Mechanics Physics 151

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1 Mechanics Physics 151 Lectue 5 Cental Foce Poblem (Chapte 3)

2 What We Did Last Time Intoduced Hamilton s Pinciple Action integal is stationay fo the actual path Deived Lagange s Equations Used calculus of vaiation Discussed consevation laws Genealized (conugate) momentum Symmety Invaiance Momentum consevation We ae almost done with the basic concepts One moe thing to cove

3 Goals fo Today Enegy consevation Define enegy function Subtle diffeence fom the Newtonian vesion Cental foce poblem Fist application Motion of a paticle unde a cental foce Simplify the poblem using angula momentum consevation Discuss qualitative behavio of the solution Use enegy consevation Distinguish bounded/unbounded obits Actual solution Thusday

4 Enegy Consevation Conside time deivative of Lagangian dl( q, q, t) L dq L dq L = + + dt q dt q dt t Using Lagange s equation one can deive d L L q L + = dt q t Conseved if Lagangian does not depend explicitly on t L d L = q dt q Define this as enegy function hqqt (,, )

5 Enegy Function? Does enegy function epesent the total enegy? Let s ty an easy example fist Single paticle moving along x axis mx L= V( x) h= mx L mx = + V( x) = T + V How geneal is this? L hqqt (,, ) q L q Total enegy

6 Enegy Function Suppose L can be witten as L( q, q, t) = L ( q, t) + L ( q, q, t) + L ( q, q, t) Tue in most cases of inteest Deivatives satisfy L q = q 1 L q 1 = L 1 1 st ode in q q L q nd ode in q = L L hqqt (,, ) q L= L L q L hqqt (,, ) q L q Eule s theoem

7 Enegy Function hqqt (,, ) = L L L= T V Enegy function equals to the total enegy T + V if T L and = V = L 1 st condition is satisfied if tansfomation fom i to q is time-independent nd condition holds if the potential is velocity-independent No fictions Fiction would dissipate enegy Let s look into the 1 st condition

8 Kinetic Enegy mi T = i i Using the chain ule This wouldn t wok if = ( q,..., q, t) because d i dt = ( q,..., q ) i i 1 n d = q dt i = q i i 1 n i q i q i + t Time-independent m m m = qq = qq i i i i i i i i k k i i k, q qk k, i q qk nd ode homogeneous No q

9 Enegy Consevation L hqqt (,, ) q L q Enegy function equals to the total enegy if Constaints ae time-independent Kinetic enegy T is nd ode homogeneous function of the velocities Potential V is velocity-independent Enegy function is conseved if Lagangian does not depend explicitly on time These ae estatement of the enegy consevation theoem in a moe geneal famewok Conditions ae clealy defined

10 Cental Foce Poblem Conside a paticle unde a cental foce Foce F paallel to Assume F is consevative V is function of if F is cental Such systems ae quite common Planet aound the Sun Satellite aound the Eath Electon aound a nucleus F = V() These examples assume the body at the cente is heavy and does not move O F m

11 Two-Body Poblem Conside two paticles without extenal foce 1 and elative to cente of mass Lagangian is L Motion of CoM ( m + m ) R m 1 i i = + i= 1 Motion of paticles aound CoM V() m 1 O R 1 CoM m Potential is function of = 1 Stong law of action and eaction m m ( ) 1 1 = = ( m1+ m) m1+ m mi i 1 mm 1 = i= 1 m1+ m ( )

12 Two-Body Cental Foce L ( m + m ) R 1 mm ( m + m ) 1 1 = + R is cyclic 1 V() CoM moves at a constant velocity Move O to CoM and foget about it m 1 O R CoM m L 1 mm ( m + m ) = 1 1 V() Relative motion of two paticles is identical to the motion of one paticle in a cental-foce potential mm 1 Reduced mass µ = o ( m m ) µ = m + m 1+ 1

13 Hydogen and Positonium Positonium is a bound state of a positon and an electon Simila to hydogen except m(p) >> m(e + ) Potential V() is identical Tun them into cental foce poblem µ = mm e e me positonium ( me + me) = µ = mm p e hydogen me ( m + m ) p Spectum of positonium identical to hydogen with m e m e / e e e + e p q V() =

14 Spheical Symmety Cental-foce system is spheically symmetic It can be otated aound any axis though the oigin Lagangian L= T( ) V( ) doesn t depend on the diection L= p=const Diection of L is fixed L by definition is always in a plane Angula momentum is conseved Choose pola coodinates Pola axis = diection of L = (, θ, ψ) = (, θ) Azimuth Zenith = 1/π L O

