mm-382 === Subject: : 1/x + 1/y + 1/z = 1/nGiven integer n>=1 and integers x,y,z such that :1/x + 1/y + 1/z = 1/n and 1<=x<=y<=z , what is the maximal possible value of z ?=== === Subject: : Re: 1/x + 1/y + 1/z = 1/n> Given integer n>=1 and integers x,y,z such that :> 1/x + 1/y + 1/z = 1/n and 1<=x<=y<=z ,> what is the maximal possible value of z ?Consider Pythagorean quartets - for example:a^2+b^2+c^2=d^2 Divide both sides by (abcd)^2As there exist infinite many Pythagorean quartets, what you can say aboutthe maximal possible value of z, which can be because of the symmetry :(bcd)^2 or (acd)^2 or (abd)^2 and - of course - n=(abc)^2 >=1=== === Subject: : Re: 1/x + 1/y + 1/z = 1/n>Given integer n>=1 and integers x,y,z such that :>1/x + 1/y + 1/z = 1/n and 1<=x<=y<=z ,>what is the maximal possible value of z ?It's natural to conjecture that z <= n(n+1)(n^2+n+1) based on1/(n+1) + 1/(n^2+n+1) + 1/n(n+1)(n^2+n+1) = 1/n, and I thinkthis is in fact optimal. Here's a proposed proof:Let a/b = 1/n - 1/x = (x-n)/nx. I claim that z <= (b^2 + b)/a.To see this, rewrite 1/y + 1/z = a/b as (ay-b)(az-b) = b^2 > 0.Since all quantities are positive integers, we have az-b <= b^2from which the bound follows.Now we know n < x <= 3n (in fact if x = 3n we only get the trivialsolution x = y = z = 3n). Taking a = x-n, b = nx (note that a/b isnot necessarily in lowest terms), the upper bound is nx(nx+1)/(x-n).Rewriting in terms of a = x-n >= 1, the upper bound becomes: z <= 2 n^3 + n + n^2 ( a + (n^2+1)/a ).The right-hand side, as a function of a, has a unique local minimumat a = sqrt(n^2+1), and thus achieves its maximum value at one ofthe endpoints, a=1 or a=2n. Since n^2+1 >= 2n, it's easy to seethat a=1 is the higher of the two, which finally gives us: z <= 2 n^3 + n + n^2 (n^2 + 2) = n(n+1)(n^2+n+1).Does the result generalise to longer sums of Egyptian fractions? === === Subject: : 1 + 2^z + ... + n^zProve using Hadamard`s factorization Theorem that f(z) = 1 + 2^z + ... + n^zhas a infinity of zeros.Show that if w is a zero of f then the conjugate of w is a zero of f too=== === Subject: : Re: 1 + 2^z + ... + n^z> Prove using Hadamard`s factorization Theorem that f(z) = 1 + 2^z + ... + n^z> has a infinity of zeros.> Show that if w is a zero of f then the conjugate of w is a zero of f too> [Homework?][Assumed: n>=2 integer?]Even without Hadamard (you would have to remind me what it saysanyway:-)=:(Hint:) g(z) = 1+n^z has easily computable roots, and thedifference f(z)-g(z) is a relatively small perturbation of f forsufficiently large abs(z).And the conjugates: use basics of complex algebra.Cheers, ZVK(Slavek)=== === Subject: : Re: 1 + 2^z + ... + n^z>> Prove using Hadamard`s factorization Theorem that f(z) = 1 + 2^z + ... + n^z>> has a infinity of zeros.>> Show that if w is a zero of f then the conjugate of w is a zero of f too>[Homework?]>[Assumed: n>=2 integer?]>Even without Hadamard (you would have to remind me what it says>anyway:-)=:>(Hint:) g(z) = 1+n^z has easily computable roots, and the>difference f(z)-g(z) is a relatively small perturbation of f for>sufficiently large abs(z).Why is that? If z = x + iy then |n^z| = n^x; I don't see whyf - g can't be much larger than f for some very large z(with y large but small x, which doesn't seem like z'swe can avoid, since that's where f vanishes...)>And the conjugates: use basics of complex algebra.>Cheers, ZVK(Slavek)=== === Subject: : Re: 1 + 2^z + ... + n^z,> Prove using Hadamard`s factorization Theorem that f(z) = 1 + 2^z + ... + n^z> has a infinity of zeros.> Show that if w is a zero of f then the conjugate of w is a zero of f too>> [Homework?]> [Assumed: n>=2 integer?]Even without Hadamard (you would have to remind me what it says> anyway:-)=:(Hint:) g(z) = 1+n^z has easily computable roots, and the> difference f(z)-g(z) is a relatively small perturbation of f for> sufficiently large abs(z).Suppose n = 3 and z = [(2m+1)(Pi)/ln(3)]*i for a large positive integers m?=== === Subject: : Re: 1 + 2^z + ... + n^z > has a infinity of zeros.>> Show that if w is a zero of f then the conjugate of w is a zero of f too[Homework?]>[Assumed: n>=2 integer?]Even without Hadamard (you would have to remind me what it says>anyway:-)=:(Hint:) g(z) = 1+n^z has easily computable roots, and the>difference f(z)-g(z) is a relatively small perturbation of f for>sufficiently large abs(z).> Why is that? If z = x + iy then |n^z| = n^x; I don't see why> f - g can't be much larger than f for some very large z> (with y large but small x, which doesn't seem like z's> we can avoid, since that's where f vanishes...)[...]> I see. So, back to Hadamard...Cheers, ZVK(Slavek).=== === Subject: : Re: 1 + 2^z + ... + n^zWe define de order of f as (are equivalent):ord(f) = lim_r{log(log(M(r)))/log(r)}, where M(r) = max{|f(z)| : |z| = r}ord(f) = lim_n{n*log(n)/-log(|a_n|) if f(z) = a_0 + a_1*z + ... + a_n*z^n +...ord(f) = inf{a>0 : |f(z)| oo . Theexponent f convergence of the sequence is defined bymu = inf{k>0 : sum_n{1/|a_n|^k} a_n = pi^n/n!Then, ord(sin(pi*z)) = lim_n{n*log(n)/(log(n!)-nlog(pi))} = ... = 1sin(pi*z) = exp(a*z + b) * z * Prod_{n=-oo...oo}((1-z/n)exp(z/n)) == exp (a*z +b) * z * Prod_{n=1...oo}(1-z^2/n^2) == exp(a*z + b)*z*(1-z^2)(1-z^2/2^2)(1-z^2/3^2)....Zdislav V. Kovarik escribi.97 en el mensaje Prove using Hadamard`s factorization Theorem that f(z) = 1 + 2^z + ... +n^z> has a infinity of zeros.> Show that if w is a zero of f then the conjugate of w is a zero of f too>[Homework?]> [Assumed: n>=2 integer?]> Even without Hadamard (you would have to remind me what it says> anyway:-)=:> (Hint:) g(z) = 1+n^z has easily computable roots, and the> difference f(z)-g(z) is a relatively small perturbation of f for> sufficiently large abs(z).> And the conjugates: use basics of complex algebra.> Cheers, ZVK(Slavek)=== === Subject: : 3D pyramid is a mapping of periodic tableThe Periodic Table arranges itself into a square pyramid if the quantumnumbers are slightly modified. Two quantum numbers (n, l ) are combined asan Aufbau quantum number. A new quantum number for precession (similar tospin) is also introduced.This format of the Periodic Table may also be viewed as a series of squarematrices.The website has been updated to include force vectors. Please see;www.members.shaw.ca/doulting/=== === Subject: : Re: A (Non)Prime Conjecture> One way to approach the prime numbers is to look at a number and see> if it has any factors other than itself. Another way to approach the> problem is to look at all of the numbers that are NOT prime and> conclude that anything left over IS prime. This latter approach is the> basis of my conjecture.I can't complain about this as I found that an ouput of composite numberscan be faster than trial division.(I was using implementation of a trial division algorithm as a means ofevaluating other programming languages. Then the shareware version of thetrial division algorithm was partially demonstration of a one form GUIapplication in place of console applications.)Another poster on this newsgroup propagated the idea of a composite outputthat output prime numbers directly based unassigned array elements of thecomposite output. The point to the poster was that thinking in terms of agraphical structure is okay but modern implementation should be in the formof a memory structure...http://pages.prodigy.net/halsteadinvest/ kbh-prm.htm=== === Subject: : al owensal owens says>> Um, you're certainly right about some of the>> responses. But the idea that your reply was>> infinitely superior to those that pointed out that>> R was uncountable is a little phony.I don't think I claimed it to be infinitely superior,Uh, no you didn't. What you said was>Ye, gods, people, use some sense. [...]Tailor your answers for your audience, not>your egos, or stifle the urge to respond:>the universe is growing dim much too>quickly to provide time or room for 'Net>posters to behave dimly too.My apologies for suggesting you'd said your> response was infinitely superior. All you said> was ye gods, use some sense and don't behave> dimly.When you're planning on taking people to task for> being dim this way you should _really_ get the> math right first.