The h Bar

12:01 AM

The thermodynamic potentials (enthalpy, Hemholtz free energy, Gibbs free energy, etc.) all represent energy that was used to establish a situation and therefore energy that you can have back by de-establishing that situation.

0celo7

The book seems to evaluate $h_{out}$ before anything happens.

Is that correct?

I'm just looking for a yes or no

dmckee

It looks to me like your source is using the "out" subscript to represent a transfer due to a change. In other words these might be better written $\Delta h = \Delta u + \Delta (Pv)$.

0celo7

I don't know what's going on

dmckee

but I don't actually know that.

0celo7

hmm

maybe I should average the final and initial enthalpies @dmckee ?

my heat transfer is negative 20kJ

not good!

Slereah

12:19 AM

Maybe you need a blanket

Christopher Peart

Wait how can i join this

I do not have an account on this stack exchange

That is really crazy

dmckee

@ChristopherPeart The individual chats are only weakly linked to their sites, and there is only one notion of chat membership across the whole network.

@0celo7 You need to read the class materials (text(s), notes, etc) and find out what they mean by $_{out}$ so that you know you are doing the right thing. I don't have the materials in front of me, and my sources aren't using that notation.

0celo7

@dmckee I'm reading. They don't explain.

In all of the class problems, it was obvious what $(Pv)_{out}$ was supposed to be

Christopher Peart

@dmckee Ok thanks I will leave you to the geeky stuff you do here and go do my own geeky stuff in The nineteenth byte

0celo7

12:44 AM

@dmckee Dammit. It's in the book. You're supposed to assume the flow is constant.

there goes two hours

DanielSank

I think I'm starting to understand math.

0celo7

@DanielSank what math?

DanielSank

@0celo7 all of it. How to invent it.

I already knew this, but I'm feeling it.

You make up an idea and try to express it in terms of an established idea. Then you investigate whether the new thing has the properties you want.

I'm reading about path integration.

Why's that get starred?

peterh

1:04 AM

@DanielSank I starred it, because as I know, you are a physicist and I've though you've learned it long ago, and understand it pretty well. I've found it funny that it is not only for me too hard.

@DanielSank On my current level I think I have to understand the Lagrange-formalism much better at first, so this is my current focus.

DanielSank

@peterh I am studying path integrals for diffusion. There is no Lagrangian.

@peterh yes, I learned it long ago, but there are levels of understanding.

0celo7

@ACuriousMind So what is the principal bundle in Gauge Theory, anyway?

peterh

@DanielSank ok-ok

Slereah

@0celo7 $SU(3) \times SU(2) \times U(1)$?

0celo7

1:34 AM

@Slereah that's the group

What's the total space?

DanielSank

Here's what I understand about diffusion path integrals:

Slereah

the bundle of that and the base manifold?

DanielSank

We know that the probability to wind up at $(x, t)$ is $$W(x,t) = \frac{1}{\sqrt{4 \pi D t}}\exp\left( - \frac{x^2}{4Dt} \right)$$ by solving the diffusion equation.

So then you say you want to think of this as a weighted sum over all trajectories that end at $(t, x)$, i.e. $$W(x,t) = \int_{\mathcal{C}(0,0;x,t)} \text{some thingy} \, .$$

Here $\mathcal{C}(0,0;x,t)$ means "all paths that start at $(0,0)$ and end at $(x,t)$.

The paths can be thought of as chains of shorter paths, i.e. $(0,0) \rightarrow (x_1, t_t) \rightarrow \cdots \rightarrow (x,t)$.

But of course we know that $$W(x_i, t_i; x_{i-1}, t_{i-1}) = \frac{1}{\sqrt{4 \pi D dt}} \exp \left( - \frac{(x_i - x_{i-1})^2}{4 D dt} \right) $$ where $dt \equiv t_i - t_{i-1}$.

