# Rotation as an example of symmetry in classical mechanics

I modified the question because it was confused.

On my book there is this mathematical definition of symmetry transformation:

"The equations of motion have a symmetry, if the solutions of the equations transformed by the symmetry are still solutions of the equations of motion, namely, there is a symmetry if the transformed equations of motions have the same form of the original".

I don't understand the meaning of this sentence, do you think is it a good (and easy) mathematical definition of symmetry transformation? Anyway what "equal in form" means? Then, i know rotation of an isolated system is a symmetry, can you make an easy example of an isolated system in which if we apply a rotation the mathematical definition of symmetry apply? Can you make an example of non symmetric transformation and show me why the mathematical definition doesn't apply?

• Which textbook? Commented Jun 13, 2019 at 15:59
• Title “Lezioni di fisica teorica” author: Nino Zanghì page 79. It’s in italian anyway: ge.infn.it/~zanghi/FT/ZUM.pdf Commented Jun 13, 2019 at 16:32

The definition simply says that the dynamics of a system does not change after a symmetry transformation. By "equal in form" he means that if I apply the transformation and then I rename the new variables as the old ones, the new and the old equations are identical.

The two parts of the definition are equivalent because:

• A symmetry sends a solution to another solution $$\Longrightarrow$$ A symmetry sends an equation of motion (EOM) to an equation that has the same solutions, i.e. the same equation.

• A symmetry sends an equation to the same equation $$\Longrightarrow$$ A symmetry sends a solution of the old EOM to a solution of the new EOM. But, in fact, the two EOMs are the same.

### Example of a symmetry

An isolated system has a Lagrangian $$\mathcal{L} = \sum_{i=1}^n \frac12 {\dot{\vec{q}}_i}^2 + V(|\vec{q}_k-\vec{q}_j|^2)\,.$$ Therefore the equations of motion will be of the form (recall $$\partial_{\vec{q}_i} |q_i-q_j|^2 = 2 (\vec{q}_i - \vec{q}_j)$$) $$\ddot{\vec{q}}_i = \sum_j(\vec{q}_i -\vec{q}_j)\,W_j(|\vec{q}_k-\vec{q}_j|^2)\,.$$ If I make a rotation $$\vec{q}_i \to R\cdot\vec{q}_i$$ such that $$R^TR=1$$ then the arguments of $$W$$ remain all unchanged because it's made of scalar products and the EOM reads $$R\cdot\ddot{\vec{q}}_i = \sum_j(R\cdot\vec{q}_i - R\cdot\vec{q}_j )\,W_j(|\vec{q}_k-\vec{q}_j|^2)\,.$$ Now if I just call $$R\cdot\vec{q} = \vec{q}$$ again, the equation is the same. The solutions can be as complicated as you want by choosing $$V$$ to be an ugly function. However, if I make a rotation that's the same as measuring all coordinats with a rotated frame. And since the system is isolated the frame is arbitrary anyway. Therefore if me and a friend of mine measure things with rotated frames and we look at a physical trajectory of a particle, both of us better conclude that the trajectory is a solution to the EOM.

### Example of something that is not a symmetry

Just take the example above and add an explicit vector $$\vec{X}$$ coupled to one of the $$\vec{q}$$'s, like $$\mathcal{L} \to \mathcal{L} + \vec{X} \cdot \vec{q}_1\,.$$ If you make a rotation, the old equations will look like the new ones, except for the fact that the new ones have the vector $$R^T\cdot \vec{X}$$ rather than $$\vec{X}$$. Accordingly, a solution of the original system won't be a solution of the rotated one.

And why is it physically acceptable that here rotations are not a symmetry? Because we had to choose a specific vector $$\vec{X}$$. So now that makes the choice of frame not arbitrary anymore: there is a frame that is preferred with respect to the other ones and it's the one where $$\vec{X}=(1,0,0)$$. Or, to put it in other terms, I can tell apart two configurations that differ by a rotation by checking at the direction of $$\vec{X}$$ in the two frames. This breaks the symmetry.

Suppose you keep track of the total energy in the system at hand and it remains unchanged with time. This is Conservation of energy. However you transform the time coordinate, the energy remains constant. Symmetry is in the system if some transformation leaves certain elements of the system unchanged. This includes translations in time.

Now suppose you consider the energy as it changes with position. It's important to consider the energy measurements in the SAME coordinate system. Coordinates refer to a real, singular point. You need to compare the change in energy at different points, not the same point at different coordinates. Keeping this in mind, we know that momentum remains constant in time if the space derivative of the Hamiltonian is 0 everywhere in the system.

This applies to every conjugate momentum. So If the energy is constant under rotations, then angular momentum is conserved.

Look up Hamiltonian Mechanics and conjugate coordinates for more information.