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:

\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}
1. O. Bang, Nonlinear mathematical physics: lecture notes, 2020, unpublished.