>but to be responsive to the OP's request, another claim>entirely.> Also>> the solution you gave was simply _incorrect_ ->> correct solutions are better. You said this:>> Prove that R:reals is a infinite dimensional vector space over Q:rationals.>> For proving this, I suppose NOT, i.e R is finite dimensional over Q, say n.>> Then, any r in R can be expressed as r=c_1*b_1 + ... + c_n*b_n>> where, c_i in Q and {b_1, ... , b_n} is a basis of R over Q.>> Now , I wish to find some contradiction from this.>> But, I don't know what is contradiction in this situation.>> If anyone have good idea or suggestion, please post reply.> Really, you are almost there. Since you say any r in R, look at> your right hand side. It is a product of a finite number of> rational fractions,>> This is simply wrong, because the b_j are _reals_, not rationals.That's not how _I_ read a basis over Q.I didn't say that it was inconsistent with the way you read > something. What I said was that it's _wrong_. It _is_ wrong.First, there's nothing non-standard about the meaning of> basis for R over Q - if your reading is not the same as mine> then your reading is simply not right. _Second_, with your> reading the proof doesn't make any sense at all: If we> suppose that R is a finite dimensional vector space over> Q then it follows that there exist _reals_ b_1, ... b_n as> above - it does _not_ follow that there exist rationals> b_j as above.>> And even if the above were correct, here things get a little phony:>> That well-known diagonalization proof is exactly the proof that the>> reals are _uncountable_,Nope. It is a proof that at least _one_ irrational number exists;>that the reals are uncountable doesn't have to be part of the initial>claim.> (Or maybe I'm wrong - how _does_ one give a proof that there>> exist irrational numbers by diagonalization?Sigh.The usual trick is to limit interest to the range [0,1):>The rationals in that range can _all_ be listed in standard>(reduced) form as a matrix like this, numerator | 0 1 1 2 1 3 1 2 3 4 1 5 1 2 3 4 5 6 ...> denominator | 1 2 3 3 4 4 5 5 5 5 6 6 7 7 7 7 7 7 ...> decimal expansion | *0 5 3 6 2 7 2 4 6 8 1 8 1 2 4 5 7 8 > to right of | 0 *0 3 6 5 5 0 0 0 0 6 3 4 8 2 7 1 5> decimal point | 0 0 *3 6 0 0 0 0 0 0 6 3 2 5 8 1 4 7> (read down) | 0 0 3 *6 0 0 0 0 0 0 6 3 8 7 5 4 2 1Then by the axiom of choice we can pick some digit other than the>one shown at each marked position along the diagonal; by picking>say 8 for the first one, we can guarantee that the number chosen>is not .999999999... = 1.0 and thus outside the required range,>(we couldn't pick that anyway, for some 9 is going to occur>somewhere on that diagonal, but its easier not to force ourselves>to prove that), and that decimal representation is guaranteed to>differ from every rational in that interval in at least one>position, thus not to be any of them, and since that list is by>construction exhaustive of the rationals within the interval, thus>to be some number in the stated interval but not a rational number,>thus an _ir_rational number *by definition*.Quod erat demonstrandum.Um, I didn't ask exactly the question I meant to ask. You> castigated people for taking the uncounability of R as> given, but felt free to refer to the well-known proof above> without saying how it went. The question I meant to ask> was this: How do you use a diagonalization argument to> show that an irrational exists, _other_ than a diagonalization> argument that _contains_ a proof that R is uncountable,> plus more details?Of course the answer is you don't. Hence you suggesting> that it was somehow a bad idea to simply invoke the> uncountability of R, and a better idea to invoke _that_> proof that there exists an irrational, was just silly,> because anyone who's familiar with that proof already> knows how to show R is uncountable.(Not that it matters, since your argument is simply> wrong anyway. As opposed to the argument via> the uncountability of R, which is actually correct.)>[It's messier with improper fractions with non-zero entries to the>left of the decimal point, and with signs, but it's doable, you>just have to be a bit obsessive about the details, reflecting the>digits about the decimal point to list them alternately right-next>and left-next in the vertical, and duplicating the numerator>columns beyond zero to put in a minus sign in the duplicate column,>which sign may need to be sticky, the proof is otherwise identical.]xanthian.=== === Subject: : Re: Alternative factorial calculation methods> some months ago I've seen a web page with alternative methods to calculate> factorial numbers.>....Also, I might as well mention one more result I just found a few days ago. (True?)ln(m!) = integral{0 to oo} floor(m {x})/(x^2 (1+x)) dx,where {x} is fractional part, {x} = x - floor(x).Leroy Quet=== === Subject: : Re: Alternative factorial calculation methodsLeroy's ingenious analytic methods are all very well and good.However, I recommend adopting a more empirical, probabilistic approach.n! is the mean time it takes to get all numbers in ascending order when tickets numbered 1 to n are randomly jumbled up and picked sequentially.So if you do this experiment often enough, you will have a perfect statisticalestimate of n! And note - as you are trying to estimate an INTEGER, you canbe suitably assured when you have the EXACT answer - most unusual in stats.For the really advanced, you might like to consider drawing n from m, where m > n.This makes use of the fact that n! = (P^m_n)/(C^m_n) . Impressive!-- He's the sort of fellow that uses statistics like a drunk uses a lamp-post, for support rather than illumination.------------------------------------------------- -----------------------------=== === Subject: : AN: Two new Magic Sinewave tutorials..... are newly available for free download ashttp://www.tinaja.com/glib/msintro1.pdf andhttp://www.tinaja.com/glib/deltams1.pdfPlease also see http://www.tinaja.com/magsn01.asp andhttp://www.tinaja.com/advt01.asp-- Don LancasterSynergetics 3860 West First Street Box 809 Thatcher, AZ 85552voice: (928)428-4073 email: don@tinaja.com fax 847-574-1462Please visit my GURU's LAIR web site at http://www.tinaja.com=== === Subject: : Re: A naive question about category theory> : > I wonder if this is strictly true. Who said that one cannot devise a> : > foundational theory in which both 'set' (or 'class') and 'category'> : > are primitive concepts, and of course suitable axioms are given that> : > somehow 'link' them coherently?>Well, *I* did, for starters.