Lecture 1

Alexey Boyarsky

September 8, 2015

1 Introduction

According to the traditional view, mechanics is the branch of physics that studies thebehaviour of physical bodies when subjected to forces or displacements. The first partof this course, Classical Mechanics A, has considered main principles and examples ofclassical mechanics in this traditional sense.

The main goal of theoretical physics however is to build the correspondence betweenphysical phenomena in the real world and mathematical models describing some of theirproperties. It appears that many mathematical ideas of mechanics and its two mainparts, Kinematics (that studies the possible states of physical systems and methods oftheir description) and Dynamics (that predicts how a given state of the system will evolvein the future) are very general and are very useful in a context that is much wider thanthe classical realm of mechanics.

The goal of this second part of the course is to develop this more general view onmechanics and, in particular, introduce Lagrangian and Hamiltonian formalisms, thatare very important in the description of many physical systems.

2 Kinematics. Description of the states of physical

system. Examples.

2.1 Point particles.

The simplest abstraction of a physical system is a point particle. Its state is characterisedby its position at any moment of time

~r = ~r(t)

In the Cartesian coordinate system ~r(t) = (x(t), y(t), z(t)). Depending on the move-ment of the particles, it may be convenient to use di↵erent coordinate system, for example

spherical coordinates: (see 1.12 of [1])

x =r sin ✓ cos'

y =r sin ✓ sin'

z =r cos ✓

(1)

1

and describe the state of the system by the set of functions r(t), ✓(t), '(t). Alter-natively, one can use other coordinate systems,

cylindrical coordinates: (⇢,�, Z) such that their relation to Cartesian coordinates

x = ⇢ cos�

y = ⇢ sin�

z = Z

(2)

parabolic coordinates: (�, ⌧,')x = �⌧ cos'

y = �⌧ sin'

z =1

2

�⌧

2 � �2�

(3)

but in any case three independent numbers are required to describe the position of theparticles of three independent functions of time to define its state at any moment.

2.2 Several particles

As the next simplest example consider two particles with the masses m1 and m2. We canspecify position of each particle by its own radius vector ~r1 and ~r2. Or, alternatively wecan introduce the center-of-mass position ~R

c.m.

(see 7.1 of [1])

~

R

c.m.

=m1~r1 + m2~r2

m1 + m2(4)

and the relative radius vector~r12 = ~r1 � ~r2 (5)

In either case we need 6 scalar functions to specify all positions.⌥⌃ ⌅⇧In general for the system of N particles one needs 3N scalar functions2.3 Pendulum: example of a system with constraints

Consider a pendulum – a material point of mass m moving in the xy plane, attached toa (massless) rigid rod of length l (Fig. 8). Naively, the position of this material point isdescribed by two coordinates x(t) and y(t) and two velocities: ẋ(t) and ẏ(t). However, itis clear that at any moment of time x and y are subject to a constraint:

x

2(t) + y2(t) = l2 (6)

Moreover, velocity is also constraint. As the rod is rigid (i.e. it cannot change its length)the velocity vector ~v = (ẋ, ẏ) is always directed perpendicular to the rode. This isa consequence of the constraint (73). As it is valid at all times t, di↵erentiating thisconstraint with respect to time we get:

x(t)ẋ(t) + y(t)ẏ(t) = 0 (7)

2

Thus, we see that out of two functions x(t), y(t) only one is independent. And similarly,out of two velocity components (ẋ, ẏ) only one is independent. Of course, one can chooseas such a variable x(t) and ẋ(t) and express

y(t) = ±p

l

2 � x2(t)

ẏ(t) = ⌥ x(t)ẋ(t)pl

2 � x2(t)(8)

However, one can introduces the angle ✓, such that

tan ✓(t) = �x(t)y(t)

(9)

This angle uniquely specifies the position:

x(t) = l cos ✓(t)

y(t) = l sin ✓(t)(10)

and velocity:ẋ(t) = �l✓̇ sin ✓(t)ẏ(t) = l✓̇ cos ✓(t)

(11)

We will see below (Section 6.2) that the dynamics can be expressed in terms of the samevariable ✓(t).

2.4 Rigid body

Figure 1: Proper Euler an-gles. The xyz (original)system is shown in blue,the XYZ (rotated) systemis shown in red. The lineof nodes (N) is shown ingreen.

Next example is a rigid body (see 8.1 of [1]). Naively thisis enormous collection of particles with N ⇠ 1023. However,the term rigid means that the relative positions ~r

ij

betweenparticles i and j are fixed. As a result the state of rigid bodycan be given by center-of-mass co-ordinates and by 3 anglesthat define rigid body position (for example Euler angles, seeFig. 1 and 9.6 of [1]), so we need 6 variables.

2.5 Fluid

More complex example is a fluid – a system of interactingparticles where the interaction however does not keep shapeof any volume fixed. When describing a fluid, one is con-centrated on the dynamics of macroscopic (albeit small) vol-umes of fluid, containing many molecules. The motion ofthese molecules defines the thermodynamics properties of afluid (pressure, density, temperature, etc.)

In the Lagrangian picture, individual fluid parcels are fol-lowed through time. The fluid parcels are labelled by some(time-independent) vector field ~a. (Often, ~a is chosen to be

3

Figure 2: Lagrangian fluid description a parcel ABCD moves with the flow ~v. If the flowis incompressible, the polygon ABCD deforms but its volume (area) remains constant.Mathematically incompressibility means that r · ~v(~r) = 0.

the center of mass of the parcels at some initial time t0). It ischosen in this particular manner to account for the possiblechanges of the shape over time. Therefore the center of mass is a good parametrizationof the velocity ~v of the parcel.) In the Lagrangian description, the flow is described by afunction ~X

~a

(t) giving the position of the parcel labeled ~a at the time t.In the Eulerian picture of the flow field, the flow quantities are depicted as a function

of position ~x and time t. Specifically, the flow is described by a function ~v (~x, t) givingthe flow velocity at position ~x at time t.

