Neutrinos have a lifetime which exceeds the lifetime of our universe. Therefore we measure an equal ratio of all three neutrino flavors, 1:1:1. However, lets assume that the heavier neutrinos can decay into the lightest neutrino very fast, so that only the lightest neutrino is measured in detecto...
might just need clarification, but the OP hasn't returned to offer any.
@Danu Well, hmmmm. It's like this: grad students are included in the fraternity once they're doing research. Even masters candidates. As are industry researchers, including the few doing in with only a Bachelor's degree.
But, yeah, a large fraction of that culture is either doctors or (as a post-doc I knew liked to say) "future doctors".
"Letting the radius of this cylinder approach zero, we obtain the disturbing conclusion that providing data in a, for all practical purposes, one-dimensional region determines the solution in a three-dimensional region. Such an apparent ‘free lunch’, where the solution seems to contain more information than input data, is a classical symptom of ill-posedness. "
For a question this broad, Physics Forums is probably a better place. I have my reasons for not pointing people to Physics Overflow. — David Z ♦15 mins ago
I am no geometer...however it is the vector space of derivations, where the derivation is a linear map from the real-valued smooth functions to reals that obeys the chain rule wrt to product
@0celo7 If you embed the manifold into $\mathbb{R}^n$, then the derivatives (i.e. the gradients) of functions on the manifold are really tangent to it in the naive geometric sence.
Did you also know that if you isometrically embed the manifold in $\mathbb{R}^n$ and restrict the ambient connection, you get the Levi-Civita connection?
@ACuriousMind Did you know that there is a bijective correspondence between the set of $r$-th order bundle functors on $\mathcal{M}f$ and the set of smooth left actions of the category $L^r$ on systems $\mathcal{S}=\{S_0,S_1,\dotsc\}$ of smooth manifolds?
@0celo7 : Yes of course. But you don't know the difference between curved spacetime and curved space. So spare me for the love of crap. Note that "Examples of non-trivial fiber bundles include the Möbius strip".
Go and read the mathspages Dirac's belt article: "In this sense a Mobius strip is reminiscent of spin-1/2 particles in quantum mechanics, since such particles must be rotated through two complete rotations in order to be restored to their original state".
@0celo7 : it's relevant because electromagnetic is associated with "the curvature of a fibre bundle", but there are no actual fibres bundled together. The fibre bundle is however a "space", and according to Einstein a field is a state of space. A gravitational field is inhomogeneous space, modelled as curved spacetime. An electromagnetic field isn't.
@0celo7: Just ignore JohnDuffield. There's nothing to win for you by talking to him, you should've realized that by now. @DavidZ: I don't think anything's been said that warrants "doing anything serious about it".
@0celo7 You know, one of my pet peeves is when people take something I say and interpret it to mean something else. So I don't appreciate the assumption that I'm starting suspension procedures.
@0celo7 whether it is or not, that doesn't matter for what I was saying. I was saying that I find it really irritating when people take something I say and interpret it to mean something else that I didn't say. So when you (apparently) assumed my comment about doing something serious meant I was starting suspension procedures, it doesn't help my mood. I didn't say that.
I'm already in a bad mood for other reasons too (IRL reasons). Which, incidentally, is part of the reason I'm not inclined to suspend anyone or go on a message-deleting spree or anything like that :-P
Anyways, back to "analysis" (US first year version of it at least)...how did you define reals without the notion of limits? did you do Dedekind cuts, or introduced some ad hoc axioms?
Pick set representatives for each cardinal. Take a disjoint union of these. The cardinality of the new thing is the sum of the cardinalities of the old.
@MikeMiller I immediately thought it is not good to have $\infty$ in a sum involving cardinals (that are useful as infinite numbers), so the notation was not communicating the same thing to two people
nλ=L and L=2λ compared to E0=1/2h’v
v = c/λ ⇒ λ = c/v ⇒ λ= L/2
⇒E0= 1/2h’.c.2/L= h’c/L = 1/2h’v
⇒c/L= 1/2 v ⇒ v = 2c/L ⇒L = 2c/v
⇒ E= h’c/L when L = 2c/v when v = 2c/L
Any comment?
nλ=L and L=2λ compared to E0=1/2h’v..... v = c/λ ⇒ λ = c/v ⇒ λ= L/2..... ⇒E0= 1/2h’.c.2/L= h’c/L = 1/2h’v...... ⇒c/L= 1/2 v ⇒ v = 2c/L ⇒L = 2c/v..... ⇒ E= h’c/L when L = 2c/v when v = 2c/L..... Any comment? — Riadh Boukratem39 mins ago
nλ=L and L=2λ compared to E0=1/2h’v
v = c/λ ⇒ λ = c/v ⇒ λ= L/2
⇒E0= 1/2h’.c.2/L= h’c/L = 1/2h’v
⇒c/L= 1/2 v ⇒ v = 2c/L ⇒L = 2c/v
⇒ E= h’c/L when L = 2c/v when v = 2c/L
Any comment?
