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2 files changed, 253 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/).
 
 
 
diff --git a/source/know/concept/noethers-theorem/index.md b/source/know/concept/noethers-theorem/index.md
new file mode 100644
index 0000000..5414d0d
--- /dev/null
+++ b/source/know/concept/noethers-theorem/index.md
@@ -0,0 +1,230 @@
+---
+title: "Noether's theorem"
+sort_title: "Noether's theorem"
+date: 2023-09-24
+categories:
+- Physics
+- Mathematics
+layout: "concept"
+---
+
+Consider the following general functional $$J[f]$$,
+where $$L$$ is a known Lagrangian density,
+$$f(x, t)$$ is an unknown function,
+and $$f_x$$ and $$f_t$$ are its first-order derivatives:
+
+$$\begin{aligned}
+    J[f]
+    = \iint_{(x_0, t_0)}^{(x_1, t_1)} L(f, f_x, f_t, x, t) \dd{x} \dd{t}
+\end{aligned}$$
+
+Then the [calculus of variations](/know/concept/calculus-of-variations/)
+states that the $$f$$ which minimizes or maximizes $$J[f]$$
+can be found by solving this Euler-Lagrange equation:
+
+$$\begin{aligned}
+    0
+    = \pdv{L}{f} - \dv{}{x} \Big( \pdv{L}{f_x} \Big) - \dv{}{t} \Big( \pdv{L}{f_t} \Big)
+\end{aligned}$$
+
+Now, the first steps are similar to the derivation of the
+[Beltrami identity](/know/concept/beltrami-identity/)
+(which is a special case of Noether's theorem):
+we need to find relations between the *explicit* dependence
+of $$L$$ on the free variables $$(x, t)$$,
+and its *implicit* dependence via $$f$$, $$f_x$$ and $$f_t$$.
+
+Let us start with $$x$$.
+The "hard" (explicit + implicit) derivative $$\idv{L}{x}$$
+is given by the chain rule:
+
+$$\begin{aligned}
+    \dv{L}{x}
+    &= \pdv{L}{f} \dv{f}{x} + \pdv{L}{f_x} \dv{f_x}{x} + \pdv{L}{f_t} \dv{f_t}{x} + \pdv{L}{x}
+\end{aligned}$$
+
+Inserting the Euler-Lagrange equation into the first term
+and using that $$\idv{f_{t}}{x} = \idv{f_{x}}{t}$$:
+
+$$\begin{aligned}
+    \dv{L}{x}
+    &= \bigg( \dv{}{x} \Big( \pdv{L}{f_x} \Big) + \dv{}{t} \Big( \pdv{L}{f_t} \Big) \bigg) f_x
+    + \pdv{L}{f_x} \dv{f_x}{x} + \pdv{L}{f_t} \dv{f_t}{x} + \pdv{L}{x}
+    \\
+    &= \dv{}{x} \Big( \pdv{L}{f_x} f_x \Big) + \dv{}{t} \Big( \pdv{L}{f_t} f_x \Big) + \pdv{L}{x}
+\end{aligned}$$
+
+Leading to the following expression for the "soft" (explicit only) derivative $$\ipdv{L}{x}$$:
+
+$$\begin{aligned}
+    - \pdv{L}{x}
+    &= \dv{}{x} \Big( \pdv{L}{f_x} f_x - L \Big) + \dv{}{t} \Big( \pdv{L}{f_t} f_x \Big)
+\end{aligned}$$
+
+And then by going through the same process for the other variable $$t$$,
+we arrive at:
+
+$$\begin{aligned}
+    - \pdv{L}{t}
+    &= \dv{}{x} \Big( \pdv{L}{f_x} f_{t} \Big) + \dv{}{t} \Big( \pdv{L}{f_t} f_t - L \Big)
+\end{aligned}$$
+
+Now we define the so-called **stress-energy tensor** $$T_\nu^\mu$$ as a useful abbreviation
+(the name comes from its application in relativity),
+where $$\delta_\nu^\mu$$ is the Kronecker delta:
+
+$$\begin{aligned}
+    T_\nu^\mu
+    &\equiv \pdv{L}{f_\mu} f_\nu - \delta_\nu^\mu L
+\end{aligned}$$
+
+Such that the two relations we just found can be written as follows:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            - \pdv{L}{x}
+            &= \dv{}{x} T_x^x + \dv{}{t} T_x^t
+            \\
+            - \pdv{L}{t}
+            &= \dv{}{x} T_t^x + \dv{}{t} T_t^t
+        \end{aligned}
+    }
+\end{aligned}$$
+
+And with this definition of $$T_\nu^\mu$$
+we can also rewrite the Euler-Lagrange equation in the same way,
+noting that $$f_f = \ipdv{f}{f} = 1$$:
+
+$$\begin{aligned}
+    \boxed{
+        - \pdv{L}{f}
+        = \dv{}{x} T_f^x + \dv{}{t} T_f^t
+    }
+\end{aligned}$$
+
+These three equations are the framework we need.
+What happens if $$L$$ does not explicitly contain $$x$$, so $$\ipdv{L}{x}$$ is zero?
+Then the corresponding equation clearly turns into:
+
+$$\begin{aligned}
+    \dv{}{t} T_x^t
+    &= - \dv{}{x} T_x^x
+\end{aligned}$$
+
+Such *continuity relations* are very common in physics.
