1 What is a conservation law?

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1 MATHEMATICS 7302 (Analytcal Dynamcs) YEAR , TERM 2 HANDOUT #6: MOMENTUM, ANGULAR MOMENTUM, AND ENERGY; CONSERVATION LAWS In ths handout we wll develop the concepts of momentum, angular momentum, and energy n Newtonan mechancs, and prove the fundamental denttes relatng them to force, torque, and work, respectvely. As a specal case we wll obtan the conservaton laws for momentum, angular momentum, and energy n solated systems. But let us frst clarfy what we mean by a conservaton law, and why conservaton laws are so mportant. 1 What s a conservaton law? In ordnary language, conservaton of X usually means please avod wastng X. In physcs, however, the word conservaton has a qute dfferent meanng: conservaton of X or X s conserved means that X s constant n tme,.e. dx/ = 0. More precsely, consder any physcal quantty Q that has, for each knematcally possble moton of a partcular physcal system, a defnte numercal (or vector) value at each nstant of tme. [Usually Q wll be some functon of the poston r and the velocty v = ṙ, as well as possbly havng an explct dependence on the tme t.] If for every dynamcally allowed moton of that system t happens that dq/ = 0, we then say that Q s conserved (or that Q s a constant of moton; we use the two terms synonymously) for that partcular physcal system and that partcular dynamcal law. Now, n some cases t turns out that the same quantty Q (or a very smlar quantty) s a constant of moton, not only for one or two obscure systems and dynamcal laws, but n fact for some broad and nterestng class of systems. We then ndulge ourselves and assert a conservaton theorem characterzng the stuatons n whch Q s conserved. (Indeed, much of analytcal dynamcs s devoted to answerng the queston: What can we say about physcal systems n general, wthout reference to detaled dynamcs?) Why are conservaton theorems, and more generally, conserved quanttes, so mportant and useful? Here are some reasons: 1) Conservaton theorems are general statements about the types of motons that a dynamcal law (or class of dynamcal laws) permts. In partcular, they gve mportant negatve nformaton: certan types of moton are forbdden (e.g. momentum-nonconservng collsons). 2) Conservaton theorems gve partal nformaton about the nature of a partcular moton, even f the equatons are too complcated to permt a full soluton (e.g. we can often use the conservaton of energy and/or angular momentum to fnd turnng ponts, maxmum heght reached, etc., even when we are unable to fnd explctly the full moton r(t)). We wll see lots of examples of ths! 1

2 3) Conservaton theorems ad n the full soluton of a partcular problem. A conserved quantty provdes a frst ntegral of the equatons of moton: sometmes ths s suffcent to essentally solve the problem (as n one-dmensonal systems wth poston-dependent forces); other tmes t can be used to decouple a set of ugly, coupled dfferental equatons (as n the central-force problem). We wll see lots of examples of ths knd too! 4) Conserved quanttes are ntmately connected wth symmetry propertes of the system, whch are n turn very mportant n ther own rght. We wll be able to make ths connecton more precse after we have developed the Lagrangan and Hamltonan formulatons of Newtonan mechancs. (The connecton between symmetres and conservaton laws holds also n quantum mechancs, and t forms n fact one of the central themes of modern physcs.) Let us stress that we cannot prove that momentum (for example) s conserved n the real world; ths has to be tested expermentally. What we can prove s that conservaton of momentum follows, under certan specfed conons, as a logcal consequence of our dynamcal axoms (e.g. Newton s second and thrd laws of moton); that s a theorem of pure mathematcs. It s n ths latter sense that I use the term conservaton theorem. Whether our dynamcal axoms are themselves true n the real world s, of course, a matter to be tested by experment; and one mportant ndrect method of testng those axoms s to test ther logcal consequences, such as the conservaton laws. 2 One partcle In the remander of ths handout we wll be consderng systems of partcles that obey the equatons of Newtonan mechancs. We begn by focussng on a sngle partcle (whch may be part of a larger system), and dervng the relatonshps between momentum and force, angular momentum and torque, and energy and work. Then, n Secton 3, we wll consder a system of n partcles as a whole. 2.1 Momentum and force If a partcle of mass m has velocty v, ts lnear momentum (or just momentum for short) s defned to be p = mv. (2.1) Trval calculus then gves the rate of change of momentum: dp = d(mv) = m dv = ma (2.2) where we have used the fact that m s a constant. (The mass of a partcle s assumed to be an nherent and mmutable characterstc of the partcle, and therefore constant n tme.) It follows that Newton s second law F = ma (where F s the net force actng on the partcle) can equvalently be re-expressed as: Force momentum theorem for a sngle partcle. The momentum of a partcle obeys dp/ = F, where F s the net force actng on the partcle. 2

