Let's discuss Mathematics

[AdrienIer]

Is omega max a constant or does it vary ?

 

[CKS]

I think omega max is N! divided by ((N/2)!(N/2)!) for even N and N! divided by (((N+1)/2)! ((N-1)/2)!) for odd N. (Assuming spin is either +1 or -1, giving S = N, N-2, N-4, etc.) This should help keep the e factors in the final answer.

 

[Lohrenswald]

Is omega max a constant or does it vary ?

It shouldn't depend on S, but N can probably be in there

CKS's answer might make sense in some way, but I've asked for help at university today. Got a lot of info but still stuck

so:

the e factors given by stirling's approximation should go away, and you can put the square roots into one factor, then there's the like N and S pyramid

I was told to like divide by the square root factor (looking at like an equation with omega on one side) and then do ln on both sides, which with like all the exponentials and factors and such you can do all kinds of things to like simplify that with those logarithm rules

but then my achilles heel: apparantly we're supposed to taylor expand some of those things to get it easier
problem now is I can't do that

there's a trick, so we can simplify the two taylor polynomials I need to

ln(1+x) and ln(1-x) (x=S/N, but my problem right now is more general)

I think ln(1+x) shold become x+(x^2)/2 (and stop it there because x is relatively small)

problem is I can't get it on that form

I know being stuck at that stage is like super inept of me, but I don't know how to do it

 

[Lohrenswald]

Okay I got it now

It was kinda hard to like work with expanding at x=0 for a function with arguement 1+x

I'll update if I find omega max

EDIT:

decided to do this for some reason, so this is omega max

(made a mistake in the N+S squares thing and fixed it lazily)

gotta be honest I don't like that S, but I don't see how I can get rid of it

 

[Lohrenswald]

someone said it'd probably not be that bad to estimate that N^2+S^2 is close to N^2

It feels kinda cheap, but most of the time S is fairly small compared to N (although sometimes it's a bit big), and the taylor expansion earlier kinda had to assume S/N was a fairly small number

N here is actually ten thousand, so it's kind of at the edge of those like judgements being valid, but I guess I'll have to live with it

On to the next thing though

say you want to rewrite a variable x to ly (l times y) (It's physics so it's to get rid of dimensions but whatever)

or maybe more precisely y=x/l (so l should have like opposite dimensions to x so that y is dimensionless)

now the derivate, d/dx

the problem I'm trying to solve like makes it seem that d/dx should be equal to (1/l)d/dy, but to me it seems more obvious it would be ld/dy

because like in the other notation f'(x)=f'(ly)=lf'(y) no?

Am I missing something?

 

[CKS]

df/dx = df/dy * dy/dx. Since dy/dx = 1/l, df/dx = (1/l) df/dy. So d/dx = (1/l) d/dy.

In the prime notation, you are taking the derivative with respect to something else (like t), which is why you get the opposite result.

 

[Lohrenswald]

It's quantum mechanics but it's fairly math shifted

I've thought about expressing the total S in the sum according to the Clebsch-Gordan table, but otherwise I've not really figured out much

any ideas?

I'm a bit scarce in saying what I've tried because this is a kinda desperate place to ask help, I mainly do just to not squander the oppurtunity, but I can answer questions if needed

 

[Lohrenswald]

The answer I think is 9/8 J with degeneracy 4 and -3/8 J with degeneracy 6

it only took me several days to figure this out

 

[Mouthwash]

Posts like that make me thankful I've forgotten my math beyond basic multiplication. I mean, I'll have to relearn some to get a degree, but I assure you there won't be quantum physics involved.

 

[Lohrenswald]

For the record I was wrong, and that exam went pretty bad
there were 8 eigenstates because there were 8 basis vectors for example

anyway I have another question

I've got something in the form of 1/((x^2)*(exp(1/x)-1)^2) (other constants too, of course)

I'm suppossed to evaluate this in the limit of x goes to zero and infinity (or very large), but I'm not sure how to

 

[AdrienIer]

Write the exp(x) as 1+x+x^2+x^3+... etc for the limit on 0.
For infinity there is no tension between the exp and the x^2 so just use your usual limit rules

 

[Lohrenswald]

not sure if I expressed it very well

So this is the actual expression I'm looking at, so instead of x it's T which is actually inverse of it sort of

So high T I can work with (analogous to x in the limit of zero)
but low T I'm not so sure, because then the T² goes to zero, but the exp goes to infinity
Actually it's kinda worse than that because T doesn't go to infinity, just something very big

It is a physics problem, and then it's not always like "proper math" going on

like in a previous problem in the same problem sheet, I got an answer for high T that was like hw/2+kT and simplified that to just kT, since that term dominates. It's really more stuff like that I kinda want to get

 

[AdrienIer]

Taylor expansion gets you to the end most of the time with these kinds of functions. exp(hw/kT) = 1 + hw/kT + ((hw/kT)^2)/2 + ((hw/kT)^3)/6 + etc... You then substract 1 and deal with the square in the denominator. Be careful about which term is actually the "strongest" because the parts of the sum are going to infinity when T gets lower.