+This one effectively says that if $$T_x^t$$ increases with $$t$$,
+then $$T_x^x$$ must decrease with $$x$$ by a certain amount.
+Yes, this is very abstract, but when you apply this technique
+to a specific physical problem, $$T_x^x$$ and $$T_x^t$$
+are usually quantities with a clear physical interpretation.
+
+For $$\ipdv{L}{t} = 0$$ and $$\ipdv{L}{f} = 0$$
+we get analogous continuity relations,
+so there seems to be a pattern here:
+if $$L$$ has a continuous symmetry
+(i.e. there is a continuous transformation
+with no effect on the value of $$L$$),
+then there exists a continuity relation specific to that symmetry.
+This is the qualitative version of **Noether's theorem**.
+
+In general, for $$L(f, f_x, x, t)$$,
+a continuous transformation (not necessarily a symmetry)
+consists of shifting the coordinates $$(f, x, t)$$ as follows:
+
+$$\begin{aligned}
+    f
+    \:\:\to\:\: f + \varepsilon \alpha^f
+    \qquad\qquad
+    x
+    \:\:\to\:\: x + \varepsilon \alpha^x
+    \qquad\qquad
+    t
+    \:\:\to\:\: t + \varepsilon \alpha^t
+\end{aligned}$$
+
+Where $$\varepsilon$$ is the *amount* of shift,
+and $$(\alpha^f, \alpha^x, \alpha^t)$$ are parameters
+controlling the *direction* of the shift in $$(f, x, t)$$-space.
+Given a specific $$L$$, suppose we have found a continuous symmetry,
+i.e. a direction $$(\alpha^f, \alpha^x, \alpha^t)$$
+such that the value of $$L$$ is unchanged by the shift, meaning:
+
+$$\begin{aligned}
+    0
+    &= \dv{L}{\varepsilon} \bigg|_{\varepsilon = 0}
+    = \pdv{L}{f} \dv{f}{\varepsilon} + \pdv{L}{x} \dv{x}{\varepsilon} + \pdv{L}{t} \dv{t}{\varepsilon}
+\end{aligned}$$
+
+Where we set $$\varepsilon = 0$$ to get rid of it.
+Negating and inserting our three equations yields:
+
+$$\begin{aligned}
+    0
+    &= - \pdv{L}{f} \alpha^f - \pdv{L}{x} \alpha^x - \pdv{L}{t} \alpha^t
+    \\
+    &= \bigg( \dv{}{x} T_f^x + \dv{}{t} T_f^t \bigg) \alpha^f
+    + \bigg( \dv{}{x} T_x^x + \dv{}{t} T_x^t \bigg) \alpha^x
+    + \bigg( \dv{}{x} T_t^x + \dv{}{t} T_t^t \bigg) \alpha^t
+\end{aligned}$$
+
+This is a continuity relation!
+Let us make this clearer by defining some *current densities*:
+
+$$\begin{aligned}
+    J^x
+    &\equiv T_f^x \alpha^f + T_x^x \alpha^x + T_t^x \alpha^t
+    \\
+    J^t
+    &\equiv T_f^t \alpha^f + T_x^t \alpha^x + T_t^t \alpha^t
+\end{aligned}$$
+
+So that the above equation can be written
+in the standard form of a continuity relation:
+
+$$\begin{aligned}
+    \boxed{
+        0
+        = \dv{}{x} J^x + \dv{}{t} J^t
+    }
+\end{aligned}$$
+
+This is the quantitative version of **Noether's theorem**:
+for every symmetry $$(\alpha^f, \alpha^x, \alpha^t)$$ we can find,
+Noether gives us the corresponding continuity relation.
+This result is easily generalized to more variables $$x_1, x_2, ...$$
+and/or more unknown functions $$f_1, f_2, ...$$.
+
+Continuity relations tell us about conserved quantities.
+Of the free variables $$(x, t)$$,
+we choose one as the *dynamic* coordinate (usually $$t$$)
+and then all others are *transverse* coordinates.
+Let us integrate the continuity relation over all transverse variables:
+
+$$\begin{aligned}
+    0
+    &= \int_{x_0}^{x_1} \! \bigg( \dv{}{x} J^x + \dv{}{t} J^t \bigg) \dd{x}
+    \\
+    &= \big[ J^x \big]_{x_0}^{x_1} + \dv{}{t} \int_{x_0}^{x_1} \! J^t \dd{x}
+\end{aligned}$$
+
+Usually the problem's boundary conditions ensure that $$[J^x]_{x_0}^{x_1} = 0$$,
+in which case $$\int_{x_0}^{x_1} J^t \dd{x}$$ is a conserved quantity (i.e. a constant)
+with respect to the dynamic coordinate $$t$$.
+
+In the 1D case $$L(f, f_t, t)$$
+(i.e. if $$L$$ is a *Lagrangian* rather than a *Lagrangian density*),
+the current density $$J^x$$ does not exist,
+so the conservation of the current $$J^t$$ is clearly seen:
+
+$$\begin{aligned}
+    \dv{}{t} J^t
+    &= 0
+\end{aligned}$$
+
+
+
+## References
+1.  O. Bang,
+    *Nonlinear mathematical physics: lecture notes*, 2020,
+    unpublished.