, *.answers moderators evidences ignorance and despicable netiquette.Archive-Name: physics-faq/criticism/galilean-invariance Invariant Galilean Transformations (FAQ) On All Laws (c) Eleaticus/Oren C. Webster Thnktank@concentric.netAn obvious typo or two corrected.The Brittanica section revised to lessOpussy-footingO and to more directlyanticipate the elementary measurementtheory and basic analytic geometrythat is applied to the transformationconcept.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 justi'cationof Relativity Theory as a necessary alternative to the classicalphysics of Newton and Galileo.We will prove that all laws are invariant under the Galileantransformation, rather than some being non-invariant, afterwe show you what that means.We shall also show that another primal requirement that SRexist is nonsense: Michelson-Morley and Kennedy-Thorndike doindeed 't Galilean (c+v) physics. 1. Foreword and Intent 3. The Principle of Relativity 4. The Encyclopedia Brittanica Incompetency. 5. Transformations on Generalized Coordinate Laws 6. The data scale degradation absurdity. 8. What does sci.math have to say about x0O=x0-vt? 9. But DoesnOt x.cO=x.c? 10. But IsnOt (xO-x.cO)=(x-x.c) Actually Two Transformations? 11. But DoesnOt (xO-x.c+vt) Prove The Transformation Time Dependent? 12. But IsnOt (xO-x.cO)=(x-x.c) a Tautology? 13. But IsnOt (xO-x.cO)=(x-x.c) Almost the De'nition of a Linear Transform? 14. But The Transform WonOt Work On Time Dependent Equations? 15. But The Transform WonOt Work On Wave Equations? 16. But MaxwellOs Equations ArenOt Galilean Invariant? 17. First and Second Derivative differential equations.If a law is different over there than it is here,it is not one law, but at least two, and leaves usin doubt about any third location. This is thePrinciple of Relativity: a natural law must be thesame relative to any location at which a given eventmay be perceived or measured, and whether or not theobserver is moving.The idea of location translates to a coordinatesystem, largely because any object in motion couldbe considered as having a coordinate system originmoving with it. If you perceive me moving relativeto you - who have your own coordinate system - willyour measurements of my position and velocity 'tthe same laws my own, different measurements 't?If a law has the same form in both cases it iscalled covariant. If it is identical in form, var-ables, and output values, it is called invariant.What weOre asking is that if the doesthe coordinate, xO, on some other, parallel axiswork? Speaking in terms of the axis on which x isthe coordinate, xO is the OtransformedO coordinate.The situation is complicated because weOre talkingabout coordinates - locations - but in most mean-ingful laws/equations, it is lengths/distances (andtime intervals) the equations are about, and x coord-inates that represent good, ratio scale measures ofdistances are only interval scale measures on the xOSo, if we have an x-coordinate in one system, thenwe can call the xO value that corresponds to the samepoint/location the transform of x.In particular, the Principle of Relativity is embodiedin the form of the Galilean transformation, whichrelates the original x, y, z, t to xO, yO, zO, tO bythe transform equations xO=x-vt, yO=y, zO=z, tO=t inthe simpli'ed case where attention is focused onlyon transforming the x-axis, and not y and z. In thecase of Special Relativity, the xO transform is thesame except that xO is then divided by sqrt(1-(v/c)^2),and tO=(t-xv/cc)/sqrt(1-(v/c)^2). In either case, vis 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.One example of the traditional fallacious ideathat an equation is not invariant under the galileantransformation comes from the Encyclopedia Brittanica:Before EinsteinOs special theory of relativitywas published in 1905, it was usually assumedthat the time coordinates measured in all inertialframes were identical and equal to an OabsolutetimeO. Thus, t = tO. (97)The position coordinates x and xO were thenassumed to be related by xO = x - vt. (98)The two formulas (97) and (98) are called aGalilean transformation. The laws of nonrelativ-istic mechanics take the same form in all framesrelated by Galilean transformations. This is therestricted, or Galilean, principle of relativity.The position of a light wave front speeding fromthe origin at time zero should satisfy x^2 - (ct)^2 = 0 (99)in the frame (t,x) and (xO)^2 - (ctO)^2 = 0 (100)in the frame (tO,xO). Formula (100) does nottransform into formula (99) using the transform-ations (97) and (98), however.................................................. Besides the trivially correct statement of what theGalilean OtransformO equations are, there is exactlyone 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 OstationaryO (observer S) equation. [Let observer M be the OmovingOsystem 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 xO and tO from x and t, not the other way around, and the idea that either observerOs 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 tO, xO 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 OtransformO equations are the relationships between xO and x, tO and t and have nothing to do with what one equation or the other ought to OsayO. The equationsO 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 xO=x-vt and tO=t. Solve for x=xO+vt and replace t with tO, then substitute the result in eq-99: (xO+vtO)^2 - (ctO)^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 justi'cation 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, xO+vtO is a mixed frame expression and it is xO+(-v)tO 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 OstationaryO equation. LetOs 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 isnOt, in general. Suppose you and I are both modelling the same light event and you are using EST and IOm using PST. OJust a time coordinateO 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, thatOs 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 donOt have to OunderstandO 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 anyoneOs 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 Othere is exactly one thing they got rightO, the correct M system version of eq-99a is eq-100a: (xO-x.eO)^2 - [ c(tO-t.eO) ]^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 observerOs eq-99a. Now, THATOs invariance. [You do realize that every eq-100a reduces to eq-99a, when you back substitute from the transforms, right? t.eO=t.e, x.eO=x.e-vt.]The traditional Gallilean transform is correct: tO = t xO = x - vt.But remember this: a transform of x doesnOt effectjust some values of x, but all of them, whether theyare in the formula or not. This is important if youwant to do things right. The crackpot position isstrongly against this sci.math veri'ed position, andthe apparently standard coordinate pseudo-transformationthey suggest is perhaps the result. {See Table ofLetOs use a simple equation: x^2 + y^2 = r^2, which isthe formula for a circle with radius r, centered at alocation where x=0.But what if the circle center isnOt at x=0? Well, weOdwant to use the form analytic geometry, vector algebra,and elementary measurement theory tells us to use, a formwhere 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 formsof laws/equations we want to transform every occurenceof x and x0 - by whatever name we call it: x.c, x_e,whatever.So, what is the transformed version of (x-x0)? Why,(xO-x0O); both x and x0 are x-coordinates, and everyx-coordinate has a new value on the new axis.So, what is the value of (xO-x0O) in terms of the originalx data?is also true for x0O=x0-vt: (xO-x0O)=[ (x-vt)-(x0-vt) ]=(x-x0).In other words, when we use the generalized coordinate formspeci'ed by analytic geometry, we 'nd that the value of(xO-x0O) does not depend on either time or velocity in anyway, 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 isinvariant under the galilean transforms. Just use thegeneralized coordinate form, with (x-x0)/etc, in thetransformation process, not the incompetently selectedprivileged form, with just x/etc.[The form is privileged because it assumes the circlecenter, point of emission, whatever, is at the origin ofthe axes instead at some less convenient point. Aftertransform the coordinate(s) of the circle center/originare also changed but the privileged form doesnOt makethis explicit and screws up the calculations, whichshould be based on (xO-x0O) but are calculated as (xO-0).]The value of (xO-x0O) is the same as (x-x0). That makessense.Draw a circle on a piece of paper, maybe to the rightside of the paper. On a transparent sheet, draw x and ycoordinate axes, plus x to the right, plus y at the top.Place this axis sheet so the y-axis is at the left sideof the circle sheet.Now answer two questions after noting the x-coordinate ofthe circle center and then moving the axis sheet to the right:(a) did the circle change in any way because you movedthe axis sheet (ie because you transformed the coordin-nate axis)?