>The point is simply that given that you REALLY CAN define either>in terms of the other, they CAN'T BOTH be primitive. If your>suitable axioms *correctly* define *either*, then you can prove>from them that the other can be defined in terms of it.Maybe I wasn't clear. For starters my question was deliberately naive.However you should know better than me that in (some) set theory(-ies)you define *two* symbols, say e and o (that is 'in' and 'empty') andsubject these symbols to some axioms.Now you can define in terms of some set theory, 'category' and all ofthe commonly used related concepts, right? But, at least with some ofthem, you have an unsatisfactory feeling as for the 'broadness' of theconstructions you can 'legally' get...OTOH you can take the reverse path and start from an abstract conceptof category and recover a set theory based on it that resembles/isconsistent with/matches traditional ones.Let's look at this setting from another point of view: nobody forbidsyou to use a redundant set of axioms. So, for a moment, build a jointtheory 'set+category' (in a hopefully consistent manner), OK?Let's say, for an analogy, that in the joint theory, 'set' stays onthe left and 'category' on the right (of course this has no meaning,literally). By removing unnecessary axioms (those that can be provedas a consequence of other ones) on the left and on the right you canrecover a non redundat set of axioms respectively for 'category' and'set'.But then you could substitute some subsets of axioms with equivalentones and remove unnecessary ones 'in the *middle*' thus finding anhypothetical theory in which both 'set' and 'category' are primitiveterms that *can't* be defined in terms of the other.Now you could ask me *why* to follow this approach and you may beconceptually dissatisfied with a *foundational* theory based jointlyon two primitive concepts that in *other* theories could be definedone in terms of the other. But I don't see why *a priori* thispossibility should be excluded: of course there may well be a *good*reason why it is indeed so!Michele-- > Comments should say _why_ something is being done.Oh? My comments always say what _really_ should have happened. :)- Tore Aursand on comp.lang.perl.misc=== === Subject: : Re: A naive question about equality> And here, your problem with people who dismissively attack> you for total incoherence around = is revealed by YOUR> dismissive attack on US as having merely had a basic> introduction to logic and set theory.Most people who have exhibited rational behavior in discussions with me have notexperienced dismissiveness from me to the best of my knowledge. If that is thecase, I extend my apologies to those individuals. No slight has ever beenintended. Comments that could lead to such misunderstanding tend to occur morefrequently when dealing with someone who mistakes confrontational posturing aspolemical debate.You should not include yourself among people who behave rationally, George.:-)=== === Subject: : Re: A naive question about equality : Most people who have exhibited rational behavior in discussionsYou are hardly even qualified to judge. Physician, heal thyself. : with me have not experienced dismissiveness from me to the best : of my knowledge.Your knowledge is limited.When you say I have not tried to do anything higher than 1st-order, and get rebutted,You have tried to invoke topology;that is inherently THIRD-order, and your response to the rebuttalis to just ignore it and go on talking about a bunch of superfluousbull, THAT is dismissive. : If that is the case, I extend my apologies to : those individuals. No slight has ever been intended.Pointedly ignoring something instead of engaging it isALWAYS intentional, so we can add damn liar to the listof relevant epithets here. : Comments : that could lead to such misunderstanding tend to occur more : frequently when dealing with someone who mistakes : confrontational posturing as polemical debate.I am not making any such mistakes. : You should not include yourself among people who behave rationally, George.That was not only dismissive, it was gratuitous.-- --- It's difficult ... you need to be united to have any strength, but internal issues have to be addressed. --- E. Ray Lewis, on liberalism in America=== === Subject: : Re: Angular Momentum in Rotating Bodies> I mean the position vector. The issue I'm raising is whether dr/dt> lies in the same direction as v. You say it does by definition. I say> it does by conventional interpretation and assumption.Would you please read my post where I proved that dr/dt is perpendicular to r? The Msg-ID is bdell4$ret8p$1@ID-106564.news.dfncis.de-Timo-- Timo Voipio | Helsinki, Finland | ICBM at: 60 11.800 N 024 52.760 EGeekCode ver 3: GU>CC d s-: a--- C++ UL(+)$>+++$ P+>+++ L++(+) E- W++ N++o? K? w O M- V- PS PE Y+ PGP+ t 5++ X R tv- b++(++++) DI+ D G e- h! r !y=== === Subject: : Re: Angular Momentum in Rotating Bodies>> I understand what you suggest with respect to successive approaches to>> zero time intervals of average velocity. But what actually happens in>> my estimation is that an infinitessimal dr/dt>dr/dt is *not* infinitesimal. It is a very ordinary vector.I misspoke here. You are quite correct. dr/dt has a magnitude equal tov for circular rotation but in the direction of the center of rotationalong r in my estimation. My reference to infinitessimal should havebeen with respect to dr as t --> 0. This is the reason there is nochange to r in the context of circular rotation despite dr/dt lyingalong r. >I know you're not interested but here goes for someone>who might be:Believe it or not, I am always interested. I'm not necessarilyagreeable, but I've learned through experience that many on the usenetcan shed considerable light on the problems I try to analyze. And Icontinue to be grateful for those insights.>The following can be shown:> | For each number T, there is a vector v such that:> | For each number e > 0, there is a number d > 0, such> | that for all numbers t > 0 the following implication holds:> | if> | |t| < d> | then> | distance( [r(T+t) - r(T)]/t, v ) < e>This vector v, of which the existence can be proven, is>defined as the instant velocity and *abbreviated* as dr/dt>and there is nothing you can do about it ;-)>or approaching in any way.>> directed toward the>> center of rotation combines vectorially with a tangential v to produce>> the rotation of v and r.>and v=dr/dt can be show to be always tangential to the path.>Take a cicular path> { x = k*cos(wt)> { y = k*sin(wt)>Position vector r = (x,y)> = ( k*cos(wt) , k*sin(wt) )>Getting the speeds:> { dx/dt = -k*w*sin(wt) speed in x-direction> { dy/dt = k*w*cos(wt) speed in y-direction>Velocity vector dr/dt = ( dx/dt, dy/dt )> = ( -k*w*sin(wt), k*w*cos(wt) )>Product of vector r and vector dr/dt:> r . dr/dt = - k^2*w*cos(wt) *sin(wt) + k^2*w*sin(wt) *cos(wt)> = 0 so they are penpendicular.> Consequently, the cross product> dr/dt x p would not be zero nor would dL/dt.>>hth>I don't think it helped and I don't think anything will>help, but you must be having a great trolling time...>Don't bother replying to this - I didn't write it for you.>But do enjoy the attention you are getting with this :-)Well, I wish I enjoyed the limelight quite as much as people seem tothink. Actually I do enjoy the limelight but only with good ideas. Idon't enjoy any kind of attention generated with bad ideas. And I'vealways found that provocative ideas of some merit generate fargreater interest than any kind of trolling. Rather like advertising abad product. Counterproductive and it certainly kills the market foranything else you might have to offer.I would like to comment on what you say above only to the extent thatI think I understand what the problem is and has been in my conceptualgrasp of the subject, the orientation of dr/dt with respect totangential velocity v.What you indicate above is certainly correct in the classical analysisof angular mechanics. I'm of the opinion that dr/dt operatescentripetally along r whereas classical angular mechanics considersthat it operates in the direction of tangential velocity v.However, having helped me isolate the conceptual problem, I prefer notto tax you further with peripheral arguments. You have been ofconsiderable help, and I thank you for your efforts.>remove DEL in address for email=== === Subject: : Re: Angular Momentum in Rotating Bodies>If an object has constant tangential (=linear) velocity, it is moving in a >straight line. Perhaps you meant tangential speed?>> You know, at present this remains an issue worth discussing. However,>> these are the terms in which the problem was analyzed originally. So,>> I would like to preserve them.>Velocity, as you undoubtedly know is a vector quantity. If a vector quantity >is constant, both its magnitude and *direction* stay constant. Therefore, in >circular motion there's no such thing as constant tangential velocity. If >the frame of reference is taken to be the mass m, then the velocity of >course stays constant, but then the motion wouldn't be circular (and we >would have a noninertial frame of reference, which would complicate matters).I appreciate what you are saying here. However, the original treatmentof the problem was cast in the terms indicated previously and seems tobe well understood in those terms.