Classical Mechanics Joel phần 10 pptx

But we can also thconfigura-ink of this path as a mapping t → η·, ·, ·, t of time into the infinite dimensional space of functions on ordinary space.. We see that his is the integral of

Chapter 8

Field Theory

In section 5.4 we considered the continuum limit of a chain of pointmasses on stretched string We had a situation in which the poten-

tial energy had interaction terms for particle A which depended only

on the relative displacements of particles in the neighborhood of A.

If we take our coordinates to be displacements from equilibrium, and

consider only motions for which the displacement η = η(x, y, z, t)

be-comes differentiable in the continuum limit, then the leading term inthe potential energy is proportional to second derivatives in the spacial

coordinates For our points on a string at tension τ , with mass density

as we can write the Lagrangian as an integral of a Lagrangian density

L(y, ˙y, y 0 , x, t) Actually for our string we had no y or x or t dependence, because we ignored gravity U g =R

ρgy(x, t)dx, and had a homogeneous

string whose properties were time independent In general, however,such dependence is quite possible For a three dimensional object, such

as the equations for the displacement of the atoms in a crystal, we might

have fields ~ η, the three components of the displacement of a particle,

as a function of the three coordinates (x, y, z) determining the particle,

219

as well as time Thus the generalized coordinates are the functions

η i (x, y, z, t), and the Lagrangian density will depend on these, their gradients, their time derivatives, as well as possibly on x, y, z, t Thus

The actual motion of the system will be given by a particular set

of functions η i (x, y, z, t), which are functions over the volume in tion and of t ∈ [t I , t f] The function will be determined by the laws

ques-of dynamics ques-of the system, together with boundary conditions which

depend on the initial configuration η I (x, y, z, t I) and perhaps a finalconfiguration Generally there are some boundary conditions on thespacial boundaries as well For example, our stretched string required

y = 0 at x = 0 and x = L.

Before taking the continuum limit we say that the configuration of

the system at a given t was a point in a large N dimensional tion space, and the motion of the system is a path Γ(t) in this space In the continuum limit N → ∞, so we might think of the path as a path

configura-in an configura-infconfigura-inite dimensional space But we can also thconfigura-ink of this path as

a mapping t → η(·, ·, ·, t) of time into the (infinite dimensional) space

of functions on ordinary space

Hamilton’s principal says that the actual path is an extremum of

the action If we consider small variations δη i (x, y, z, t) which vanish

on the boundaries, then

δI =

Z

dx dy dz dt δ L = 0.

Note that what is varied here are the functions η i, not the coordinates

(x, y, z, t) x, y, z do not represent the position of some atom — they

represent a label which tells us which atom it is that we are talkingabout They may well be the equilibrium position of that atom, but

Notice there is no variation of x, y, z, and t, as we discussed.

The notation is getting awkward, so we need to reintroduce the

notation A ,i = ∂A/∂r i In fact, we see that ∂/∂t enters in the same way as ∂/∂x, so we will set x0 = t and write

for µ = 0, 1, 2, 3, and write η ,µ := ∂ µ η If there are several fields η i, then

∂ µ η i = η i,µ The comma represents the beginning of differentiation, so

we must not use one to separate different ordinary indices

In this notation, we have

where we have thrown away the boundary terms which involve δη i uated on the boundary, which we assumed to be zero Inside the region

eval-of integration, the δη i are independent, so requiring δI = 0 for all functions δη i (x µ) implies

dif-in a form which looks like we are dealdif-ing with a relativistic problem,

because t and spatial coordinates are entering in the same way We

have not made any assumption of relativity, however, and our problemwill not be relativistically invariant unless the Lagrangian density isinvariant under Lorentz transformations (as well as translations).Now consider how the Lagrangian changes from one point in space-time to another, including the variation of the fields, assuming the fieldsobey the equations of motion Then the total derivative for a variation

Note that if the Lagrangian density has no explicit dependence on the coordinates x µ , the stress-energy tensor satisfies an equation ∂ ν T µν

which is a continuity equation

In dynamics of discrete systems we defined the Hamiltonian as H =

P

i p i q˙i − L(q, p, t) Considering the continuum as a limit, L = Rd3x L

is the limit of P

ijk ∆x∆y∆zL ijk , where L ijk depends on q ijk and a few

of its neighbors, and also on ˙q ijk The conjugate momentum p ijk =

∂L/∂ ˙ q ijk = ∆x∆y∆z∂L ijk /∂ ˙ q ijk which would vanish in the continuumlimit, so instead we define

π(x, y, z) = p ijk /∆x∆y∆z = ∂L ijk /∂ ˙ q ijk = δ L/δ ˙q(x, y, z).