(b) did the coordinate of the circle center change?The circle didnOt change [although SR will say it did];that means that (xO-x0O) does indeed equal (x-x0).The coordinate of the circle center did change, and itchanged at the same rate (-vt) as did every point onthe circle. That means that x0O<>x0, and the fact thecircle center didnOt change wrt the circle, means thatthe relationship of x0O with x0 is the same as that ofany xO on the circle with the corresponding x: xO=x-vt;x0O=x0-vt.This is to prepare you for the True Believer crackpots thatsay OconstantO coordinates canOt be transformed; some evensay they arenOt coordinates. These crackpots include somethat brag about how they were childhood geniuses, btw.QED: The galilean transformation for any law ongeneralized Cartesian coordinates is invariant underthe Galilean transform.The use of the privileged form explains HOW the transformedthe screwed up effect of the transform is, and how useof the generalized form corrects the screwup.The SR transforms and the Galilean transforms bothconvert good, ratio scale data to inferior intervalscale data. The effect is corrected, allowed for,when the transforms are conducted on the generalizedcoordinate forms speci'ed by analytic geometry andvector algebra.Both sets of transforms are OtranslationsO - lateralmovements of an axis, increasing over time in thesecases - but with the SR transform also involving arescaling. It is the translation term, -vt in the xtransform to xO, and -xv/cc in the t transform to tO,that degrades the ratio scale data to interval scaledata. In general, rescaling does not effect scalequality 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 saidthey were OtranslationsO (movements) - and in the caseof OgoodO rotations, ratio scale data quality is indeedpreserved, but SR violates the conditions of good ro-tations; they are not rigid rotations and they donOtappropriately rescale all the axes that must be rescaledto preserve compatibility.The proof is in the pudding, and the pudding is thecombination of simple tests of the transformations.We can tell if the transformed data are ratio scaleor interval.Ratio scale data are like absolute Kelvin. A measure-ment of zero means there is zero quantity of thestuff being measured. Ratio scale data support add-ition, subtraction, multiplication, and division.The test of a ratio scale is that if one measurelooks like twice as much as another, the stuffbeing measured is actually twice as much. Withabsolute Kelvin, 100 degrees really is twice theheat as 50 degrees. 200 degrees really is twiceas much as 100.Interval scale data are like relative Celsius, whichis why your science teacher wouldnOt let you use itin gas law problems. There is only one mathematicaloperation interval scales support, and that has tobe between two measures on the same scale: subtraction.100 degrees relative (household) Celsius is not twiceas much as 50; we have to convert the data to absoluteKelvin to tell us what the real ratio of temperaturesis.However, whether we use absolute Kelvin or relativeCelsius, the difference in the two temperature readingsis the same: 50 degrees.Thus, if we know the real quantities of the OstuffObeing measured, we can tell if two measures are ona ratio scale by seeing if the ratio of the twomeasures is the same as the ratio of the known quant-ities.If a scale passes the ratio test, the interval scale testis automatically a pass.If the scale fails the ratio test, the interval scaletest becomes the next in line.It isnOt just the bare differences on an intervalscale that provides the test, however. Differencesin two interval scale measures are ratio scale, soit is ratios of two differences that tell the tale.LetOs do some testing, and remember as we do that ourconcern is for whether or not the data are messed up,not with OreasonsO, excuses, or avoidance. Are we going to take a transformed length (difference)and see whether that length 'ts ratio or interval scalede'nitions?Of course, not. Interval scale data are ratio afterone measure is subtracted from another. That is themajor 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 theorigin, where x=0, and the other end at the coordinatethat tells us the correct lengths.Note that these x-values are ratio scale only becauseone end of each rod is at x=0. That may remind you ofthe correct way to use a ruler or yard/meter-stick:put the zero end at one end of the thing you aremeasuring. 