Notes on Mathematical Methods for Physicists Chapter3

Notes on Mathematical Methods for Physicists$\S3$

Chapter3 Vector Analysis

$\S3.1$Review of Basic Properties

(1)

(2)

(3) Vector polynomials (e.g.$\bold{A}-\bold{2B}+\bold{3C}$) are well-defined.

(4) Unit vectors$\hat{\bold{e}}_i$.

(5) Magnitude

(6) Dot product

(7) Orthogonal

(8) Projection

(9) Direction cosines

(10) Single-column matrix

(11) The addition and multiplication of vectors can also be replaced by single-column matrices.

(12) Transpose of a single-column matrix , called a$\,\,\,\textbf{row vector}$.

(13) Dot product replaced by multiplication of matrices

$\S3.2$Vectors in$3-D$Space

$\square$Vectors or Cross Product

Definition :

The direction of$\bold{C}$, i.e., that of$\hat{\bold{e}}_C$, is perpendicular to the plane of$\bold{A}$and$\bold{B}$, such that$\bold{A}$,$\bold{B}$, and$\bold{C}$form a right-handed system.

This can be equivalently represented as

Note that the cross product is a quantity specifically defined for$3-D$space.

$\square$Scalar Triple Product

Definition :

This quantity is equivalent to the volume of the space in the picture above.

There is

Example Reciprocal Lattice

Let$\bold{a}$,$\bold{b}$, and$\bold{c}$(not necessarily mutually perpendicular) represent the vectors that define a crystal lattice. The displacements from one lattice point to another may then be written

In the band theory of solids , it is useful to introduce what is called a$\,\,\,\textbf{reciprocal lattice}$$\bold{a}'$,$\bold{b}'$,$\bold{c}'$such that

and with

Then

$\square$Vector Triple Product

The equation can be proved by Levi-Civita symbols and Kronecker delta. There is

$\S3.3$Coordinate Transformations

$\square$Rotations

Rotate the axis as the picture below

So the$\,\,\,\textbf{unchanged}$vector$\bold{A}$now takes the$\,\,\,\textbf{changed}$form

which is equivalent to the matrix equation

It is easy to find that

is orthogonal.

$\square$Orthogonal Transformations

It is no accident that the transformation describing a rotation in$\R^2$was orthogonal. We can rewrite$\bold{S}$as

Summary : The transformation from one orthogonal Cartesian coordinate system to another Cartesian system is described by an$\,\,\,\textbf{orthogonal}$matrix.

$\square$Reflections

For simplicity , consider first the$\,\,\,\textbf{inversion}$operation , in which the sign of each coordinate is reversed. This will lead to

which clearly results in$\det S=-1$. The change in sign of the determinant corresponds to the change from a right-handed to a left-handed coordinate system (which cannot be accomplished by a rotation). Reflection about a plane also changes the sign of the determinant.

This formula for vector addition , multiplication by a scalar , and the dot product are unaffected by a reflection transformation of the coordinates , but this is not true of the cross product. To see this , look at one component :

The unchanged vectors are called$\,\,\,\textbf{axial vectors}$, or$\,\,\,\textbf{pseudovectors}$, while the other are called$\,\,\,\textbf{polar vectors}$, or just$\,\,\,\textbf{vectors}$. There is

It is easy to find that$\bold{A}\cdot(\bold{B}\cross\bold{C})$changes sign in the transformation , so it is called a$\,\,\,\textbf{pseudoscalar}$. And the vector$\bold{A}\cross(\bold{B}\cross\bold{C})$is a vector , so there is a general principle that a product with an odd number of pseudo quantities is "pseudo" , while those with even numbers of pseudo quantities are not.

$\square$Successive Operations

Two transformations can be carried out successively. This rule is only true for rotations and/or reflections by applying their relevant orthogonal transformations.

Notes :

1.The operations take place in right-to-left order.

2.The product of two orthogonal matrices is orthogonal , so the combined operation is an orthogonal transformation.

$\S3.4$Rotations in$\R^3$

Rotations in$\R^3$is important in practice , so we discuss now in some detail the treatment of rotations in$\R^3$. There is

The argument we made to evaluate$\hat{\bold{e}}'_\mu\cdot\hat{\bold{e}}_\nu$could as easily have been made with the roles of the two unit vectors reversed

In$\R^2$, we find that all the elements of$S$depended on a single variable , the rotation angle. In$\R^3$, the number of independent variables needed to specify a general rotation is three. The usual parameters used to specify$\R^3$rotations are the$\,\,\,\textbf{Euler angles}$. The three steps describing rotation of the coordinate axes are the following :

(1) The coordinates are rotated about the$\hat{\bold{e}}_3$axis counterclockwise (as viewed from positive$\hat{\bold{e}}_3$) through an angle$\alpha$in the range$0\leq\alpha<2\pi$, into new axes denoted$\hat{\bold{e}}'_1$,$\hat{\bold{e}}'_2$,$\hat{\bold{e}}'_3$. (The polar direction is not changed ; the$\hat{\bold{e}}_3$and$\hat{\bold{e}}'_3$axes coincide.)