 

[Michkov]

Looks similar to Planck's law, at least in the general form of the equation. It has two approximations for both short and long wavelengths, Rayleigh-Jeans and Wien respectively. Maybe looking up how you get from the general case to the long and short approximations would help.

 

[Kyriakos]

Hi, I have a general question:
why specifically use the diagonal method to show that some sets have larger cardinality than that of the set of natural numbers?

I have to assume that the point was to show that one cannot order the sequence in any way which would correspond to a tie to the natural numbers, eg 1.1, 1.2 etc would still be placed in positions (eg) 1 and 2, but then the full set would take up all of the positions used by natural numbers and still allow room for many more positions for the fractions. But why should one use a diagonal to show there are many more possible positions, instead of any other method? Is this the simplest possible to think of, or does it have any specific use in other things? I mean intuitive it would be self-evident that even a fraction of a fraction of a fraction... of something would go on in a one to one tie to the natural numbers, so is there some use in coming up with a specific and easy to iterate set which can be fed back to the original (different in the diagonal) and still allow for the new set having space for more?

To me the diagonal presentation seemed somewhat similar to the first proof of there being more prime numbers than can be accounted for, so was wondering if it was just one possible way of making the argument or something inherently valuable due to ties to other parts of set theory.

 

[Lohrenswald]

arccot(phi)=arctan(1/phi), yea?

 

[monikernemo]

Hi, I have a general question:
why specifically use the diagonal method to show that some sets have larger cardinality than that of the set of natural numbers?

I have to assume that the point was to show that one cannot order the sequence in any way which would correspond to a tie to the natural numbers, eg 1.1, 1.2 etc would still be placed in positions (eg) 1 and 2, but then the full set would take up all of the positions used by natural numbers and still allow room for many more positions for the fractions. But why should one use a diagonal to show there are many more possible positions, instead of any other method? Is this the simplest possible to think of, or does it have any specific use in other things? I mean intuitive it would be self-evident that even a fraction of a fraction of a fraction... of something would go on in a one to one tie to the natural numbers, so is there some use in coming up with a specific and easy to iterate set which can be fed back to the original (different in the diagonal) and still allow for the new set having space for more?

To me the diagonal presentation seemed somewhat similar to the first proof of there being more prime numbers than can be accounted for, so was wondering if it was just one possible way of making the argument or something inherently valuable due to ties to other parts of set theory.

I believe there are many variants of Cantor's diagonal argument to show that R is strictly larger than N, so perhaps you might want to explicitly state which version of the proof you are talking about?

Most of the proof is similar to the ''general diagonalisation arguement'' where one example is from Cantor's Theorem, which says that card P(S) (collection of subsets of S) is strictly larger than that of card S.

First you assume there is a bijection f: S -> P(S), then you look at T subset of S that contains all the elements t in S such that t not in f(t). So obviously, this subset of S cannot be in the image of f since if there exists s such that f(s) = T, then we get a contradiction because

1. If s in T, then by definition of T, s not in T, which is a contradiction
2. If s not in T, then by definition of T, s in T, which is yet another contradiction.

So f cannot be a bijection.

And the ''diagonal argument'' to show R is strictly larger than N is similar. Its different ways of encoding the same idea.

 

[Lohrenswald]

I'm kind of feeling insane right now

it's about scalar products, so these are vectors.
Quoting P.C. Matthews book "Vector Calculus" page 4
assume * is a dot and T is theta lol

a*b=|a||b|cosT
[...]
the dot product is commutative, i.e. a*b=b*a
[...]
since the quantity |b|cosT represents the component of the vector b in the direction of the vedtor a, the scalar a*b kan be thought of as the magnitude of a multiplied by the component of b in the direction of a

What I'm wondering now is like, why isn't it the other way around? it seems pretty clear from this figure for example that the component of a in b is larger than the one of b in a
shouldn't it like be "symmetrical"?

what am I missing?

 

[monikernemo]

I'm kind of feeling insane right now

it's about scalar products, so these are vectors.
Quoting P.C. Matthews book "Vector Calculus" page 4
assume * is a dot and T is theta lol

What I'm wondering now is like, why isn't it the other way around? it seems pretty clear from this figure for example that the component of a in b is larger than the one of b in a
shouldn't it like be "symmetrical"?

what am I missing?

Yes, you can totally relabel a as b and b as a. Like you have pointed out the situation is symmetric.

 

[CKS]

What I'm wondering now is like, why isn't it the other way around? it seems pretty clear from this figure for example that the component of a in b is larger than the one of b in a
shouldn't it like be "symmetrical"?

what am I missing?

The component of a in the direction of b is bigger than the component of b in the direction of a, but this is balanced out by a being bigger than b:
(a cosT)b = a(b cosT) and both are equal to a b cosT.

a cosT > b cosT, but b < a by the same ratio, so it balances out.