Mathematics

Semiclassical

1:02 PM

What’s nice is that $D^n(xy)= n D^{n-1}y+ xD^n y\implies D^n x -x D^n = n D^{n-1}$

Akiva Weinberger

$D^n(xy)=\sum D^kxD^{n-k}y$, no? Unless I misunderstand you

Semiclassical

Which isn’t so different from $Dx*n -x^n D=nx^{n-1}$

@AkivaWeinberger $D=d/dx$

Akiva Weinberger

Yes, and?

Semiclassical

So what’s $D^2 x$ ?

Akiva Weinberger

Oh, I forgot the binomial coefficient

Oh, right, sorry

Semiclassical

1:06 PM

That too

Akiva Weinberger

I was thinking of $D^n(fg)$

Semiclassical

well, same here. I just had the special case of that :)

Akiva Weinberger

Pincherle derivative

In mathematics, the Pincherle derivative T’ of a linear operator T:K[x] → K[x] on the vector space of polynomials in the variable x over a field K is the commutator of T with the multiplication by x in the algebra of endomorphisms End(K[x]). That is, T’ is another linear operator T’:K[x] → K[x] T ′ := [ T , x ] = T x − x T = − ad ⁡ ( x ) T , {\displaystyle...

$(D^n)'=nD^{n-1}$

philmcole

@Semiclassical Can I ask you a physics question about potential? As I understand, a potential in the mathematical sense is just a scalar function whose derivative gives a conservative field. From the physics standpoint the potential at a point $x$ gives the work done when moving from the reference point $x_0$ to $x$. I was wondering can the $x_0$ be arbitrarily chosen or does it need to be a certain point like usually one assumes it's infinity?

Akiva Weinberger

It's a consequence of $D'=1$ and $(TS)'=T'S+TS'$

Semiclassical

1:10 PM

@philmcole (up to an annoying minus sign)

mercio

anyone has a scalar function whose derivative doesn't give a conservative field ?

philmcole

@Semiclassical Okay so it's just a convenience to choose $x_0=\infty$. Where does the minus sign come in?

Semiclassical

@philmcole it’s arbitrary and we usually take it to be at infinity for convenience

The minus sign comes from the fact that, if you want total energy to be conserved, then increasing KE means that PE should decrease

And since a net force acting on an object at rest increases KE, then that same force must decrease PE in that scenario

There are cases where x0=infinity isn’t allowed, though

An obvious one being U=mgy

philmcole

Does that mean if we choose $x_0$ to not be infinity than the potential has a maximum at $x_0$ always and decreases in every direction?

Semiclassical

Depends on the scenario.

philmcole

1:16 PM

okay so not generally because you mentioned the decreasing PE..

Semiclassical

For instance, if you’ve got two particles of the same charge, then PE increases without bound as the two particles are brought together

philmcole

Yeah so why would one choose a reference point $x_0$ other than infinity?

Semiclassical

Well, take a uniform gravitational field $F_y=-mg$

How much work does it take to move a particle to y=infinity?

philmcole

okay yeah if it's constant then it takes infinite work

mercio

depends in which direction

philmcole

1:25 PM

or negative infinite work

skull

does infinity =?= infinity

Semiclassical

Well, it’s still infinite work in either direction. Just ambiguous whether it’s infinite work by the field or against the field

A more subtle example is provided by the case of an infinite line charge of uniform density

The force law is F ~ 1/r in that case, which goes to zero at infinity so you might think you’re safe

philmcole

why you're not?

Semiclassical

Because $\int (1/r)dr=\ln r+C$, and ln(r) blows up at infinity

What’s more, it also blows up at zero

philmcole

cool so there is somewhere a minimum potential

Semiclassical

1:29 PM

Nope.

philmcole

:(

Semiclassical

ln(r) just keeps going up

Secret

Will lookup what is $(\int () - \lambda)^{-1}$ later

Semiclassical

What you have to do in that case is pick some finite reference distance $R$

And then the potential energy is $\ln r-\ln R =\ln(r/R)$

philmcole

okay

It is kind of weird for me to think about having different reference points for potential energy

Semiclassical

1:32 PM

otoh, this isn’t an issue if you take the line current to really have some finite radius

philmcole

since my natural feeling would be potential energy is absolute (but it isn't)

Semiclassical

@philmcole it’s even worse if you want to get some notion of potential that’s appropriate for a magnetic field

philmcole

is it different there?

Semiclassical

Yeah. The issue is that the magnetic field of a current-carrying wire will circulate around said wire

philmcole

oh so basically it's not conservative...

and has no potential in the mathematical sense

Semiclassical

1:36 PM

Well, you have two options

One is to look for a scalar potential. In that case, you’ll end up finding: if you wind around a wire, then your initial potential won’t be the same as your final potential even if you start and end at the same point

So any scalar potential would have to be a multivalued function, which isn’t fun to deal with

philmcole

so the constructed scalar potential replacement is discontinuous?

Semiclassical

In the sense that the definition of your scalar potential would have to change along the way, yes

(The potential itself still needs to change continuously and smoothly)

philmcole

okay what's the second option?

