path: root/source/know/concept/noethers-theorem/index.md

---
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.