The h Bar

12:39 AM

@ACuriousMind ah dang

new font looks weeird

at least on mac

I'm not a fan but...eh, might just be not being used to it. Changing the font on anything you're used to in a different font looks weird

More Anonymous

7:21 AM

Hey all! is it possible to measure the Riemann Curvature using this setup?

Where the grey diagonal line is a one waymirror

(not sure how its done in LIGO)

Prateek Mourya

8:02 AM

Does anyone know a good book on binomial series

123

8:14 AM

Hi All..

123

11:36 AM

What is exact and inexact differential in terms of thermodynamic coordinates? Pls explain in easy to understand example.

ACuriousMind

see physics.stackexchange.com/q/278742/50583 and its linked questions

123

@ACuriousMind Thanks, In book it is clearly explained how work is path function. But how how heat is path function? Any exmaple

ACuriousMind

Since $\mathrm{d}U = \delta Q - \delta W$, if heat were a path function and the differential $\mathrm{d}Q$ of a function $Q$, then $\delta W = \mathrm{d}Q - \mathrm{d}U$ would also be exact - it would be the differential of $Q-U$ since $\mathrm{d}$ is linear.

More Anonymous

12:23 PM

@ACuriousMind Is a co-chain just a linear combination of differential forms multiplied with a prefactor

?

Srijan Suryansh

is this graph correct for Electric field vs R

More Anonymous

And does this prefactor change at every point?

Srijan Suryansh

ACuriousMind

@MoreAnonymous a linear combination of differential forms is just a differential form :P

More Anonymous

@ACuriousMind True!

:P

ACuriousMind

12:25 PM

a co-chain is just a functional on chains

More Anonymous

trying to wrap my head on this

ACuriousMind

the answer by joshphysics shows how every differential form is equivalent to a co-chain on the smooth chains (i.e. "paths" over which you can integrate)

this equivalence is also at the heart of why the deRham cohomology is the same as the ordinary cohomology of a smooth manifold

More Anonymous

I'm not really good at cohomology

ACuriousMind

then just ignore my remark :P

More Anonymous

why is he mapping $S_k \to Z$ ??

where $S_k$ the set of all singular k-cubes in $R^n$ ?

ACuriousMind

12:35 PM

it's just a definition; another way to phrase this is to say that chains are formal linear combinations of his "cubes" with integral coefficients (which he shows later is equivalent to the definition in terms of functions)

More Anonymous

thanks this helps^

ACuriousMind

I think his answer is pretty good at explaining what the intuition behind this is - "gluing" cubes together and assigning a "weight" to each cube

note that most usual math texts will talk about simplices, not cubes, but that's a non-essential difference

More Anonymous

Yea ... I think I get it now. Just one quibble. How does he know that that heat should be mapping: cubes to a number and not say a cube infinitesimal?

ACuriousMind

I don't know what a "cube infinitesimal" is.

A 1-cube is just a path. Thermodynamics tells us there's a way to assign a quantity "heat" to any process, and processes are paths in thermodynamic state space.

you shouldn't try to mix rigorous approaches (like joshphysics' approach here) with non-rigorous ideas like "infinitesimals", it rarely goes well

(there are rigorous approaches that use something like infinitesimals but they require different machinery)

More Anonymous

So here's one way of thinking of it (for one dimension):

$ \int_0^\infty f(x) dx = c $ ... Taking $x \to 2x$:

$ \int_0^infty f(2x) d x = c/2$

This is the same as saying in Riemann integration as count only the even strips. Then the question becomes how does one find the integral for an arbitrary sequence?

13

Q: The Definite Integral Problem (with a twist)?

The Definite Integral Problem (with a twist) In the Riemann integral one essentially calculates the area by splitting the area into $N$ rectangular strips and then taking $N \to \infty$. Here's something I asked myself related to the Riemann integral. Let's say I split the area into say $3$ st...