(2) The coordinates are rotated about the$\hat{\bold{e}}'_2$axis counterclockwise (as viewed from positive$\hat{\bold{e}}'_2$) through an angle$\beta$in the range$0\leq\beta\leq\pi$, into new axes denoted$\hat{\bold{e}}''_1$,$\hat{\bold{e}}''_2$,$\hat{\bold{e}}''_3$. (This tilts the polar direction toward the$\hat{\bold{e}}'_1$direction , but leaves$\hat{\bold{e}}'_2$unchanged.)

(3) The coordinates are now rotated about the$\hat{\bold{e}}''_3$axis counterclockwise (as viewed from positive$\hat{\bold{e}}''_3$) through an angle$\gamma$in the range$0\leq\gamma<2\pi$, into the final axes , denoted$\hat{\bold{e}}'''_1$,$\hat{\bold{e}}'''_2$,$\hat{\bold{e}}'''_3$. (This rotation leaves the polar direction ,$\hat{\bold{e}}''_3$, unchanged.)

The matrices that represent those rotations are

 

Note the order of$S$.

Note that those matrices are orthogonal with determinant$+1$.

$\S3.5$Differential Vector Operators

We are going to discuss$\,\,\,\textbf{vector field}$. Physicists need to be able to characterize the rate at which the values of vectors (and also scalars) change with position , and this is most effectively done by introducing differential vector operator concepts. It turns out that there are a large number of relations between these differential operators , and it is our current objective to identify such relations and learn how to use them.

$\square$Gradient ,$\nabla$

The$\,\,\,\textbf{gradient}$characterizes the change of a scalar field , here$\varphi$, with position. There is

which is of the form corresponding to the dot product of

We have given the$3\cross1$matrix of derivatives the name$\nabla\varphi$(often referred to in speech as "del phi" or "grad phi") ; we give the differential of position its customary name$d\bold{r}$.

The$\nabla\varphi$is a vector. To prove this , we must show that it transforms under rotation of the coordinate system according to$(\nabla\varphi)'=S(\nabla\varphi)$. We have

Having now established the legitimacy of the form$\nabla\varphi$, we proceed to give$\nabla$a life of its own. We therefore define

Note that$\nabla$is a$\,\,\,\textbf{vector differential operator}$.

In physics , we have$\bold{F}(\bold{r})=-\nabla V(\bold{r})$.

There is a simple geometric interpretation of the gradient :$\nabla V$is the direction of most rapid increase in$V$, and its magnitude is associated with the speed of increasing.

Example Gradient of$r^n$

$\square$Divergence ,$\nabla\cdot$

The$\,\,\,\textbf{divergence}$of a vector is defined as the operation

Example Divergence of Central Force Field

Consider$\nabla\cdot f(r)\hat{\bold{r}}$. We write

Using

We can simplify the equation :

Of course , I have already learnt about the origin of divergence ---- the representative of flux in a differential form. If the physical problem being described is one in which fluid (molecules) are neither created or destroyed , we will also have an$\,\,\,\textbf{equation of continuity}$, of the form

If the divergence of a vector field is zero everywhere , its lines of force will entirely of closed loops ; such vector fields are termed$\,\,\,\textbf{solenoidal}$.

$\square$Curl ,$\nabla\cross$

This vector is called the$\,\,\,\textbf{curl}$of$\bold{V}$. Note that when the determinant is evaluated , it must be expanded in a way that causes the derivatives in the second row to be applied to the functions in the third row (and not to anything in the top row) ; we will encounter this situation repeatedly , and will identify the evaluation as being$\,\,\,\textbf{from the top down}$.

Example Curl of a Central Force Field

Consider$\nabla\cross[f(r)\hat{\bold{r}}]$. Writing

so the result is 0.

Noting that a vector whose curl is zero everywhere is termed$\,\,\,\textbf{irrotational}$.

$\S3.6$Differential Vector Operators : Further Properties

$\square$Successive Applications of$\nabla$

The possible results include the following :

All five of these expressions involve second derivatives , and all five appear in the second-order differential equations of mathematical physics , particularly in electromagnetic theory.