>> As noted in my reply to Dirk van der Moortel, I misstated this point.>> Assuming you're referring above to dr/dt>No. I am referring to dv/dt.>> rather than dv/dt, I should>> have indicated a centripetal acceleration of dr/dt towards the center>> of rotation.>Centripetal acceleration of dr/dt... WtF? You mean the centripetal >acceleration of the object, d^2/(dt)^2 r?This notation is somewhat unclear. What I meant is what I said. I wascommenting on a point raised by DvM where I inadvertently indicatedthat dr/dt lay in the direction of r whereas what I meant was thatdr/dt in my estimation is centripetally directed along r.>> My contention is that dr should be taken along r towards the center of>After being proven wrong several times over with differring methods, why do >you still insist? As this is x-posted to mathematics I think that you know >that if something has been proven algebraically with no errors in the method >and premises, the result has conclusively proven true. Period. End of debate.Well, I gues that's the end of the debate. I wasn't aware we weredebating. I presented an alternative opinion as to the direction inwhich dr/dt should be taken to support the idea that circular rotationat constant angular velocity represents a constant change in angularmomentum rather than constant angular momentum. And this does notrepresent an algebraic problem. It's a problem of mechanical insightand interpretation.As to why I persist, that's my business. Period. End of debate.>...just out of curiosity, how did you pass your Physics 101?With flying colors.>-- >Timo Voipio | Helsinki, Finland | ICBM at: 60 11.800 N 024 52.760 E>GeekCode ver 3: GU>CC d s-: a--- C++ UL(+)$>+++$ P+>+++ L++(+) E- W++ N++>o? K? w O M- V- PS PE Y+ PGP+ t 5++ X R tv- b++(++++) DI+ D G e- h! r !yremove DEL in address for email=== === Subject: : Re: Angular Momentum in Rotating Bodies> However, having helped me isolate the conceptual problem, I prefer not> to tax you further with peripheral arguments. You have been of> considerable help, and I thank you for your efforts.You are welcome.I hope you learned something after all :-)=== === Subject: : Re: Angular Momentum in Rotating Bodies> Now, I consider that this - in my terms - interpretive assumption is>> worth dicussing because it seems incorrect. Any centripetal force>> would produce a centripetal dr/dt independent of v and would continue>> to do so in the presence or absence of v. In the absence of v, dr/dt>> would result in some velocity in the direction of the accelerating>> force. And in the presence of v it combines vectorially with v to>> produce rotation.>Lester, I preferred your style on> http://groups.google.com/groups?&threadm=3ef325d5.60167653@ netnews.att.net>But I must admit, this thread is vastly more successful ;-)>You know, Dirk, if you're referring to the Transverse Doppler thread,yes we had some rather agreeable fun with it once I understood whereyou were going. And it's certainly the kind of comic relief we can alluse with the kinds of things we think and write about.The unfortunate thing is that it was intended to be taken quiteseriously. But somewhere along the line I began to realize that it waspretty unintelligible for practical purposes. Too bad. I still thinkit's correct if we could just figure out what it says.And it's too bad as well about the follow up request for any kind ofmechanical interpretation for isotropic transverse doppler. Old Manshowers me with formulas for angular momentum which were undoubtedlyhoary with age when he was a youth - if he ever was. But neither henor anyone else bothers to reply to a relatively simple request for amechanical explanation. Most knowlegeable thinkers are sure it's therebut no one seems to be able to explain how it is mechanized.I don't mind being provocative. I do mind being stupid or caught offguard. And that's exactly what happened with Meron's analysis ofdL/dt. It was directly to the point, and I'm afraid I reacted badly toit. But I haven't forgotten it or the point I was trying to make, andthe present effort is just directed toward correcting that problem andmaking clear the problem I see with conventional angular mechanicalconcepts.Perhaps one day we can have some more fun somewhere along the line. Italways helps to keep the obtuse expression of relatively simple ideasremove DEL in address for email=== === Subject: : Re: Angular Momentum in Rotating Bodies> I mean the position vector. The issue I'm raising is whether dr/dtIf r means position vector, then dr/dt means vectorchange of position vector per unit time.If you want to see what that is, plot two positions differingby a short time, and look at their vector difference.You'll see it's the same direction as v. Not surprisingly,since v also means vector change of position per unit time.> lies in the same direction as v. You say it does by definition. I say> it does by conventional interpretation and assumption.Then you'll need to provide a different definition of v. Isee that point has been made in this thread already.I was trying without success to establish a new perspective on the> issue. Of course, if as you suggest dr/dt = v as a matter of> definition, then there is no issue and no basis for discussion.What's your alternate definition?> What puzzles me is why the definition is taken the way it is. We have> one definition in the form of L = r x p and we want to investigate the> implications and properties of that definition.This is a definition of angular momentum.Then we have dL/dt = dr/dt x p + r x dp/dt as one of the properties.> You say that we merely define dr/dt as lying in the direction of v.No, we define v as being another name for dr/dt. It's notin the direction. It's the same thing.> And the only reason that I can see for that is to make the cross> product zero and L constant.No, it's defined that way because that's how it's defined.That definition leads to certain properties like theabove, but it's defined that way because it's convenientto have a name for first derivative of position.> So, I really have to ask why the direction of dr/dt is a matter of> definition and not interpretationBecause derivative has a specific meaning which will notallow you to choose direction arbitrarily.Look at r(t). Look at r(t+h) for small h. Those twopositions differ by a vector. As h gets smaller andsmaller, that vector will converge to a specificdirection. The quantity (r(t+h)-r(h))/h will convergeto a specific magnitude and direction. There's nochoice in this. It follows directly from the processof evaluating the derivative, from the definition ofderivative.What you are doing seems to be like this process:1. Define dog to refer to a certain kind of animal.2. Examine the properties of the animals which fallin the agreed upon class dog.3. In walks Lester, who says, I want to examinethese properties, but first we'll define dog to meanvehicle powered by internal combustion engine.The properties follow from the meaning assigned to theword dog. If you change the meaning of dog tocar, its properties become the properties of cars.If you change velocity to mean something other thantime derivative of position, then the properties weassociate with time derivative of position will nolonger be properties of the thing you are now callingvelocity. - Randy=== === Subject: : Re: Angular Momentum in Rotating Bodies>> I mean the position vector. The issue I'm raising is whether dr/dt>> lies in the same direction as v. You say it does by definition. I say>> it does by conventional interpretation and assumption.> Have you considered the radius->infinity case? That makes >it obvious (to me).