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 xO at v=.5, t=10.xO=x-vt.A B C a b c- --10 20 40 5 15 35- --B/A = 2 b/a = 3C/A = 4 c/a = 7C/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 = 10C-A = 30 c-a = 30C-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 SRresults is that the differences do not correspond soneatly 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 valuesdo correspond neatly and exactly to the ratio scaleresults.Using the generalized coordinate form, such as (x-x0),the transform produces an interval scale xO and aninterval scale x0O. That gives us a ratio scale (xO-x0O),just like - and equal to - (x-x0).It has become apparent - whether misleading or not -that the crackpot responses to the obvious derive froma common source, whether it be bandwagoning or theirSR instructors.Below, in the sci.math subject, we see that all sci.mathrespondents agree with the basic controversial positionof this faq: every coordinate is transformed, whether asupposed constant or not.Think about it, the generalized coordinate of a circlecenter, x0, applies to in'nities upon in'nities ofcircle locations (given y and z, too); it is a constantonly for a given circle, and even then only on a givencoordinate axis.And even variables are often held OconstantO duringeither integration or differentiation.The utility of a variable is that you can discuss allpossible particular values without having to single outjust one. That utility does not make particular - singledout - values on the variableOs axis not values of thevariable just because they have become named values.In any case, all that is preamble to the incompetent ideathey have proposed for a transform of coordinates. It isbased 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: (xO)^2 - (ictO)^2 = 0. ThatOs the one thing theBrittanica got right. Note that the leading crackpot justcriticized 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 xO and t to tO 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 toanother is NOT a substitution of the/a de'nition of xfor itself; that is not a coordinate transformation.The most that can said for such a substitution is thatit is a change of variable.But the crackpots are calling this a coordinate trans-form of the original equation: (xO+vt)^2 - (ictO)^2 = 0.It is not a coordinate transform, of course, exceptaccidentally. (xO+vt) is not the primed systemcoordinate, it is another form/expression of x. Theyget that substitution by solving xO=x-vt for x; x=xO+vt.So, by incompetent misnomer, they accomplish what theyhave been railing against all along.It has been the generalized coordinate form in question allthis time: (x-x0)^2 - (ict)^2 = 0.Here they substitute for x instead of transforming to theprimed frame: (xO+vt-x0)^2 - (ictO)^2. -- ^ | ^ |It is still x ^ but see what they have accomplishedby their mis/malfeasance: [xO+vt-x0]=[xO+(vt-x0)]=[xO-(x0 -vt)]. =[xO-x0O]The crackpots have been bragging about how you donOthave to transform the circle centerOs coordinate totransform the circle centerOs coordinate. Braggingthat what they were doing was not what they saidthey were doing.This does give us insight as to some of the crackpotvariations on their x0O<>x0-vt theme, which in all thevariations will be discussed in later sections..They are used to seeing the mixed coordinate form,(xO+vt-x0) without realizing what it respresented,so - accompanied with a lack of understanding ofthe term OdependentO - they are used to seeing justthe one vt term, and not the one hidden in the de'-nition of xO and are used to imagining it makes thewhole expression time dependent and thus not invariant.About which, let x=10, let, x0=20, v=10, and tvariously 10 and 23:(x-x0)=-10. Using their (xO+vt-x0):For t=10, we have (xO+vt-x0) = [ (10-10*10) + (10*10) - (20) ] = -90 + 100 - 20 = -10 = (x-x0)For t=23, we have (xO+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 tjust so you can see that the crackpots not only donot understand the obvious logic of the algebra{ (xO-x0O)=[ (-vt)-(x0-vt) ]=(x-x0) } - which showsthat the transform has no possible time term effect -but they donOt understand even a simple arithmeticdemonstration of the facts.Oh. Their (xO+vt-x0) or (xO+vtO-x0) reduces the sameway since tO=t: (x-vt+vt-x0)=(x-x0).Their process, which says (xO+vtO) is the transformof x, says that (xO+vtO) is the moving system locationof x, but it canOt