Newton s Laws: What is Their Operational Meaning?

Newto s Laws: What is Their Operatioal Meaig? Olga Kosheleva ad Vladik Kreiovich Uiversity of Texas at El Paso 500 W. Uiversity El Paso, TX 79968, USA olgak@utep.edu, vladik@utep.edu Abstract Newto s mechaics is oe of the most successful theories i the history of sciece; its success is based o three Newto s laws. At first glace, the Newto s laws that describe the relatio betwee masses, forces, ad acceleratios are very clear ad straightforward. However, the situatio becomes more ambiguous if we take ito accout that the otios of mass ad force are ot operatioally defied. I this paper, we describe the operatioal meaig of Newto s laws. 1 It Is Importat to Reformulate Newto s Laws i Operatioal Terms: Formulatio of the Problem Origial formulatio of Newto s laws: remider [1, 5]. The first Newto s law law of iertia states that if o force is actig o a body, this body retais its speed ad directio of motio. The secod law states that the force F is equal to the product of mass m ad acceleratio a: F = m a. The third law states that if a body A acts o a body B with a force F, the the body B acts o the body A with the force F. Pedagogical problem: we eed a operatioal reformulatio. Of course, we have a ituitive otio of what is a mass ad what is a force. However, for most people, these ituitive otios are somewhat vague, ad to uderstad Newto s laws, we eed to be able to provide a precise umerical meaig of these terms. Without such operatioal meaig, Newto s laws souds very abstract: there exist some precise otios of mass ad force for which the above three laws hold. This is probably how some studets uderstad these laws. If all we have is such a abstract formulatio, it is o woder that some studets have trouble applyig these laws to real-life problems. 1

Foudatioal problem: we eed a operatioal reformulatio. Operatioal reformulatio is eeded also because Newto s laws aim at describig the physical world. How do we kow that these laws are valid? How ca we check that these laws are ot valid? For example, what do physicists mea whe they claim that Newto s laws are ot valid i relativistic mechaics? Whe Newto s laws are formulated i the above abstract form, without providig ay operatioal meaig for mass ad force, the it is ot clear how to check whether the give experimetal data supports these laws or ot. To be able to do that, we eed to reformulate Newto s laws i operatioal terms, i.e., i terms of observatios. 2 Reformulatig Newto s Laws i Operatioal Terms: A Straightforward Approach First Newto s law: a straightforward reformulatio. The first Newto s law was actually first formulated by Galileo [3]. This law has a straightforward operatioal iterpretatio: if we have oly oe body A, the its acceleratio is zero: a A = 0. Of course, i reality, we always have some other bodies i the Uiverse, but if these bodies are sufficietly far away, we ca safely assume that their ifluece is egligible. We ca therefore reformulate this law i the followig form: whe we move a body A further ad further away from all other bodies, its acceleratio gets closer ad closer to 0. Commet. This reformulatio assumes that the force betwee the bodies decreases as the distace betwee them icreases. This is defiitely true for usual forces such as gravity or electromagetic forces, but it is worth metioig that ot all forces are like that: for example, the force actig betwee the two quarks icreases whe the distace betwee them icreases; see, e.g., [1]. Secod Newto s law: does it mea aythig? By itself, the secod Newto s law ca be simply viewed as a defiitio of the force: oce we kow how to defie masses, we ca the defie the force F as the product m a. Thus, o matter how bodies move, the secod law is always satisfied, if we simply take F def = m a. From this viewpoit, the secod law does ot tell us aythig at all. Okey, there is a implicit assumptio of determiism here, that if we place the same bodies at same locatios with same iitial velocities, the we will observe the same acceleratios, but from the secod law itself, we caot coclude aythig beyod that. Commets. It is also usually implicitly assumed that a fiite umber of parameters is sufficiet to describe a body, its positio, velocity, ad orietatio, ad 2

