# This Quantum World/Appendix/Fields

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## Contents

### 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).

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

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

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 $z$:

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

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.

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

If we take the limit $dx\rightarrow0$ (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

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

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

• the gradient $\frac{\partial f}{\partial\mathbf{r}}$ 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 $\frac{d\mathbf{r}}{d\lambda},$ which is tangent on $\mathcal{C}.$

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

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

The gradient of $f$ is another input-output device: pop in $d\mathbf{r},$ and get the difference

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

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

#### Curl

The curl of a vector field $\mathbf{A}$ is defined by

$\hbox{curl}\,\mathbf{A}=\frac{\partial}{\partial\mathbf{r}}\times\mathbf{A}=\left(\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z}\right)\mathbf{\hat x}+ \left(\frac{\partial A_x}{\partial z}-\frac{\partial A_z}{\partial x}\right)\mathbf{\hat y}+ \left(\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y}\right)\mathbf{\hat z}.$

To see what this definition is good for, let us calculate the integral $\oint\mathbf{A}\cdot d\mathbf{r}$ over a closed curve $\mathcal{C}.$ (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 $\mathbf{A}$ along $\mathcal{C}$ (or around the surface enclosed by $\mathcal{C}$). 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,

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

$(^*{}^*{}^*)\quad\left[\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z}\right]dy\,dz=(\hbox{curl}\,\mathbf{A})_x\,dy\,dz.$

Let us represent this infinitesimal rectangle of area $dy\,dz$ (lying in the $y$-$z$ plane) by a vector $d\mathbf{\Sigma}$ whose magnitude equals $d\Sigma=dy\,dz,$ 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 $d\mathbf{\Sigma}$ is related to the direction of circulation.) This allows us to write (***) as a scalar (product) $\hbox{curl}\,\mathbf{A}\cdot d\mathbf{\Sigma}.$ 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 $\Sigma$ with infinitesimal rectangles and add up their circulations, we get $\int_\Sigma\hbox{curl}\,\mathbf{A}\cdot d\mathbf{\Sigma}.$

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 $\partial\Sigma$ of $\Sigma$ survive.

The bottom line: $\oint_{\partial\Sigma}\mathbf{A}\cdot d\mathbf{r} =\int_\Sigma\hbox{curl}\,\mathbf{A}\cdot d\mathbf{\Sigma}.$

This is Stokes' theorem. Note that the left-hand side depends solely on the boundary $\partial\Sigma$ of $\Sigma.$ 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 $\mathbf{A}$ is the gradient of a scalar field $f,$ and if $\mathcal{C}$ is a curve from $\mathbf{A}$ to $\mathbf{b},$ then

$\int_\mathcal{C} \mathbf{A}\cdot d\mathbf{r}=\int_\mathcal{C} df=f(\mathbf{b})-f(\mathbf{A}).$

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

$\int_\Sigma\left(\hbox{curl}\,\frac{\partial f}{\partial\mathbf{r}}\right)\cdot d\mathbf{\Sigma}=\oint_{\partial\Sigma}\frac{\partial f}{\partial\mathbf{r}}\cdot d\mathbf{r} =0.$

#### Divergence

The divergence of a vector field $\mathbf{A}$ is defined by

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

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

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

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

$\left[\frac{\partial A_x}{\partial x}+\frac{\partial A_y}{\partial y}+\frac{\partial A_z}{\partial z}\right]\,dx\,dy\,dz=\hbox{div}\,\mathbf{A}\,d^3r.$

If we fill up a region $R$ with infinitesimal parallelepipeds and add up their net outward fluxes, we get $\int_R\hbox{div}\,\mathbf{A}\,d^3r.$ 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:

$\int_{\partial R}\mathbf{A}\cdot d\mathbf{\Sigma} =\int_R\hbox{div}\,\mathbf{A}\,d^3r.$

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

$\oint_{\partial\partial R}\mathbf{A}\cdot d\mathbf{r} =\int_{\partial R}\hbox{curl}\,\mathbf{A}\cdot d\mathbf{\Sigma}=\int_R\hbox{div curl}\,\mathbf{A}\,d^3r.$

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:

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

#### Some useful identities

$d\mathbf{r}\times\left(\frac{\partial}{\partial\mathbf{r}}\times\mathbf{A}\right)\frac{\partial}{\partial\mathbf{r}}(\mathbf{A}\cdot d\mathbf{r})-\left(d\mathbf{r}\cdot\frac{\partial}{\partial\mathbf{r}}\right)\mathbf{A}$
$\frac{\partial}{\partial\mathbf{r}}\times\left(\frac{\partial}{\partial\mathbf{r}}\times\mathbf{A}\right)= \frac{\partial}{\partial\mathbf{r}}\left(\frac{\partial}{\partial\mathbf{r}}\cdot\mathbf{A}\right) -\left(\frac{\partial}{\partial\mathbf{r}}\cdot\frac{\partial}{\partial\mathbf{r}}\right)\mathbf{A}.$