# 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).

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

#### Curl

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

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

To see what this definition is good for, let us calculate the integral ${\displaystyle \oint \mathbf {A} \cdot d\mathbf {r} }$  over a closed curve ${\displaystyle {\mathcal {C}}.}$  (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 ${\displaystyle \mathbf {A} }$  along ${\displaystyle {\mathcal {C}}}$  (or around the surface enclosed by ${\displaystyle {\mathcal {C}}}$ ). Let's start with the boundary of an infinitesimal rectangle with corners ${\displaystyle A=(0,0,0),}$  ${\displaystyle B=(0,dy,0),}$  ${\displaystyle C=(0,dy,dz),}$  and ${\displaystyle D=(0,0,dz).}$

The contributions from the four sides are, respectively,

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

${\displaystyle (^{*}{}^{*}{}^{*})\quad \left[{\frac {\partial A_{z}}{\partial y}}-{\frac {\partial A_{y}}{\partial z}}\right]dy\,dz=({\hbox{curl}}\,\mathbf {A} )_{x}\,dy\,dz.}$

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

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 ${\displaystyle \partial \Sigma }$  of ${\displaystyle \Sigma }$  survive.

The bottom line: ${\displaystyle \oint _{\partial \Sigma }\mathbf {A} \cdot d\mathbf {r} =\int _{\Sigma }{\hbox{curl}}\,\mathbf {A} \cdot d\mathbf {\Sigma } .}$

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

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

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

${\displaystyle \int _{\Sigma }\left({\hbox{curl}}\,{\frac {\partial f}{\partial \mathbf {r} }}\right)\cdot d\mathbf {\Sigma } =\oint _{\partial \Sigma }{\frac {\partial f}{\partial \mathbf {r} }}\cdot d\mathbf {r} =0.}$

#### Divergence

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

${\displaystyle {\hbox{div}}\,\mathbf {A} ={\frac {\partial }{\partial \mathbf {r} }}\cdot \mathbf {A} ={\frac {\partial A_{x}}{\partial x}}+{\frac {\partial A_{y}}{\partial y}}+{\frac {\partial A_{z}}{\partial z}}.}$

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

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

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

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

If we fill up a region ${\displaystyle R}$  with infinitesimal parallelepipeds and add up their net outward fluxes, we get ${\displaystyle \int _{R}{\hbox{div}}\,\mathbf {A} \,d^{3}r.}$  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 ${\displaystyle \partial R}$  of ${\displaystyle R}$  survive. The bottom line:

${\displaystyle \int _{\partial R}\mathbf {A} \cdot d\mathbf {\Sigma } =\int _{R}{\hbox{div}}\,\mathbf {A} \,d^{3}r.}$

This is Gauss' law. Note that the left-hand side depends solely on the boundary ${\displaystyle \partial R}$  of ${\displaystyle R.}$  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 ${\displaystyle \Sigma }$  is a closed surface — and thus the boundary ${\displaystyle \partial R}$  or a region of space ${\displaystyle R}$  — then ${\displaystyle \Sigma }$  itself has no boundary (symbolically, ${\displaystyle \partial \Sigma =0}$ ). Combining Stokes' theorem with Gauss' law we have that

${\displaystyle \oint _{\partial \partial R}\mathbf {A} \cdot d\mathbf {r} =\int _{\partial R}{\hbox{curl}}\,\mathbf {A} \cdot d\mathbf {\Sigma } =\int _{R}{\hbox{div curl}}\,\mathbf {A} \,d^{3}r.}$

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: ${\displaystyle \partial \partial =0.}$  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:

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

#### Some useful identities

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