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 $$\mathcal{C}$$ 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,


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

These add up to



(^*{}^*{}^*)\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}. $$