Chapter 6
Lagrangian Mechanics
6.1Generalized Coordinates
A t of generalized coordinates q 1,...,q n completely describes the positions of all particles in a mechanical system.In a system with d f degrees of freedom and k constraints,n =d f −k independent generalized coordinates are needed to completely specify all the positions.A constraint is a relation among coordinates,such as x 2+y 2+z 2=a 2for a particle moving on a sphere of radius a .In this ca,d f =3and k =1.In this ca,we could eliminate z in favor of x and y ,i.e.by writing z =± a 2−x 2−y 2,or we could choo as coordinates the polar and azimuthal angles θand φ.
For the moment we will assume that n =d f −k ,and that the generalized coordinates are independent,satisfying no additional constraints among them.Later on we will learn how to deal with any remaining constraints among the {q 1,...,q n }.
The generalized coordinates may have units of length,or angle,or perhaps something totally different.In the theory of small oscillations,the normal coordinates are convention-ally chon to have units of (mas
s)1/2×(length).However,once a choice of generalized coordinate is made,with a concomitant t of units,the units of the conjugate momentum and force are determined:
p σ =ML 2T ·1 q σ , F σ =ML 2T 2·1 q σ ,(6.1)
where A means ‘the units of A ’,and where M ,L ,and T stand for mass,length,and time,respectively.Thus,if q σhas dimensions of length,then p σhas dimensions of momentum and F σhas dimensions of force.If q σis dimensionless,as is the ca for an angle,p σhas dimensions of angular momentum (ML 2/T )and F σhas dimensions of torque (ML 2/T 2).
1
2CHAPTER 6.LAGRANGIAN MECHANICS
6.2Hamilton’s Principle
The equations of motion of classical mechanics are embodied in a variational principle,called Hamilton’s principle.Hamilton’s principle states that the motion of a system is such that the action functional S q (t ) =t 2 t 1
论文翻译公司dt L (q,˙q,
t )(6.2)is an δS =0.Here,q ={q 1,...,q n }is a complete t of generalized coordinates for our mechanical system,and
L =T −U (6.3)
is the Lagrangian ,where T is the kinetic energy and U is the potential energy.Setting the first variation of the action to zero gives the Euler-Lagrange equations,d dt momentum p σ ∂L ∂˙q σ =force F σ ∂L ∂q σ.(6.4)
Thus,we have the familiar ˙p σ=F σ,also known as Newton’s cond law.Note,however,that the {q σ}are generalized coordinates ,so p σmay not have dimensions of momentum,nor F σof force.For example,if the generalized coordinate in question is an angle φ,then the corresponding generalized momentum is the angular momentum about the axis of φ’s rotation,and the generalized force is the torque.
6.2.1Momentum conrvation
Whenever L is independent of a generalized coordinate q σ,the conjugate force F σ=∂L ∂q σvanishes and therefore the conjugate momentum p σ=∂L ∂˙q σis conrved.This is an example of a deep result known as Noether’s theorem which we will explore more fully next week.
Noether’s theorem guarantees that to every continuous symmetry of L there corresponds an associated conrved quantity.
6.2.2
Invariance of the equations of motion Suppo ˜L (q,˙q,t )=L (q,˙q,t )+d dt
G (q,t ).(6.5)Then
˜S [q (t )]=S [q (t )]+G (q b ,t b )−G (q a ,t a ).(6.6)Since the difference ˜S −S is a function only of the endpoint values {q a ,q b },their variations are identical:δ˜S =δS .This means that L and ˜L result in the same equations of motion.
6.2.HAMILTON’S PRINCIPLE 3Thus,the equations of motion are invariant under a shift of L by a total time derivative of a function of coordinates and time.
6.2.3Remarks on the order of the equations of motion
The equations of motion are cond order in time.This follows from the fact that L =L (q,˙q,t ).Using the chain rule,d dt ∂L ∂˙q σ =∂2L ∂˙q σ∂˙q σ ¨q σ +∂2L ∂˙q σ∂q σ ˙q σ +∂2L ∂˙q σ∂t
.(6.7)That the equations are cond order in time can be regarded as an empirical fact.It follows,as we have just en,from the fact that L depends on q and on ˙q ,but on no higher time derivative terms.Suppo the Lagrangian did depend on the generalized accelerations ¨q as well.What would the equations of motion look like?
Taking the variation of S ,
δt b t a
dt L (q,˙q,¨q ,t )= ∂L ∂˙q σδq σ+∂L ∂¨q σδ˙q σ−d dt ∂L ∂¨q σ δq σ t b t a +t b t a dt ∂L ∂q σ−d dt ∂L ∂˙q σ +d 2dt 2 ∂L ∂¨q σ δq σ.(6.8)
The boundary term vanishes if we require δq σ(t a )=δq σ(t b )=δ˙q σ(t a )=δ˙q σ(t b )=0∀σ.The equations of motion would then be fourth order in time.
6.2.4Lagrangian for a free particle
For a free particle,we can u Cartesian coordinates for each particle as our system of generalized coordinates.For a single particle,the Lagrangian L (x ,v ,t )must be a function solely of v 2.This is becau homogeneity with respect to space and time preclude any dependence of L on x or on t ,an
d isotropy of space means L must depend on v 2.We next invoke Galilean relativity,which says that the equations of motion are invariant under transformation to a reference frame moving with constant velocity.Let V be the velocity of the new reference frame K relative to our initial reference frame K .Then x =x −V t ,and v =v −V .In order that the equations of motion be invariant under the change in reference frame,we demand
L (v )=L (v )+d dt G (x ,t ).(6.9)
The only possibility is L =12m v 2,where the constant m is the mass of the particle.Note:
L =12m (v −V )2=12m v 2+d dt 12m V 2t −m V ·x =L +dG dt
.(6.10)
恣意4CHAPTER6.LAGRANGIAN MECHANICS For N interacting particles,
L=1
2
N
a=1
m a
d x
a
dt
2
−U
{x a},{˙x a}
hhb.(6.11)
Here,U is the potential energy.Generally,U is of the form
U=groovy
a U1(x a)+
a<a
v(x a−x a ),(6.12)
however,as we shall e,velocity-dependent potentials appear in the ca of charged parti-cles interacting with electromagneticfields.In general,though,
L=T−U,(6.13) where T is the kinetic energy,and U is the potential energy.
