@Relativisticcucumber anyway, your latest question about BEC, im not sure what you want more than the fact that Bose-Einstein statistics favour a LOT of particles sharing the same state.
$r = x^2 + y^2 + z^2$ as measured from the center of the earth (most likely but it is hard to tell because i am not really certain what sort of context you are working in)
@123 if you are writing down this formula then it sounds like you are doing a problem in which you are considering the earth and another object in space and considering the force one mass exerts on the other
alternatively, you can be at the surface of the earth and considering the gravitational force acting on an object near the surface of the earth like you do when you consider a baseball flying through the air
if you are looking at earth in 3D space, it has gravitational field $\vec{g} = -\frac{GM_e}{r^3}\vec{r}$ where $r^3 = (x^2 + y^2 + z^2)^{3/2}$ and $\vec{r} = (x, y, z)$. If you are looking at just a plane in space (so specializing to 2 of the 3 dimensions), just set $z = 0$ and let $\vec{r} = (x, y)$.
the militaries of modern countries are something out of sci-fi when compared to the greatest militaries from hundreds of years ago
air force alone makes it a completely different thing
but in the movie Avatar, they show the old war techniques of horses and arrows winning over modern military tech in a war
but they didnt use nukes in the first movie maybe because of public backlash back on earth. in the beginning of the second Avatar, they use a nuke and win the war
let $M$ be a smooth manifold. let $x \in M$. i want to define the tangent space $T_xM$ as the space of all derivations over smooth functions at $x$ (with appropriate addition and scalar multiplication defined on the derivations). I am a bit confused because wikipedia states that a derivation as $D: C^\infty(M) \to \mathbb{R}$, so it sends a smooth function $f: M \to \mathbb{R}$ to a real number. Is that right? I would have thought that it should send the smooth function evaluated at the point $x$ to a real number, so $f(x = x') \mapsto Df(x) \lvert_{x = x'}$
u can think of $D_{x'}$ as the map $D_{x'}(f(x))= Df(x)|_{x=x'}$. the input has to be the function $f(x)$, not the function evaluated at the point $f(x')$ becuz the latter loses the neighborhood information required to differentiate @SillyGoose
to be more precise, the input is a germ of functions
becuz the equivalence class of all functions, which hav the same neighborhood information around $x'$, get mapped to the same real number
@SillyGoose it is in fact wrong. You should see it as a map $D: C^\infty_{x^0}(M) \longrightarrow \mathbb{R}$. $C^\infty_{x_0}(M)$ is usually called the germ of $C^\infty(M)$ functions at $x_0$. It is the quotient of $C^\infty(M)$ by the relation $f \sim g$ if there exists a neighbourhood $U$ of $x_0$ such that $f|_U = g|_U$
this is just telling you that the derivations at $x_0$ should only depend on the local behaviour of the functions at $x_0$
@naturallyInconsistent sorry, stumbled about this conversation and am very confused. I know who you're talking about in MIT, but who is "miao miao" here??
> “Meow” is an established spelling in English. Rossini spells it “miau” in his famous duet for two cats - but he would; he was Italian. “Miao” is the preferred French spelling.
@SillyGoose all the big language of bundles and manifolds and derivations wasn't very helpful when it came to actually computing a simple derivative, there is a very deep lesson in this
only disadvantage is $\int (\dots)+(\dots)dx$ looks like it's enclosed by the integral and dx whereas $\int dx (\dots) + (\dots)$ need to use a bracket/parentheses to enclose what you're trying to integrate
If one is to believe beauty is immutable and the way one perceives it doesn't matter 🤔
when one "understands" something in mathematics is like any other evocation :O
I was kinda tongue in cheek about it because I don't really think of math as "art" like most mathematicians do but now Idk
like how a piece of art can evoke deep feelings in some regions of the brain, math can do the same just in a different way..but one is emotional and the other is logical. I'm not convinced you can call them the same thing (calling math art)
@bolbteppa I realized I am a man by now when yesterday I wrote an integral with the measure after the function and I barred it to have the measure first :P
It's also much easier to read, so you know the variable first. It's similar to the Spanish ¿ (although we don't put the measure at the end too :P)
I can't let myself do that now though, not until I take measure theory. My prof would definitely fail me if I wrote $\int d\mu f$ :P
@SirCumference that's because in classical mechanics there is always a number $J$ - the QM intro text is emphasizing that while the operator $\vec J^2$ is a scalar operator, there is no scalar operator $J$ that you could square
@Mr.Feynman for me it's more that I rag on physicists for not doing math properly all the time, so I feel some need to defend them when they're just being a bit silly about notation :P
@bolbteppa true it’s not so useful for computing things…but one doesn’t really need to know much at all to compute things. The “abstract nonsense” is my preferred way of organizing information in my head :P
It is hard to visualize the abstract holonomy of a connection…and i prefer to think in words anyways. “Abstract nonsense” also lets you recognize things in the future with little additional effort
it also just answers questions that people maybe are okay with glossing over but which seem pretty fundamental. like why should the connection 1-form be lie algebra valued? and the lie algebra of the structure group no less. well the answer is answered immediately if one just constructs it, and the construction is quite natural
@SirCumference i like it :D it makes me not forget to add the measure in the first place lol
a lot of stuff even in the abstract nonsense just seems to be linear algebra anyways...
let $P$ be a principal bundle. define a connection on $P$. i am trying to show the equivalence of the horizontal subspaces defined by the corresponding connection one-form and the horizontal subspaces defined by the definition of a connection.
I should be able to ignore definition 10.4 and from the listed properties in the definitions show that $X \in H_pP$ iff $X \in \ker\omega_p$
Nakahara proves one direction in the text, so I am attempting the other direction and got to this point.
However, I am not sure how to show that $(10)$ vanishes
