This Quantum World/Appendix/Fields

Fields

As you will remember, a function is a machine that accepts a number and returns a number. A field is a function that accepts the three coordinates of a point or the four coordinates of a spacetime point and returns a scalar, a vector, or a tensor (either of the spatial variety or of the 4-dimensional spacetime variety).

Gradient

Imagine a curve $𝒞$ in 3-dimensional space. If we label the points of this curve by some parameter $\lambda ,$ then $𝒞$ can be represented by a 3-vector function $𝐫(\lambda ).$ We are interested in how much the value of a scalar field $f(x,y,z)$ changes as we go from a point $𝐫(\lambda )$ of $𝒞$ to the point $𝐫(\lambda +d\lambda )$ of $𝒞.$ By how much $f$ changes will depend on how much the coordinates $(x,y,z)$ of $𝐫$ change, which are themselves functions of $\lambda .$ The changes in the coordinates are evidently given by

${(}^{*})dx=\frac{dx}{d\lambda }d\lambda ,dy=\frac{dy}{d\lambda }d\lambda ,dz=\frac{dz}{d\lambda }d\lambda ,$

while the change in $f$ is a compound of three changes, one due to the change in $x,$ one due to the change in $y,$ and one due to the change in $z$:

${(}^{*}{}^{*})df=\frac{df}{dx}dx+\frac{df}{dy}dy+\frac{df}{dz}dz.$

The first term tells us by how much $f$ changes as we go from $(x,y,z)$ to $(x+dx,y,z),$ the second tells us by how much $f$ changes as we go from $(x,y,z)$ to $(x,y+dy,z),$ and the third tells us by how much $f$ changes as we go from $(x,y,z)$ to $(x,y,z+dz).$

Shouldn't we add the changes in $f$ that occur as we go first from $(x,y,z)$ to $(x+dx,y,z),$ then from $(x+dx,y,z)$ to $(x+dx,y+dy,z),$ and then from $(x+dx,y+dy,z)$ to $(x+dx,y+dy,z+dz)$? Let's calculate.

$\frac{\partial f(x+dx,y,z)}{\partial y}=\frac{\partial [f(x,y,z)+\frac{\partial f}{\partial x}dx]}{\partial y}=\frac{\partial f(x,y,z)}{\partial y}+\frac{{\partial }^{2}f}{\partial y\partial x}dx.$

If we take the limit $dx\to 0$ (as we mean to whenever we use $dx$), the last term vanishes. Hence we may as well use $\frac{\partial f(x,y,z)}{\partial y}$ in place of $\frac{\partial f(x+dx,y,z)}{\partial y}.$ Plugging (*) into (**), we obtain

$df=(\frac{\partial f}{\partial x}\frac{dx}{d\lambda }+\frac{\partial f}{\partial y}\frac{dy}{d\lambda }+\frac{\partial f}{\partial z}\frac{dz}{d\lambda })d\lambda .$

Think of the expression in brackets as the dot product of two vectors:

• the gradient $\frac{\partial f}{\partial 𝐫}$ of the scalar field $f,$ which is a vector field with components $\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z},$
• the vector $\frac{d𝐫}{d\lambda },$ which is tangent on $𝒞.$

If we think of $\lambda$ as the time at which an object moving along $𝒞$ is at $𝐫(\lambda ),$ then the magnitude of $\frac{d𝐫}{d\lambda }$ is this object's speed.

$\frac{\partial }{\partial 𝐫}$ is a differential operator that accepts a function $f(𝐫)$ and returns its gradient $\frac{\partial f}{\partial 𝐫}.$

The gradient of $f$ is another input-output device: pop in $d𝐫,$ and get the difference

$\frac{\partial f}{\partial 𝐫}\cdot d𝐫=df=f(𝐫+d𝐫)-f(𝐫).$

The differential operator $\frac{\partial }{\partial 𝐫}$ is also used in conjunction with the dot and cross products.