3 Remark. In Ensten s specal relatvty, momentum s no longer p = mv; rather, t s p = mv/ 1 v 2 /c 2. Then dp/ s no longer equal to ma, so that F = dp/ s no longer equvalent to F = ma. It turns out that, n specal relatvty, F = ma s false but F = dp/ s true. As a specal case of the force momentum theorem we obtan: Conservaton of momentum for a sngle partcle. The momentum p s a constant of moton f and only f F = 0 at all tmes. Of course, the fact that F = 0 mples the constancy of p (or equvalently of v) s smply Newton s frst law. 1 Note also the weaker but broader result: Let e be any fxed (.e. constant) vector; then d dp (p e) = e = F e (2.3) so that f the component of the force n some fxed drecton e vanshes, then the component of momentum n that same drecton s a constant of moton. (Ths s useful, for nstance, n analyzng projectle moton near the Earth s surface, where the horzontal component of the force vanshes.) Thus far ths s all farly trval. 2.2 Angular momentum and torque If a partcle of mass m has poston r (relatve to the chosen orgn of coordnates) and velocty v, ts angular momentum s defned to be L = r p = mr v. (2.4) (Note that ths depends on the choce of orgn.) Let us now compute the rate of change of angular momentum by usng Newton s second law: dl = d ( dr ) ( (r p) = p + r dp ) = r F, (2.5) where n the fnal equalty we used the fact that dr p = v p = 0 snce v and p are collnear, and the just-derved fact (equvalent to Newton s second law) that dp = F. Ths suggests that we should defne the torque 2 We have therefore proven: τ = r F. (2.6) 1 Actually, ths s not qute so: Newton s frst law apples only to partcles subject to no force, whle the just-stated corollary of Newton s second law apples to partcles on whch the net force s zero, even f ths zero net force arses as the sum of two or more nonzero forces. 2 From the Latn torquere, to twst (compare modern Spansh torcer and modern Italan torcere). The torque s also sometmes called the moment of force a termnology that I fnd needlessly confusng. 3

4 Torque angular momentum theorem for a sngle partcle. The angular momentum of a partcle obeys dl/ = τ, where τ s the net torque actng on the partcle. As a specal case we obtan: Conservaton of angular momentum for a sngle partcle. The angular momentum L s a constant of moton f and only f τ = 0 at all tmes. Note also the weaker but broader result: Let e be any fxed vector; then d dl (L e) = e = τ e (2.7) so that f the component of the torque n some fxed drecton e vanshes, then the component of angular momentum n that same drecton s a constant of moton. 3 Moreover, f L s a constant of moton, then the partcle s path les entrely n some fxed plane through the orgn (except for one degenerate possblty). To show ths, we must consder two cases: Case 1: L 0. We have r L = r (r p) = 0 by the propertes of the cross product. Hence r les n the plane through the orgn that s perpendcular to the fxed drecton L. Case 2: L = 0. In ths case we have L = mr dr = 0, so that r and dr/ are parallel whenever they are nonzero. It follows that the moton les on some fxed lne passng through the orgn, except for one degenerate possblty: namely, f the partcle reaches zero velocty at the orgn (.e., r = 0 and dr/ = 0 occur smultaneously), then the partcle can re-emerge from the orgn along a dfferent lne. Cauton! All the theory developed here concerns the angular momentum, torque, etc. wth respect to a pont (the orgn) that s fxed wth respect to an nertal frame. Later (n Secton 3.2 below) we wll be a bt more general. 2.3 Energy and work If a partcle of mass m has velocty v, ts knetc energy s defned to be K = 1 2 mv2 = 1 mv v. (2.8) 2 We can then compute the rate of change of knetc energy by usng Newton s second law: dk = d (1 mv v) = mdv v = ma v = F v (2.9) 2 where F s the net force actng on the partcle. We call F v the power (or rate of dong work) done on the partcle by the net force. We have therefore proven: 3 Because of the geometrcal nterpretaton of angular momenta and torques, we often refer n ths context to the angular momentum and torque around some fxed axs e, as a synonym for n some fxed drecton e. 4

