Mechanics Physics 151 Lectue 5 Cental Foce Poblem (Chapte 3)
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
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
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 (,, )
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
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
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
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
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
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
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 ( )
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 ) 1 1 1 µ = m + m 1+ 1
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() =
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
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 =
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
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
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! (?)
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
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 ()
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
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
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
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 ()
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
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
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
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?