If a muon in the upper atmosphere has a constant velocity of $v=.950c$ with an internal clock of $2.2E-6 s$ and I'm asked to find how far it travels w.r.t. an observer on earth that's just the velocity * the dilated time?

I thought the velocity was transformed as well

@Obliv No, the speeds have to be equal in magnitude. I discuss that here: physics.stackexchange.com/a/562271/123208 & physics.stackexchange.com/a/554417/123208

@Earman "The Physics Detective" is a character called John Duffield. He is well known to members of this site, and not for good reasons.

Thanks @all. I think I understand the reason why it didn’t get well-enough attention.

Hi All...

Hello @JohnRennie Sir

@Earman It got all the attention it deserved :-)

@123 Hi :-)

@123 Sorry for the delay, I had to dash of for a few minutes. Let's continue in the problem solving room.

Aye Sir

Moved

I am trying to understadn why 3.265 implies conservation fo angular momentum. My understanding is that we are saying the expectation of energy will be the same whether you apply the hamiltonian first and then rotate your state or if you rotate your state and then apply the hamiltonian

Is that on the right track? I am most confused about interpreting what applying the hamiltonian means since i feel like at least abstractly there is a natural interpretation on L as it rotates your state

@SillyGoose The Hamiltonian generates time evolution, so for any $O$ with $[O,H] = 0$ we have also that $[O,U(t)] = 0$ and therefore if you feed an initial state $\lvert \psi\rangle$ into the Schrödinger equation that is an eigenstate of $O$ it will remain an eigenstate of $O$ at all times

that's what we mean by "conservation", right - you start with a fixed value at time $t=0$ and it never changes

oh I see