Consider the case where L does not depend explicitly on (~x, t), so

This is a continuity equation, similar to the equation from fluid

me-chanics, ∂ρ/∂t + ~ ∇ · (ρ~v) = 0, which expresses the conservation of

mass That equation has the interpretation that the change in themass contained in some volume is equal to the flux into the volume,

because ρ~v is the flow of mass past a unit surface area In the current case, we have four conservation equations, indexed by µ Each of these

can be integrated over space to tell us about the rate of change of the

“charge” Q µ (t) =R

d3V T µ0 (~ x, t), d

dt Q µ (t) =

Z

d3V ∂

∂x i T µi (~ x, t).

We see that his is the integral of the divergence of a vector current

( ~ J µ)i = T µi, which by Gauss’ law becomes a surface integral of the flux

of J µ out of the volume of our system We have been sloppy about ourboundary conditions, but in many cases it is reasonable to assume there

is no flux out of the volume In this case the right hand side vanishes,and we find four conserved quantities

L(η, ˙η, ∇η) depends on spatial derivates of η as well, and we may ask

whether we need to require absense of dependence on ∇η for a

coordi-nate to be cyclic Independence of both η and ∇η implies independence

on an infinite number of discrete coordinates, the values of η(~ r) at

ev-ery point ~ r, which is too restrictive a condition for our discussion We

will call a coordinate field η i cyclic ifL does not depend directly on η i,although it may depend on its derivatives ˙η i and ∇η i

The Lagrange equation then states

then the derivative dΠ/dt involves the integral of a divergence, which

by Gauss’ law is a surface term

We want to discuss the relationship between symmetries and conserved

quantities which is known as Noether’s theorem It concerns

in-finitesimal tranformations of the degrees of freedom η i (x µ) which mayrelate these to degrees of freedom at a changed point That is, the

new fields η 0 (x 0 ) is related to η(x) rather than η(x 0 ), where x µ → x 0

µ=

x µ + δx µ is some infinitesimal transformation of the coordinates ratherthan of the degrees of freedom For a scalar field, like temperature,under a rotation, we would define the new field

η 0 (x 0 ) = η(x),

but more generally the field may also change, in a way that may depend

on other fields,

η i 0 (x 0 ) = η i (x) + δη i (x; η k (x)).

This is what you would expect for a vector field ~ E under rotations,

because the new E x 0 gets a component from the old E y

The Lagrangian is a given function of the old fields L(η i , η i,µ , x µ)

If we substitute in the values of η(x) in terms of η 0 (x 0) we get a newfunction L 0, defined by

L 0 (η 0

i , η 0 i,µ , x 0 µ) = L(η i , η i,µ , x µ ).

The symmetries we wish to discuss are transformations of this typeunder which the form of the Lagrangian density does not change, sothat L 0 is the same functional form as L, or

L 0 (η 0

i , η 0 i,µ , x 0 µ) = L(η 0

i , η i,µ 0 , x 0 µ ).

In considering the action, we integrate the Lagrangian density over

a region of space-time between two spacial slices corresponding to aninitial time and a final time We may, however, consider an arbitraryregion of spacetime Ω ⊂ R4 The corresponding four dimensional vol-

ume in the transformed coordinates is the region x 0 ∈ Ω 0 The actionfor a given field configuration η

∂x 0

∂x

L(η, η ,µ , x)d4x.

The Jacobian is

det (δ µν + ∂ ν δx µ) = 1 + Tr∂δx µ

∂x ν = 1 + ∂ µ δx µ .

It makes little sense to assume the Lagrangian density is invariant unless

the volume element is as well, so we will require the Jacobian to be

identically 1, or ∂ µ δx µ = 0 So then δS = 0 for the symmetries we wish

This differs from S(η) by S 0 (η 0)− S(η) = δ1S + δ2S, because

1 the Lagrangian is evaluated with the field η 0 rather than η,

8.1 NOETHER’S THEOREM 227where

If we define dΣ µ to be an element of the three dimensional surface

Σ = ∂Ω of Ω, with outward-pointing normal in the direction of dΣ µ,the difference in the regions of integration may be written as an integralover the surface,

by Gauss’ Law (in four dimensions)

As ¯δ is a difference of two functions at the same values of x, this

operator commutes with partial differentiation, so ¯δη i,µ = ∂ µ δη¯ i Using

this in the second term of δ1S and the equations of motion in the first,

which holds for arbitrary volumes Ω Thus we have a conservationequation

∂ µ J µ = 0.

The infinitesimal variations may be thought of as proportional to an

infinitesimal parameter , which is often in fact a component of a vector The variations in x µ and η i are then

four-δx µ =  dx µ

dη i d ,

Appendix A

These are some notes on the use of the antisymmetric symbol  ijk forexpressing cross products This is an extremely powerful tool for manip-ulating cross products and their generalizations in higher dimensions,

and although many low level courses avoid the use of , I think this is

a mistake and I want you to become proficient with it

In a cartesian coordinate system a vector ~ V has components V ialongeach of the three orthonormal basis vectors ˆe i , or ~ V =P

where the Kronecker delta δ ij is defined to be 1 if i = j and 0 otherwise.