$\square$Laplacian

Laplacian is the divergence of the gradient. We have

When$\varphi$is the electrostatic potential , we have$\nabla\cdot\nabla\varphi=0$in vacuum. This is called the$\,\,\,\textbf{Laplacian equation}$.

Often the Laplacian is written as$\nabla^2$, or$\Delta$in the older European literature.

Example Laplacian of a Central Field Potential

Calculate$\laplacian\varphi(r)$.

$\square$Irrotational and Solenoidal Vector Fields

The curl of the divergence :

The divergence of the curl :

We can thus see that the divergence is$\,\,\,\textbf{irrotational}$, and the curl is$\,\,\,\textbf{solenoidal}$. Therefore ,

$\square$Maxwell's Equations

We who are learning physics all know this well.

$\square$Vector Laplacian

There is

(This can be proved by Levi-Civita symbols.) The term$\nabla\cdot\nabla\bold{V}$is called the$\,\,\,\textbf{vector Laplacian}$, written as$\laplacian\bold{V}$.

Example Electromagnetic Wave Equation

In vacuum , we can derive the Maxwell equation , and thus we have

Then the equation is decided only by$\bold{E}$. In vacuum , there is$\nabla\cdot\bold{E}=0$. The result is the vector electromagnetic wave equation for$\bold{E}$,

The components are three scalar wave equations.

$\square$Miscellaneous Vector Identities

These are further examples of usages of$\nabla$.

Example Divergence and Curl of a Product

First , simplify$\nabla\cdot(f\,\bold{V})$,

Now , simplify$\nabla\cross(f\,\bold{V})$. Consider the$x$-component :

This is the$x$-component of$f(\nabla\cross\bold{V})+(\nabla f)\cross\bold{V}$, so we have

Example Gradient of a Dot Product

We can prove the equation by applying$\nabla_A$and$\nabla_B$, which operate only on$\bold{A}$or$\bold{B}$.

$\S3.7$Vector Integration

$\square$Line Integrals

Possible forms :

These integrals are calculated over some path$C$.

$\square$Surface Integrals

Possible forms :

Here ,$d\bold{\sigma}$is$\hat{\bold{n}}dA$, where$\hat{\bold{n}}$is a unit vector indicating the normal direction.

$\square$Volume Integrals

$\S3.8$Integral Theorems

$\square$Gauss' Theorem

This theorem connects a surface integral of a vector to a volume integral of the of the divergence of the vector. Assume a vector$\bold{A}$and it is$C^1$over a$\,\,\,\textbf{simply connected}$region of$\R^3$. Then Gauss' theorem states that

Notice that if the region of interest is the complete$\R^3$, and the volume integral converges , the surface integral must vanish , giving that

$\square$Green's Theorem

A frequently useful corollary of Gauss' theorem is a relation known as$\,\,\,\textbf{Green's theorem}$. If$u$and$v$are two scalar functions , we have

Also we have

So there is

Thus , applying Gauss' theorem , we obtain

This is Green's theorem. An alternative of Green's theorem is

These results we have already obtained are of great importance. For instance , consider a vector field$\bold{B}(x,y,z)$which satisfies that$\bold{B}(x,y,z)=B(x,y,z)\bold{a}$, in which$\bold{a}$is a constant vector to an arbitrary direction. Applying the equation above , we can get

In a similar manner , using$\bold{B}=\bold{a}\cross\bold{P}$in which$\bold{a}$is a constant vector , we may show

These last two forms of Gauss' theorem are used in the vector form of Kirchhoff diffraction theory.

$\square$Stokes' Theorem

This theorem relates a surface integral of a derivative of a function to the line integral of the function , with the path of integration being the perimeter bounding the surface.

The theorem states that

These can be applied in Faraday's law and Oersted's law.

$\S3.9$Potential Theory

In some fields of physics , we introduce$\,\,\,\textbf{potential}$to simplify our problems.

$\square$Scalar Potential

If in a simply connected region of space , an irrotational vector field can be expressed as the negative gradient of a scalar function$\varphi$:

We call$\varphi$a$\,\,\,\textbf{scalar potential}$, and this simplifies the$\bold{F}$(three functions) into one function. Notice that$\bold{F}$is a derivative , so the potential is only determined up to a additive constant , which can be used to adjust its value at infinity (usually zero) or at some other reference point.

Consider

Thus we can conclude that$\bold{F}$is irrotational , or to say , is$\,\,\,\textbf{conservative}$, because

$\square$Vector Potential

A solenoidal force field can be written as$\bold{B}=\nabla\cross\bold{A}$.