The two specifications are related as follows:

~v(~x, t) = ~v⇣~

X

~a

(t)⌘

=@

~

X

~a

(t)

@t

(12)

because both sides describe the velocity of the parcel labeled ~a at the time t. The numberof needed variables is formally infinite, one vector for each point r, but locally there areonly there numbers:

~v

~x

(t) = ~v(~x, t).

2.6 Electromagnetic field

This is the first example of a system that can not be defined as a collection of (interacting)particles. This also also infinitely-dimensional system: a configuration of electromagneticfield is described by 2 vector fields: electric field ~E

~r

(t) = ~E(~r, t) and magnetic ~B~r

(t) =~

B(~r, t) or 6 numbers for each point in space (6 scalar functions). However, as we will seebelow, not all of these functions are independent !

4

3 Equations of motion and degrees of freedom

Already in the simplest examples we see, however, that it is not enough to know onlypositions of particles or configuration of the the fields. For a point particle we need toknow also the velocity of the particle. The same position but di↵erent velocity at theinitial state will result in very di↵erent evolution of the the same particle. To definea state of a system completely, we need to specify enough information to be able todetermine this state at any moment in the future. Therefore, the definition of state ofa system is defined by its dynamics or equations of motion. For a wide class of physicalsystem the equation of motion are local di↵erential equation, second order in time. Tofind a specific solution of such equations (and define evolution of the physical system athand) we need specify the initial conditions containing the both the value of the functionand its time derivative at the initial moment of time, the analog of the initial positionand initial velocity. For the systems that can be understood as collections of (interacting)particles, equations of motion can in principle be derived from the Newtons equations forpoint particles (see 2.1 of [1]):

m~̈r =X

i

~

F

i

. (13)

In the case of electric field:m~̈r = q ~E (14)

If we want also to include magnetic field we need to write a term q[~v⇥ ~B]/c in right-handside, where c is the speed of light:

~

F

L

= q

✓~

E +1

c

[~v ⇥ ~B]◆

(15)

This term is proportional to v/c and vanishes in non-relativistic limit. Therefore, let usconsider dynamics of relativistic particle.

3.1 Relativistic particles

For relativistic particles the equations of motion are very similar, e.g. for a particle inexternal electromagnetic field:

8><

>:

m

du

0

ds

= q(~u · ~E)

m

d~u

ds

= q⇣u

0~

E + [~u⇥ ~B]⌘ (16)

where uµ = (u0, ~u) = dxµ/ds is a 4-vector of velocity, s — is an interval, q — is a particle’scharge. Here the coordinates are xµ and velocities are uµ:

u

µ =

cp

1� v2/c2,

~vp1� v2/c2

!(17)

5

In terms of the the field strength F µ⌫ with Ei = F 0i and Bi = 12✏ijk

F

jk

this can berewritten in a more compact form

m

d

2x

µ

ds

2= qF µ⌫u

⌫

(18)

3.2 Particles, connected by springs

Let us consider equations of motion of a system consisting of many interacting particles,for example a system of equal particles connected by springs (see 11.4 of [1]): We will

Figure 3: System of large amount of particles connected by springs

denote by xi

the displacement of the ith particle from its equilibrium position (i runsfrom 1 to N , and x

N+1 = x1). So the equation on ith particle is:

mẍ

i

= k(xi�1 � xi)� k(xi � xi+1) = k(xi+1 + xi�1 � 2xi) (19)

Let we denote equilibrium distance between particles as a and use continius functionfor deviation from equilibrium state u(x), such that x

i

! u(x), xi�1 ! u(x � a) and

x

i+1 ! u(x + a). In limit a! 0 we have:

lima!0

k

m

(u(x + a) + u(x� a)� 2u(x)) = d2u

dx

2lima!0

ka

2

m

(20)

(to prove it use taylor series). Denoting lima!0

ka

2

m

as we obtain equation of string motion

(see 11.5 of [1]):@

2u

@t

2=

@

2u

@x

2(21)

⌫

�

�Let us introduce the number of degrees of freedom defined by (one half of) the numberof functions one needs to specify at time t = 0 to fully determine future evolution of thesystem at t > 0.

In all the examples above, that equations of motion are second order di↵erential equations,and for solving them we need to specify the position or configuration function of thesystem and its derivative in the initial moment of time. It is because all these systemsare di↵erent generalizations of particle dynamics and Newton’s second law. However, thisanalogy is not always the case!

6

Figure 4: Random walk of the particle on 1D lattice.

3.3 The di↵usion equation

Let us consider a very small particle in a medium ( e.g. gas). The particles forming themedium hit our test particle from time to time ( as they are involved in the random ther-mal motion) and displace it. Let us find a mathematical model describing the evolution ofsuch a test particle in a medium. For simplicity, let us assume that the particle can moveonly along a line ( one-dimensional motion). We can measure the position of the particleat discrete moments of time separated by equal time intervals ⌧ : t = 0, ⌧, 2⌧, ..N⌧, . . . .Let us assume that during this time intervals our particle is displaced by equal distance lto the left or to the right. The direction is random ( as it the displacement is caused bycollisions with randomly moving thermal particles). The probability to move the particleone step right is equal to p and the probability to move one step left is q = 1� p.