that oce we kow the values of all these parameters, we ca uiquely determie all the forces. Determiism is what distiguishes Newto s mechaics from quatum physics, where we ca oly predict probabilities of differet measuremet results, but ot the measuremet results themselves. What if we also take ito accout the third law? If we also take the third law ito accout, the the situatio chages. Literally, the third law says that for every two bodies A ad B, the force F A B with which the body B acts o the body A ad the force F B A describig the ifluece of the body A o the body B are related by the formula F B A = F A B. If we substitute the defiitio F = m a ito this formula, we coclude that i the situatio whe we oly have two bodies A ad B, the followig is true: m B a B A = m A a A B, where m A ad m B are the masses of the bodies A ad B, ad a A B ad a B A are their acceleratios. We still do ot have a operatioal defiitio of mass, so the above rule ca be reformulated as follows: it is possible to assig, to every body A, a umber m A so that i every situatio i which there are oly two bodies A ad B, we have m B a B A = m A a A B. How ca check this possibility experimetally? How to check the third law? Oe thig we ca check right away: that the vectors a A B ad a B A have the same directio. Sice these two vectors have the same directio, we ca defie their ratio r A B def = a B A a B A as a real umber for which a B A = r A B a A B. Accordig to the above formula, this (observable) ratio has the form r A B = m A m B. So, the questio of how to reformulate the third law i operatioal terms ca be described as follows: for every two bodies A ad B, we ca experimetally determie the ratios r A B ; we wat to check whether there exist values m A for which r A B = m A m B for all pairs (A, B). Oe ca easily see that if such values m A exist, the for every three bodies A, B, ad C, we have r A C = r A B r B C. Vice versa, if this property is satisfied, the we ca fid appropriate m A : for example, we ca fix some object A 0 ad the take m A = r A A0. Ideed, i this case, for C = A 0, we have r A A0 = r A B r B A0, i.e., m A = r A B m B ad thus, r A B = m A. m B Commet. A additioal implicit assumptio behid Newto s physics is that i geeral, the body mass does ot chage with time. To be more precise, it may chage e.g., for a rocket flyig to the Moo but this is because the 3

origial rocket cosisted of two parts: the rocket itself ad the fuel. Each part retais its mass, but the parts become separated as the fuel flies away. This costacy of mass is what separates Newto s mechaics from special relativity, where a body s mass chages with the body s speed v as m 0 m =, where c is the speed of light. 1 v2 c 2 Straightforward iterpretatio of Newto s laws. Thus, the straightforward iterpretatio of Newto s laws is as follows. If the body A is the oly body i the world, the its acceleratio is equal to 0: a A = 0. For every body A, its acceleratio a A is uiquely determied by the positios, velocities, ad orietatios of this body A ad of all other bodies. Let a A B deote the acceleratio of the body A i the situatio whe the oly other body preset is body B. I this case: for every two bodies A ad B, the vectors a A B ad a B A are colliear, i.e., a B A = r A B a A b for some scalar r A B ; for every three bodies A, B, ad C, we have r A C = r A B r B C. 3 Reformulatig Newto s Laws i Operatioal Terms: A Additioal Property Additivity of Forces The above reformulatio is rather weak. Oe ca see that i this reformulatio, the first ad the third laws are meaigful, while the secod law which is usually portrayed as the mai law of Newto s physics practically disappears: it is reduced simply to determiism. So, how did Newto make predictios? If this is the case, if the secod law does ot have ay serious meaig, the how come Newto succeeded i gettig so may observable predictios out of his laws? Yes, he used a specific formula for the gravity force, but this is ot sufficiet: this would be sufficiet for situatios whe we have oly two bodies, but Newto also aalyzed situatios with three or more bodies. How did he do it? Eter additivity of forces. I his aalysis, Newto also also used aother property, a property which he did ot explicitly formulate as oe of his laws, but which is very importat for makig predictios: the implicit property of additivity of forces. Namely, he assumes that i the presece of several bodies, a force actig o a give body A is equal to the sum of the forces comig from all these bodies. 4