Kenshin

Hello

DanielSank

So $$W(x,t) = \prod_{\tau=0}^t \frac{1}{\sqrt{4 \pi D d\tau}} \exp \left\{ - \frac{1}{4D} \int_0^t \dot{x}(\tau)^2 \, d\tau \right\}\, .$$

Kenshin

1:43 AM

For the some thingy integral, shouldn't C represent a specific path?

how can you integrate over many different paths at the same time

shouldn't C be a specific path, and instead have a summation of many integrals over each possible specific path C

i.e. put an extra summation sign in front of your integral and include some weighting constants

DanielSank

Ooops, should have been $$W(x,t) = \int_{\mathcal{C}(0,0;x,t)} \prod_{\tau=0}^t \frac{1}{\sqrt{4 \pi D d\tau}} \exp \left\{ - \frac{1}{4D} \int_0^t \dot{x}(\tau)^2 d\tau \right\} \, .$$

So therefore $$\text{some thingy} = \prod_{\tau=0}^t \frac{1}{\sqrt{4 \pi D d \tau}} \exp \left \{ - \frac{1}{4D} \int_0^t \dot{x}(\tau)^2 d \tau \right\} \equiv d_W x(\tau) \, . $$

I think from here the trick is going to be how to integrate against functionals of $x(\tau)$, but I think that can probably be defined in some kind of algebraic way relative to known cases.

In fact, the known cases are those functionals defined as $$F[x(\tau)] = 1 \text{ if the path goes through some certain set of points and }0 \text{ otherwise}$$

Ha! Yes. We can define any other functional as linear combinations of those!

@0celo7 I just figured out a thing!

0celo7

@Slereah how do you know it's unique

Slereah

Because every manifold is special in its own way

0celo7

@Slereah you're right, it is

If spacetime is R^4, at least.

Slereah

1:58 AM

If you have the fiber and group can you have non-unique bundles?

What would be an example

Ideally a 2D example

DanielSank

Hmmm, this book could be a lot more concise.

I think I just summarized the first half of chapter 1.

oh well

heather

i just had a large facepalm, "i need to comment my code", "i am an idiot" moment.

DanielSank

Nice, nice.

heather

what the heck was I thinking a month ago?

anyone have some tachyons handy? =P

DanielSank

@heather I spent mine on candy.

sorry

@heather Probably not much.

heather

2:10 AM

@DanielSank what shop were you at that accepts tachyons as currency? I want to visit it =D

@DanielSank true

i'm commenting it now

as best I can

0celo7

@Slereah mobius band vs. cylinder over $S^1$.

Slereah

Different group tho

Identity versus Z2

0celo7

Both are vector bundles.

@Slereah what?

Slereah

The local trivializations have different group structures

0celo7

@Slereah just take any principal bundle that's not trivial

then take the base space and product with the group

Hopf fibration will work

DanielSank

2:17 AM

@heather The best possible situation is to write the code so that comments aren't as necessary.

For example, use good variable names and structure things logically.

@heather Santa Barbara Intergalactic Candy and Soda Company.

heather

@DanielSank yeah...i thought i did that =/

0celo7

@DanielSank I gave heather an integral problem. Do you want to see if you can crack it?

DanielSank

@0celo7 Just give the problem.

I'm not particularly good at integrals but I'll take a look.

0celo7

@DanielSank Let $g:[0,1]\to\Bbb R$ be continuous. Show that there exists a unique continuous function $f:[0,1]\to\Bbb R$ such that $g(x)-\int_0^x f(x-t)e^{-t^2}\, dt=f(x)$.

DanielSank

@0celo7 No.

0celo7

2:28 AM

@DanielSank Did you take a look?

DanielSank

@0celo7 I looked at it for five seconds.

0celo7

Good enough I suppose.

DanielSank

Actually the more I look at it the more interesting it looks...

hmmmmmmm

0celo7

@heather Have you learned any interesting math lately?

heather

@0celo7 not really - I've been reading about quantum error correction and trying to fix my terrible python code

i kind of need to learn how to balance my activities.

DanielSank

2:30 AM

Quantum error correction involves some pretty interesting math.

@0celo7 look up the "toric code" and see if you can understand it.

0celo7

@heather Your python code requires quantum corrections?

heather

@0celo7 lol, no

those were two separate things

0celo7

@DanielSank The wiki page?

heather

::facepalms::

i can be so idiotic

::goes off to fix half the functions in the program::

actually, wait, no...

@DanielSank, dictionaries number starting from 1, right?

0celo7

a

heather

2:36 AM

ah, I see the problem

fixed!

=D

DanielSank

@heather Dictionaries aren't indexed.

@0celo7 The Kitaev article.

heather

@DanielSank yeah...ignore me =)

DanielSank

k

0celo7

@DanielSank "Topological phases and quantum computation"?

