The h Bar

6:06 AM

I have a difficult choice to make (buying a physical copy of): GSW or Polchinski?

any opinions are welcome

123

6:25 AM

HelloooOoo Gyus..

Hi @JohnRennie

Sir

John Rennie

Hi :-)

123

Unfortunately no question for today.. But always happy to discuss with this family of physics.

B. Brekke

7:03 AM

"The relaxation time approximation assumes constant broadening of the energy spectrum by $\Gamma$ that corresponds to the characteristic relaxation time $\tau$ = $\hbar$/$2\Gamma$." I do not understand what this phrase means. Especially the "broadening of the energy spectrum"

John Rennie

Isn't it just lifetime broadening due to the energy time uncertainty?

B. Brekke

It is in the context of an application of an external electric field to a simple material. So some excitation is expected to "live" longer due to this perturbation?

1 hour later…

Curious youth

8:25 AM

How to be a power ranger?

Slereah

Get one of those magic rocks

Curious youth

I think physics is so hard that I need them. How do you find them?

This is my Undergraduate assignment for this week.

2 hours later…

10:15 AM

@NiharKarve I would recommend GSW but maybe you would prefer Polchinski given your cft comments and it being more 'modern'

10:44 AM

@bolbteppa well, I do have supplemental CFT resources so the string theory textbook doesn't have to be self-contained. I was actually gravitating towards GSW myself (which'll probably be a better investment since online versions of Polchinski are more easily available)

GSW is just the canonical text, a book like Kiritsis' one has even more modern/recent stuff in it

I'd forgotten about that (I have Kiritsis' lecture notes from 1997, pre-In a Nutshell, they're quite good)

11:28 AM

Anyone knows if the following identity is derived somewhere in Griffith's?

ACuriousMind

@schn No idea about Griffith, but it's not hard to find on the 'net, e.g. math.stackexchange.com/q/2076653/143136. The right key phrase is "Green's function for the Laplacian".

@ACuriousMind nice, thanks

It's derived in chapter 1

Yeah, I found it

What is the problem with $\nabla \cdot \mathbf{A} = 0$ implying $\mathbf{k} \cdot \mathbf{A}_{\mathbf{k}} = 0$

11:36 AM

@bolbteppa I was just going to ask :)

So you take the divergence of the sum

i.e. $\nabla \cdot (\mathbf{a}f)$, where $f$ is some scalar (the exponential). This is a well known identity right?

$$0 = \nabla \cdot \mathbf{A} = \nabla \cdot \sum \mathbf{A}_{\mathbf{k}} e^{i \mathbf{k} \cdot \mathbf{r}} = \sum \mathbf{A}_{\mathbf{k}} \cdot \nabla e^{i \mathbf{k} \cdot \mathbf{r}} = i \sum \mathbf{A}_{\mathbf{k}} \cdot \mathbf{k} e^{i \mathbf{k} \cdot \mathbf{r}} \to \mathbf{A}_{\mathbf{k}} \cdot \mathbf{k} = 0 $$

How does the third equality follow? Why is $\nabla \cdot \mathbf A_k =0$?

It's constant

Ah, a constant vector

Yeah

11:40 AM

makes sense

@bolbteppa Neat, thanks!

@bolbteppa Have you read both Jackson and Griffith's?

Parts of both

Basic question. Why is $\nabla \times \mathbf a_1$ sometimes called the longitudinal part and the divergence $\nabla \cdot \mathbf a_2$ the transverse of a given vector field $\mathbf a$?

I'm familiar with the terms irrotational and solenoidal.

Helmholtz decomposition

In physics and mathematics, in the area of vector calculus, Helmholtz's theorem, also known as the fundamental theorem of vector calculus, states that any sufficiently smooth, rapidly decaying vector field in three dimensions can be resolved into the sum of an irrotational (curl-free) vector field and a solenoidal (divergence-free) vector field; this is known as the Helmholtz decomposition or Helmholtz representation. It is named after Hermann von Helmholtz.As an irrotational vector field has a scalar potential and a solenoidal vector field has a vector potential, the Helmholtz decomposition states...

11:56 AM

@bolbteppa It is what we just talked about I guess? :)

With $\mathbf A_k \cdot \mathbf k=0$?

Or $\nabla \cdot \mathbf{A} = 0$ implying $\mathbf{k} \cdot \mathbf{A}_{\mathbf{k}}$.

Or even $\nabla \cdot \mathbf{A} = 0$ implying $\mathbf{k} \cdot \mathbf{A}=0$?

3 hours later…

2:37 PM

The link just explains why they use those terms

2 hours later…

Shing

4:11 PM

p-adic number sounds pretty sexy

but I dont quite get it, would anyone be kind enough to explain this to me?

does it have anything to do with Casimir effect?

ACuriousMind

p-adic numbers don't really have anything to do with mainstream physics

there's some speculation, but nothing well-established that I know of

why would you think it has something to do with the Casimir effect?

Slereah

4:24 PM

I remember that there was some attempt to base QM on p-adic numbers

It didn't do much