I precise terms, the force F A B,...,C that bodies B,..., C exert o body A is equal to the sum of the forces F A B,..., F A C that the body A would experiece i the presece of oly oe other body B,..., or C: F A B,...,C = F A B +... + F A C. Commet. This additivity property is sometimes explicitly metioed as a importat part of the secod Newto s Law for example, it is listed as such o the Wikipedia page o Newto s laws but Newto ever explicitly formulated this property. Let us reformulate additivity i operatioal terms. Accordig to the secod Newto s law, FA B,...,C = m A a A B,...,C, FA B = m A a A B,..., F A C = m A a A C, where a are correspodig acceleratios. Substitutig these expressios ito the above formula ad dividig both sides by the commo factor m A, we get the desired reformulatio. Operatioal reformulatio of additivity of forces. The acceleratio a A B,...,C that bodies B,..., C exert o body A is equal to the sum of the acceleratios a A B,..., a A C that the body A would experiece i the presece of oly oe other body B,..., or C: a A B,...,C = a A B +... + a A C. Commet. I the appedix, we describe how to tell whe a fuctio of may variables ca be represeted as a sum of such pairwise expressios. 4 What We Ca Coclude Based o Additivity of Forces First coclusio: mometum is preserved. For each body A, due to additivity, we have m A A = F A B. If we add up all both sides correspodig B A to all the bodies A, we will be able to coclude that m A a A = F A B. A A B I the right-had side of this formula, each pair of objects (A, B) occurs twice: as F A B ad as F B A. Due to the third Newto s law, F A B + F B A = 0. Thus, F A B = 0 ad, therefore, m A a A = 0. Each acceleratio a A is a time A A B derivative of the correspodig velocity v A. Thus, ( ) d m A v A = 0. dt A 5

I other words, the mometum A m A v A does ot chage with time. Commets. A alterative derivatio of the mometum preservatio property is give i [2]. It is worth metioig that the mometum is preserved i special relativity as well, the differece is that i special relativity theory, as we have metioed earlier, the mass chages whe velocity chages. Secod coclusio: additivity of mass. Let us assume that we have two bodies A ad B travellig together, with the same acceleratio a. We ca view this situatio i two differet ways: as two differet bodies A ad B each travellig with the acceleratio a, or as a sigle composite body AB travellig with a acceleratio a. I the first case, for the body A, the secod Newto s law has the form m A a = F A, where, due to additivity, the force F is the sum of two compoets: the force F A B comig from the body B ad the force F A X comig from all other bodies X: m A a = F A B + F A X. Similarly, we have m B a = F B A + F B X. By addig these two formulas ad by takig ito accout that, due to the third Newto s law, F A B + F B A = 0, we coclude that (m A + m B ) a = F A X + F B X. O the other had, i the secod iterpretatio, we have a sigle composite body AB of some mass m AB which is acceleratig due to forces F A X ad F B X actig o this composite body. Due to additivity of forces, the overall, force actig o the composite body AB is equal to F A X + F B X. Thus, for this composite body, the secod Newto s Law takes the form m AB a = F A X + F B X. By comparig the formulas correspodig to the two possible iterpretatio of this situatio, we coclude that m AB = m A + m B. I other words, mass is additive i the sese that the mass of the composite body is equal to the sum of the masses of its compoets. Commet. This argumet is similar to the oe give i [2]. Derivig laws of gravity (almost). Sice oe of the mai origial successes of Newto s physics was the descriptio of the motio caused by the gravitatioal forces, it is worth metioed that the formula for the gravitatioal force ca be almost uiquely determied based o additivity. 6