be because x is moving further inthe negative direction from the moving viewpoint.That formula will only work out with v<0 which is indeedthe velocity the primed system sees the other moving at.However, that formula cannot be derived from xO=x-vt,the formula for transformation of the coordinates fromthe unprimed to the primed,The crackpotsO positions/arguments were put to sci.mathin such a way that at least two or three who posted re-sponses thought it was your faq-er who was on the idiotOsside of the questions.Their responses: -I. x0O = x0. In other words: x0O <> x0-vt, or constant values on the x-axis are not subject to the transform.AA: de'ne constant values on the x-axis, butin the context of the question that is not relevant. The relevant fact isthat if the unprimed observer holds an object at point x0, then theprimed observer assigns to that object a coordinate x0O which isnumerically related to x0 by x0O= x0 -vt.AA: xO=x-v*t=x0-vtO, so if x0Ois to give the coordinate in the (xO,tO,)-system, it will be given byx0O=x0-v*tO: ie., it is not given by a constant. Thus, being at rest(constant of the point X0 isthe number x0 in the unprimed system, and x0O in the primed system.Clearly x0 and x0O are different, if vt is not zero. However one may saythat (though it sounds/is stupid) the point X0 itself is the samethroughout the transformation. However that expression soundsmeaningless, since a transform (ok, maybe we should call it a change ofbasis) is only a function that takes the pointOs representation in onesystem into the same pointOs representation in another system. It ispreferrable to use three notations: X0 for the point itself and x0 andx0O for the pointsO representations in some coordinate systems.GG: most idiotic to come up, and it doesso frequently. And in a number of guises.The idea being that x.cO <> x.c-vt, with x.c being whatwe have called x0 above; the notation makes no difference.Some crackpots have managed to maintain that position evenafter graphs have illustrated that such an idea means thatafter a while a circle center represented by x.cO could beoutside the circle.The leading crackpot just make that explicit, as far asone can tell from his befuddled post in response to a lineabout active transforms, which are actually moving bodysituations, 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.cO=x.c,the circle center would be at (0,0) still! The circle is herebut its center is way, way over there! Indeed, although a changeof coordinate systems is not movement of any object described inthe coordinates, the x.cO=x.c crackpottery is tantamount to thecircle staying put but the center moving away. Or vice versa.One crackpot puts the (xO-x.cO)=(x-vt - x.c+vt) relationshiplike this: (x-vt+vt - x.c).See, he says, that is transforming x (with x-vt - x.c) and thenreversing the transform (x-vt+vt - x.c).ThatOs just another crackpot form of the idiocy thatx.cO <> x.c-vt. YouOll have noticed the implicationis that there is no transform vt term relating to x.c. Time Dependent?That particular crackpottery is perhaps more corrupt thanmoronic, since it includes deliberately hiding a vt term fromview, and pretending it isnOt there. [However, we have seenabove that it is a familiar incompetency, and not likely anoriginal.]Look, the crackpots say, there is a time term in thetransformed (xO - x.c+vt). The transform isnOt invariant!ItOs time dependent!Just put xO in its original axis form, also, which revealsthe other time term, the one they hide: (xO-x.c+vt) = (x-vt - x.c+vt) = (x-x.c).So, at any and all times, the transform reduces to theoriginal expression, with no time term on which to bedependent.Then there is the fact that if you leave the equationin any of the various notation forms - with or withoutreducing them algebraicly - the arithmetic always comesdown to the same as (x-x.c). That means nothing to crack-pots, but may mean something to you.My dictionary relates OtautologyO to needless repetition.ThatOs another form of the x.cO <> x.c-vt idiocy.The repetition involved is the vt transformation term.Apply the -vt term to the x term, and it is needlessrepetition to apply it anywhere again? The OagainO isto the x.c term. The x.cO = x.c crackpot idiocy.The repetition of the vt terms is required by the presenceof two x values to be transformed.Be sure to note the next section. a Linear Transform?Now, how on earth can we relate a tautology to a basicde'nition in math?we get this de'nition: --A