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Khallil Benyattou
Khallil Benyattou
23:09
You can derive an atom, @Semiclassical?
Semiclassical
Semiclassical
the equations for the bohr model, yes
assume linear momentum is quantized as $\hbar$ i.e. $L=mvr=n\hbar$
then there's a cute mnemonic: $\alpha=\beta$, where $\alpha$ is the fine-structure constant and $\beta=v/c$
Khallil Benyattou
Khallil Benyattou
Ah, that's all gibberish to me right now. :-P
What do you mean by the term quantised?
Semiclassical
Semiclassical
can only occur in integer multiples
that's the main assumption of the bohr model. it's ad hoc, but it works for hydrogen
out for now
Khallil Benyattou
Khallil Benyattou
See ya!
Daniel Fischer
Daniel Fischer
23:24
Today was a good day, I found two interesting questions to think about :-D
Khallil Benyattou
Khallil Benyattou
Which fields are they in, @DanielFischer?
Daniel Fischer
Daniel Fischer
@KhallilBenyattou Analysis
Khallil Benyattou
Khallil Benyattou
Ah, real or complex, @DanielFischer?
Daniel Fischer
Daniel Fischer
@KhallilBenyattou One real analysis, one functional analysis.
iluso
iluso
Is there a notation for "n is a k-th power" ?
Khallil Benyattou
Khallil Benyattou
23:26
@DanielFischer Oohh, what's functional analysis about?
Daniel Fischer
Daniel Fischer
@KhallilBenyattou Topological vector spaces and continuous [or discontinuous, but I'm not into that] linear [or non-linear, but I'm neither into that] operators between them.
iluso
iluso
:22679296 $\exists m\in\mathbb{N}, n=m^k$
Khallil Benyattou
Khallil Benyattou
That sounds pretty cool, @DanielFischer! There's a module at my uni to do with that and I might take it!
Daniel Fischer
Daniel Fischer
@iluso "$\sqrt[k]{n} \in \mathbb{N}$" is not something you'd like, I suppose?
iluso
iluso
@DanielFischer I was looking for something that can fit nicely as a $\sum$ condition
Daniel Fischer
Daniel Fischer
23:31
@iluso Just write $m^k$ instead of $n$?
iluso
iluso
@DanielFischer Is something like $\sum_{m^k \in\mathbb{N}}$ readable ?
Khallil Benyattou
Khallil Benyattou
$$ \sum_{m^k \in \mathbb{N}}$$
On a new line, it's not so bad I guess. ^_^"
Daniel Fischer
Daniel Fischer
@iluso Well, so-so, but with $m\in \mathbb{N}$, you always have $m^k \in \mathbb{N}$.
Khallil Benyattou
Khallil Benyattou
The kth root notation seems more explicit!
iluso
iluso
$$ H(k+1, n) = k! \sum_{j=0}^k \frac{(-1)^j}{(k-1-j)!} \sum_{\substack{d|n \\ d\geq 2 \\ d^{1/(j+1)} \in \mathbb{N}}} H(k-j, n/d) $$
Arg
I think I'll go for a sentence like "the inside sum is take over the d such that..." ;)
r9m
r9m
23:45
@DanielFischer This was awesome!! :D (+1)
Daniel Fischer
Daniel Fischer
@r9m Heh :-) That was a really fun problem. (One of the two I mentioned above.)
r9m
r9m
@DanielFischer link to the other one please then :D
Daniel Fischer
Daniel Fischer
@r9m This one
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