Hi, i've read your answer and there are many things that i don't understand. If i don't bother you i'd like to ask you some details

No problem. Tried to be as clear as possible :)

ok first thing that i don't understand is

You can invoke invariance by rotation which is enough to get the usual solid angle measure, but this extra assumptions that you have implicitly taken is arbitrary and does not even come from the symmetries of the Hamiltonian (merely rotation symmetric about..

If the hamiltonian of the system is $H=-\mu cos\theta B$ where $\theta$ is the angle between magnetic field and dipole, the system is invariant under rotation right?

Not if you can’t change $\vec B$. You only have rotation invarince about the axis it defines

Intuitively, the presence of the magnetic field defines a preferred direction (the direction of the magnetic field), so breaks rotational invariance

I thought it was invariant under rotation because if i rotate the frame of reference the hamiltonian is always the same

Granted a full model encompassing the em field (eg QED) would be rotation invariant, but in your model it is treated as an external source

ok, so it is $\phi$ invariant?

Well that is true and is precisely why the hamiltonian is not rotation invariant

Yes

@lpz what do you mean? this seems interesting

Are you familiar with qm or hamiltonian mechanics?

yes, but today i feel confused

I'm naively thinking that if the hamiltonian is the same after a transformation then the equation of motion are also the same and so the system appears identical

Take in qm. You have the anticommutation rules of angular momentum. $H$ is invariant bu rotation iff it commutes with all the $L_i$. However you have $H=BL_z$ which does not commute with $L_x,L_y$, so no full rotation invariance. However, it commutes with $L_z$ hence the remaining $\phi$ invariance.

Replace the anticommutation by poisson brackets and you get the same argument for classical hamiltonian mechanics

i'm not familiar with poisson brackets, however the Hamiltonian is $H=-\mu cos\theta B$ in every frame right? what change?

B changes so it can’t be the same theta

ok this is weird i think i'm confusing on something stupid :D

B is the amplitude, it does not change

what is wrong with that?

I was lazy, I meant $\vec B$

but $\vec B$ is not there in the hamiltonian

Take for example a particle in aplane under the influence of a uniform force

Example, i put a B field and the magnetic dipole is parallel to it. Then, $theta=0$. If i rotate my frame of reference i always have $theta=0$

You have $H=p_x^2/2+p_y^2/2-Fx$

I’ll just finish my example and i’ll come back to that

yes, sorry. didn't want to interrupt

the kinetic part is rotation invariant but the potential breaks rotation invariance

No prob, i’m just a bit slow to type math

you could write it as $\vec F \cdot \vec r$ written in a rotation invariant form

But this does not mean that it doesn’t break rotation invariance.

i agree

Same thing here

now i'm confused because i also agree with my example

your external magnetic field is fixed and creates a preferred direction

well just need to apply the mathematical criterion

in the first example, let’s rotate a bit:

Example, i put a B field and the magnetic dipole is parallel to it. Then, $theta=0$. If i rotate my frame of reference i always have $theta=0$

ah ok you are already answering sorry again

go on

i'm reading (interested)

Ince again in your example you are cheating because you are rotating$\vec B$ as well

so for the example you have the following inititesimal transformations

aaaah

wait maybe i get what you mean

Got it?

if i rotate the frame of reference is like the system rotates (opposite direction)

if the B field rotates with the system is part of the system

while i should consider it something external?