> Mark L. FergersonHi Mark. It's good to hear from you again. I don't know if you caughtany of the original Meron-Heymann-Dogfrey-Poe-me thread, but thequestion you ask was actually part of the genesis. It was entitledLinear and Angular Momentum as I recall and represented an attempt toget this general idea off the ground.Basically I was trying to get some conceptual basis established forthe analysis of angular momentum in linear terms. I think what you aresuggesting is that an infinite radius represents motion in a straightline and that dr/dt thus has to take place in the direction of v.(Please correct me if I'm wrong.) And I actually agree. But let me seeif I can explain what I'm suggesting.Given a constant force f acting on a point mass m, we produce aconstantly changing velocity v in the direction of the force. And ifwe call that direction r then dr/dt certainly acts along v. This iswhat I would consider longitudinal linear acceleration and velocity.However, if we then impose some constant velocity v2 across thedirection of v of a magnitude equal to dr/dt but normal to thedirection of the accelerating force we then produce circular rotation.And this is all I see happening in the context of angular momentum incircular rotational situations.The secondary orthogonal velocity simply becomes the tangentialvelocity in rotation and dr/dt acts centripetally and combines withthe secondary orthogonal velocity to produce rotation through thevector addition of dr/dt and v2. There is no net change in r becausedr is infinitessimal and r is finite.Thus as r --> oo v2 --> 0 and linear acceleration in the direction off and v effectively resumes just as it did prior to the imposition ofv2.I don't see this as any kind of remarkable consideration except thatit is different from the classical interpretation and alters thenature of dL/dt in the context of circular rotation. It makes dL/dtnon zero in such cases and indicates that circular rotation actuallyrepresents a constant change in angular momentum rather than aconstant angular momentum as is supposed in the conventionalinterpretation.Sometimes it is difficult to get my meaning across clearly. I startedoff with Meron's critique of dL/dt but with the intention of bringingup my revisionist interpretation of the direction of dr/dt in order tojustify certain other interpretations of physics problems.All I'm really trying to do is to get to some common conceptualfoundation for linear and angular mechanics. And if thisinterpretation is allowed then I think we have a very realisticpossibility of addressing linear, macro, and micro angular mechanicson a common analytical framework.remove DEL in address for email=== === Subject: : Re: Angular Momentum in Rotating Bodies>> On Nov 15, 2002 in a discussion with Mati Meron the following points>> were raised in connection with angular momentum and the circular>> rotation of a point mass at distance r and constant tangential>> velocity v.>>> If an object has constant tangential (=linear) velocity, it is moving in a >> straight line. Perhaps you meant tangential speed?>If an object has constant linear velocity, in other words it is moving>in a straight line at constant speed, it also has constant angular momentum.>SocksGood point. It also has what I call constant angular momentum inuniversal terms, that is constant angular momentum with respect toevery point in space. The most that can be said for circular rotationis that it represents a kind of locally constant angular momentum,that is with respect to only one point in space.remove DEL in address for email=== === Subject: : Re: Angular Momentum in Rotating Bodies>[not understanding highschool physics]>Would it be possible to entince you into reading a book before>you cross post to various news groups?>SocksI'm not sure what you're after. I thought that considering the use ofa time derivative of a vector cross product, the inclusion of sci.mathwould be appropriate. As for the parenthetic not understanding highschool physics notation, if you're applying it to me you mightreconsider.remove DEL in address for email=== === Subject: : Re: Angular Momentum in Rotating Bodies>>On Nov 15, 2002 in a discussion with Mati Meron the following points>were raised in connection with angular momentum and the circular>rotation of a point mass at distance r and constant tangential>velocity v.>My contention was and is that this situation represents constantly>changing angular momentum and his response was that in this situation>angular momentum is constant because dL/dt = 0.>His reasoning was that angular momentum represents >[1] L = r x p by definition and that>[2] dL/dt = dr/dt x p + r x dp/dt.>He further states that dr/dt = v so that the cross product >[3] dr/dt x p = 0>because v and p are co linear.>>What do you think the definition of p is? p = mv, where m is a scalar.>>p and v are therefore collinear vectors, hence v x p = 0 by definition.>>> Well, of course, v x p = 0. There's never been any question about>> that. The question I'm asking is whether dr/dt = v? Everyone seems to>> think so.>It is -- by definition. Let r be the vector which runs from the>or such. The velocity vector v is simply defined to be dr/dt.My reply to Edward Green this morning addressed this problem. It iscertainly taken to be a matter of definition. However, if we areinvestigating the properties and mechanical implications of onedefined quantify, that is L = r x p, I don't think we are justified insimply adopting subordinate definitions which bear directly on thatissue without compromising that analysis.>> And I agree that they have the same scalar magnitude for>> circular rotation. What I don't agree on is that their direction is>> the same. I think dr/dt operates centripetally along r and not in the>> direction of tangential velocity v.>You are free to do this by making your own definition for what v is.>In this case, I think it is counterproductive to take this direction.>Velocity is an established concept and a novel definition impedes>other people understanding what you are talking about. It is better>to call your concept by a new name. Well, there's glory^1 for you.Ah, yes, I've had this suggestion before, and it is not without merit.However, I would like to reiterate that at this juncture what I'maddressing is just the physical interpretation of a well understooddefinition in the form of L = r x p and dL/dt.>[1] a nice knockdown argument as per Humpty-Dumpty in Alice in>_Through the Looking Glass_.>-- remove DEL in address for email=== === Subject: : Re: Angular Momentum in Rotating Bodies>You are making the same mistake I made: dr/dt and v _are both>>vectors_. v = dr/dt by definition, and of course, in magnitude in>>direction (how else would two vectors be equal?). End of story,>>pretty much.I'm not sure dr/dt = v by definition.Oish. Good thing Mati doesn't seem to be paying attention this time!Let me interpret that remark in the most favorable light: The>question is, by r do we mean the position vector, or the scalar>length of the radius. r is often used for the position vector, and>in this case, the velocity is the first time derivative of the>position vector. End of story.> I mean the position vector. The issue I'm raising is whether dr/dt> lies in the same direction as v. You say it does by definition. I say> it does by conventional interpretation and assumption.Do it by orthogonal components then:x = |r|*cos(theta), y = |r|*sin(theta)|r| is constant for circular motion, andw = d(theta)/dt = constant (w is a scaler)dx/dt = -|r|*w*sin(theta), dy/dt = |r|*w*cos(theta)v = dr / dt = i*(dx / dt) + j*(dy/dt) = |r|*w*[-i*sin(theta) + j*cos(theta)]the component of v along r is given by the dot product, v*r:v*r = [|r|^2]*w*[i*cos(theta) + j*sin(theta)]*[-i*sin(theta) + j*cos(theta)] = [|r|^2]*[cos(theta)*sin(theta) - sin(theta)*cos(theta)] = 0Therefore, for circular motion, the radial component of the velocity,v = dr/dt, is always zero. [Old Man]=== === Subject: : Re: Angular Momentum in Rotating Bodies>You are making the same mistake I made: dr/dt and v _are both>>vectors_. v = dr/dt by definition, and of course, in magnitude in>>direction (how else would two vectors be equal?). End of story,>>pretty much.I'm not sure dr/dt = v by definition.Oish. Good thing Mati doesn't seem to be paying attention this time!Let me interpret that remark in the most favorable light: The>question is, by r do we mean the position vector, or the scalar>length of the radius. r is often used for the position vector, and>in this case, the velocity is the first time derivative of the>position vector. End of story.