5 Work energy theorem for a sngle partcle. The knetc energy of a partcle obeys dk/ = F v, where F s the net force actng on the partcle. The work energy theorem can also be stated n ntegral form, as follows: The work done on the partcle by a force F durng an nfntesmal dsplacement r s by defnton F r; then the change n knetc energy between tme t 1 and tme t 2 equals the total work done between tme t 1 and tme t 2,.e. K(t 2 ) K(t 1 ) = snce v = dr. Here are two specal cases: t 2 t 1 dk = t 2 t 1 F dr = r(t 2 ) r(t 1 ) F dr (2.10) 1. If F = 0 (free partcle), then the knetc energy K s a constant of moton. Ths s, of course, trval, snce for a free partcle the velocty vector v (and not just ts magntude) s a constant of moton, by Newton s frst (or second) law. A slghtly more general (and less trval) case s: 2. If F v = 0 (.e., the force s always perpendcular to the velocty), then the knetc energy K s a constant of moton. Here are some cases where ths occurs: (a) For the magnetc force F mag = qv B (why?). (b) For the normal force assocated to a fxed (.e., non-movng) constrant, such as a partcle sldng frctonlessly on a fxed curve or a fxed surface: the normal force s perpendcular to the curve or surface, whle the velocty s tangental to the curve or surface. 4 (c) For crcular orbts n a central force (why?). The work energy theorem always holds, no matter the nature of the force. But f the force s conservatve, then we can rephrase the work energy theorem by a change of accountng : nstead of talkng about knetc energy and work, we can alternatvely talk about knetc energy and potental energy. Let us recall the defntons: A force law F = F(r,v, t) s called conservatve f (a) F depends only on the poston r, not on the velocty v or explctly on the tme t; and (b) the vector feld F(r) s conservatve,.e. there exsts a scalar feld U(r) such that F = U. 4 Notce, by contrast, that a movng constrant (such as a movng nclned plane or a rotatng wre) can do work, snce the object s velocty s no longer tangental to the curve or surface. 5

6 (In both MATH1302 and MATH1402 you have studed necessary and suffcent conons for a vector feld n R 3 to be conservatve.) In ths stuaton we call U(r) the potental energy assocated to the force feld F(r). Recall that U s unque up to an arbtrary adve constant, and that t can be defned by the lne ntegral r 1 U(r 1 ) = r 0 F dr (2.11) where r 0 s an arbtrarly chosen pont (the pont where the potental energy s defned to be zero) and the ntegral s taken over an arbtrary curve runnng from r 0 to r 1 (snce the vector feld F s conservatve, the lne ntegral takes the same value for all choces of such a curve). By the chan rule we have d dr U(r(t)) = ( U) (please make sure you understand n detal the reasonng here), and hence dk = F v = F dr It follows that the total energy = ( U) dr s a constant of moton. We have therefore proven: (2.12) = d U(r(t)). (2.13) E = K + U = 1 2 mv2 + U(r) (2.14) Conservaton of energy for a sngle partcle movng n a conservatve force feld. For a partcle of mass m subject to a conservatve force F = F(r) = ( U)(r), the total energy E = K + U = 1 2 mv2 + U(r) s a constant of moton. Of course, ths s nothng other than a rephrasng of the work energy theorem n whch we refer to potental energy rather than to work. 3 Systems of partcles Let us now consder a system of n partcles, whch we number from 1 to n. We denote by m the mass of the th partcle and by r,v,a the poston, velocty and acceleraton of the th partcle. Then p = m v s the momentum of the th partcle, K = 1m 2 v 2 s the knetc energy of the th partcle, and so forth. We defne the total mass n M = m (3.1) and the center-of-mass poston vector n m r r cm = =1 =1 = n m =1 6 n m r =1 M. (3.2)

