@Feynman_00 I think the issue is that a group is, at its core, a set $G$ with a composition operation $G\times G\rightarrow G$ which obeys the group composition rules. Ultimately the group elements have to actually be something - matrices, functions, sets, etc. If two groups are isomorphic to one another (in whatever sense is relevant in a given context), then they can be the same from the perspective of group theory even if their elements are different types of object.
The most common concrete definitions of the 3D rotation group use either the set of orientation-preserving isometries of $\mathbb R^3$ which fix the origin or more concretely the set of $3\times 3$ orthogonal matrices with determinant +1, which follows from the former set via an arbitrary choice of orthonormal basis. These are, strictly speaking, different groups because their carrier sets are not the same, but they are isomorphic in an obvious way.
When people talk about the "abstract" rotation group, they are essentially referring to the group structure while "forgetting" about the concrete carrier set which underlies the group. In that sense, I tend to think of the abstract rotation group essentially as an equivalence class of groups all related by (Lie group) isomorphism.
I remember being taught to use these in maths and science classes in the 1970s, but now it has been mentioned I haven't seen them used in a paper for ages.
@Obliv You can select any point to use as the origin, but if the point isn't at the COM the resulting frame will be non-inertial.
You can work in a non-inertial frame of course, but it greatly complicates the equations of motion.
I tried to ask this question but it got closed: physics.stackexchange.com/questions/740326/…. I think it counts as a conceptual question, rather than a check-my-work question, so I'm not sure how to edit it to make it follow the rules. Can anyone help point me in the right direction?
hm from my minimal understanding of finite group theory. If you have a finite group G acting on a non-empty set. Then, there exists a homomorphism from G into the symmetric group (by definition of a group action), giving you a permutation representation of the action. Then, there exists a homomorphism from the permutation representation into the (linear) representation. Can you go backwards to find the original, most abstract group?
why does taking the cartesian product of subsystems and then pulling a state out of that space imply that the state will be a product of states of the subsystem?
to me a cartesian product does not output anything but an ordered pair
If you consider some surface with a total area A then we can divide it up into tiny parts with area dA. Then we add up all these tiny parts to get the total area: A = dA₁ + dA₂ + dA₃ + ...
Now we can measure the force on each area. The force may not be constant i.e. the forces on different areas may be different, but because each dA is so small the force can be taken as constant over the area dA.
does anyone have a moderate level resource to get enough information abt quantum entanglement to start reading some papers on the subject? I'm looking at this rn and it's a bit dense xD arxiv.org/pdf/quant-ph/0702225.pdf
@Feynman_00 I mean, distinguishing between isomorphic groups is a bit pointless: We usually say a Lie group is a matrix Lie group if it is isomorphic to a group of matrices. Whether it is defined in terms of matrices or not doesn't really matter
@Slereah Could you help me understand the difference between an outer product, a dyadic product, and a tensor product of two vectors?
In particular, in the context of looking at an element of the tensor product of two hilbert spaces in QM
gah nevermind okay i think i got it figured out lol... looking at google images when you search tensor product of two vectors is quite deceiving. half the pictures are of other products
@ACuriousMind Oh ok, so the problem was only in the statement "$\mathrm{SO}(3)$ is a representation". In that case, I'd only make sense of this as "it is the image of the representation", as the representation is a map $G\rightarrow\mathrm{GL}(V)$
What is meant with cosmic ray flux? i am trying to understand this graph https://www.researchgate.net/figure/The-flux-of-cosmic-rays-as-a-function-of-the-energy-Over-a-very-broad-energy-interval-it_fig2_236635142
@Mad it just means particles per unit solid angle per unit time. But since it's in a log-log graph they've just changed the units to give an example at what each point means