6.3Remarks on the Choice of Generalized Coordinates
Any choice of generalized coordinates will yield an equivalent t of equations of motion. However,some choices result in an apparently simpler t than others.This is often true with respect to the form of the potential energy.Additionally,certain constraints that may be prent are more amenable to treatment using a particular t of generalized coordinates. The kinetic energy T is always simple to write in Cartesian coordinates,and it is good practice,at least when one isfirst learning the method,to write T in Cartesian coordinates and then convert to generalized coordinates.In Cartesian coordinates,the kinetic energy of a single particle of mass m is
T=1
2m
˙x2+˙y2+˙x2
.(6.14)
If the motion is two-dimensional,and confined to the plane z=const.,one of cour has
T=1
2m
˙x2+˙y2
.
Two other commonly ud coordinate systems are the cylindrical and spherical systems. In cylindrical coordinates(ρ,φ,z),ρis the radial coordinate in the(x,y)plane andφis the azimuthal angle:
x=ρcosφ˙x=cosφ˙ρ−ρsinφ˙φ(6.15)
y=ρsinφ˙y=sinφ˙ρ+ρcosφ˙φ,(6.16) and the third,orthogonal coordinate is of cour z.The kinetic energy is
T=1
2m
˙x2+˙y2+˙x2
=1
2m
˙ρ2+ρ2˙φ2+˙z2
.(6.17)
6.4.HOW TO SOLVE MECHANICS PROBLEMS5 When the motion is confined to a plane with z=const.,this coordinate system is often
referred to as‘two-dimensional polar’coordinates.
In spherical coordinates(r,θ,φ),r is the radius,θis the polar angle,andφis the azimuthal
angle.On the globe,θwould be the‘colatitude’,which isθ=π
2−λ,whereλis the latitude.
<θ=0at the north pole.In spherical polar coordinates,
x=r sinθcosφ˙x=sinθcosφ˙r+r cosθcosφ˙θ−r sinθsinφ˙φ(6.18) y=r sinθsinφ˙y=sinθsinφ˙r+r cosθsinφ˙θ+r sinθcosφ˙φ(6.19) z=r cosθ˙z=cosθ˙r−r sinθ˙θ.(6.20) The kinetic energy is
T=1
2m
˙x2+˙y2+˙z2
=1
deem是什么意思2m
˙r2+r2˙θ2+r2sin2θ˙φ2
.(6.21)
traceability6.4How to Solve Mechanics Problems
Here are some simple steps you can follow toward obtaining the equations of motion:
1.Choo a t of generalized coordinates{q1,...,q n}.
2.Find the kinetic energy T(q,˙q,t),the potential energy U(q,t),and the Lagrangian
L(q,˙q,t)=T−U.It is often helpful tofirst write the kinetic energy in Cartesian coordinates for each particle before converting to generalized coordinates.
3.Find the canonical momenta pσ=∂L
∂˙qσand the generalized forces Fσ=∂L
∂qσ
.
4.Evaluate the time derivatives˙pσand write the equations of motion˙pσ=Fσ.Be careful
to differentiate properly,using the chain rule and the Leibniz rule where appropriate.
实力英文>cost5.Identify any conrved quantities(more about this later).
西瓜的英文
6.5Examples
6.5.1One-dimensional motion
克服困难的英语作文For a one-dimensional mechanical system with potential energy U(x),
L=T−U=1
2
m˙x2−U(x).(6.22)
The canonical momentum is
p=∂L
∂˙x
=m˙x(6.23)
6CHAPTER 6.LAGRANGIAN MECHANICS and the equation of motion is d dt ∂L ∂˙x =∂L ∂x
⇒m ¨x =−U (x ),(6.24)
which is of cour F =ma .Note that we can multiply the equation of motion by ˙x to get
0=˙x m ¨x +U (x ) =d dt 12m ˙x 2+U (x ) =dE dt
,(6.25)
where E =T +U .6.5.2Central force in two dimensions
Consider next a particle of mass m moving in two dimensions under the influence of a potential U (ρ)which is a function of the distance from the origin ρ= x 2+y 2.Clearly cylindrical (2d polar)coordinates are called for:
L =12m ˙ρ2+ρ2˙φ
2 −U (ρ).(6.26)The equations of motion are d dt ∂L ∂˙
ρ =∂L ∂ρ⇒m ¨ρ=mρ˙φ2−U (ρ)(6.27)d dt ∂L ∂˙φ =∂L ∂φ⇒d dt mρ2˙φ =0.(6.28)
Note that the canonical momentum conjugate to φ,which is to say the angular momentum,is conrved:p φ=mρ2˙φ=const .(6.29)
We can u this to eliminate ˙φ
from the first Euler-Lagrange equation,obtaining m ¨ρ=
p 2φmρ3−U (ρ).(6.30)We can also write the total energy as
E =12
m ˙ρ2+ρ2˙φ2 +U (ρ)=12m ˙ρ2+p 2φ
2mρ2+U (ρ),(6.31)
from which it may be shown that E is also a constant:dE dt = m ¨ρ−p 2φmρ3
+U (ρ) ˙ρ=0.(6.32)We shall discuss this ca in much greater detail in the coming weeks.