We are interested in the probability to find a particle in the point x at the timemoment t. In our discrete model x = ml and t = ⌧N . Therefore, we are looking for thefunction on the lattice, P

N

(m). To be in at the point m at the time moment N + 1, ourparticle had either to come from the left (if it was at the moment N at the point m� 1and moved to the right) or from the right ( if it was at the point m + 1 and moved left),see Fig.4. Therefore, we have the following equation for the evolution of probability:

P

N+1(m) = pPN(m� 1) + qPN(m + 1) (22)

The most interacting case of this theory appears when p = q = 1/2 (there is no directedmotion in average). Subtracting P

N

(m) from both sides of (22):

P

N+1(m)� PN(m) =1

2(P

N

(m� 1) + PN

(m + 1)� 2PN

(m)) (23)

Taking limit ⌧ ! 0, l ! 0 we can write:

P

N+1(m)� PN(m) =⌧

⌧

(PN+1(m)� PN(m)) ⇡ ⌧

@P

@t

(24)

l

2

2l2(P

N

(m� 1) + PN

(m + 1)� 2PN

(m)) ⇡ l2

2

@

2P

@x

2(25)

7

Introducing a constant D = liml,⌧!0

l

2

2⌧ we have di↵usion equation which describes evolution

of probability distribution:@P

@t

= D@

2P

@x

2(26)

If one starts with a very sharp (�-function shaped) peak at t = 0, the evolution of thispeak with time will be given by the following expression (see Section 3.3.1 for details):

P (x, t) =1

2⇡

1Z

�1

e

ikx

e

�Dk2tdk =

1

2p⇡Dt

e

� x2

4D2t2 (27)

This is a Gaussian distribution with variance equal to �2 = 2Dt. Time dependence ofthis function is shown on Fig. 5.

Figure 5: Di↵usion evolution of initial point-like distribution.

For the mixture of gases di↵usion is also described by equation of the same form:

@n

A

@t

= D@

2n

A

@x

2(28)

where nA

is a number density of particles of some type A. This is the first order di↵erentialequation in time on particles concentration n

A

(x, t). So the only degree of freedom is afunction n

A

(x, t).

3.3.1 A general solution of the di↵usion equation

This is an additional information for those, not needed for the main courseIt’s not di�cult to find a general solution of this equation which is not grow on infinity. Suppose

P (x, t) = X(x)T (t), where X and T are some unknown functions. Substituting in (26):

Ṫ

T

= DX

00

X

(29)

8

Left-hand side depends only on t, whereas right-hand side depends only on x. The only possibility isthat both sides are equal to some constant �.

1. Let � = Dk2 > 0. Then for X we have:

X

00 = k2X (30)

and solutions are X(x)1,2 = C1,2e±kx. One of this solutions grow on +1, other on �1, so theyare not interesting for us.

2. If � = 0 we have:X

00 = 0 (31)

with solution X = Ax + B. From condition on infinities A = 0, so X is some constant. For T wealso have Ṫ = 0, so T is a constant also.

3. In case � = �Dk2 < 0:X

00 + k2X = 0 (32)

and solutions are X(x)1,2 = C1,2e±ikx, which don’t grow on infinity. For T we have:

Ṫ = �Dk2T (33)

with solution T (t) = Ce�Dk2t.

This equation is linear, so the full solution is a sum of particular ones. Taking all cases together we have:

P (x, t) =X

k

C

k

e

ikx

e

�Dk2t =1Z

�1

C(k)eikxe�Dk2t

dk (34)

because k is continuous variable. Coe�cients C(k) are determined by initial conditions.Suppose initial conditions in form P (x, 0) = �(x), which means that particle has probability 1 to be

in the point x = 0. So, for t = 0:

�(x) =1Z

�1

C(k)eikxdk (35)

So it’s easy to find C(k):

C(k) =12⇡

1Z

�1

�(x)e�ikxdx =12⇡

(36)

3.4 Euler equation for fluid

Let’s look at equations of motion of ideal fluid.The first one is the continuity equation. The mass in some (small but macroscopic)

volume V is M =R

V

⇢dV and its change per unit time is equal to total mass flow (given

by in the integral of ⇢~v over the surface of this volume S):

dM

dt

= �Z

S

⇢~v · ~ndS, (37)

where ~n is a unit vector normal to the surface and oriented out from it. Using Gausstheorem we have:

d

dt

Z

V

⇢dV = �Z

V

~r · (⇢~v)dV (38)

9

so we arrive to the continuity equation:

@⇢

@t

+ ~r · (⇢~v) = 0 (39)

If div ~v = r · ~v = 0 (incompressible fluid !) this equation can be rewritten as the

d⇢

dt

⌘ @⇢@t

+ ~v ·r⇢ = 0 (40)

It is the first equation. For second one let’s use Newton’s second law to our volume:Z

V

dv

dt

dm =

Z

V

g dm�I

S

p dS (41)

where g is force per unit volume of fluid and p is pressure. Using dm = ⇢ dV and Gausstheorem again we have:

⇢

dv

dt

= ⇢g �rp (42)

using the identity dv(r,t)dt

⌘ @v@t

+ @v@x

i

@x

i

@t

= @v@t

+ (v ·r)v and we finally have:

@v

@t

+ (v ·r)v = g � 1⇢

rp (43)

So we have 2 equations of first order on field of velocities and density, they are degrees offreedom. The pressure is defined from the equation of fluid state. Although Eq. (43) isjust the second Newton’s law, the equation is first order in time because it is velocity ~v

~x

(t)is the coordinate in this case. The variable ~x is just an index, marking “fluid particle”.

3.5 Systems with constraints

3.5.1 Pendulum

m

�

x

y

l

Figure 6: The pendulum.