Ideed, the value of the gravitatioal force F A B, by defiitio, is determied oly by the masses of the bodies m A ad m B ad by the mutual locatio r of these two bodies: FA B = F (m A, m B, r) for some vector-valued fuctio F. If the body B cosists of two parts B 1 ad B 2 of masses, correspodigly, m 1 ad m 2, the we ca view this situatio i two differet ways: we ca treat the body B as two differet bodies B 1 ad B 2 each affectig the body A, or we ca treat B as a sigle body affectig the body A. I the first case, due to additivity of forces, the force actig o the body A is equal to the sum F A B = F A B1 + F A B2 = F (m A, m 1, r) + F (m A, m 2, r). O the other had, i the secod iterpretatio, due to the additivity of masses m B = m 1 + m 2, this same force has the form F A B = F (m A, m B, r) = F (m A, m 1 + m 2, r). By comparig the formulas correspodig to the two possible iterpretatio of this situatio, we coclude that F (m A, m 1 + m 2, r) = F (m A, m 1, r) + F (m A, m 2, r). I other words, for every m A ad r ad for each spatial compoet i, the fuctio f(m) def = F i (m A, m, r) satisfies the additivity property f(m 1 + m 2 ) = f(m 1 ) + f(m 2 ). Oe ca easily see that the oly cotiuous fuctio with this property is a fuctio f(m) = k m, where k def = f(1). Ideed, this is trivially true for m = 1. For m = 1, we have so f(1) = f so f(1) = f ad f For ratioal m = p, we have ( p f(m) = f = f ) ( p f(m) = f = p f ) +... + f = 1 f(1) = k m. +... + f ( times), (p times), = p 1 k = p k = k m. 7

Sice every real umber ca be represeted as a limit of ratioal umbers, ad f(m) = k m for all ratioal umbers, cotiuity implies that f(m) = k m for all values m. Thus, F (m A, m B, r) = f(m A, r) m B, where we deoted f(m A, r) def = F (m A, 1, r). Similarly, if the body A cosists of two parts A 1 ad A 2 with masses m 1 ad m 2, the we ca view this situatio i two differet ways: as two differet bodies A 1 ad A 2 both affected by B, or as a sigle composite body A affected by the body B. I the first case, due to the additivity of forces, the overall force actig o the body A is equal to F (m 1, m B, r) + F (m 2, m B, r). I the secod case, this force is equal to F (m A, m B, r) = F (m 1 + m 2, m B, r). By comparig these two expressios for the same force, we coclude that F (m 1 + m 2, m B, r) = F (m 1, m B, r) + F (m 2, m B, r). Substitutig the expressio F (m A, m B, r) = f(m A, r) m B ito this formula ad dividig both sides of the resultig equality by m B, we coclude that f(m 1 + m 2, r) = f(m 1, r) + f(m 2, r). Thus, similar argumets lead to f(m A, r) = m A g( r). Hece, for some fuctio g( r). F (m A, m B, r) = m A m B g( r) Commet. It is worth metioig that for this formula, the first Newto s law is automatically satisfied: whe m B = 0, we have F = 0. Derivig laws of gravity (cot-d). If we require that this expressio be rotatio-ivariat, we ca the coclude that g( r) = r h(r) for some fuctio h(r), where r def = r is the distace betwee the two bodies. Commet. For this formula, the third Newto s law is also automatically satisfied, sice here, F B A = m B m A ( r) h(r) = F A B. Derivig laws of gravity (fial part). Fially, if we require that the depedece be scale-ivariat, i.e., that a re-scalig of distaces r λ r (e.g., chagig from meters to cetimeters) will lead to the same formula for the force, but maybe after a appropriate re-scalig of force. I precise terms, this meas that for every λ, there exists a value a(λ) for which h(λ r) = a(λ) h(r). If we first re-scale by a factor of λ 1 (i.e., go from r to r = λ 1 r), ad the by a factor of λ 2 (i.e., go from r to r = λ 2 r = λ r, where λ def = λ 1 λ 2 ), the we get h(λ r) = h(λ 2 (λ 1 r)) = a(λ 2 ) h(λ 1 r) = a(λ 2 ) a(λ 1 ) h(r). 8