7 We also defne the center-of-mass velocty vector v cm = dr cm / and the center-of-mass acceleraton vector a cm = d 2 r cm / 2. Fnally, we defne the poston of the th partcle relatve to the center of mass: r (cm) = r r cm (3.3) and the correspondng veloctes v (cm) = dr (cm) / and acceleratons a (cm) Note that the weghted sum of these relatve postons satsfes m r (cm) = = d 2 r (cm) / 2. m (r r cm ) = Mr cm Mr cm = 0. (3.4) The same therefore holds also for the weghted sum of the relatve veloctes or acceleratons (why?). We now make the followng fundamental assumpton about the nature of the forces actng on our partcles: The net force F actng on the th partcle s the vector sum of two-body forces exerted by the other partcles of the system, plus possbly an external force. That s, we assume that F j + F (ext) (3.5) F = j where F j s the force exerted on the th partcle by the jth partcle (we refer collectvely to all these forces as nternal forces), and F (ext) s the external force actng on the th partcle (.e., a force exerted by somethng outsde our system). We can also wrte F = j F j + F (ext) (3.6) (summng over all j nstead of just j ) f we make the conventon that F = 0. 5 We furthermore assume that the nternal forces obey Newton s thrd law: F j = F j for all pars j. (3.7) (Note that our conventon F = 0 s equvalent to sayng that ths relaton holds also for = j.) 3.1 Momentum and force The total momentum of the system s, by defnton, P = p (3.8) where p = m v s the momentum of the th partcle. 5 Whenever we wrte we mean, of course, n. =1 7

8 3.1.1 Knematc dentty We have a smple knematc dentty: P = p = m v = m dr = d ( ) m r = d (Mr cm) = Mv cm. (3.9) That s, the total momentum of the system s the same as the momentum that a sngle partcle would have f t were located at the center of mass r cm and had a mass equal to the total mass M. (Ths s good: t s what justfes treatng composte partcles, such as the Earth, for some purposes as f they were pont masses.) Dynamcal theorems Let us now consder Newtonan dynamcs. As we have seen, Newton s second law for the th partcle can be expressed as dp = F = j F j + F (ext). (3.10) Let us now sum these equatons over : we obtan dp = dp =,j F j + F (ext). (3.11) But,j F j = 0 by Newton s thrd law (3.7) because the forces cancel n pars. (You should make sure that you understand ths reasonng; t mght be useful to wrte out explctly the cases n = 2 and n = 3.) We have therefore proven: Force momentum theorem for a system of partcles. The total momentum of the system obeys dp/ = F (ext), where F (ext) = F (ext) s the total external force actng on the system. Snce by the knematc dentty (3.9) we have dp/ = Ma cm, the force momentum theorem can equvalently be rephrased as: Center-of-mass theorem for a system of partcles. The center of mass of the system moves as f t were a sngle partcle whose mass equals the total mass M and whch s acted on by a force equal to the total external force F (ext) : that s, F (ext) = Ma cm. (Once agan, ths s what justfes treatng a composte partcle, such as the Earth, for some purposes as f t were a pont mass.) As an mportant specal case we obtan: Conservaton of momentum for a system of partcles. The total momentum P s a constant of moton f and only f the total external force F (ext) = s zero at all tmes. F (ext) 8

9 In partcular, an solated system (.e. one subject to no external forces) has F (ext) = 0 for all, so that we have: Conservaton of momentum for an solated system of partcles. The total momentum P of an solated system s a constant of moton. Ths holds no matter what the nternal forces are, provded only that they obey Newton s thrd law. Ths s therefore an extremely general and mportant result. 3.2 Angular momentum and torque Let us now be a bt more general than we were prevously: nstead of consderng angular momenta and torques wth respect to the orgn only, let us consder them wth respect to an arbtrarly movng pont Q whose poston s gven by r Q (t). We therefore defne the poston of the th partcle wth respect to Q, r (Q) = r r Q, (3.12) and the angular momentum of the th partcle wth respect to Q, L (Q) = m r (Q) dr(q). (3.13) [Note that L (Q) s not n general equal to r (Q) p do you see why?] The total angular momentum of the system (wth respect to Q) s then L (Q) = L (Q). (3.14) Knematc dentty The total angular momentum L (Q) admts a very smple decomposton: L (Q) = m r (Q) dr(q) (3.15a) = m (r (cm) ( (cm) dr + r cm (Q) ) ) + dr(q) cm (3.15b) = m r (cm) dr(cm) + ( ) m r (cm) dr(q) cm ( + r cm (Q) dr (cm) ) ( ) m + m r (Q) cm dr(q) cm (3.15c) = L (cm) + Mr (Q) cm dr(q) cm (3.15d) 9

