The h Bar

@Sanjana just in case you havent seen a bazillion versions of that integral, it is commonly appearing in intro to QFT as similar to a part of the correlation functions. This integral infamously is troublesome. It should be a Bessel K function, and you can reason things out from there. You can also show its asymptotic behaviour, approximate it in various ways, etc. If the argument in the paper is just nonsense, you might prefer to look at the rigorous derivations instead.

Relativisticcucumber

5:07 AM

GUYS i think spinor might have had kittens im not sure. spinor was surrounded by 2 kittens and possibly more it was hard to see but i got a picture of one of them

3

they are a lot smaller than the picture makes them look. SO CUTE. the other one was moving too fast, but it was all orange, no white

naturallyInconsistent

6:15 AM

awwwwww~~~

Mr. Feynman

7:31 AM

@Relativisticcucumber looks like the rep was not an irrep and spinor is not a spinor

Unless, spinor was a Dirac spinor from the very beginning, so she's always been reducible

Yeah, these lil potatoes must be Weyl spinors

Sanjana

8:35 AM

@naturallyInconsistent Sorry for this late reply but yes, I have seen this integral. This isnt usually worked out in books, and that was the reason I found this QFT notes from a friend at a particular IIT.

Indeed from the last equation in the snapshot, they make a complex scaling of coordinates to convert that final integral into a "well known" integral form for a Bessel function.

So maybe the argument is wrong, but atleast the results are not...

This is how the complete solution looks like

Relativisticcucumber

@Mr.Feynman omg

Nairit Sahoo

8:50 AM

Hii everyone

I am unable to do one very trivial computation from Jackson

The relevant part is attached as a png file here.

I am unable to get the 2nd term in the square brackets of 3.114 of Jackson from 2.16 of the same book.

John Rennie

9:05 AM

@NairitSahoo I've uploaded the picture for you.

Nairit Sahoo

9:33 AM

@JohnRennie Thank you so much. I am new here and actually dont know how to do that

naturallyInconsistent

@NairitSahoo how far have you gotten?

Nairit Sahoo

2.16 has two terms. The first term follows directly from 3.70
The 2nd term of 2.16 I can't understand at all

naturallyInconsistent

The 2nd term of Equation (2.16) is the image charge term.

Nairit Sahoo

Yes, that's what I thought and said. But I am unable to get the exact functional form

What modification should I do to 3.70 to get that term?

naturallyInconsistent

You are getting this competely wrong. You have Equation (2.16) first and foremost. Then use Equation (3.70) twice to get Equation (3.114). It is clear that Equation (2.16) is in an earlier chapter and is not found by appealing to the later chapter. There must be some other way to get Equation (2.16)

Nairit Sahoo

10:03 AM

@naturallyInconsistent Oh, I mispoke... I understood eqn. 2.16 fully. When I said "The 2nd term of 2.16 I can't understand at all", I meant that I couldn't understand the term coming in eqn 3.114 arising from the 2nd term in 2.16.

naturallyInconsistent

@NairitSahoo you should get something from what we have already talked about, so where have you gotten?

Nairit Sahoo

@naturallyInconsistent Nowhere, you didn't say anything new. I am unable to get the 2nd term in the square brackets of 3.114

naturallyInconsistent

@NairitSahoo and im asking you what you have tried

Nairit Sahoo

Nothing in particular. I have no idea, where to begin. For example, I feel that I have to modify $\vec{x'} \to \frac{a^2}{x'^2} \vec{x'}$ But I don't think it will bring any changes in the RHS of 3.70 apart from something in the spherical harmonics which I cannot figure out

So I am guessing even the approach is wrong

Can you give some hints?

naturallyInconsistent

10:24 AM

@NairitSahoo This is a good first step.

But you should try something along that lines but in a systematic way

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