> I mean the position vector. The issue I'm raising is whether dr/dt> lies in the same direction as v. You say it does by definition. I say> it does by conventional interpretation and assumption.Do it by orthogonal components then:x = |r|*cos(theta), y = |r|*sin(theta)|r| is constant for circular motion, andw = d(theta)/dt = constant (w is a scaler)dx/dt = -|r|*w*sin(theta), dy/dt = |r|*w*cos(theta)v = dr / dt = i*(dx/dt) + j*(dy/dt) <---- vector derivative = |r|*w*[-i*sin(theta) + j*cos(theta)]the component of v along r is given by the dot product, v*r:v*r = [|r|^2]*w*[i*cos(theta) + j*sin(theta)]*[-i*sin(theta) + j*cos(theta)] = [|r|^2]*[cos(theta)*sin(theta) - sin(theta)*cos(theta)] = 0Therefore, for circular motion, the radial component of the velocity,v = dr/dt, is always zero. [Old Man]=== === Subject: : Re: Angular Momentum in Rotating Bodies>I suggest you start a religion. Your way of reasoning is much>more similar to religious argument than to physics argument.Arf!>Arfur> Ah, yes, I remember it well. This time 'round I'm trying for a civil> discussion of a contentious point rather than ranting and raving. So> far reasonably successfully.Vector tine derivative:v = dr/dt = i*(dx/dt) + j*(dy/dt)where r = i*x + j*yfor circular motion, |r| is a constat, and d(theta)/dt is also constant.Show that the dot product, v*r = 0, thus demonstrating that theradial component of velocity is zero for all angles and for all times.Get it? [Old Man]=== === Subject: : Re: Angular Momentum in Rotating Bodies>On Nov 15, 2002 in a discussion with Mati Meron the following points>were raised in connection with angular momentum and the circular>rotation of a point mass at distance r and constant tangential>velocity v.>My contention was and is that this situation represents constantly>changing angular momentum and his response was that in this situation>angular momentum is constant because dL/dt = 0.>His reasoning was that angular momentum represents >[1] L = r x p by definition and that>[2] dL/dt = dr/dt x p + r x dp/dt.>He further states that dr/dt = v so that the cross product >[3] dr/dt x p = 0>because v and p are co linear.>>What do you think the definition of p is? p = mv, where m is a scalar.>>p and v are therefore collinear vectors, hence v x p = 0 by definition.>Well, of course, v x p = 0. There's never been any question about>that. The question I'm asking is whether dr/dt = v? Everyone seems to>think so. And I agree that they have the same scalar magnitude for>circular rotation. What I don't agree on is that their direction is>the same. I think dr/dt operates centripetally along r and not in the>direction of tangential velocity v.Let me give it one last try and then I'll give up. 1. If your symbol r represents a (position) vector, then you apparentlydo not understand the definition of a derivative. In this casev = dr/dt, by definition of a derivative.2. If your symbol r represents the magnitude of the position vector,then dr/dt is equal to the radial component of the velocity vector.But for circular motion, the magnitude of the radius is constant, so theradial component of velocity is zero, but the magnitude of the velocityvector is nonzero. So dr/dt and v do not have the same magnitude, one iszero and the other is not. Again this indicates a lack of understandingof what a derivative is.A derivative represents a rate of change of a variable. Your continued obtuseness is what leads people to think you are a simpletroll looking for attention. You can claim you are being treated poorly,but what about your treatment of us?I suggest you read up on what the derivative of a scalar is and thenwhat the derivative of a vector is. And here's a hint: a vector in aplane is represented by two scalars, which can vary independently.Education is not having everything spoon fed to you. Sometimes you need tolearn things for yourself. The net is a great resource, but it shouldn'treplace thinking and learning on your own.>remove DEL in address for email--John E. PrussingUniversity of Illinois at Urbana-ChampaignDepartment of Aerospace Engineeringhttp://www.uiuc.edu/~prussing=== === Subject: : Re: Angular Momentum in Rotating Bodies>> I mean the position vector. The issue I'm raising is whether dr/dt>If r means position vector, then dr/dt means vector>change of position vector per unit time.>If you want to see what that is, plot two positions differing>by a short time, and look at their vector difference.>You'll see it's the same direction as v. Not surprisingly,>since v also means vector change of position per unit time.I think I see what you're getting at but would like to ask a questionon just this much of the post to be sure. If we have just any oldvector r and it changes magnitude without changing direction, is dr/dtzero?Perhaps I am making a mistake or misconstruing what you say, but Ialways learned that a vector is composed of both magnitude anddirection. So, I would have thought that changes to either wouldchange the vector itself for analytical purposes. ====== === Subject: : Re: Angular Momentum in Rotating Bodies As I> understand the concept, the constant angular revolution of one> celestial body around another considerably more massive body is> explained in terms of the orbiting body's falling toward the other at> exactly the same velocity as it orbits tangentially, resulting in a> circular orbit of constant radius, period, and dimension. > This is true only if you take the average velocity between points on the orbit separated by 1/4 of a revolution. Don't confuse this with the instantaneous velocity at a single point on the orbit.-- === === Subject: : Re: Angular Momentum in Rotating Bodies> I think I see what you're getting at but would like to ask a question> on just this much of the post to be sure. If we have just any old> vector r and it changes magnitude without changing direction, is dr/dt> zero?Let's define a vector r such that its magnitude varies with time:r = i v t cos theta + j v t sin theta,where theta and v are (scalar) constants.Then dr/dt will bedr-- = i v cos theta + j v sin theta,dtand if v != 0 then dr/dt = constant and != 0.-Timo-- Timo Voipio | Helsinki, Finland | ICBM at: 60 11.800 N 024 52.760 EGeekCode ver 3: GU>CC d s-: a--- C++ UL(+)$>+++$ P+>+++ L++(+) E- W++ N++o? K? w O M- V- PS PE Y+ PGP+ t 5++ X R tv- b++(++++) DI+ D G e- h! r !y=== === Subject: : Re: Angular Momentum in Rotating BodiesMime-version: 1.0Content-type: text/plain; charset=US-ASCIIContent-transfer-encoding: 7bit> The classical definition of dL/dt regards dr/dt as lying in the> direction of v. I regard dr/dt as lying centripetally along r. I think> this represents a reasonable point for discussion.But...but...there's no I regard... up for discussion here !!v is the velocity VECTOR...it's the rate of change of positionr is the position VECTOR...dr/dt is (by def'n) the rate of change of position...two vectors are equal iff they have the same magnitude and direction...The thing that you're saying lies centripetally along r is just NOT therate of change of position...maybe it's a physically interesting vector(cough), but you really need to give it a different name.