O the other had, we have h(λ r) = a(λ) h(r). By comparig these two formulas, we coclude that a(λ) = a(λ 1 λ 2 ) = a(λ 1 ) a(λ 2 ). This equatio is similar to the oe that we had before, except that ow we have multiplicatios istead of additios. We ca use l(x) to reduce multiplicatio to additio. By takig logarithms of both sides, we get l(λ 1 λ 2 ) = l(λ 1 ) + l(λ 2 ), where we deoted l(x) def = l(a(x)). For the fuctio A(X) def = l(exp(x)) = l(a(exp(x))), we have l(x) = A(l(x)), so the above formula takes the form A(l(λ 1 λ 2 )) = A(l(λ 1 )) + A(l(λ 2 )). Here, l(λ 1 λ 2 ) = x 1 +x 2, where x i def = l(λ i ), so the formula takes the additivity form A(x 1 + x 2 ) = A(x 1 ) + A(x 2 ). We already kow that i this case, A(x) = α x for some x. Thus, l(x) = A(l(x)) = α l(x), ad a(x) = exp(l(x)) = exp(α l(x)) = x α. Now, from h(λ r) = a(λ) h(r), whe r = 1, we get h(x) = h(1) x α, i.e., h(r) = C r α for some values C ad α. Therefore, F (m A, m B, r) = C m A m B r r α. Commets. This is almost Newto s law describig gravity. To get exactly the Newto s law, we eed to specify α = 3. Similarly, if we defie electrostatic forces as depedig oly o the additive charges q A ad q B, the we get F A B = D q A q B r r β for some values D ad β. Ackowledgmets. This work was supported i part by the Natioal Sciece Foudatio grats HRD-0734825 ad HRD-1242122 (Cyber-ShARE Ceter of Excellece), ad DUE-0926721. The authors are thakful to all the participats of the IEEE Symposium o Computatioal Itelligece for Egieerig Solutios CIES 2014 (Orlado, Florida, December 9 12, 2014) for valuable discussios. Refereces [1] R. Feyma, R. Leighto, ad M. Sads, The Feyma Lectures o Physics, Addiso Wesley, Bosto, Massachusetts, 2005. 9

[2] E. Freudethal, E. Hagedor, ad O. Kosheleva, Coservatio of eergy implies coservatio of mometum: how we ca explai coservatio of mometum to before-calculus studets, Joural of Ucertai Systems, 2014, Vol. 8, No. 3, pp. 169 172. [3] Galileo Galilei, Dialogue Cocerig the Two Chief World Systems, Eglish traslatio of the 1632 book, Moder Library, New York, 2001. [4] V. Kreiovich, Astroomical tests of Relativity: beyod Parameterized Post-Newtoia Formalism (PPN), to testig fudametal priciples, I: S. Klioer, P. K. Seidelma, ad M. H. Soffel (eds.), Relativity i Fudametal Astroomy, Proceedigs of IAU Symposium No. 261, Cambridge Uiversity Press, Cambridge, UK, 2009, pp. 56 61. [5] I. Netwo, The Pricipia: Mathematical Priciples of Natural Philosophy, Eglish traslatio of 1687 book, Uiversity of Califoria Press, Berkeley, Califoria, 1999. A Formalizig Additivity of Forces: How to Tell Whe a Fuctio of Several Variables is Equal to the Sum of Pairwise Fuctios Defiitio. Assume that the itegers from 1 to are divided ito several groups A,..., B. For a tuple x 1,..., x ad for a group A, by x A, we deote a sub-tuple cosistig of all the values x i with i A. We say that a fuctio f(x 1,..., x ) is a sum of pairwise fuctios if f(x 1,..., x ) = A,B f AB (x A, x B ) for some fuctios f AB. Propositio. Whe a fuctio f is three times differetiable, the f is a sum 3 f of pairwise fuctios if ad oly if = 0 wheever i, j, ad k belog x i x j x k to differet groups. Proof. Let us first prove that if f is a sum of pairwise fuctios, the the correspodig third order derivatives are equal to 0. Without losig geerality, let us assume that i A, j B, ad k C. Let us first differetiate the fuctio f with respect to x i ad x j. The derivative of the sum is equal to the sum of the derivatives. Of all the pairwise terms formig f, oly the term f AB (x A, x B ) ca deped both o x i ad x j : all other terms either do ot deped o x i for i A 2 or do ot deped o x j for j B, ad thus, the secod derivatives of all x i x j other terms are equal to 0. Thus, 2 f x i x j = 2 f AB x i x j. The fuctio f AB depeds 10