DanielSank

@0celo7 Not sure. Link?

I can tell if I look at it.

heather

2:40 AM

imagine a graph of the sine function. the y axis is my mood and the x axis is time. The graph is my mood during coding =P

0celo7

arxiv.org/pdf/0904.2771.pdf

DanielSank

That's not the one I was thinking of.

One sec...

hmmmm, not finding it...

well ok fine start with the Wiki article.

0celo7

@DanielSank what's wrong with what's above?

DanielSank

@0celo7 Maybe nothing. I just don't know it well.

0celo7

@DanielSank I know literally nothing about quantum computing

What's going on in this article

DanielSank

2:50 AM

@0celo7 Too broad.

0celo7

@DanielSank What is a "code?"

What is "error correction?"

DanielSank

You have a vector (quantum state).

It is a superposition of tensor products of other vectors (States of individual qubits).

Somehow, one of the constituent vector's states changes (error).

You need to figure out that an error happened, but you're not allowed to check the vectors to see which one changed. You can only measure things like parity of multiple vectors.

Physics Meta

0

Q: Diagram Drawing Tool in Editor

It would be very nice if this editor had a diagram drawer, don't yall think? Like with easy drag-and-drop tools to place boxes and vectors for free-body diagrams, springs, circuit components, etc. etc.

feature-request

DanielSank

This is a "code" because the group of vectors is used to represent the information of only one logical vector. This is essential why it's ok for one constituent to suffer an error.

0celo7

3:04 AM

@DanielSank Huh

How does one measure parity of multiple vectors?

heather

3:15 AM

well, i fixed several problems =D

and then i found some problems =(

Anonymous

4:29 AM

Anonymous

@YashasSamaga

Anonymous

I found this in an old article

Anonymous

Here's the link

Anonymous

dtic.mil/dtic/tr/fulltext/u2/727198.pdf

Anonymous

It was released by the Foreign Technology Division it seems ( the scanned pages aren't a good read though! )

Anonymous

4:32 AM

(Page 72)

Leonhard Euler

5:51 AM

Hello.

Yashas Samaga

So we actually don't know that photons are massless?

is it just that our experimental apparatus is not able to measure it if it did have some tiny tiny tiny mass?

Leonhard Euler

@YashasSamaga, hi

Yashas Samaga

@LeonhardEuler hi

Leonhard Euler

@YashasSamaga, how do you convert 760 mm of Hg into Pascal?

Yashas Samaga

1Pa = 1N/m^2

760mm of Hg = the pressure exerted by 760mm of mercury

P = density * g * h

substitute the numbers in SI units and you'll get your answer in terms of Pa

P = 13600 * 10 * 0.76

If I remember correctly, it is nearly 101325 Pa

Leonhard Euler

5:58 AM

10mm=1cm,?@YashasSamaga

Yashas Samaga

:|

yes

Anonymous

@YashasSamaga We don't have sufficient experimental evidence to say that rest mass of photon is zero although most modern theories assume it to be 0. Mathematically, assuming its rest mass to be zero explains well a large number of phenomena. Check out the experiment carried out by Roderic Lakes in 1998.

Anonymous

Studies of galactic magnetic fields suggest a limit of less than $3 × 10^{−27}$ eV.

Anonymous

Here is the paper: silver.neep.wisc.edu/~lakes/mu.pdf

Leonhard Euler

@YashasSamaga, is it fine?

Yashas Samaga

6:06 AM

yes

DanielSank

@0celo7 Suppose each vector is 2D. The states are $|\uparrow \rangle$ and $|\downarrow \rangle$. The parity is even if there is an even number of vectors in the $|\uparrow\rangle$ state.

Yashas Samaga

@AccidentalFourierTransform After 30 days, I figured out what you meant by ""We all agree that photons travel at the speed of light." Not necessarily. If photons happen to have mass, then vγ<c and your argument about 0/0 is invalid. You only proved that mγ=0⇔vγ=c. But we could have, in principle, mγ=0 and vγ<c. "

Balarka Sen

7:50 AM

@Slereah Yeah, totally. Vector bundles over $X$ with structure group $G$ are in fact classified by maps $X \to BG$.

*upto homotopy

Leonhard Euler

8:22 AM

Hello @JohnRennie

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