*** empty log message ***

vector-curl.tex

@@ -1,83 +1,13 @@
-\section*{Vector curl}
+\section*{Curl in tensor form}
 The curl of a vector function can be expressed in tensor form as
 $$\mathop{\rm curl}{\bf F}=\epsilon_{ijk}\,{\partial F_k\over\partial x_j}$$
 where $\epsilon_{ijk}$ is the Levi-Civita tensor.
 The following script demonstrates that this formula is equivalent
 to computing curl the old fashioned way.
-First, define $\epsilon_{ijk}$.
 
 \medskip
-{\tt
-epsilon=zero(3,3,3)
+\verb$-- Define epsilon.$
 
-epsilon[1,2,3]=1
-
-epsilon[2,3,1]=1
-
-epsilon[3,1,2]=1
-
-epsilon[3,2,1]=-1
-
-epsilon[1,3,2]=-1
-
-epsilon[2,1,3]=-1
-}
-
-\medskip
-\noindent
-Next, define a generic vector function $\bf F$ and
-then compute $A=\epsilon_{ijk}\,\partial F_k/\partial x_j$.
-The first summation is over $k$ which corresponds to indices 3 and 4.
-The second summation is over $j$ which (with $k$ out of the way)
-corresponds to indices 2 and 3.
-
-\medskip
-{\tt
-F=(FX(),FY(),FZ())
-
-A=outer(epsilon,d(F,(x,y,z)))
-
-A=contract(A,3,4)
-
-A=contract(A,2,3)
-}
-
-\medskip
-\noindent
-Now compute curl the old fashioned way and check for equality.
-
-\medskip
-{\tt
-B=(
-
-\ d(F[3],y)-d(F[2],z),
-
-\ d(F[1],z)-d(F[3],x),
-
-\ d(F[2],x)-d(F[1],y)
-
-)
-
-\medskip
-A-B
-}
-
-$$\left(\matrix{0\cr0\cr0}\right)$$
-
-\newpage
-
-\noindent
-The following is a variation on the previous script.
-The product $\epsilon_{ijk}\,\partial F_k/\partial x_j$
-is computed in just one line of code.
-In addition, there is an optimization.
-The outer product and the first contraction have been replaced with a
-dot product.
-
-\medskip
-\verb$F=(FX(),FY(),FZ())$
-
-\medskip
 \verb$epsilon=zero(3,3,3)$
 
 \verb$epsilon[1,2,3]=1$
@@ -92,23 +22,73 @@ dot product.
 
 \verb$epsilon[2,1,3]=-1$
 
-\medskip
-\verb$A=contract(dot(epsilon,d(F,(x,y,z))),2,3)$
+\verb$-- F is a generic vector function.$
 
-\medskip
-\verb$B=($
+\verb$F=(FX(),FY(),FZ())$
 
-\verb$ d(F[3],y)-d(F[2],z),$
+\verb$-- A is the curl of F.$
 
-\verb$ d(F[1],z)-d(F[3],x),$
+\verb$A=outer(epsilon,d(F,(x,y,z)))$
 
-\verb$ d(F[2],x)-d(F[1],y)$
+\verb$A=contract(A,3,4) --contract across k$
 
-\verb$)$
+\verb$A=contract(A,2,3) --contract across j$
 
-\medskip
-\verb$--Are A and B equal? Subtract to find out.$
+\verb$-- B is the curl of F computed the old fashioned way.$
+
+\verb$BX=d(F[3],y)-d(F[2],z)$
+
+\verb$BY=d(F[1],z)-d(F[3],x)$
+
+\verb$BZ=d(F[2],x)-d(F[1],y)$
+
+\verb$B=(BX,BY,BZ)$
+
+\verb$-- Are A and B equal? Subtract to find out.$
 
-\medskip
 \verb$A-B$
 
+$$\left(\matrix{0\cr0\cr0}\right)$$
+
+\newpage
+
+\noindent
+The following is a variation on the previous script.
+The product $\epsilon_{ijk}\,\partial F_k/\partial x_j$
+is computed in just one line of code.
+In addition, the outer product and the contraction across $k$
+are now computed with a dot product.
+
+\medskip
+\verb$F=(FX(),FY(),FZ())$
+
+\verb$epsilon=zero(3,3,3)$
+
+\verb$epsilon[1,2,3]=1$
+
+\verb$epsilon[2,3,1]=1$
+
+\verb$epsilon[3,1,2]=1$
+
+\verb$epsilon[3,2,1]=-1$
+
+\verb$epsilon[1,3,2]=-1$
+
+\verb$epsilon[2,1,3]=-1$
+
+\verb$A=contract(dot(epsilon,d(F,(x,y,z))),2,3)$
+
+\verb$BX=d(F[3],y)-d(F[2],z)$
+
+\verb$BY=d(F[1],z)-d(F[3],x)$
+
+\verb$BZ=d(F[2],x)-d(F[1],y)$
+
+\verb$B=(BX,BY,BZ)$
+
+\verb$-- Are A and B equal? Subtract to find out.$
+
+\verb$A-B$
+
+$$\left(\matrix{0\cr0\cr0}\right)$$
+