Let us come back to the example of pendulum, whose kinematics we have alreadyconsidered in Section 2.3. Let us write its equations of motion. The main problem here

10

is that to describe the dynamics of the material point m in x, y coordinates we need tointroduce some explicit forces. In this case this is the tension of the rope, T = (T

x

, T

y

).So the laws of its motion is:

mẍ = �Tx

(44)

mÿ = Ty

�mg (45)

We have 4 unknown functions (x y, Tx

, T

y

) and only 2 equations. But we also haveconstraints — relations between quantities that describe state of the system. Constraintsare some functions of form F (q1, . . . , q̇1, . . . ) = 0. For pendulum we have constraints (73)

x

2 + y2 = l2 (46)

(47)

and its time derivative, saying that velocity is orthogonal to the rod (Eq. (7)).The problem can be simplified if we introduce coordinates that take into account

constraints. It’s easy to see that position of the system is fully determined by an angle✓, Eq. (9):

x(t) = l cos ✓(t)

y(t) = l sin ✓(t)

ẋ(t) = �l✓̇ sin ✓(t)ẏ(t) = l✓̇ cos ✓(t)

(48)

and write the 2nd Newton’s law in the coordinates along and orthogonal to the rod (radialcoordinates). The radial movement (movement along the rod) is trivial:

r(t) =p

x

2(t) + y2(t) = l = const (49)

while the equation, projected to the direction orthogonal to the rod, does not contain theunknown and redundant tension force ~T and can be written as

ml

2✓̈ = �mgl sin ✓ (50)

One recognizes in Eq. (50) the equation for the angular momentum with the torque dueto gravitational force in the right hand side.

3.5.2 Pendulum on the movable support

Consider the system of 2 particles in the gravitational field of mass M and m. The firstparticle moves along horizontal wire, and the second is connected to in through the ropeof length l, see Fig. 7. Using Newton’s laws we can write equations on system’s dynamics:

Mẍ

M

= Tx

(51)

Mÿ

M

= N �Mg � Ty

(52)

mẍ

m

= �Tx

(53)

mÿ

m

= Ty

�mg (54)

11

M

g

T

N

l

T

mg

m

Mg

Thetay

x

Figure 7: Pendulum on the movable support.

In this system we have 7 unknown functions of time: xM

, yM

, xm

, ym

, Tx

, T

y

(tension ofthe rope), N (reaction of the wire). Again, the situation is the same as in the previousSection 6.2: we have 3 unknown forces (T

x

, T

y

, N) that should be found if we are workingin the redundant coordinates x

m

, x

M

, y

m

, y

M

.We have only 4 equations (51)–(54), so where do we get other 3? Needed relations

come from constraints. In our case we have 2 constraints:

y

M

= const (55)

(xM

� xm

)2 + (yM

� ym

)2 = l2 (56)

We can introduce an angle ✓(t) – between the rod and the vertical line:

tan ✓ = �xM � xmy

M

� ym

(57)

Di↵erentiating constraints we can, in principle, find relations between accelerations andangle ✓ and plug it into equation of motions. It’s easy to see, that position of thesystem is fully determined by 2 quantities: the x coordinate of the center of mass xc.m. =(Mx

M

+ mxm

)/(M + m) and ✓.

12

3.6 Maxwell’s equations

Electromagnetic field is even more complicated:

~r · ~E = 4⇡⇢ (58)~r · ~B = 0 (59)

~r⇥ ~E = �1c

@

~

B

@t

(60)

~r⇥ ~B = 1c

4⇡~j +

@

~

E

@t

!(61)

From the point of view of dynamics, the second pair of equations, (60)–(61) are the firstorder di↵erential equations for quantities ~E and ~B. Each of them requites one initialcondition: ~E(x, t = 0) and ~B(x, t = 0). For any initial values of ~E and ~B we can findtime derivatives such that the equations are satisfied.

But the first pair of equations (equations (58)–(59)) have no time derivative at all!They have to be valid at all moments, including the initial moment t = 0. If we choosean initial condition (58) and (59) may be satisfied or not. Moreover, if we define 2components of ~E at t = 0, we can find the third one from (58). This same is truefor (59) and the components of ~B at t = 0 We can say therefore that (58) and (59)constrain the space of possible initial conditions and, therefore, reduce the number ofdegrees of freedom. We see that we can specify only four scalar functions (arbitraryinitial conditions for 2 components of ~E and for 2 components of ~B at each point ofspace). There is a way to interpret these 4 functions as 2 local degrees of freedom.1

It should be stressed that this is just a di↵erent example of the system with constraintsthat involves derivatives. In the examples of pendulum (Secs. 6.22.3) we solved thisproblem by introducing new variables that solve trivially the constraints (e.g. switchingfrom Cartesian to polar coordinates). Let us see how this can be done in this case.

Let use the following property of vector analysis:

~r · [~r⇥ ~A] = 0 (62)

that holds for any vector field ~A.2 We can therefore solve equation (59) by the followinganzatz :

~r · ~B = 0 =) ~B = ~r⇥ ~A (63)

Then substituting this solution to the equations (60), we get

~r⇥ ~E = �1c

@(~r⇥ ~A)@t

(64)

1Recall that a mechanical degree of freedom is a pair: a coordinate and its time derivative that canbe freely chosen at initial moment, local degree of freedom means that we have such data at every point.

2It is easy to demonstrate by explicit computation that ~r · (~r ⇥ ~A) = 0 for any vector field ~A.Interestingly, the converse is also true: if divergence of a vector ~Z is equal to zero, this vector can berepresented as a curl of some other vector ~Z = ~r⇥ ~W . This is known as Poincaré lemma.