As the basis vectors ˆe k are orthonormal, i.e orthogonal to each other

and of unit length, we have ˆe i · ˆe j = δ ij

229

Doing a sum over an index j of an expression involving a δ ij isvery simple, because the only term in the sum which contributes is

the one with j = i Thus P

j F (i, j)δ ij = F (i, i), which is to say, one just replaces j with i in all the other factors, and drops the δ ij and the

summation over j So we have ~ A · ~ B =P

i A i B i, the standard expressionfor the dot product1

We now consider the cross product of two vectors, ~ A × ~B, which

is also a bilinear expression, so we must have ~ A × ~ B = (P

i A iˆe i)×

(P

j B j eˆj) =P

iP

j A i B j(ˆe i × ˆe j) The cross product ˆe i × ˆe j is a vector,

which can therefore be written as ~ V = P

k V k eˆk But the vector result

depends also on the two input vectors, so the coefficients V k really

depend on i and j as well Define them to be  ijk, so

ˆ

e i × ˆe j =X

k

 kij eˆk

It is easy to evaluate the 27 coefficients  kij, because the cross product

of two orthogonal unit vectors is a unit vector orthogonal to both ofthem Thus ˆe1׈e2 = ˆe3, so 312 = 1 and  k12 = 0 if k = 1 or 2 Applying

the same argument to ˆe2× ˆe3 and ˆe3× ˆe1, and using the antisymmetry

of the cross product, ~ A × ~ B = − ~ B × ~ A, we see that

123 = 231 = 312 = 1; 132 = 213 = 321 =−1,

and  ijk = 0 for all other values of the indices, i.e  ijk = 0 whenever any

two of the indices are equal Note that  changes sign not only when the

last two indices are interchanged (a consequence of the antisymmetry of

the cross product), but whenever any two of its indices are interchanged Thus  ijk is zero unless (1, 2, 3) → (i, j, k) is a permutation, and is equal

to the sign of the permutation if it exists

Now that we have an expression for ˆe i × ˆe j, we can evaluate

Much of the usefulness of expressing cross products in terms of ’s

comes from the identity

X

k

 kij  k`m = δ i` δ jm − δ im δ j` , (A.5)

1Note that this only holds because we have expressed our vectors in terms of

orthonormal basis vectors.

A.1 VECTOR OPERATIONS 231

which can be shown as follows To get a contribution to the sum, k must be different from the unequal indices i and j, and also different from ` and m Thus we get 0 unless the pair (i, j) and the pair (`, m)

are the same pair of different indices There are only two ways thatcan happen, as given by the two terms, and we only need to verify the

coefficients If i = ` and j = m, the two ’s are equal and the square

is 1, so the first term has the proper coefficient of 1 The second termdiffers by one transposition of two indices on one epsilon, so it musthave the opposite sign

We now turn to some applications Let us first evaluate

Note that ~ A · ( ~ B × ~C) is, up to sign, the volume of the parallelopiped

formed by the vectors ~ A, ~ B, and ~ C From the fact that the  changes

sign under transpositions of any two indices, we see that the same istrue for transposing the vectors, so that

~

A × ( ~ B × ~C) = ~ B ~ A · ~C − ~C ~ A · ~ B. (A.7)

This is sometimes known as the bac-cab formula.

Exercise: Using (A.5) for the manipulation of cross products,show that

From the second definition, we see that the determinant is the volume

of the parallelopiped formed from the images under the linear map A

of the three unit vectors ˆe i, as

(Aˆ e1)· ((Aˆe2)× (Aˆe3)) = det A.

In higher dimensions, the cross product is not a vector, but there

is a generalization of  which remains very useful In an n-dimensional space,  i1i2 i n has n indices and is defined as the sign of the permuta- tion (1, 2, , n) → (i1 2 i n), if the indices are all unequal, and zero

otherwise The analog of (A.5) has (n − 1)! terms from all the

permu-tations of the unsummed indices on the second  The determinant of

Appendix B

The gradient operator

We can define the gradient operator

While this looks like an ordinary vector, the coefficients are not

num-bers V i but are operators, which do not commute with functions of

the coordinates x i We can still write out the components wardly, but we must be careful to keep the order of the operators andthe fields correct

straightfor-The gradient of a scalar field Φ(~ r) is simply evaluated by distributing

the gradient operator

The general application of the gradient operator ~ ∇ to a vector ~ A

gives an object with coefficients with two indices, a tensor Some parts

of this tensor, however, can be simplified The first (which is the trace

233

of the tensor) is called the divergence of the vector, written and defined

This is expressible in terms of the curls of ~ A and ~ B.

The curl is like a cross product with the first vector replaced by the

differential operator, so we may write the i’th component as

where the sign which changed did so due to the transpositions in the

indices on the , which we have done in order to put things in the form

of the definition of the curl Thus

~

∇ · ( ~ A × ~ B) = (~ ∇ × ~ A) · ~ B − ~ A · (~∇ × ~ B). (B.9)