Categories: Mathematics, Physics.

# Noether’s theorem

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 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
(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:

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:

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*:

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:

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:

## References

- O. Bang,
*Nonlinear mathematical physics: lecture notes*, 2020, unpublished.