Our construction is

Checking the$y$- and$z$-components of$\nabla\cross\bold{A}$first , noting that$A_x=0$,

The$x$-component :

$\bold{B}$is solenoidal , so$\nabla\cdot\bold{B}=0$. Then

So

Actually ,$\bold{A}$is far from unique , as we can add to it the gradient of$\,\,\,\textbf{any}$scalar function. Moreover , our verification of$\bold{A}$was independent of the values of$x_0$and$y_0$. In addition , we can derive another formula for$\bold{A}$in which the roles of$x$and$y$are interchanged :

Thus , we may introduce different kinds of gauge to restrict the vector potential. In electromagnetic theories , there are$\,\,\,\textbf{Lorentz gauge}$:

and$\,\,\,\textbf{Coulomb gauge}$:

and transformations of$\bold{A}$and$\varphi$to satisfy them or other legitimate gauge condition are called$\,\,\,\textbf{gauge transformations}$.

$\square$Gauss' Law

The Gauss' law states that for an arbitrary volume$V$,

$q$is the charge in the volume.

To prove this law , notice that the case that$\partial V$does not enclose$q$is easy to handle. If there is no charge in the volume , then$\nabla\cdot\bold{E}=0$is satisfied throughout the entire volume$V$. Thus ,

If$q$is within the volume$V$, we must be more devious. We can use a volume as below.

We now consider$\oint\bold{E}\cdot d\vec{\sigma}$in the surface of this modified volume. The contribution from the connecting tube will become negligible in the limit that it shrinks toward zero cross section , as$\bold{E}$is finite everywhere on the tube's surface. The integral over the modified$\partial V'$will thus be that of the original$\partial V$(over the outer boundary , which we designate$S$) , plus that of the inner spherical surface ($S'$).

But note that the "outward" direction for$S'$is toward smaller$r$, so$d\vec{\sigma}'=-\hat{\bold{r}}dA$. We have

There is$dA=\delta^2d\Omega$, so

Then

We can also use its differential form

Thus we can see that Gauss' law is the integral form of one of Maxwell's equations.

$\square$Poisson's Equation

Assuming a situation independent of time , write$\bold{E}=-\nabla\varphi$, we obtain

This kind of equation is called$\,\,\,\textbf{Poisson's equation}$. For a more special situation ,$\nabla^2\varphi=0$is called$\,\,\,\textbf{Laplace's equation}$.

Here , we are going to consider a particular situation. To make Poisson's equation apply to a point chare$q$, the Dirac delta function is what we need. Thus , we write

For charge$q$at$\bold{r}=0$. Inserting the point-charge potential for$\varphi$, we have

This means the derivatives of$1/r$is meaningless at the point$\bold{r}=0$, and only if applied in a integral can it be meaningful. Also ,

$\square$Helmholtz's Theorem

We now turn to two important theorems which establish conditions for the existence and uniqueness of solutions to time-independent problems in electromagnetic theory.

Claim 1

A vector field is uniquely specified by giving its divergence and its curl within a simply connected region and its normal component on the boundary.

Let$\bold{P}$be a vector field satisfying the conditions

where$s$may be interpreted as a given source (charge) density and$\bold{c}$as a given circulation (current) density. Assuming that the normal component$P_n$on the boundary is also given , we want to show that$\bold{P}$is unique.

The proof goes by assuming the existence of a different vector field$\bold{P}'$, which also satisfies those criteria. Consider$\bold{Q}=\bold{P}-\bold{P}'$, which must have$\nabla\cdot\bold{Q}=0$,$\nabla\cross\bold{Q}=0$, and$Q_n=0$identically. So there is$\bold{Q}=-\nabla\varphi$,$\nabla^2\varphi=0$.

Now apply Green's theorem ,

But$Q_n=0$, so$\nabla\varphi\cdot d\vec{\sigma}=0$everywhere on$\partial V$. Thus we obtain

This can only be satisfied if$\bold{Q}$is identically zero throughout the region. Therefore$\bold{P}=\bold{P}'$.

$Q.E.D\quad\square$

Claim 2

A vector$\bold{P}$with both source and circulation densities vanishing at infinity may be written as the sum of two parts , one of which is irrotational , the other of which is solenoidal.

Then$\bold{P}$can be written as$\bold{P}=-\nabla\varphi+\nabla\cross\bold{A}$. There is also$s=\nabla\cdot\bold{P}$and$\bold{c}=\nabla\cross\bold{P}$defined well.