linear transformation, A, on the space is a method of corr-esponding to each vector of the space another vector of thespace such that for any vectors U and V, and any scalarsa 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= ( ).and aAX+bAC = ( x-x.c, y-y.c, z-z.c, 0 ).Need it be said?Sure: QED. On the galilean transform thede'nition of a linear transform, A(aU+bV)=aAU + bAV,is completely satis'ed.The generalized form transforms exactly andnon-redundantly - with ONE TRANSFORM, not atransform and reverse transform - and non-tautologically, just as the very de'nitionof a linear transform says it should.And does so with absolute invariance, with thisgalilean transformation.The main crackpot that has asserted such a thing was referringequation; for which we have shown repeatedly elsewhere that thenumerical calculations are identical for any primed values asfor the unprimed values.The presence - before transformation - of a velocity termseems to confuse the crackpots. It turns out there is ex-treme historical reason for this, as you will see in thesubject on MaxwellOs equations.forms and the galilean transforms.Oh? Just what is the magical term in them that prevents(xO-x.cO)=(x-vt - x.c+vt)=(x-x.c) from holding true?It turns out not to be magic, but reality, that interfereswith 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 simpledemonstration that the galilean transform ongeneralized cartesian coordinates is invariant,their 'rst defense is usually an incredibly stupidx0O=x0, because the coordinate of a circle center,or point of emission, etc, is a constant and canOtbe transformed.The last defense is but MaxwellOs equations are notinvariant under that coordinate transform. Whenasked just what magic occurs in Maxwell that wouldprevent the simple algebra (xO-x0O)=[ (x-vt)-(x0-vt) ]=(x-x0)from working, and when asked them for a demonstration,they will never do so, however many hundreds oftimes their defense is asserted.The reason may help you understand part of EinsteinOs1905 paper in which he gave us his absurd SpecialRelativity 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 theforce components in the direction of the x axis areX and L, Y and M are in the y direction, Z and N inthe z direction.Those values are not, however, coordinates, but valuesvery much like acceleration values.BTW, the current fad is that E and B are O'eldsO, havingbeen Oforce 'eldsO for a while, after being OforcesO.So, when Einstein says he is applying his coordinatetransforms to the Maxwell form he presented, he iseither 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): XO=X. LO=L. YO=B(Y-(v/c)N). MO=B(M+(v/c)Z). ZO=B(Z+(v/c)M). NO=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; EinsteinOs transform v is the velocity of a coordinate axis, not the velocity he touched it.(c) if they were honest Einsteinian transforms, theyOd be 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 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 Einsteinianblunders, this one transcends error and is an incredibleexample 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 donOt know what coordinates are. Thevarious 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 speci'ed. 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 speci'ed. 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 speci'ed. 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.EinsteinOs - and MaxwellOs - E and B arenot coordinate vectors. thatmisinforms the idea that Maxwell isnOt invariant under thegalilean transform: confusions about velocities.Velocities With Respect to Coordinate Systems.--Aaron Bergman supplied the background in a post to a sci.physics.*newsgroup: with a current I in each.Now, according to simple E&M, each current generates a magnetic'eld and this causes either a repulsion or attraction betweenthe wires due to the interaction of the magnetic 'eld and thecurrent. LetOs just use the case where the currents are parallel.Now, suppose you are running at the speed of the current betweenthe 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 acontradiction. galilean transform (xO-x.cO)=(x-x.c), insures that it is an error to imagine there is anydifference between the data and law in one frame and in another;the usual, convenient rest frame is the best frame and only framerequired for universal analysis. [Well, (xO<>x, x,cO<>x.c, but(xO-x.cO)=(x-x.c).]Second, given that you decide unnecessarily to adapt a law tozz