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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. |