The formulas proposed for$\varphi$and$\bold{A}$are as following ,

Then ,

What we need to do is just check these equation. First , check that$-\nabla^2\varphi=s$, we examine

Now check that$\nabla\cross(\nabla\cross\bold{A})=\bold{c}$. There is

We look first at

Now , integrate by parts , we have an equation (from example 3.7.3)

Apply on the problem , we get

Thus , we only need to check that$-\nabla^2\bold{A}=\bold{c}$. For component$j$,

$\S3.10$Curvilinear Coordinates

$\square$Orthogonal Coordinates in$\R^3$

In Cartesian coordinates , a point$(x_0,y_0,z_0)$can be identified as the intersection of three planes ,$x=x_0$,$y=y_0$and$z=z_0$. A change in$x$is a displacement$\,\,\,\textbf{normal}$to the plane$x=x_0$, so as the changes in$y$and$z$. Then those changes are always along the same directions marked as$\hat{\bold{e}}_x$,$\hat{\bold{e}}_y$and$\hat{\bold{e}}_z$.

Now consider a spherical polar coordinate system as an example of a curvilinear coordinate system , with the picture below.

(The definition is well-known and I am not going to repeat it here.)

Important observations :

(1) General coordinates need not be lengths ;

(2) A surface of constant coordinate value may have a normal whose direction depends on position ;

(3) Surfaces with different constant values of a same coordinate need to be parallel ;

(4) Changes in the value of a coordinate may move$\bold{r}$in both an amount and a direction that depends on position.

It is obvious that these unit vectors are mutually perpendicular , meaning that a small displacement in one of those directions will not change the values of the other two coordinates. But , a "large" displacement would bring changes to the values , because the normals are position-dependent.

A coordinate system is said to be$\,\,\,\textbf{orthogonal}$if its unit vectors are mutually perpendicular.

It is significant to realize that

because there is not a constant unit vector. So , the component formulas for$\bold{V}=V_r\hat{\bold{e}}_r+V_\theta\hat{\bold{e}}_\theta+V_\varphi\hat{\bold{e}}_\varphi$describe decompositions applicable to the point at which the vector is specified.

When we deal a question with an arbitrary curvilinear system , with coordinates labeled$(q_1,q_2,q_3)$, we naturally consider how$q_i$changes with the Cartesian coordinates. In this case ,$x$can be thought as a function of$q_i$, thus we have

But

Thus

where

Spaces with a measure of distance given as above are called$\,\,\,\textbf{metric}$or$\,\,\,\textbf{Riemannian}$.

The equation above can be seen as the dot product of two vectors , which are in the$dq_i$and$dq_j$direction. Thus , if the coordinate is perpendicular , the coefficients will vanish when$i\neq j$.

Rewrite the equation as

Note that$h_i$may be position-dependent. We call$h_i$the scale factors or$\,\,\,\text{Lam}\acute{\text{e}}$coefficients.

$\square$Integrals in Curvilinear Coordinates

After we introduced the definition of the scale factors , we can use them to set up formulas for integration in the curvilinear coordinates. There is

Also surface integrals

For volume integrals , they take the form as

$\square$Differential Operators in Curvilinear Coordinates

Gradient

Just replace$dr_i$by$h_idq_i$:

Thus ,

Divergence

The value of divergence must be the same as the value calculated in Cartesian coordinates , but the volume element is totally different from the former coordinate.

Laplacian

Curl

$\square$Circular Cylindrical Coordinates

This coordinate uses$(\rho,\varphi,z)$to describe positions. The ranges are

But for$\rho=0$,$\varphi$is not well defined.

There is

$\square$Spherical Polar Coordinates

Thus ,

$\square$Rotation and Reflection in Spherical Coordinates

Rotation

Consider a rotation identified by Euler angles$(\alpha,\beta,\gamma)$converts the coordinates of a point from$(r,\theta,\varphi)$to$(r,\theta',\varphi')$.

The first rotation is by an angle$\alpha$around$z$axis , thus only changes$\varphi$into$\varphi-\alpha$. It does not change the components of$\bold{A}$, either.

The second rotation changes the the values of both$\theta$and$\varphi$, and also changes the directions of$\hat{\bold{e}}_\theta$and$\hat{\bold{e}}_\varphi$.

Referring to the picture above , we can summarize that

Carrying out the spherical trigonometry (though this book does not tell me how to do it) , there is

The third rotation leaves the components of$\bold{A}$unchanged but requires the replacement of$\varphi'$by$\varphi'-\gamma$.

Summarizing ,

and

Reflection