A question regarding Noether's theorem. Does external force and torque "break the symmetry? How this is possible since symmetry is related to space?

Question

I always wanted to physicist but I choose mechanical engineering. Now I am trying to get accepted in a graduate program in physics so I am learning classical mechanics. So I am new to physics, I appreciate any help I can get.

So we learned that when external torque is zero, angular momentum is conserved, external force is zero, momentum is conserved and in a closed system when there is no heat and work transfer across the boundary (just a one case where energy is conserved) energy is conserved.

I assume that that still holds to be correct.

Then I was introduced to Noether's Theorem really packing all conservation laws in one theorem.

It says when space is homogeneous (Lagrangian is invariant to translation), linear momentum is conserved.

When space is isotropic (Lagrangian is invariant to rotation), angular momentum is conserved.

Time is homogeneous, energy is conserved.

So I am trying to put what I learned and this together but failing to do so.

I understand that Lagrangian changes when there is an external force etc. I totally understand the mathematical derivations showing symmetries leading to conservation laws.

My question is,

1. Is symmetries "broken" when external force, torque applied do work on the system?

2. If that is correct how do we able to "break" symmetries of the space, that should be impossible? Shouldn't space be independent of which object we apply force which we don't?

Some AI model says, we break symmetries of the space by exerting force in one direction and making that favorable, but i don't trust that.

So it is not the math that I don't understand it is the two questions that I labeled. Any help of any kind is appreciated.

Answer 1

Perhaps a simple example is in order:

1. Consider a non-relativistic 1D free particle with Lagrangian $$ L_0(q,\dot{q},t)~=~\frac{m}{2}\dot{q}^2. \tag{1}$$ It is well-known that the momentum $p=m\dot{q}$ is conserved since $q$ is a cyclic variable, i.e. $L_0$ has a translation symmetry $q\to q+a$.

2. Next introduce an external force $F_{\rm ext}(t)$ in the Lagrangian $$ L(q,\dot{q},t)~=~\frac{m}{2}\dot{q}^2+ qF_{\rm ext}(t), \tag{2}$$ cf. e.g. this Phys.SE post. Now the translation symmetry $q\to q+a$ is broken, and the momentum $p$ is no longer conserved.

Answer 2

We usually consider potential energy rather than force in analytical mechanics and quantum mechanics. For any operator that commutes with the Hamiltonian $H$ in form $[A, H] = 0$, we can say $A$ connects a conserved operator. While when we apply an extra force, or say, an extra potential, for example $-\alpha/r$ in the hydrogen atom, this does break the origin symmetry.

The origin symmetry of an electron is $\mathrm{SO}(3)$, which means 3-dimensional rotational invariance. After applying Coulomb interaction, the symmetry becomes $\mathrm{SO}(4) = \mathrm{SO}(3) \times \mathrm{SO}(3)$. Again, when we apply a magnetic field, the symmetry breaks from $\mathrm{SO}(4)$ to $SO(2)$, this is how symmetry breaks happen.

Answer 3

I prefer to think of Noether's theorem as a way of examining the properties of an equation (or set of equations) that describe a particular phenomenon.

Example:
Newtonian mechanics: gravitational interaction, giving rise to orbital motion.

The gravitational attraction by a spherically symmetric celestial body is itself spherically symmetric.

In the Principia Newton gave a derivation of Kepler's law of areas. (More precisely: Newton derived a more general area law, showing that it obtains for any central force.) Link to an answer that I posted in 2022: Newton and the area law

As we know, in retrospect we recognize this area law as a way of expressing conservation of angular momentum.

The reasoning that arrives at this area law has multiple elements, but one element among them is unique to it: the property that the force that is exerted is symmetrical under change of orientation.

I argue that the notions, symmetry-of-the-force and symmetry-of-the-space, should be considered individually.

The only way to be able to formulate classical mechanics at all is to assume that space is uniform.

With uniformity of space assumed:
Then application of Noether's theorem is specifically to the force that is being exerted.

Example: A case where a force of attraction has cilindrical symmetry, but not spherical symmetry.

The Earth has an equatorial bulge (equatorial radius about 21 kilometer larger than polar radius). The gravitational attraction from the Earth is correspondingly not completely spherically symmetric. Orbits in the equatorial plane are not affected. The effect on orbits that are tilted with respect to the equatorial plane is as follows: the orbit is not quite planar (but that averages out), and there is a precession of the plane of the orbit, around the Earth's axis. There is a type of orbit, called Sun-synchronous orbit, that takes advantage of that precession effect. A Sun-synchronous orbit is a low Earth orbit that is close to a polar orbit, but an an inclination of 98.7 degrees. At that inclination the precession of the plane of orbit has a period of one year.

So that is an asymmetry of the force having as result that the momentum of the orbiting object is affected (depending on inclination and orbital altitude).

Historically: up until the revolution of relativistic physics there had always been an implicit assumption that space is euclidean.

As we know: in order to formulate the special theory of relativity it must be assumed (either implicitly or explicitly) that spacetime is Minkowski spacetime.

As profound of a change that was:
Minkowski spacetime accomodates the notions of angular momentum and linear momentum, and most importantly: Minkowski spacetime accomodates symmetry under velocity translation.

As we know, the Lorentz transformations were recognized in the context of exploring symmetries of solutions to Maxwell's equations.

I regard that as supporting evidence for the case that symmetry-of-the-force and symmetry-of-the-space should be considered individually.

When a physicist formulates a hypothesis, formulated as an equation, how do you assess whether what that equation describes has the conservation properties that you want? According to Noether's theorem you have the option of establishing that the equations feature the corresponding symmetry.

Answer 4

In general consider a Lagrangian system $$\frac{d}{dt}\frac{\partial L}{\partial \dot x}-\frac{\partial L}{\partial x}=Q(x,\dot x),\quad L=L(x,\dot x),$$ here $x=(x^1,\ldots,x^m)^T$ are local coordinates on a configuration manifold $M$. Here $Q$ are non potential generalized forces. They can be usual forces torques or something else. Notice that the word "external" is senseless here.

Assume in addition that the Lagrangian $L$ possesses a symmetry group $g^s:M\to M$ generated by a vector field $v=v(x):$ $$\frac{d}{ds}g^s(x)=v(g^s(x)),\quad g^0(x)=x,\quad L\Big(g^s(x),\frac{\partial g^s(x)}{\partial x}\dot x\Big)=L(x,\dot x)\quad\forall s.$$ Then the Noether type theorem is read as follows $$\frac{d}{dt}\Big(\frac{\partial L}{\partial \dot x}v\Big)=Qv.$$ In particular when $Qv=0$ then the expression $\frac{\partial L}{\partial \dot x}v$ turns into the standard Noether integral.

Answer 5

To answer that question how that is possible: By leaving out stuff.

To flesh this out a bit: We have good reason to believe that fundamentally, space is isotropic and thus momentum is conserved. However, in many real-life examples, we consider roughly human-sized objects close to the surface of the earth, and we model that by a gravitational potential $\phi=mgz$. Then, vertical momentum is not conserved -- when you throw something up, it falls back down.

Of course, really, there is a compensating momentum of the earth, so that overall momnetum is conserved. But since the earth is much much heavier, it does not move in any measurable way whenyou throw a ball or shoot a cannonball.