oly o the variables x l with l A or l B. Thus, its secod derivative also oly depeds o these variables, ad caot deped o x k for k C (for which 3 f k A ad k B). So, ideed, = 0. x i x j x k Let us ow prove that, vice versa, if all the correspodig third derivatives of the fuctio f(x 1,..., x ) are equal to 0, the the fuctio f(x 1,..., x ) is a sum of pairwise fuctios. This proof is based o the fact that if we kow the partial derivative g, of a fuctio g(x 1,..., x ), the we ca represet the x 1 fuctio g(x 1,..., x ) as g(x 1, x 2,..., x ) = g(0, x 2,..., x ) + x1 0 g x 1 (t, x 2,..., x ) dt. Similarly, if we kow the partial derivatives with respect to x 1,..., x k, the we ca write g(x 1,..., x k, x k+1,..., x ) = g(0,..., 0, x k+1,..., x )+ (g(x 1, 0,..., 0, x k+1,..., x ) g(0, 0,..., 0, x k+1,..., x ))+ (g(x 1, x 2, 0,..., 0, x k+1,..., x ) g(x 1, 0, 0,..., 0, x k+1,..., x )) +... + (g(x 1,..., x k 1, x k, x k+1,..., x ) g(x 1,..., x k 1, 0, x k+1,..., x )) = g(0,..., 0, x k+1,..., x ) + x2 0 xk 0 x1 0 g x 1 (t, 0,..., 0, x k+1,..., x ) dt+ g x 2 (x 1, t, 0,..., 0, x k+1,..., x ) dt +... + g x k (x 1, x 2,..., x k 1, t, x k+1,..., x ) dt. x i x j We have already metioed that from the fact that f is a sum of pairwise fuctios, it follows that for all i A ad all j B, the partial derivative 2 f depeds oly o the variables x A ad x B. This secod partial derivative has the form g x i, where g def = f x j. Thus, we ca get the above itegral represetatio of the fuctio g = f x j. I this represetatio, the first term g(0,..., 0, x k+1,..., x ) does ot deped o the variables x A, while all other terms deped oly o x A ad x B. Thus, for every j B, we have f x j = f 1 (x B, x C,...) + f 2 (x A, x B ) for appropriate fuctios f 1 ad f 2. Now that we have this iformatio about the partial derivatives of the fuctio f with respect to variables x B, we ca apply the itegral formula oce agai ad get f(x A, x B, x C,...) = F 1 (x A, x C,...) + F 2 (x B, x C,...) + F 3 (x A, x B ) 11

for appropriate fuctios F i. Whe we oly have three groups of variables, we have the desired represetatio of the fuctio f as a sum of pairwise fuctios. Whe we have more tha three groups of variables, we ca cotiue our decompositio. For the fuctios F 2 ad F 3, the secod order derivatives with 2 f respect to x A ad x C are equal to 0, so = 2 F 1. The left-had x A x C x A x C side depeds oly o x A ad x C, thus the right-had side also oly depeds o x A ad x C. Thus, similarly to the above, we ca coclude that F 1 (x A, x C, x D,...) = F 11 (x A, x D,...) + F 12 (x C, x D,...) + F 13 (x A, x C ). A similar represetatio is possible for F 2, so we have f(x A, x B, x C, x D,...) = F 11 (x A, x D,...) + F 12 (x C, x D,...) + F 13 (x A, x C )+ F 21 (x B, x D,...) + F 22 (x C, x D,...) + F 23 (x B, x C ) + F 3 (x A, x B ). By combiig F 12 ad F 22 together ito a sigle fuctio F 4, we get f(x A, x B, x C, x D,...) = F 11 (x A, x D,...) + F 4 (x C, x D,...) + F 13 (x A, x C )+ F 21 (x B, x D,...) + F 23 (x B, x C ) + F 3 (x A, x B ). If we have four groups of variables, the the propositio is prove, otherwise we ca use the same reductio oce agai, etc. After each reductio, we have fuctios depedig o oe fewer groups of variables, so evetually, this reductio will stop ad we will get the desired represetatio. The propositio is prove. 12