=== === Subject: : Re: Angular Momentum in Rotating Bodies>>> As I>> understand the concept, the constant angular revolution of one>> celestial body around another considerably more massive body is>> explained in terms of the orbiting body's falling toward the other at>> exactly the same velocity as it orbits tangentially, resulting in a>> circular orbit of constant radius, period, and dimension. >>This is true only if you take the average velocity between points on the >orbit separated by 1/4 of a revolution. Don't confuse this with the >instantaneous velocity at a single point on the orbit.>-- >Actually, I think it is true for instanteous velocity because v and rchange direction continuously throughout the orbit. However, this maybe grist for another discussion.remove DEL in address for email=== === Subject: : Re: Angular Momentum in Rotating Bodies>> The classical definition of dL/dt regards dr/dt as lying in the>> direction of v. I regard dr/dt as lying centripetally along r. I think>> this represents a reasonable point for discussion.>But...but...there's no I regard... up for discussion here !!the next time someone wants to consider a point for discussion I'll besure to consult it or you.If this seems a trifle sharp, I apologize. But I'm not aware that indiscussing the properties, implications, and characteristics ofexpressions such as L = r x p and dL/dt we are bound by anyconsideration except mechanical plausibility.If you want define dr/dt in the direction of v, then you could moreeasily just start out by defining dL/dt as zero and investigate theimplications of that definition instead.If on the other hand you would claim that by taking dr/dt in thedirection of the center of rotation, I am prejudicing dL/dt bydefinition, all I can suggest is that we try to determine whichapproach makes more sense.I don't regard - if you'll allow - the matter as one of definition. Iconsider the direction of dr/dt as a problematic but criticalconsideration with respect to judging dL/dt. And I consider that dr/dtis centripetal in orientation because that direction coincides withthe centripetal force needed to keep the point mass m in circularrotation. At least it does in celestial mechanics and I suspect inordinary angular mechanics as well.And if that is not a suitable subject for discussion then I don't knowwhat could be. Since the whole purpose and subject of discussion isthe evaluation of dL/dt.[. . .]remove DEL in address for email=== === Subject: : Re: Angular Momentum in Rotating Bodies> Now, I consider that this - in my terms - interpretive assumption is> worth dicussing because it seems incorrect. Any centripetal force> would produce a centripetal dr/dt independent of v and would continue> to do so in the presence or absence of v. In the absence of v, dr/dt> would result in some velocity in the direction of the accelerating> force. And in the presence of v it combines vectorially with v to> produce rotation.>Lester, I preferred your style on> http://groups.google.com/groups?&threadm=3ef325d5.60167653@ netnews.att.net> But I must admit, this thread is vastly more successful ;-)I'm not _sure_, but I think it might be bad protocol to bring uppossible past indiscretions, unless the victim is running for publicoffice.=== === Subject: : Re: Angular Momentum in Rotating Bodies>The classical definition of dL/dt regards dr/dt as lying in the>direction of v. I regard dr/dt as lying centripetally along r. I think>this represents a reasonable point for discussion.>>But...but...there's no I regard... up for discussion here !!> the next time someone wants to consider a point for discussion I'll be> sure to consult it or you.If this seems a trifle sharp, I apologize. But I'm not aware that in> discussing the properties, implications, and characteristics of> expressions such as L = r x p and dL/dt we are bound by any> consideration except mechanical plausibility.> No, you are incorrect. Mechanical plausibility is certainly necessary,but it is simply insufficient. You are engaged in an intellectualexercise of sorts, and that exercise involves calculations thatmake use of techniques from calculus to derive relationships amongvarious quantities. Given that, those techniques, and the rulesof operation with those techniques, play a key role. Certainly itis possible to use calculus and at the same time disregard thoserules; calculus students do it all the time, and those of us whoteach them have a technical term for those methods: ERRORS.Can you define a vector q that serves your purposes (that you'resuggesting ought to be dr/dr)? Sure: just let q be the product ofspeed |v| with the unit normal of the curve traced out by r(t). Canyou define a vector K that serves your purposes for L (that is, asr x p, where you have the vector p pointing along the direction ofthe vector q above)? Again, sure: just define it: K = r x mqNow, you can have your little discussion and eat it too. Yes, youcan show that dK/dt does indeed change with time. So what? Of whatvalue is an arbitrarily-defined vector that changes with time. Thebig deal about invariants in classical mechanics has to do withthem being.... INVARIANT! The presence of a conserved quantityenables equations to be simplified, and complex problems to bereduced to less-complex problems (along with the inclusion of aconstraint that states the invariance property).> If you want define dr/dt in the direction of v, then you could more> easily just start out by defining dL/dt as zero and investigate the> implications of that definition instead.> It's not up to him, or you, or me, for that matter, to want todefine dr/dt in any fashion. If you're using calculus (and fromall appearances, you do seem to be using calculus), then dr/dt isalready defined. Similarly, dL/dt, dv/dt, and any other d*/dt thatyou care to discuss.> If on the other hand you would claim that by taking dr/dt in the> direction of the center of rotation, I am prejudicing dL/dt by> definition, all I can suggest is that we try to determine which> approach makes more sense.I don't regard - if you'll allow - the matter as one of definition. I> consider the direction of dr/dt as a problematic but critical> consideration with respect to judging dL/dt. And I consider that dr/dt> is centripetal in orientation because that direction coincides with> the centripetal force needed to keep the point mass m in circular> rotation. At least it does in celestial mechanics and I suspect in> ordinary angular mechanics as well.> However, dr/dt is not associated (directly) with centripetalacceleration, nor (by Newton's Second Law of Motion) withcentripetal force. No doubt you are aware of the formula F = ma,identifying force with the product of mass with acceleration. Further,acceleration is dv/dt, or d^2 r/dt^2. Under the assumption of constantspeed |v|, dv/dt is easily shown to be perpendicular to v, and isdirected along the radius of curvature of whatever curve is beingtraced out by r(t).> And if that is not a suitable subject for discussion then I don't know> what could be. Since the whole purpose and subject of discussion is> the evaluation of dL/dt.> You want to interpret dL/dt as a rate of change wrt time. If so, and todo this correctly, you need to use the machinery that is already inplace, i.e., differential calculus.In short, whether it's a suitable subject for discussion, or not, theidentification of dr/dt with a vector along the radial direction amountsto a rejection of the methods of calculus for computing thesequantities. Perhaps you should address that as the topic, rather thanusing calculus-derived terminology to deny the applicability ofcalculus. Since it appears that 300+ years of work points to therejection of your claim, you may have a tough row to hoe in thatpursuit.