13

Removing curl from both sides of (64) we find that

~

E = �1c

@

~

A

@t

+ ~U = �1c

@

~

A

@t

� ~rA0 (65)

where ~U is a vector field such that ~r⇥ ~U = 0 and , therefore, it can be represented as agradient of a function A0.3

As a result, we have expressed 6 functions ~E, ~B via 4 functions (A0, ~A):8>><

>>:

~

E = �~rA0 � 1c

@

~

A

@t

~

B = ~r⇥ ~A(66)

Is this relation unique? Using the property

~r⇥ ~rf = 0 (67)

for any function f we see that the magnetic field ~B does not change if we change:

~

A! ~A + ~rf (68)

However, ~E changes under the transformation (68). However, we along with the trans-formation (68) we do a transformation

A

0 ! A0 � 1c

ḟ (69)

then both ~E, ~B will not change, i.e. the pair (A0 � 1c

ḟ ,

~

A + ~rf) corresponds to the same~

E, ~B for any scalar function f(t, x). This property can be seen in the Lorentz-convariantnotations. If we introduce a 4-vector of electromagnetic field Aµ = (A0, ~A) and recallthat ~E, ~B are related to the field strength tensor F

µ⌫

via

E

i

= F0i

B

i

=1

2✏

ijk

F

jk

(70)

Then we can rewrite the relation (64) as

F

µ⌫

=@A

µ

@x

⌫

� @A⌫@x

µ

(71)

It is then obvious that if we change

A

µ

! Aµ

+@f

@x

µ

(72)

(keep in mind that x0 = ct) then the field-strength Fµ⌫

(and ~E, ~B as a consequence) doesnot change.

3Similarly to the comment above, if ~r⇥ ~U = 0 , ~U can be represented as a gradient of some scalarfunction.

14

⌫

�

�The freedom to choose Aµ for given ~E and ~B is called gauge invariance. One needsadditional conditions to uniquely relate Aµ and ~E, ~B. Such a condition is called gaugecondition or gauge fixing.

In terms of the field strength Fµ⌫

the Maxwell’s equations have the form (Eqs. (58)(61)):

@F

µ⌫

@x

⌫

= jµ (73)

where 4-vector jµ = (c⇢,~|), and the pair of equations (59)–(60) gets rewritten as theso-called Bianchi identity:

✏

µ⌫�⇢

@

⌫

F

�⇢

= 0 (74)

Notice that the ansatz (71) solves (74) identically. If we plug the ansatz (71) into equa-tion (73), we find ✓

�� 1c

2

@

2

@t

2

◆A

µ

+ @µ

⇣@

⌫

A

⌫

⌘= jµ (75)

Let us choose a simple way to fix this freedom by imposing the condition

A

0 = 0 (76)

Plugging this into the Maxwell’s equations, we get

✓�� 1

c

2

@

2

@t

2

◆A

i

+ @i

⇣@

j

A

j

⌘= ji (77)

We are then with only three functions ~A. However, this is not the final answer –our gauge freedom is not fully fixed. Indeed if we do gauge transformation (68) witha function f that does not depend on time, this will not change the condition A0 = 0.Therefore, we expect another component of A

µ

to be unphysical (possible to eliminateby a an additional gauge transformation. It is convenient to choose the function f(x

i

)such that

@

i

Ã

i = @i

(Ai + @if) = 0 (78)

This implicitly reduces the number of independent components of Ai

( one of them can befound using the above equation if the other two are known) and simplifies the equationsof motion. Let us stress however that if A0 = 0, then

~

E = �1c

@

~

A

@t

~r · ~E = �1c

@(~r · ~A)@t

(A0 = 0)

(79)

Therefore, we can impose both conditions A0 = 0 and ~r · ~A = 0 at the same timeonly if ~r · ~E = 0 and, therefore, if ⇢ = j0 = 0. Therefore, in the empty space Maxwellequations reduce to

15

8<

:� ~A� 1

c

2

@

2~

A

@t

2= 0

~r · ~A = 0(80)

The initial field Aµ had 4 degrees of freedom and to describe the solution of Eq. (75)one needed to specify 4 components of field and 4 time derivatives. However, A

µ

are notdirectly observable, we need to know only they combinations that give us ~E and ~B. Onthe other hand, the 6 components of ~E and ~B are directly observable, however are notindependent. Solving explicitly the constraints on ~E and ~B with the help of Aµ we seethat the system has only two degrees of freedom and we need to specify two independentfunctions and two independent time derivatives of these functions ( ”velocities”) at theinitial moment t = 0.

3.7 Quantum mechanics. Schrödinger equation

In quantum mechanics is described by the complex-valued wave function (x, t) and itsevolution is described by the 1st order equation

i~@ @t

= �~2 � 2m

+ U (81)

The state of the system is (~x) and there is no analog of velocity, as the evolution isdefined by the the equation that is first order in time.

4 Lagrange mechanics

In all the examples we considered above the we have seen that the state of the systemwas defined as a minimal set of data required to determine the subsequent evolution ofthe system. For many systems this set consists of a configuration ( the analog of theposition of a particle) and its time derivative ( the analog of velocity), however there arealso deviations from this picture defined by the dynamics of the system ( its equationsof motion). The nature of configuration may be di↵erent: this can be a discret set ofvariables or a number of continuous functions (e.g. vector field). More over, the samesystem can be rewritten in di↵erent coordinate systems (di↵erent parametrisations ofconfigurations), depending for example on the external forces acting on this system.

In this Section we will try to find common feature of all the above examples and finda description of the evolution of a dynamical system that will have the same form forall these systems. As the particles’ dynamics is the basic example for mechanics, we willstart with rewriting the dynamics of such a system of particles in an arbitrary system ofcoordinates and will then generalize the resulting description for a wider class of systems.

Let qi

, i = 1 . . . n, are new co-ordinates and f is some function of coordinates andtime f = f(q1, . . . , qn, t).

