# Geometrical intuition for Noether's Theorem

I have been reading some questions about the relation between Noether's Theorem and Lie Algebras and I wanted to get some intuition on it, but I didn't find what I really wanted. Also, the majority of questions I found on this topic were in the context of field theory or QM.

The theorem states that, if the Lagrangian is invariant under the one-parameter group of diffeomorphisms $$\{\mathbf{f}^s:\mathcal{M}\to\mathcal{M}, s\in\mathbb{R}\}$$, which is a Lie Group, then there's a conserved quantity given by $$\mathcal{N}=\sum_{i=1}^np_i(t)\frac{\partial\mathbf{f}^s(q(t))}{\partial s}\bigg|_{s=0}\equiv\sum_{i=1}^np_i(t)\psi_i(t)$$

As far as I understood, the functions $$\psi_i(t)$$ are the infinitesimal generators of the Lie Group and form a basis for the tangent space (defined as the Lie Algebra) of this Group at the identity. Is this correct?

Also, if $$\psi(t)=\{\psi_1(t),\dots,\psi_n(t)\}$$ form a basis of a vector space, can we see $$\mathcal{N}$$ as a linear combination of this basis, and hence an element of the vector space? If so, is there some geometrical intuition, motive or meaning for which $$\mathcal{N}$$ has to be conserved?

The most basic intuitive explanation of Noether's theorem is that it is the extension to generalised coordinates of the principle that since force is defined as the rate of change of momentum, if there is no force in a particular direction, the component of momentum in that direction is a constant.

A symmetry of the Lagrangian in configuration space means that the Langrangian doesn't change as you move along that direction, so its gradient along that direction is zero. The partial derivatives $$f_i=\partial \mathcal{L}/\partial q_i$$ are the generalised forces, and the partial derivatives $$p_i=\partial \mathcal{L}/\partial \dot{q_i}$$ are the corresponding generalised momenta. Thus, if the Lagrangian is constant along the direction $$q_i$$, the force in that direction is zero. If the Langrangian is constant along another direction that is some linear combination of the basis of generalised coordinates $$\sum \psi_i q_i$$ then the corresponding linear combination of forces is zero: $$\sum \psi_i f_i=0$$. And the corresponding linear combination of generalised momenta $$\sum \psi_i p_i$$ is constant.

Thus, $$\psi=(\psi_1,\psi_2,\ldots,\psi_n)$$ is the vector pointing in the direction where the symmetry tells us there is no force. It is pointing in the direction along which the symmetry operation pushes points.

Some of your confusion might arise from there being two Lie groups involved here, one inside the other. You talk about the one-parameter group of diffeomorphisms being "the Lie Group". Having one parameter, it has only one infinitesimal generator. But then you also refer to the $$\psi_i(t)$$ being a set of $$n$$ infinitesimal generators to "the Lie Group". They can't be the same group! One is the Lie group of the symmetry, the other is the Lie group of the configuration space. The $$\psi_i(t)$$ are not just generators of the Lie algebra of the configuration space, they are the components of one vector in the Lie algebra of the configuration space $$\sum \psi_i(t)$$, that generates the one-parameter Lie group of the symmetry. The coefficients of the linear combination we have to use are incorporated into the components.

It's like if we break down the vector $$\psi=(1,2,3)$$ into components $$\psi_1=e_x$$, $$\psi_2=2e_y$$, $$\psi_3=3e_z$$. The vector is the sum of its components: $$\psi=\sum \psi_i$$. Although these three component vectors between them can generate the full $$xyz$$ space by taking weighted combinations, we are not free to choose any element of $$xyz$$-space, only the simple unweighted sum of these three components. The choice of $$\mathcal{N}$$ is baked in to the $$\psi_i$$.

• I think I got what's happening. There's a conserved quantity along the direction in which there are no forces by the symmetry in our system. The vector $\psi(t)$ that points in this direction has components $\{\psi_1(t),\dots,\psi_n(t)\}$ and this single vector generates the Lie Group of symmetries. The conserved quantity is then a linear combination of these components weighted by each momentum $p_i$, because $p_i$ would be the conserved quantity if $\psi_i$ was the only non-zero component of $\psi$. Mar 11 at 17:41
• Yes, that's right. For example, consider the rotational symmetry of a central force. Being a central force, the component of force in a tangential direction is zero. At a point $(x,y)$ the force is zero in the direction $\psi=(-y,x)$. Thus the momentum $(m\dot{x},m\dot{y})$ has a component in this direction $mx\dot{y}-my\dot{x}$, which is the formula for the angular momentum. $\psi=(-y,x)$ is the single vector at each point always pointing in the direction of no force. Mar 11 at 18:04
• Your answer is more than I was looking for when I wrote the question! Thanks! Mar 11 at 18:07