Noether's theorem for space translational symmetry

Question

Imagine a ramp potential of the form $U(x) = a*x + b$ in 1D space. This corresponds to a constant force field over $x$. If I do a classical mechanics experiment with a particle, the particle behaves in the "same way" no matter what the initial position of the particle is. This should give rise to space translational symmetry.

Now, consider Newton's equations, $\dot{p} = -U'(x)$. For the potential above, $U'(x) = a \neq 0$. Therefore, $\dot{p} \neq 0$. This is not in accordance with Noether's theorem for translational symmtery. What am I missing here?

Answer 1

Translational symmetry in the sense of the standard formulation of Noether theorems means that the Lagrangian is invariant under the action of the group of spatial translations. This is not the case in your example because $U$ does not admit such invariance.

However there is another, more physical, version of the idea of translational invariance for a physical system:

The class of solutions of the equation of motion is invariant under spatial displacements.

In other words, if $$x=x(t)$$ is a solution with initial conditions $$x(0)=x_0\:,\quad \frac{dx}{dt}|_{t=0}=\dot{x}_0\:,$$ the solution with initial conditions $$x(0)=x_0 + s\:,\quad \frac{dx}{dt}|_{t=0}=\dot{x}_0$$ must be $$x=x(t)+s$$ that is, the initial solution changed by the same initial given translation at each time $t$. This fact is by no means trivial.

This invariance requirement is valid for your example as you can directly check. However, since this invariance requirement is weaker than the one used in the standard version of Noether theorem, it does not imply that the momentum is conserved.

In Lagrangian formulation the two notions of invariance are not equivalent. The Noetherian one implies the second one but the converse implication is false. In Hamiltonian formulation they are equivalent provided we restrict ourselves to deal with canonical transformations.

The natural question however arises whether this weaker notion of invariance of your system implies the existence of a conserved quantity (diferent from the momentum). The answer is positive in our case. There is in fact another, weaker, version of Noether theorem stating that, if the Lagrangian is not invariant under the one-parameter ($\epsilon$) group of transformations $$x \to x_\epsilon\:, \quad \dot{x} \to \dot{x}_\epsilon = \frac{d}{dt}x_\epsilon$$ but, at first order in the parameter $\epsilon$, the transformed Lagrangian differs from the initial one just due to a total derivative $$\frac{d}{dt}f(t,x) = \frac{\partial f}{\partial x}\dot{x} + \frac{\partial f}{\partial t}$$ then there is a conserved quantity along the solution of the motion equations: $$I(t,x, \dot{x}) = \frac{\partial L}{\partial \dot{x}} \partial_\epsilon x_\epsilon|_{\epsilon=0} - f(t,x)\:.$$ The proof is a trivial generalization of the know classical one. In the considered case $$L(t,x, \dot{x}) = \frac{m}{2}\dot{x}^2 - ax-b\:.$$ Thus, since our group of transformations is $$x \to x+\epsilon\:, \quad \dot{x} \to \dot{x}\:,$$ we have $$\partial_{\epsilon}|_{\epsilon=0} L(t,x_\epsilon, \dot{x}_\epsilon)= -a = \frac{d}{dt} (-at)\:.$$ We conclude that there exists a conserved quantity. This is $$I(t,x, \dot{x}) = m \dot{x} + at\:.$$ A posteriori this is obvious from the equations of motion themselves, but it also arises form a weak symmetry of the Lagrangian.

Answer 2

The Lagrangian is

\begin{equation} L = \frac{1}{2} \dot{x}^2 - ax-b. \end{equation}

Introducing spatial translation $x \rightarrow x+\Delta$ for constant $\Delta$ we see that

\begin{equation} L \rightarrow L' = \frac{1}{2} \dot{x}^2 - ax - a\Delta -b. \end{equation}

Therefore the action changes as

\begin{equation} \delta S = \int{dxdt \; (L'-L)} = \int{dx dt \; (-a\Delta)} \neq 0. \end{equation}

Therefore $U(x)$ has broken the translational symmetry. There can be no associated conserved current.

Answer 3

I) Let the Lagrangian be

$$\tag{1} L~=~\frac{m}{2}v^2-U(x), \qquad v~:=~\dot{x}.$$

Let the force

$$\tag{2} F~=~-U'(x) $$

be a constant.

II) Infinitesimal translations

$$\tag{3} \delta x~=~\varepsilon $$

is a quasi-symmetry

$$\tag{4} \delta L ~=~\varepsilon \frac{df}{dt}, \qquad f~:=~Ft $$

of the Lagrangian (1). Here $\varepsilon$ is an infinitesimal parameter. The corresponding bare Noether charge is

$$\tag{5} Q^0 ~:=~p, \qquad p~:=~\frac{\partial L}{\partial v}~=~mv,$$

so the corresponding full Noether charge is

$$\tag{6} Q~:=~Q^0-f~=~mv - Ft. $$

Noether's theorem states that the quantity (6) is conserved in time, as one can easily verify.

III) The system also possesses other quasi-symmetries (and thereby conservation laws by Noether's theorem). E.g. the following infinitesimal transformation

$$\tag{7} \delta x~=~\varepsilon t $$

is a quasi-symmetry

$$\tag{8} \delta L ~=~\varepsilon \frac{df}{dt}, \qquad f~:=~mx+\frac{F}{2}t^2 $$

of the Lagrangian (1). The corresponding bare Noether charge is

$$\tag{9} Q^0 ~:=~p t, $$

so the corresponding full Noether charge is

$$\tag{10} Q~:=~Q^0-f~=~mvt- mx -\frac{F}{2}t^2. $$

IV) If one think a bit more, one can probably construct a quasi-symmetry whose corresponding Noether charge is the acceleration itself.

Answer 4

Noether's theorem tells us that a conserved quantity is related to a symmetry of the action, where the action $S$ is given by:

$$ S = \int L dt $$

where $L$ is the Lagrangian given by:

$$ L = T - V $$

Since the potential $V$ is a function of position the Lagrangian and hence the action is not symmetric under displacements in space.