We consider a system of particles with Cartesian co-ordinates ~r1, . . . ,~rN and we wantto change them to independent generalized co-ordinates q1, . . . , qn, n 3N . The relationsbetween old and new co-ordinates are:

~r

i

= ~ri

(q1, . . . , qn, t) (82)

16

For each particle we have:

m

i

d~v

i

dt

= ~Fi

(83)

4.1 Transformation of 2nd Newton’s law under the change of

variables (82)

Let we have a particle in the 3-dimensional oscillator potential. The equations of motionare:

m~̈r = �k~r (84)

Let we make a general co-ordinate change:

x

i

= fi

(q1, q2, q3, t) (85)

ẋ

i

=X

j

@f

i

@q

j

@q

j

@t

+@f

i

@t

(86)

ẍ

i

=X

j,k

@

2f

i

@q

j

@q

k

@q

j

@t

@q

k

@t

+X

j

@f

i

@q

j

@

2q

j

@t

2+X

j

@

2f

i

@t@q

j

@q

j

@t

+@

2f

i

@t

2(87)

and equation of motion will be in form:

m

✓@f

i

@q1

@

2q1

@t

2+@f

i

@q2

@

2q2

@t

2+@f

i

@q3

@

2q3

@t

2

◆=

� kfi

�m X

j,k

@

2f

i

@q

j

@q

k

@q

j

@t

@q

k

@t

+X

j

@

2f

i

@t@q

j

@q

j

@t

+@

2f

i

@t

2

!(88)

For example, for Galilei’s transformation xi

= qi

+ vi

t equations don’t become toodi�cult:

mq̈

i

= �kqi

� kvi

t (89)

But for rotating co-ordinate system with frequency ! (see 5.2 of [1]):

8><

>:

x =x0 cos!t + y0 sin!t

y =� x0 sin!t + y0 cos!tz =z0

(90)

we get:8><

>:

m (ẍ0 cos!t + ÿ0 sin!t) =(m!2 � k)(x0 cos!t + y0 sin!t)�m!(�ẋ0 sin!t + ẏ0 cos!t)m (�ẍ0 sin!t + ÿ0 cos!t) =(m!2 � k)(�x0 sin!t + y0 cos!t) + m!(ẋ0 cos!t + ẏ0 sin!t)

mz̈

0 =� kz0(91)

17

4.2 Covariant form

Let’s multiply every of this equations by @~ri

/@q

j

and add results:

NX

i=1

m

i

d~v

i

dt

· @~ri@q

j

=NX

i=1

~

F

i

· @~ri@q

j

(92)

As j changes from 1 to n we have n such equations.

NX

i=1

m

i

d~v

i

dt

· @~ri@q

j

=d

dt

NX

i=1

m

i

~v

i

· @~ri@q

j

�NX

i=1

m

i

~v

i

· ddt

@~r

i

@q

j

(93)

We can further rewrite this expression using two identities

@~v

i

@q̇

j

=@~r

i

@q

j

(94)

and@~v

i

@q

k

=d

dt

@~r

i

@q

k

(95)

we haved

dt

NX

i=1

m

i

~v

i

· @~vi@q̇

j

�NX

i=1

m

i

~v

i

· @~vi@q

j

= Qj

(96)

where we have introduced notation Qj

(called generalized forces):

Q

j

⌘NX

i=1

~

F

i

· @~ri@q

j

(97)

The meaning of these functions is work under infinitesimal displacement �qj

. Indeed, thework A is given by

A =X

i

~

F

i

�~r

i

=X

j

X

i

~

F

i

@~r

i

@q

j

!�q

j

=X

j

Q

J

�q

j

(98)

Let’s rewrite left part of (96) as

d

dt

@

@q̇

j

NX

i=1

m

i

v

2i

2| {z }

⌘T

� @@q

j

NX

i=1

m

i

v

2i

2= Q

j

(99)

By definition, the sum in left hand side is a kinetic energy of the system T . So, finally:

d

dt

@T

@q̇

j

� @T@q

j

= Qj

(100)

These n equations are called Lagrange equations. Qi

are called generalized forces. Themain profit of this approach is that Lagrange equations are covariant — they doesn’tchange form after change of co-ordinate system.

18

Lagrange equations contains n + 1 functions — n generalized forces Qi

and kineticenergy T . But they can be simplified for the case of potential forces ~F

i

. The potentialityof force means, that there exists scalar function of co-ordinates and time V

i

(~r, t), that~

F

i

= �~rVi

. So, if all forces are potential ones:

Q

j

= �NX

i=1

@

~

V

i

@x

i

@x

i

@q

j

+@

~

V

i

@y

i

@y

i

@q

j

+@

~

V

i

@z

i

@z

i

@q

j

!= �@U

@q

j

(101)

where U =Pi

V

i

— potential energy of the system. Therefore:

d

dt

@T

@q̇

j

� @T@q

j

= �@U@q

j

(102)

and, using fact that @U/@q̇j

= 0, we finally have:

d

dt

@L

@q̇

j

� @L@q

j

= 0 (103)

whereL = T � U (104)

Function L(q, q̇, t) is called Lagrange function, or just Lagrangian of the system (see 10.4of [1]).

5 The principle of least action

It turns out that Lagrange equations of motion (103) can be obtained from conceptioncalled the principle of least action or Hamilton’s principle (see 10.1 of [1]).

Namely, we consider a system whose state is characterized by a set of discrete or con-tinuous variables q

j

. Its dynamics is determined via the function L(q1, q2, . . . , qs, q̇1, q̇2, . . . , q̇n, t),or briefly L(q, q̇, t).

Let the system occupy, at the instants t1 and t2, positions defined by two sets of valuesof the co-ordinates, q(1) and q(2). Then the condition is that the system moves betweenthese positions in such a way that the integral

S =

t2Z

t1

L(q, q̇, t)dt (105)

takes the least possible value4. The function L is Lagrangian of the system concerned,and the integral (105) is called the action.

The fact that the Lagrangian contains only q and q̇, but not the higher derivatives q̈,...q , etc., expresses the result already mentioned, that the mechanical state of the systemis completely defined when the co-ordinates and velocities are given.