Semiclassical

The other route is to look for a vector potential instead of a scalar potential

Namely, you look for another vector field $\mathbf{A}$ such that $\mathbf{B}=\nabla \times \mathbf{A}$

And that’s a strategy that works generically, so it’s what people usually do

philmcole

oh nice finally it makes sense to me

Semiclassical

1:45 PM

The main annoyance off the bar is that it’s a vector field, so there’s three scalar components

So in principle three times as complicated

philmcole

I was so confused why it was suddenly different. Nobody told me that it is since the usual definition of potential doesn't work anymore.

Thank you I really understand it now!

Semiclassical

Well, the Lorentz force law for a charged particle in a magnetic field is weird in that it’s velocity-dependent

Hence the framework of KE+PE=E isn’t very useful for it

The work-energy Theorem still makes sense, but now the notion of potential energy isn’t really sensible

Hence one usually defers such discussions to an E&M course

The other issue is that, for scalar potentials, it’s enough to pick a reference point if you want to specify the potential uniquely

philmcole

Why would energy conservation not make sense anymore if the force is velocity dependent?

or I mean the notion of potential energy

Semiclassical

Because then the potential energy isn’t just a function of spatial location

philmcole

Potential energy is still defined as the work done?

Oh and since the work is the integral of the force which now depends also on $v$ it is no good

Semiclassical

1:56 PM

Right

philmcole

Cool I really learned something today :D

thanks again

Semiclassical

this all leads to the notion of gauge invariance which is confusing

philmcole

We didn't have this yet

Semiclassical

You won’t for a bit I imagine

philmcole

Hopefully I'll get on the top of the hill at some time where all those things finally tie together and make sense to me :P

Semiclassical

2:01 PM

You end up with scenarios where two different vector fields can look very different and yet still have the same curl

So both of them can be valid vector potentials for the same magnetic field, despite them looking so different

philmcole

This leads to problems I guess

Semiclassical

That’s like a vectorial version of the usual “you can shift a scalar potential by a constant without changing the physics”

And it leads to headaches

philmcole

Do you have to deal with them right now?

sounds like it :P

Semiclassical

not so much myself

user193319

If $X$ is a metric space that isn't totally bounded, how can I show there is an $r > 0$ and a collection of points $\{x_n\}$ such that $ \{B(x_n,r)\}_{n=1}^\infty$ is a collection of pairwise disjoint sets?

Semiclassical

2:04 PM

But if you do grad level physics you definitely run into this eventually

philmcole

Thanks for the warning!

philmcole

2:16 PM

@mercio I think if you take any differentiable scalar function you get a conservative field, since one definition of conservativity is "a field $f$ is conservative iff there exists $F$ with $f=\nabla F$". But you could take any non-differentiable scalar function which has no associated conservative field.

Peter Wildemann

The situation is even worse for more general gauge theories :)

But actually, although this "gauge freedom" is a bit annoying for some reasons it gives you an extra degree of freedom to play with, which is for example useful, when trying to show existence of equations involving these "gauge fields".

Semiclassical

It makes doing QFT a headache

Especially if you do stuff like QCD. (QED still has subtleties with gauge freedom, but QCD even more so)

philmcole

I just recently visited CERN and heard a talk about QCD. Crazy stuff.

Semiclassical

Nice!

philmcole

I didn't understand a thing.

But the talk was great and gave me an idea of what's happening right now in Physics.

Semiclassical

2:32 PM

The trouble with QCD (beyond just being inherently more complicated than QED) is that the coupling scale is different

With QED, the relevant number is the so-called fine structure constant, which is roughly 1/137

That’s small enough that doing perturbation theory is very effective, despite the fact that the series you get are actually not convergent

Same can’t be said for QCD in most cases of interest

So that approach will not tell you what you want to know, and in general it’s a lot of work to get stuff out of QCD

philmcole

Sounds really difficult. All I understood is that they need to do a looot of simulations in QCD. Basically most of what they talked about were based on numeric results.

Semiclassical

2:49 PM

probably lattice QCD stuff

which is hard

philmcole

yeah this was it!

Akiva Weinberger

3:29 PM

Hmm... JEE ? JEE is well known for being dumb. — Alex K Chen 1 min ago

2

@AbhasKumarSinha Thoughts?

@BalarkaSen

(Source)

Abcd

3:57 PM

f(x)= 1 , x is rational
= 0 , x i s irrational

Is this periodic

Akiva Weinberger

Should be

though it's confusing, 'cause it doesn't have a smallest period

Abcd

@AkivaWeinberger how :/

Akiva Weinberger

$f(x+1)=f(x)$, so $1$ is a period (not "the" period but "a" period)

Every rational number is a period

Abcd

ok

Akiva Weinberger

(We know that, if $p$ is a period, then $kp$ is a period for integer $k$. So every periodic function has infinitely many periods. This is strange in that there's no smallest one.)

Abcd

4:08 PM

@AkivaWeinberger I think you should see the JEE Maths section, I'd like to know your thoughts.

Links:

drive.google.com/file/d/0BxIR7Kj71SMKd0V2UlBMYVJpRHM/view

drive.google.com/file/d/0BxIR7Kj71SMKT2YtaFBPajRaOE0/view

(scroll a bit down for maths section)

Akiva Weinberger

4:43 PM

I just got distracted by Q39 in the first link (p22)

How do you approach that

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$Mathematics$ Mathematics