10 where the mddle two terms n (3.15c) vansh by vrtue of (3.4). Therefore, the total angular momentum about Q s the sum of two terms: the total angular momentum of the system about ts center of mass, plus the angular momentum about Q that the total mass would have f t were concentrated at the center of mass Dynamcal theorems We defne the torque on the th partcle wth respect to Q: τ (Q) Usng (3.6) ths decomposes nto an nternal and an external part: = r (Q) F. (3.16) τ (Q) = τ (Q)(nt) + τ (Q)(ext) (3.17) where τ (Q)(nt) = r (Q) j F j (3.18) and τ (Q)(ext) The total external torque s, by defnton, = r (Q) F (ext). (3.19) τ (Q)(ext) = τ (Q)(ext) = r (Q) F (ext). (3.20) The total nternal torque s, by defnton, τ (Q)(nt) = τ (Q)(nt) =,j r (Q) F j. (3.21) Usng Newton s thrd law F j = F j, we can equvalently rewrte ths as τ (Q)(nt) =,j r (Q) ( F j ) =,j r (Q) j ( F j ) (3.22) where n the second equalty we have smply nterchanged the summaton labels and j. Takng the half-sum of (3.21) and (3.22), we obtan τ (Q)(nt) = 1 (r (Q) 2 r (Q) j ) F j (3.23a) From (3.23b) we conclude two thngs: = 1 2,j (r r j ) F j. (3.23b),j 1. The total nternal torque s ndependent of the choce of the reference pont Q. (Ths s a very strong fact, gven that we have allowed Q to move n a totally arbtrary way. It s a consequence of Newton s thrd law.) 10

11 2. If the strong form of Newton s thrd law holds recall that ths says that the force F j s drected along the lne jonng to j we have (r r j ) F j = 0 and hence the total nternal torque vanshes. Let us now evaluate the tme dervatve of the angular momentum of the th partcle. Snce we have dl (Q) L (Q) = m r (Q) d ( = m r (Q) ( (Q) dr = m = m r (Q) dr(q) ) dr(q) dr(q) d2 r (Q) 2 + r (Q), (3.24) d2 r (Q) ) 2 (3.25a) (3.25b) (3.25c) = m r (Q) (a a Q ) (3.25d) = r (Q) F m r (Q) a Q (3.25e) = τ (Q) m (r r Q ) a Q (3.25f) where we have wrtten a Q = d 2 r Q / 2. Summng ths now over we obtan the rate of change of the total angular momentum: dl (Q) = τ (Q)(ext) + τ (Q)(nt) m (r r Q ) a Q (3.26a) = τ (Q)(ext) + τ (Q)(nt) M (r cm r Q ) a Q [why?] (3.26b) = τ (Q)(ext) M (r cm r Q ) a Q (3.26c) where n the last step we assumed the valy of the strong form of Newton s thrd law (so that the total nternal torque s zero). The dentty (3.26) holds for an arbtrary moton of the reference pont Q, but because of the last term (the one nvolvng a Q ) t s not very useful n general. However, f the acceleraton vector a Q ponts along the lne from Q to the center of mass, then ths last term vanshes (why?). In partcular ths happens f or (a) Q s unaccelerated (wth respect to an nertal frame), so that a Q = 0 (b) Q s the center of mass, so that r cm r Q = 0. (Luckly, these nce stuatons are the ones occurrng most often n practce.) We have therefore proven: 11

12 Torque angular momentum theorem for a system of partcles (assumng the strong form of Newton s thrd law). If the acceleraton vector a Q ponts along the lne from Q to the center of mass, then the total angular momentum of the system obeys dl (Q) / = τ (Q)(ext), where τ (Q)(ext) s the total external torque on the system. As a specal case we obtan: Conservaton of angular momentum for system of partcles (assumng the strong form of Newton s thrd law). If the acceleraton vector a Q ponts along the lne from Q to the center of mass, then the total angular momentum L (Q) s a constant of moton f and only f the total external torque τ (Q)(ext) s zero at all tmes. In partcular, for an solated system (.e. one subject to no external forces) we obtan: Conservaton of angular momentum for an solated system of partcles (assumng the strong form of Newton s thrd law). If the acceleraton vector a Q ponts along the lne from Q to the center of mass, then the total angular momentum L (Q) of an solated system s a constant of moton. Ths holds no matter what the nternal forces are, provded only that they obey the strong form of Newton s thrd law. Ths s therefore an extremely general and mportant result. 3.3 Energy and work Fnally, let us consder the relatons nvolvng energy and work for a system of partcles Knematc dentty The total knetc energy of the system s, by defnton, the sum of the knetc energes of the ndvdual partcles: K = K = 1 2 m v 2 = 1 m 2 v v. (3.27) Let us rewrte ths n terms of the center of mass, as follows: K = 1 m 2 v v = 1 2 = 1 2 (3.28a) m (v cm + v (cm) ) (v cm + v (cm) ) (3.28b) m (v 2 cm + 2v cm v (cm) ( ) = 1 2 Mv2 cm + v cm m v (cm) = 1 2 Mv2 cm m v (cm) v (cm) 2 ) (3.28c) m v (cm) 2 (3.28d) (3.28e)