4It should be mentioned that this formulation of the principle of least action is not always valid for theentire path of the system but only for any su�ciently short segment of the path. The integral (105) forentire path must be extremum, but not necessarily a minimum. This fact, however, is of no importanceas regards the derivation of the equations of motion, since only the extremum condition is used.

19

Let us now derive the di↵erential equations which solve the problem of minimizingthe integral (105). For simplicity, we shall at first assume that the system has only onedegree of freedom, so that only one function q(t) has to be determined.

Let q = q(t) be the function for which S is a minimum. This means that S is increasedwhen q(t) is replaced by any function of the form

q(t) + �q(t) (106)

where �q(t) is a function which is small everywhere in the interval of time from t1to t2; it has no connection to virtual displacement from previous section and is called avariation of the function q(t). Since, for t = t1 and for t = t2, all the functions (106)must take the values q(1) and q(2) respectively, it follows that

�q(t1) = �q(t2) = 0 (107)

The change in S when q is replaced by q + �q is

t2Z

t1

L(q + �q, q̇ + �q̇, t)dt�t2Z

t1

L(q, q̇, t)dt

When this di↵erence is expanded in powers of �q and �q̇ in the integrand, the leadingterms are of the first order. The necessary condition for S to have a minimum is thatthese terms (called the first variation, or simply the variation, of the integral) should bezero. Thus the principle of least action may be written in the form

�S = �

t2Z

t1

L(q, q̇, t)dt = 0 (108)

or, e↵ecting the variation,t2Z

t1

✓@L

@q

�q +@L

@q̇

�q̇

◆dt = 0

Since �q = d�q/dt, we obtain, on integrating the second term by parts,

�S =@L

@q̇

�q

����t2

t1

+

t2Z

t1

✓@L

@q

� ddt

@L

@q̇

◆�qdt (109)

The conditions (107) show that the integrated term in 109) is zero. There remains anintegral which must vanish for all values of �q. This can be so only if the integrand iszero identically. Thus we have

d

dt

@L

@q̇

� @L@q

= 0

When the system has more than one degree of freedom, the n di↵erent functions qi

(t)must be varied independently in the principle of least action. We then evidently obtainsequations of the form

d

dt

@L

@q̇

i

� @L@q

i

= 0 (110)

20

We see, that principle of the least action gives the right equations of motion for thesystem of particles. In fact, the scope of it’s application is much more wider and liesfrom elementary particle physics up to gravity and cosmology. So we will postulate it asa fundamental principle of mechanics. Writing Lagrangian is equivalent to determiningall properties of the physical system.

One further general remark should be made. Let us consider two functions L0(q, q̇, t)and L(q, q̇, t), di↵ering by the total derivative with respect to time of some function f(q, t)of co-ordinates and time:

L

0(q, q̇, t) = L(q, q̇, t) +df(q, t)

dt

(111)

The integrals (105) calculated from these two functions are such that

S

0 =

t2Z

t1

L

0(q, q̇, t)dt =

t2Z

t1

L(q, q̇, t)dt +

t2Z

t1

df(q, t)

dt

dt = S + f(q(2), t2)� f(q(1), t1)

i.e. they di↵er by a quantity which gives zero on variation, so that the conditions �S 0 = 0and �S = 0 are equivalent, and the form of the equations of motion is unchanged. Thusthe Lagrangian is defined only to within an additive total time derivative of any functionof co-ordinates and time.

Note that we have derived the Lagrangian equations for arbitrary generalized coordi-nates q

i

and therefore their form will remain the same under arbitrary change of coordi-nates.

✏�

��

From now on we will describe the dynamical system via its Lagrangian, introducinggeneralized coordinates, as dictated by symmetries of the problem.

6 Examples of Lagrangians

6.1 Lagrangian of a free particle and Gallilean invariance

Consider the Lagrangian of a free particle:

L =1

2m~v

2 (112)

Let us perform a Gallilean transformation ~v0 = ~v + ~V . Naively, the form of the La-grangian 112 changes:

L(v0) =1

2m

⇣~v

2 + ~V 2 + 2(~v · ~V )⌘

(113)

Notice, however that the additional terms form a total derivative:

~

V

2 + 2(~v · ~V ) = ddt

⇣~

V

2t + 2(~r · ~V )

⌘(114)

and therefore, as discussed around Eq. (111) does not change the extremum of the action.

21

m

�

x

y

l

Figure 8: The pendulum.

6.2 Pendulum

So far it looks like Hamilton’s principle is a very complicated way of obtaining the equationof motion, that one can write down immediately. For more complicated systems thatcontain certain constraints, however, such a framework is extremely useful. To give anexample: Consider a pendulum, a mass m attached to a massless rod of length l that issuspended from a pivot at position (x, y) = (0, 0) around which it can swing freely. Thepotential of the mass in the gravitational field is given by mgy. The Lagrange functionof the pendulum is thus given by

L (x, y, ẋ, ẏ) =m

2

�ẋ

2 + ẏ2��mgy. (115)

The Euler-Lagrange equations for the x- and y-coordinates lead to two equations ofmotion, ẍ = 0 and ÿ = �g.

Unfortunately these equations are completely wrong. What we found are the equa-tions of motion of a free particle in 2 dimensions in a gravitational field. Solutions are e.g.trajectories of rain drops or of cannon balls but certainly not the motion of a pendulum.What went wrong? We forgot to take into account the presence of the rod that imposesa constraint, namely that x2 + y2 = l2. A better approach would be to use a coordinatesystem that accounts automatically for this constraint, namely to describe the state ofthe pendulum by the angle ✓ (t) between the pendulum and the y-direction, see Fig. 8.But how does the equation of motion look in terms of this angle?