This is more precisely what I'm talking about

^

Also please use @MoreAnonymous? (it helps when Im multitasking)

ACuriousMind

1:14 PM

@MoreAnonymous I don't see what that question has to do with what we are discussing

if you want to talk about infinitesimal paths, then the rigorous notion of an infinitesimal path is just a tangent vector (equivalence classes of paths is one of the definitions for tangent vectors). A functional on tangent vectors at every point is an assignment of a cotangent vector to every point, which is another possible definition of what a differential form is (given the assignment is smooth in the appropriate sense)

More Anonymous

@ACuriousMind He's putting a weight (a number) on a finite cube. My question was not do it for the limiting case

an infinitesimal cube

ACuriousMind

the infinitesimal notion of a $k$-cube is just a collection of $k$ tangent vectors

a functional on these is a $k$-form

More Anonymous

Agreed

Balarka Sen

@MoreAnonymous Morally your idea is the idea of differential forms, yes. You put weights on small cubes.

More Anonymous

But my point is I can always multiply a coefficient $a$ to the say a one form at point $P$ $dx|_P $ and get $a dx|_P$ then at point $dx |_{P +\delta}$ I can multiply $b$ and get $b dx|_{P + \delta}$. I can continue doing this game for $N$ points and take the limit $N \delta = c$ where $N \to \infty$ and $\delta \to 0$

Balarka Sen

1:22 PM

I am not sure if there is a deeper connection

More Anonymous

@BalarkaSen Neither am I

:P

ACuriousMind

@MoreAnonymous "at a point" you don't have a 1-form, you just have cotangent vectors

More Anonymous

@ACuriousMind Can't I go to a one form using the musical isomorphism ?

ACuriousMind

no, I mean, a 1-form is an assignment of a cotangent vector to each point

"a 1-form at point $p$" = "the cotangent vector the 1-form assigns to $p$"

on a contractible domain - such as $\mathbb{R}^n$, you can write every differential form as linear combination of basis forms $\mathrm{d}x^i$ as $\omega = \sum_i f_i\mathrm{d}x^i$ where the $f_i$ are functions

there's no need to do this on some discretization and then take a limit or anything, it just works out of the box in the smooth case

More Anonymous

@ACuriousMind Yes but how do we know heat is smooth then?

ACuriousMind

1:26 PM

that's the usual phyiscs assumption - everything is smooth until it isn'T :P

note that "smooth" does not mean you can't integrate it over things like a zig-zag path - there's a natural way to integrate a smooth form over piecewise smooth paths

More Anonymous

@ACuriousMind yes thy physicists everything is nice convergent smooth etc etc XD

@ACuriousMind You can integrate my thingy too :P

Depending on how open minded you are to extend the definition of integration

ACuriousMind

@MoreAnonymous I haven't fully understood what you're proposing since you have to really define what your limits are and why they should converge, but I don't understand why you're even trying to do this.

More Anonymous

@ACuriousMind I believe the answerer does provide a proof of convergence?

6

A: The Definite Integral Problem (with a twist)?

Let $b_r = \sum_{d \mid r} a_d\mu(\frac{r}{d})$. We prove that if the $b_r$'s are small enough, the result is true. Claim: If $\lim_{n \to \infty} \frac{\log^2(n)}{n}\sum_{r=1}^n |b_r| = 0$ and $f$ is smooth, then $$\lim_{k \to \infty} \lim_{n \to \infty} \sum_{r=1}^n a_rf\left(\frac{kr}{n}\righ...

@ACuriousMind On why?

ACuriousMind

I also don't understand what that question has to do with differential forms or heat

rigorous axiomatizations of physics assume sufficient differentiability for everything to work out - there's no benefit to not assuming it

if you want to ask "how do we know heat is smooth?", you have to first explain what your axiomatization of heat is that does not already include that

and then you have to explain why anyone should use that axiomatization

More Anonymous

True

I guess to get something like I'm proposing one would need a fundamentally different form of process. I don't think you would get it from something nice like $F = ma$