13 where the mddle term n (3.28d) vanshes because of (3.4). Thus, the total knetc energy can be nterpreted as the sum of two terms: the knetc energy of the moton of the center of mass, and the sum of the knetc energes of the ndvdual partcles wth respect to the center of mass Dynamcal theorems Let us now consder Newtonan dynamcs. For each partcle, the work energy theorem stll holds: dk ( ) = F v = F j + F (ext) v. (3.29) j In order to express ths n terms of potental energes, let us assume that both the external and the nternal forces are conservatve. For the external forces, ths means that for each there s a potental energy U (ext) such that F (ext) = ( U (ext) )(r ). (3.30) For the nternal forces, ths means that for each par {, j} of dstnct partcles there s a potental energy U {,j} (r,r j ) such that F j = U {,j} (r,r j ) F j = j U {,j} (r,r j ) (3.31a) (3.31b) where means the gradent wth respect to r when r j s held fxed, and j means the reverse. Remark. Usually U {,j} (r,r j ) s just a functon of the nter-partcle separaton vector r r j (ths expresses the translaton-nvarance of the potental energy,.e. the fact that t does not depend on the choce of orgn of coordnates). Note that f U {,j} (r,r j ) s of ths form, then Newton s thrd law F j = F j holds do you see why? Indeed, most often U {,j} (r,r j ) s just a functon of the nter-partcle dstance r r j (ths expresses the translaton-nvarance and rotaton-nvarance of the potental energy,.e. the fact that t does not depend on the choce of orgn of coordnates or the orentaton of the coordnate axes). Note that f U {,j} (r,r j ) s of ths form, then the strong form of Newton s thrd law holds do you see why? And recall, fnally, that Newton s thrd law mples the conservaton of momentum, whle the strong form of Newton s thrd law mples the conservaton of angular momentum. Here we have just seen a frst hnt of the deep connecton between symmetres (= nvarances) and conservaton laws, whch plays a central role n modern physcs and to whch we wll return later n ths course when we study the Lagrangan and Hamltonan formulatons of Newtonan mechancs. It follows that f we defne the total potental energy (external plus nternal) to be U(r 1,...,r n ) = n =1 U (ext) (r ) <j n U {,j} (r,r j ), (3.32)

14 we have F = U(r 1,...,r n ) (3.33) (why?). Summng the work energy theorem (3.29) over and nsertng (3.33), we obtan dk = v U(r 1,...,r n ) = by the chan rule. We have therefore proven: dr U(r 1,...,r n ) = d U(r 1(t),...,r n (t)) Conservaton of energy for a system of partcles. For a system of partcles subject to conservatve nternal and external forces, the total energy E = K + U = 1 m 2 v 2 + U(r 1,...,r n ) (3.35) s a constant of moton. (3.34) Ths s wonderful but please note that both the nternal and external potental energes must be ncluded n U. Alas, the nternal potental energy s often naccessble n practce unless you know the detals of the nternal dynamcs (consder, for nstance, a gas consstng of nteractng molecules). From the pont of vew of the external forces alone, energy sometmes seems to dsappear untl you realze that t went nto nternal potental energy (as e.g. n nelastc collsons). 6 Note, fnally, that the conservaton-of-energy theorem holds whenever (3.33) holds; t s not necessary for the potental energy to have the specal form (3.32) nvolvng external forces plus two-body forces. (For nstance, mult-body forces are also allowable.) 6 Feynman gves a brllant explanaton of ths pont n vol. I, secton 4.1 of the Feynman Lectures: see 04.html 14