This Quantum World/Appendix/Fields

From Wikibooks, open books for an open world
< This Quantum World(Redirected from This quantum world/Appendix/Fields)
Jump to: navigation, search

Fields[edit]

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[edit]

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[edit]

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).

Dydz.png

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.
Right hand rule simple.png

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}.

Stokes.png

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[edit]

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[edit]


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}.