--- 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.