Timeline for Group actions confusion
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 21 at 18:41 | vote | accept | Geigercounter | ||
| Aug 20 at 23:04 | comment | added | Geigercounter | Moreover, what is then the $D_0$ group element introduced in (2.70)? Where does this come into play if we're simply solving the equations in my previous comment? | |
| Aug 20 at 22:33 | comment | added | Geigercounter | I think I am getting more confused. You said above that the group action is not the Poisson bracket, yet here you say that the group action is give by $\partial_\phi w = \{w, H_0\}$ (with an initial condition)? Are you simply saying that the group action is the same as the action of the generator and solving this equation? I then don't see how this constrains the constant shift of $w \rightarrow w+c$ to $w \rightarrow w+1$ | |
| Aug 20 at 20:38 | comment | added | ACuriousMind♦ | @Geigercounter That's for linear operators on a vector space. In this case you have a phase space (which does not necessarily have a vector space structure) and you compute the flow of the vector field by solving - as I already said - the differential equation $\partial_\phi w = \{ w, H_0 \}$ with initial condition $w(0) = w$. Note that solving the equations of motion $\partial_t f = \{f, H\}$ for $H$ the Hamiltonian is just a special case of it and you don't have any "particular formula" to solve arbitrary equations of motion either, do you? | |
| Aug 20 at 18:42 | comment | added | Geigercounter | Yes but what is the particular formula to compute this? In some references I have found for example $$ e^{-\xi A}B e^{\xi A} \equiv \sum_n \frac{(-\xi)^n}{n!}\text{ad}^n_A B, \qquad \text{ad}_A B \equiv -\{A,B\}$$ for the group action of $A$ on $B$ | |
| Aug 20 at 18:21 | comment | added | ACuriousMind♦ | @Geigercounter The group action is not the Poisson bracket, the Poisson bracket is the infinitesimal version of the group action. The group action has a parameter $\phi$ so that $w(\phi)$ is what it does to $w$, and $\partial_\phi w(\phi) = \{w,H_0\}$ (this is more or less the definition of the flow/group action). The solution to the differential equation $\partial_\phi w(\phi) = 1$ with initial condition $w(0) = w$ is $w(\phi) = w + \phi$. | |
| Aug 20 at 17:23 | comment | added | Geigercounter | But then I do not see how the group action leads to $w \rightarrow w+1$ if the group action is simply the Poisson bracket... I'm probably misunderstanding though | |
| Aug 20 at 16:43 | history | answered | ACuriousMind♦ | CC BY-SA 4.0 |