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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?

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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.

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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.

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  • $\begingroup$ Can you please explain how you arrived at the expression for $\delta S$ from the definition of action, especially with regards to $dx$ $\endgroup$ – IanDsouza Apr 2 '16 at 9:27
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    $\begingroup$ I'm using $L$ to mean the Lagrangian density rather than Lagrangian, so the action would be defined as $\int{d^4x L}$ in 4D spacetime. Since you have only one spatial dimension, $d^4x \rightarrow dx dt$. Then $\delta S = \int{dx dt \delta L}$ which gives the expression for $\delta S$ I have written. $\endgroup$ – Orca Apr 2 '16 at 9:49
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    $\begingroup$ In response to your statement about doing experiments at different points in space and getting the same result, I would say that for this potential $U(x)$, you do not actually get the same result for your experiment. Consider $U(1)$ and $U(2)$ at different $x=1$ and $x=2$. These are different values, therefore you experience different potentials at different spacetime points and cannot get the same result for your experiment. An analogy would be doing the same experiment in different gravitational fields, and expecting the same result. $\endgroup$ – Orca Apr 2 '16 at 9:55
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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.

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  • $\begingroup$ As often happens our answers are very similar :) $\endgroup$ – Valter Moretti Apr 2 '16 at 12:24
  • $\begingroup$ @ValterMoretti: True :) $\endgroup$ – Qmechanic Apr 2 '16 at 12:26
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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.

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  • $\begingroup$ Okay.. I see that the consideration is the symmetry of the action. But now I am confused about some physical intuition. I was lead to believe that translational symmetry in a system is intricately related to the fact that if I do an experiment at different points in space and get similar results (up to the initial translation), the system is said to be translationally symmetric. This seems to be true for the above setup. But now I assume that this notion of translational symmetry is not important for Noether's theorem. Is this correct? $\endgroup$ – IanDsouza Apr 2 '16 at 8:17
  • $\begingroup$ @IanDsouza: correct. This is a common misunderstanding of what Noether's theorem states. $\endgroup$ – John Rennie Apr 2 '16 at 9:13

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