1 files changed, 23 insertions, 70 deletions
diff --git a/source/know/concept/beltrami-identity/index.md b/source/know/concept/beltrami-identity/index.md
index be9a344..fe93e28 100644
--- a/source/know/concept/beltrami-identity/index.md
+++ b/source/know/concept/beltrami-identity/index.md
@@ -26,9 +26,11 @@ $$\begin{aligned}
     = \pdv{L}{f} - \dv{}{x} \Big( \pdv{L}{f'} \Big)
 \end{aligned}$$
 
-We now want to know exactly how $$L$$ depends on the free variable $$x$$,
-since it is a function of $$x$$, $$f(x)$$ and $$f'(x)$$.
-Using the chain rule:
+We now want to know exactly how $$L$$ depends on the free variable $$x$$.
+Of course, $$x$$ may appear explicitly in $$L$$,
+but usually $$L$$ also has an *implicit* dependence on $$x$$ via $$f(x)$$ and $$f'(x)$$.
+To find a relation between this implicit and explicit dependence,
+we start by using the chain rule:
 
 $$\begin{aligned}
     \dv{L}{x}
@@ -45,16 +47,17 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Although we started from the "hard" derivative $$\idv{L}{x}$$,
-we arrive at an expression for the "soft" derivative $$\ipdv{L}{x}$$,
-describing the *explicit* dependence of $$L$$ on $$x$$:
+we arrive at an expression for the "soft" derivative $$\ipdv{L}{x}$$
+describing only the *explicit* dependence of $$L$$ on $$x$$:
 
 $$\begin{aligned}
     - \pdv{L}{x}
     = \dv{}{x} \bigg( f' \pdv{L}{f'} - L \bigg)
 \end{aligned}$$
 
-What if $$L$$ does not explicitly depend on $$x$$, i.e. $$\ipdv{L}{x} = 0$$?
-In that case, the equation can be integrated to give the **Beltrami identity**:
+What if $$L$$ does not explicitly depend on $$x$$ at all, i.e. $$\ipdv{L}{x} = 0$$?
+In that case, the equation can be integrated to give the **Beltrami identity**,
+where $$C$$ is a constant:
 
 $$\begin{aligned}
     \boxed{
@@ -63,69 +66,19 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Where $$C$$ is a constant.
-This says that the left-hand side is a conserved quantity in $$x$$,
-which could be useful to know.
-If we insert a concrete expression for $$L$$,
-the Beltrami identity might be easier to solve for $$f$$ than the full Euler-Lagrange equation.
-The assumption $$\ipdv{L}{x} = 0$$ is justified;
-for example, if $$x$$ is time, it means that the potential is time-independent.
-
-
-## Higher dimensions
-
-Above, a 1D problem was considered, i.e. $$f$$ depended only on a single variable $$x$$.
-Consider now a 2D problem, such that $$J[f]$$ is given by:
-
-$$\begin{aligned}
-    J[f] = \iint_{(x_0, y_0)}^{(x_1, y_1)} L(f, f_x, f_y, x, y) \dd{x} \dd{y}
-\end{aligned}$$
-
-In which case the Euler-Lagrange equation takes the following form:
-
-$$\begin{aligned}
-    0 = \pdv{L}{f} - \dv{}{x} \Big( \pdv{L}{f_x} \Big) - \dv{}{y} \Big( \pdv{L}{f_y} \Big)
-\end{aligned}$$
-
-Once again, we calculate the hard $$x$$-derivative of $$L$$ (the $$y$$-derivative is analogous):
-
-$$\begin{aligned}
-    \dv{L}{x}
-    &= \pdv{L}{f} \dv{f}{x} + \pdv{L}{f_x} \dv{f_x}{x} + \pdv{L}{f_y} \dv{f_y}{x} + \pdv{L}{x}
-    \\
-    &= \dv{f}{x} \bigg( \dv{}{x} \Big( \pdv{L}{f_x} \Big) + \dv{}{y} \Big( \pdv{L}{f_y} \Big) \bigg)
-    + \pdv{L}{f_x} \dv{f_x}{x} + \pdv{L}{f_y} \dv{f_y}{x} + \pdv{L}{x}
-    \\
-    &= \dv{}{x} \Big( f_x \pdv{L}{f_x} \Big) + \dv{}{y} \Big( f_x \pdv{L}{f_y} \Big) + \pdv{L}{x}
-\end{aligned}$$
-
-This time, we arrive at the following expression for the soft derivative $$\ipdv{L}{x}$$:
-
-$$\begin{aligned}
-    - \pdv{L}{x}
-    &= \dv{}{x} \Big( f_x \pdv{L}{f_x} - L \Big) + \dv{}{y} \Big( f_x \pdv{L}{f_y} \Big)
-\end{aligned}$$
-
-Due to the derivatives, this cannot be cleanly turned into an analogue of the 1D Beltrami identity,
-and therefore we use that name only in the 1D case.
-
-However, if $$\ipdv{L}{x} = 0$$, this equation is still useful.
-For an off-topic demonstration of this fact,
-let us choose $$x$$ as the transverse coordinate, and integrate over it to get:
-
-$$\begin{aligned}
-    0
-    &= - \int_{x_0}^{x_1} \pdv{L}{x} \dd{x}
-    \\
-    &= \int_{x_0}^{x_1} \dv{}{x} \Big( f_x \pdv{L}{f_x} - L \Big) + \dv{}{y} \Big( f_x \pdv{L}{f_y} \Big) \dd{x}
-    \\
-    &= \Big[ f_x \pdv{L}{f_x} - L \Big]_{x_0}^{x_1} + \dv{}{y} \int_{x_0}^{x_1} \Big( f_x \pdv{L}{f_y} \Big) \dd{x}
-\end{aligned}$$
-
-If our boundary conditions cause the boundary term to vanish (as is often the case),
-then the integral on the right is a conserved quantity with respect to $$y$$.
-While not as elegant as the 1D Beltrami identity,
-the above 2D counterpart still fulfills the same role.
+This says that the left-hand side is a conserved quantity
+with respect to $$x$$, which could be useful to know.
+Furthermore, for some Lagrangians $$L$$,
+the Beltrami identity is easier to solve for $$f$$ than the full Euler-Lagrange equation.
+The condition $$\ipdv{L}{x} = 0$$ is often justified:
+for example, if $$x$$ is time, it simply means that the potential is time-independent.
+
+When we add more dimensions, e.g. for $$L(f, f_x, f_y, x, y)$$,
+the above derivation no longer works due to the final integration step,
+so the name *Beltrami identity* is only used in 1D.
+Nevertheless, a generalization does exist
+that can handle more dimensions:
+[Noether's theorem](/know/concept/noethers-theorem/).