> [. . .]remove DEL in address for email> === === Subject: : Re: Angular Momentum in Rotating BodiesI mean the position vector. The issue I'm raising is whether dr/dt> lies in the same direction as v. You say it does by definition. I say> it does by conventional interpretation and assumption. Have you considered the radius->infinity case? That makes > it obvious (to me).You seem to me to be feeding the misconception that there is somethingto get here. There may be such things sitting down at the sametable, but the chief thing to get about dr/dt lies in the samedirection as v is that there is nothing to get!Saying that dr/dt, which is v, lies in the same direction as v, islike saying that women have the same gender as human females. Thenwhen somebody wonders isn't that a matter of conventionalinterpretation you bait him by inviting him to consider the limitingcase of tall women. :-) Given that by r we mean the position vector,as stated, dr/dt lies in the same direction as v because it *is* v;circular motion, rectilinear motion, or any motion at all.=== === Subject: : Re: Angular Momentum in Rotating Bodies> Now, I consider that this - in my terms - interpretive assumption is> worth dicussing because it seems incorrect. Any centripetal force> would produce a centripetal dr/dt independent of v and would continue> to do so in the presence or absence of v. In the absence of v, dr/dt> would result in some velocity in the direction of the accelerating> force. And in the presence of v it combines vectorially with v to> produce rotation.> Lester, I preferred your style on> http://groups.google.com/groups?&threadm=3ef325d5.60167653@ netnews.att.net> But I must admit, this thread is vastly more successful ;-)> I'm not _sure_, but I think it might be bad protocol to bring up> possible past indiscretions, unless the victim is running for public> office.Nah, Lester and I really enjoyed that exchange :-)Besides, aren't we all running for public office on Usenet?Dirk -Bad Protocol- Vdm=== === Subject: : Re: Angular Momentum in Rotating Bodies>I'm not _sure_, but I think it might be bad protocol to bring up>possible past indiscretions, unless the victim is running for public>office.Usenet is a Markov process; no memory.kmuldrew@ucalgary.ca=== === Subject: : Re: Angular Momentum in Rotating Bodies>I'm not _sure_, but I think it might be bad protocol to bring up>possible past indiscretions, unless the victim is running for public>office.> Usenet is a Markov process; no memory.Andrei Markov was known for his excellent memory: http://groups.google.com/groups?&q=author%3Aken+author% 3Amuldrew;-)kmuldrew@ucalgary.ca=== === Subject: : Re: Angular Momentum in Rotating Bodies>> I mean the position vector. The issue I'm raising is whether dr/dtIf r means position vector, then dr/dt means vector>change of position vector per unit time.If you want to see what that is, plot two positions differing>by a short time, and look at their vector difference.>You'll see it's the same direction as v. Not surprisingly,>since v also means vector change of position per unit time.I think I see what you're getting at but would like to ask a question> on just this much of the post to be sure. If we have just any old> vector r and it changes magnitude without changing direction, is dr/dt> zero?No.dr/dt is a vector whose components are (dx/dt, dy/dt, dz/dt).If any of those changes (as they must if the magnitude ofr is changing), then this is a nonzero vector.> Perhaps I am making a mistake or misconstruing what you say, but I> always learned that a vector is composed of both magnitude and> direction. So, I would have thought that changes to either would> change the vector itself for analytical purposes.That's correct. - Randy=== === Subject: : Re: Angular Momentum in Rotating Bodies>I mean the position vector. The issue I'm raising is whether dr/dt>lies in the same direction as v. You say it does by definition. I say>it does by conventional interpretation and assumption.>> Have you considered the radius->infinity case? That makes >>it obvious (to me).>> Mark L. Fergerson> Hi Mark. It's good to hear from you again. I don't know if you caught> any of the original Meron-Heymann-Dogfrey-Poe-me thread, but the> question you ask was actually part of the genesis. It was entitled> Linear and Angular Momentum as I recall and represented an attempt to> get this general idea off the ground. Sorry, I've been so busy I usually have time only for small interjections in other people's threads. And no, I didn't keep up with the thread you refer to, so I'm glad to get back to the beginning as it were. (I have until saturday evening to unscrew this; hope I can be of help)> Basically I was trying to get some conceptual basis established for> the analysis of angular momentum in linear terms. I think what you are> suggesting is that an infinite radius represents motion in a straight> line and that dr/dt thus has to take place in the direction of v. Correct.> (Please correct me if I'm wrong.) And I actually agree. But let me see> if I can explain what I'm suggesting.Given a constant force f acting on a point mass m, we produce a> constantly changing velocity v in the direction of the force. And if> we call that direction r then dr/dt certainly acts along v. This is> what I would consider longitudinal linear acceleration and velocity. With you so far, with the caveat that infinite radius is a somewhat fuzzy concept; if you scale your dr and dt small enough, any radius is effectively infinite (but that's something of a red herring). Also we must note that constant velocity only exists when we do not have an accelerating force (this is important in the following).> However, if we then impose some constant velocity v2 across the> direction of v of a magnitude equal to dr/dt but normal to the> direction of the accelerating force we then produce circular rotation. In masses not connected with other masses, fine. Of course you're mixing increments of velocity and time with constant forces, but I see what you mean. Now try applying identical incremental forces repeatedly and see what happens. First, note that orthogonal changes direction every time you change position. You'll note that you get incremental changes in the orthogonal velocity components with the result that you have no net velocity along either axis, but instead the incremental velocity changes are smoothed into a constantly increasing rotational velocity (rpm) about the axis. This is what most of us mean by angular acceleration (note that acceleration requires energy input to change velocity). Now, stop applying the incremental forces and note that the rpm stops changing. This is what most of us mean by angular momentum. (but you knew that)> And this is all I see happening in the context of angular momentum in> circular rotational situations. When you try to apply this thinking to parts of an extended solid body, you get into trouble. The transverse force is hidden in the radial tensile forces and the longitudinal force is hidden in the circumferential tensile forces. In the constant rpm case the forces are bced. If you unbce them, rpm changes (or the flywheel frags), but that costs energy.> The secondary orthogonal velocity simply becomes the tangential> velocity in rotation and dr/dt acts centripetally and combines with> the secondary orthogonal velocity to produce rotation through the> vector addition of dr/dt and v2. There is no net change in r because> dr is infinitessimal and r is finite. Your terminology is a bit confusing; you mean by v2 the velocity of an element of a rotating body along the circumference, and dr to mean change in radius? In one sense dr goes from a maximum through zero and to a negative maximum and repeats, when seen in the coordinates that contain the center of rotation. Are you switching from Cartesian coordinates (seen by the element) to cylindrical coordinates (seen by the system) without saying so? That may be the problem.> Thus as r --> oo v2 --> 0 and linear acceleration in the direction of> f and v effectively resumes just as it did prior to the imposition of> v2. So you're sa