15 Moe Fomally Lagangian in pola coodinates θ is cyclic, but ψ is not We can choose the pola axis so that the initial condition is Now ψ is constant. We can foget about it = (, θ, ψ ) m L= T V = + + V d L L dt ψ ψ ( sin ψθ ψ ) ( ) ( sin cos ) = m ψ ψ ψθ = ψ = π, ψ = nd tem vanishes ψ =

16 Angula Momentum m L= T V = + V θ is cyclic. Conugate momentum p θ conseves Altenatively ( θ ) ( ) L = = = θ pθ m θ const l Aeal velocity Keple s nd law da dt 1 = θ = Tue fo any cental foce const Magnitude of angula momentum d da

17 Radial Motion m L = T V = + V ( θ ) ( ) Lagange s equation fo Deivative of V is the foce m = m θ + f () Centifugal foce Using the angula momentum l l m = + f () 3 m d dt V() ( m ) m θ + = Cental foce V() f() = l = We know how to integate this. But we also know what we ll get by integating this m θ

18 Enegy Consevation 1 E = T + V = m + + V = m + l + V = m ( θ ) ( ) ( ) const l = E V() m m One can solve this (in pinciple) by t d t = dt = = t() l E V() m m Then invet t() (t) Then calculate θ(t) by integating 1 st ode diffeential equation of θ = l m NB: This neve goes negative Done! (?)

19 Degees of Feedom A paticle has 3 degees of feedom Eqn of motion is nd ode diffeential 6 constants Each consevation law educes one diffeentiation By saying time-deivative equals zeo We used L and E 4 conseved quantities Left with constants of integation = and θ We don t have to use consevation laws It s ust easie than solving all of Lagange s equations

20 Qualitative Behavio Integating the adial motion isn t always easy Moe often impossible You can still tell geneal behavio by looking at E V () V() + l m Enegy E is conseved, and E V must be positive m = + V () m Plot V () and see how it intesects with E l = E V() m m Quasi potential including the centifugal foce = E V () > E > V ()

21 Invese-Squae Foce Conside an attactive 1/ foce k k f() = V() = Gavity o electostatic foce k l V () = + m 1/ foce dominates at lage Centifugal foce dominates at small A dip foms in the middle V () l m k

22 Unbounded Motion Take V simila to 1/ case Only geneal featues ae elevant E = E 1 > min E = V ( ) 1 min Paticle can go infinitely fa E 1 V () 1 m Aive fom = E Tuning point E = V = Go towad = E 3 A 1/ foce would make a hypebola

23 Bounded Motion E = E min < < max Paticle is confined between two cicles E 1 V () Goes back and foth between two adii E 1 m E 3 Obit may o may not be closed. (This one isn t) A 1/ foce would make an ellipse

24 Cicula Motion E = E 3 = (fixed) Only one adius is allowed Stays on a cicle Classification into unbounded, bounded and cicula motion depends on the geneal shape of V Not on the details (1/ o othewise) E = V ( ) = = const = E 1 E E 3 V ()

25 Anothe Example V a = f = 3 4 Attactive -4 foce V has a bump 3a Paticle with enegy E may be eithe bounded o unbounded, depending on the initial a V = + l m 3 E V l m V

26 Stable Cicula Obit Cicula obit occus at the bottom of a dip of V m = E V = Top of a bump woks in theoy, but it is unstable Initial condition must be exactly = and = = const Stable cicula obit equies dv m = = d dv d > E E stable unstable

27 Powe Law Foce V () V() + l m dv d = l = f( ) = 3 m dv df 3l = + > d d m 4 = = df d = < 3 f( ) Suppose the foce has a fom k > fo attactive foce Condition fo stable cicula obit is n 1 n 1 kn < 3k n > 3 f = k Powe-law foces with n > 3 can make stable cicula obit n

28 Summay Stated discussing Cental Foce Poblems Reduced -body poblem into cental foce poblem l m = + f () 3 m Poblem is educed to one equation Used angula momentum consevation l V () V() + m Unbounded, bounded, and cicula obits Condition fo stable cicula obits Qualitative behavio depends on Next step: Can we actually solve fo the obit?

Mechanics Physics 151

Mechanics Physics 151 Mechanics Physics 5 Lectue 5 Centa Foce Pobem (Chapte 3) What We Did Last Time Intoduced Hamiton s Pincipe Action intega is stationay fo the actua path Deived Lagange s Equations Used cacuus of vaiation