2

Consider one parameter transformation: $y = y ( \tilde{y}, \alpha)$ such that lagrangian satisfies: $\tilde{L}(\tilde{y}, \alpha) = L(y ( \tilde{y}, \alpha))$. We say that equation is invariant under this transformation if: $$\tilde{L}(\tilde{y}, \alpha) = L(\tilde{y}) + \frac{dF(\tilde{y},\alpha)}{dt}$$ Show that: $$\int \frac{\delta L}{\delta y_t} \frac{\partial y}{\partial \alpha} dx - \frac{\partial F}{\partial \alpha}$$ is a constant of motion.

Here $\tilde{y}$ is an old field before transformation, so is lagrangian $\tilde{L}$. Space is 2-dimensional $(x, t)$ and the exercise is listed in section concerning KdV equation.

Any hints?

qoqosz
  • 817

1 Answers1

1

Since OP's problem looks like a homework assignment, we will only provide OP with a series of hints rather than a full solution.

I) The Lagrangian $L~=~\int \! dx ~{\cal L}$ is the spatial integral of the Lagrangian density ${\cal L}$. Let us call the position field $q(x,t)$ and the corresponding velocity field $v(x,t)$. The Lagrangian $L[q(\cdot,t),v(\cdot,t)]$ is a functional of the position and the velocity fields, cf. e.g. this Phys.SE answer. The momentum field is the functional derivative

$$ \tag{1} p(x,t)~:=~\frac{\delta L[q(\cdot,t),v(\cdot,t)]}{\delta v(x,t)}.$$

An infinitesimal variation $\delta L$ of the Lagrangian functional $L$ is given by $$\tag{2} \delta L~=~\int \! dx~\left( \frac{\delta L}{\delta q}\delta q+\frac{\delta L}{\delta v}\delta v \right). $$ The exercise basically asks OP to derive a functional version of Noether's Theorem, which he probably knows for ordinary point mechanics.

II) Let $$\tag{3} \delta q~=~\varepsilon Y$$ be an infinitesimal variation of the position field, where $\varepsilon$ is an infinitesimal constant, and where $Y$ is the generator. Correspondingly, the velocity field transforms as $$\tag{4}\delta v~=~\varepsilon \dot{Y}.$$ (The transformation (3) is a so-called vertical transformation. In general, one could also allow horizontal contributions from variation of $x$ and $t$.)

III) Assume that the transformation (2) is a so-called quasi-symmetry $$\tag{5} \delta\left(\left. L \right|_{v=\dot{q}}\right)~=~ \varepsilon\frac{d}{dt} \left(\left. F\right|_{v=\dot{q}}\right),$$ where $F[q(\cdot,t),v(\cdot,t)]$ is some functional. (If $F=0$ is zero, then the quasi-symmetry becomes a symmetry of the Lagrangian $L$.)

IV) The (temporal component of the) bare Noether current $j^0$ for a vertical transformation (3) is simply the momentum field $p$ times the vertical generator $Y$,

$$\tag{6} j^0 ~:=~p Y. $$

The bare Noether charge $Q^0$ is the spatial integral

$$\tag{7} Q^0 ~:=~ \int \! dx~j^0. $$

The full Noether charge $Q$ is improved with (minus) the $F$-functional,

$$\tag{8} Q ~:=~Q^0-F. $$

Formula (8) corresponds to OP's last formula (v1). Noether's Theorem states that $Q$ is a conserved quantity on-shell, i.e. if the equation of motion

$$\tag{9} \left.\frac{d}{dt} \left(\frac{\delta L}{\delta v}\right|_{v=\dot{q}}\right)~=~\left. \frac{\delta L}{\delta q}\right|_{v=\dot{q}} $$

is satisfied. The proof in the functional setting is very similar to the usual proof given in ordinary point mechanics.

Community
  • 1
Qmechanic
  • 201,751
  • Thank you for your answer. My main problem is how does the partial derivative $\partial F / \partial \alpha$ appears? I can't see this one. – qoqosz Nov 25 '12 at 18:36
  • @qoqosz: The first order contribution is $\alpha~\partial F / \partial \alpha$ in the notation of the exercise. This should be identified with $\varepsilon F$ in my notation. – Qmechanic Nov 25 '12 at 19:35
  • 1
    Thanks a lot! I must think on this a little more but it's definitely clearer now. – qoqosz Nov 25 '12 at 23:07