Curl

The curl of a vector field $𝐀$ is defined by

$\text{curl}𝐀=\frac{\partial }{\partial 𝐫}\times 𝐀=(\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z})\widehat{𝐱}+(\frac{\partial A_x}{\partial z}-\frac{\partial A_z}{\partial x})\widehat{𝐲}+(\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y})\widehat{𝐳}.$

To see what this definition is good for, let us calculate the integral $\oint 𝐀\cdot d𝐫$ over a closed curve $𝒞.$ (An integral over a curve is called a line integral, and if the curve is closed it is called a loop integral.) This integral is called the circulation of $𝐀$ along $𝒞$ (or around the surface enclosed by $𝒞$). Let's start with the boundary of an infinitesimal rectangle with corners $A=(0,0,0),$ $B=(0,dy,0),$ $C=(0,dy,dz),$ and $D=(0,0,dz).$

The contributions from the four sides are, respectively,

• $\overline{AB}:A_y(0,dy/2,0)dy,$
• $\overline{BC}:A_z(0,dy,dz/2)dz=[A_z(0,0,dz/2)+\frac{\partial A_z}{\partial y}dy]dz,$
• $\overline{CD}:-A_y(0,dy/2,dz)dy=-[A_y(0,dy/2,0)+\frac{\partial A_y}{\partial z}dz]dy,$
• $\overline{DA}:-A_z(0,0,dz/2)dz.$

These add up to

${(}^{*}{}^{*}{}^{*})[\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z}]dydz=(\text{curl}𝐀{)}_{x}dydz.$

Let us represent this infinitesimal rectangle of area $dydz$ (lying in the $y$-$z$ plane) by a vector $d\mathrm{\Sigma }$ whose magnitude equals $d\mathrm{\Sigma }=dydz,$ and which is perpendicular to the rectangle. (There are two possible directions. The right-hand rule illustrated on the right indicates how the direction of $d\mathrm{\Sigma }$ is related to the direction of circulation.) This allows us to write (***) as a scalar (product) $\text{curl}𝐀\cdot d\mathrm{\Sigma }.$ Being a scalar, it it is invariant under rotations either of the coordinate axes or of the infinitesimal rectangle. Hence if we cover a surface $\mathrm{\Sigma }$ with infinitesimal rectangles and add up their circulations, we get $\underset{\mathrm{\Sigma }}{\int }\text{curl}𝐀\cdot d\mathrm{\Sigma }.$

Observe that the common sides of all neighboring rectangles are integrated over twice in opposite directions. Their contributions cancel out and only the contributions from the boundary $\partial \mathrm{\Sigma }$ of $\mathrm{\Sigma }$ survive.

The bottom line: ${\oint }_{\partial \mathrm{\Sigma }}𝐀\cdot d𝐫=\underset{\mathrm{\Sigma }}{\int }\text{curl}𝐀\cdot d\mathrm{\Sigma }.$

This is Stokes' theorem. Note that the left-hand side depends solely on the boundary $\partial \mathrm{\Sigma }$ of $\mathrm{\Sigma }.$ So, therefore, does the right-hand side. The value of the surface integral of the curl of a vector field depends solely on the values of the vector field at the boundary of the surface integrated over.

If the vector field $𝐀$ is the gradient of a scalar field $f,$ and if $𝒞$ is a curve from $𝐀$ to $𝐛,$ then

$\underset{𝒞}{\int }𝐀\cdot d𝐫=\underset{𝒞}{\int }df=f(𝐛)-f(𝐀).$

The line integral of a gradient thus is the same for all curves having identical end points. If $𝐛=𝐀$ then $𝒞$ is a loop and $\underset{𝒞}{\int }𝐀\cdot d𝐫$ vanishes. By Stokes' theorem it follows that the curl of a gradient vanishes identically:

$\underset{\mathrm{\Sigma }}{\int }\left(\text{curl}\frac{\partial f}{\partial 𝐫}\right)\cdot d\mathrm{\Sigma }={\oint }_{\partial \mathrm{\Sigma }}\frac{\partial f}{\partial 𝐫}\cdot d𝐫=0.$

Divergence

The divergence of a vector field $𝐀$ is defined by

$\text{div}𝐀=\frac{\partial }{\partial 𝐫}\cdot 𝐀=\frac{\partial A_x}{\partial x}+\frac{\partial A_y}{\partial y}+\frac{\partial A_z}{\partial z}.$

To see what this definition is good for, consider an infinitesimal volume element $d^{3}r$ with sides $dx,dy,dz.$ Let us calculate the net (outward) flux of a vector field $𝐀$ through the surface of $d^{3}r.$ There are three pairs of opposite sides. The net flux through the surfaces perpendicular to the $x$ axis is

$A_x(x+dx,y,z)dydz-A_x(x,y,z)dydz=\frac{\partial A_x}{\partial x}dxdydz.$

It is obvious what the net flux through the remaining surfaces will be. The net flux of $𝐀$ out of $d^{3}r$ thus equals

$[\frac{\partial A_x}{\partial x}+\frac{\partial A_y}{\partial y}+\frac{\partial A_z}{\partial z}]dxdydz=\text{div}𝐀d^{3}r.$

If we fill up a region $R$ with infinitesimal parallelepipeds and add up their net outward fluxes, we get $\underset{R}{\int }\text{div}𝐀d^{3}r.$ Observe that the common sides of all neighboring parallelepipeds are integrated over twice with opposite signs — the flux out of one equals the flux into the other. Hence their contributions cancel out and only the contributions from the surface $\partial R$ of $R$ survive. The bottom line:

$\underset{\partial R}{\int }𝐀\cdot d\mathrm{\Sigma }=\underset{R}{\int }\text{div}𝐀d^{3}r.$

This is Gauss' law. Note that the left-hand side depends solely on the boundary $\partial R$ of $R.$ So, therefore, does the right-hand side. The value of the volume integral of the divergence of a vector field depends solely on the values of the vector field at the boundary of the region integrated over.

If $\mathrm{\Sigma }$ is a closed surface — and thus the boundary $\partial R$ or a region of space $R$ — then $\mathrm{\Sigma }$ itself has no boundary (symbolically, $\partial \mathrm{\Sigma }=0$). Combining Stokes' theorem with Gauss' law we have that

${\oint }_{\partial \partial R}𝐀\cdot d𝐫=\underset{\partial R}{\int }\text{curl}𝐀\cdot d\mathrm{\Sigma }=\underset{R}{\int }\text{div curl}𝐀d^{3}r.$

The left-hand side is an integral over the boundary of a boundary. But a boundary has no boundary! The boundary of a boundary is zero: $\partial \partial =0.$ It follows, in particular, that the right-hand side is zero. Thus not only the curl of a gradient but also the divergence of a curl vanishes identically:

$\frac{\partial }{\partial 𝐫}\times \frac{\partial f}{\partial 𝐫}=0,\frac{\partial }{\partial 𝐫}\cdot \frac{\partial }{\partial 𝐫}\times 𝐀=0.$

Some useful identities

$d𝐫\times (\frac{\partial }{\partial 𝐫}\times 𝐀)\frac{\partial }{\partial 𝐫}(𝐀\cdot d𝐫)-(d𝐫\cdot \frac{\partial }{\partial 𝐫})𝐀$

$\frac{\partial }{\partial 𝐫}\times (\frac{\partial }{\partial 𝐫}\times 𝐀)=\frac{\partial }{\partial 𝐫}(\frac{\partial }{\partial 𝐫}\cdot 𝐀)-(\frac{\partial }{\partial 𝐫}\cdot \frac{\partial }{\partial 𝐫})𝐀.$