Here comes into play a great advantage of Hamilton’s principle: it is independent ofthe coordinate system that one chooses. Suppose one goes from one coordinate systemx1, x2,..., xN to another coordinate system q1, q2,..., qf via the transformations q = q (x)and x = x (q). The trajectory x (t) becomes then q (x (t)). The action functional canthen be rewritten as

S [x] =

t2Z

t1

L (x (t) , ẋ (t)) dt =

t2Z

t1

L

x (q (t)) ,

fX

i=1

@x (q (t))

@q

i

q̇

i

!dt. (116)

22

The rhs of Eq. 116 is again of the form

S [q] =

t2Z

t1

L̃ (q (t) , q̇ (t)) dt (117)

with a new Lagrangian L̃. Also here Hamilton’s principle must hold, i.e., the dynamicevolution of the system follows from the Euler-Lagrange equations

@L̃

@q

i

� ddt

@L̃

@q̇

i

= 0 (118)

for i = 1, ..., f .If we have a system with constraints we can sometimes introduce coordinates that

automatically fulfill those constraints. The equations of motion are then simply given bythe Euler-Lagrange equations in these coordinates. Let us go back to the pendulum. Wedescribe now the configuration of the pendulum by the angle ✓ (t), see Fig. 8. In termsof this angle the kinetic energy of the pendulum is given by ml2✓̇2/2 and the potentialenergy by �mlg cos ✓. This leads to the following Lagrange function:

L

⇣✓, ✓̇

⌘=

ml

2

2✓̇

2 + mgl cos ✓. (119)

The corresponding Euler-Lagrange equation is given by

✓̈ (t) = �gl

sin ✓ (t) , (120)

which is indeed the equation of motion of the pendulum.

6.3 Pendulum on a movable support

Consider a mass M that can move freely along a horizonal line without friction. Attachedto the mass M is a pendulum of mass m via a massless connection of length l (Fig. 9).We calculate now the Lagrange equations for this system.

We first need to find a suitable coordinate system. The system has 2 degrees offreedom (you can find this number by subtracting the two constraints from the 4 degreesof freedom of the unconstrained masses). Practical coordinates are the position X ofthe mass M along the line and the angle ✓ between the pendulum and the direction ofgravity. The position of the pendulum body is then given by

x = X + l sin ✓ and z = �l cos ✓.

The kinetic energy is then given by:

T =1

2MẊ

2 +1

2m

�ẋ

2 + ż2�

=1

2MẊ

2 +1

2m

⇣Ẋ + l✓̇ cos ✓

⌘2+⇣l✓̇ sin ✓

⌘2�.

This simplifies to

T =1

2(m + M) Ẋ2 +

1

2m

2lẊ ✓̇ cos ✓ +

⇣l✓̇

⌘2�.

23

M

m

�

x

y

Figure 9: Example: a pendulum on a movable support.

The potential energy is given by

V = �mgl cos ✓.

We can now obtain the equations of motions by taking derivatives of the Lagrangianwith respect to the coordinates and to their time derivatives. This is done separately forthe two coordinates. The Lagrange equation 118 for the coordinate X is given by:

d

dt

@ (T � V )

@Ẋ

�� @ (T � V )

@X

= 0,

leading to

(m + M) Ẍ + mld

dt

⇣✓̇ cos ✓

⌘= 0

or(m + M) Ẍ = ml

⇣✓̇

2 sin ✓ � ✓̈ cos ✓⌘

.

Note that the partial derivative with respect to Ẋ is only taken on those places wherethis variable occurs but that the derivative with respect to the time t acts on all variablesincluding ✓ en ✓̇. Another point to note here is that quantity @ (T � V ) /@Ẋ is conserved(i.e. does not change with time). This follows always immediately if the Lagrangian doesnot depend on one of the coordinates (here X). You can check easily that this quantityis here the total momentum in the X-direction.

For the other coordinate, ✓, we obtain:

d

dt

@ (T � V )

@✓̇

�� @ (T � V )

@✓

= 0,

leading to

ml

⇣l✓̈ + Ẍ cos ✓ � Ẋ ✓̇ sin ✓

⌘+ mlẊ ✓̇ sin ✓ + mgl sin ✓ = 0

or

✓̈ +Ẍ

l

cos ✓ +g

l

sin ✓ = 0.

This example shows how straightforward the equations of motion can be derived withthe Lagrange formalism as compared to deriving them from Newton’s formalism whichinvolves force vectors.

24

6.4 Particle in a central force field

For a particle in a central field the motion takes place in a plane (see 6.1 of [1]). Wechoose polar coordinates. The velocity has a radial and a tangential component. Thekinetic and the potential energies are given by

T =m

2

⇣ṙ

2 + r2✓̇2⌘

and V = V (r) .

The Lagrange equation for r is

d

dt

✓@L

@ṙ

◆= mr̈ =

@L

@r

= mr✓̇2 � dVdr

and for ✓:d

dt

✓@L

@✓̇

◆= m

d

dt

⇣r

2✓̇

⌘=@L

@✓

= 0.

Here we find that mr2✓̇, the angular momentum, is conserved as the Lagrangian does notdepend on ✓.

References

[1] Analytical Mechanics, G.R. Fowles and G.L. Cassiday 6th or 7th edition (ThomsonLearning, inc., 1999), ISBN 9780534408138.

25

Introduction Kinematics. Description of the states of physical system. Examples. Point particles. Several particlesPendulum: example of a system with constraintsRigid bodyFluidElectromagnetic field

Equations of motion and degrees of freedomRelativistic particlesParticles, connected by springsThe diffusion equationA general solution of the diffusion equation

Euler equation for fluidSystems with constraintsPendulumPendulum on the movable support

Maxwell's equationsQuantum mechanics. Schrödinger equation

Lagrange mechanicsCovariant form

The principle of least actionExamples of LagrangiansLagrangian of a free particle and Gallilean invariancePendulumPendulum on a movable support Particle in a central force field