Calculate an Integral: Intro to Superconductivity by M. Tinkham

[vatlychatran]

In the book Introduction to Superconductivity by M. Tinkham, there is an integral as shown below. Could you help me to derive the result? Thanks a lot!

"3.2 ORIGIN OF THE ATTRACTIVE INTERACTION
We now must examine the origin of the negative V_{kk'} needed for superconductivity. If we take the bare Coulomb interaction V(\vec{r})= e^2/r and carry out the computation of V(\vec{q})
V(\vec{q}) = V(\vec{k}-\vec{k'}) = V_{\vec{k}\vec{k'}} = \Omega^{-1}\int V(\vec{r})e^{i\vec{q}.\vec{r}}d\vec{r}
we find
V(\vec{q}) =\frac{4\pi e^2}{\Omega q^2}"

 

[Steve Drake]

not 100% sure but if its a volume integral, that's where the 4pi comes from
(dhpi from 0 to 2pi = 2pi and sin(theta) from 0 to pi = 2. so somehow the other bit must equal e^2/q^2?

 

[vatlychatran]

You are right about the value 4pi. When I change to spherical coordinates, I get this value from theta and phi, but I had to assume that cos(\vec{q},\vec{r}) (this is from dot product) is not a function of theta and phi.
In fact, I'm not clear with the calculation of M. Tinkham in some points. There is no information about the angle between q and r. He doesn't say about the interval of the integral, and I assume that r runs from 0 to infinite then I will have a result in which terms relating to the upper limit do not vanish. I can set upper limit to certain value because there is something like screening phenomenon but the result is not vanish. For the other term, It seems to be -\frac{1}{q^2}.
I like the way leading to this problem because it is pedagogic but solving it makes me annoy. Hope someone can help me.

 

[Bill_K]

V(k) ≡ Ω^-1 ∫ (e²/r) e^ik·x d³x

Integrate by parts twice:

= Ω^-1 ∫ ∇² (e²/r) (-1/k²)e^ik·x d³x

and since ∇²(1/r) = 4π δ³(x),

V(k) = 4πe²/(Ωk²)

 

[vatlychatran]

V(k) ≡ Ω^-1 ∫ (e²/r) e^ik·x d³x

Integrate by parts twice:

= Ω^-1 ∫ ∇² (e²/r) (-1/k²)e^ik·x d³x

and since ∇²(1/r) = 4π δ³(x),

V(k) = 4πe²/(Ωk²)

thank you so much!

 

[tiny-tim]

hi vatlychatran!

alternatively, convert to spherical coordinates, with the "North Pole" in the q direction, and it's a straightforward integration (no parts)

 

[vatlychatran]

hi vatlychatran!

alternatively, convert to spherical coordinates, with the "North Pole" in the q direction, and it's a straightforward integration (no parts)

Hi tiny-tim, from your instructions I get this
V(\textbf{q})=\frac{4\pi e^2}{\Omega q^2}(1-\cos{qR})

in which R is upper limit of r. I don't know how remove the term cos(qR) but at least I get one important thing that V(\textbf{q})\geq 0.
Thanks a lot because from your instructions I understand another thing.

 

[tiny-tim]

hi vatlychatran!

… I don't know how remove the term cos(qR) …

shouldn't it be something like qRcosθ, with the cos disappearing on taking the θ limits?

can you show how you got that? ​

 

[vatlychatran]

hi vatlychatran!


shouldn't it be something like qRcosθ, with the cos disappearing on taking the θ limits?

can you show how you got that? ​

Of course, wait me, cos(qR) is from the integral of (exp(-iqr)-exp(iqr)) or (-2isin(qr)). Wait me for details

V(\textbf{q})=\frac{1}{\Omega}\int \frac{e^2}{r} e^{i\textbf{q.r}}d\textbf{r}
=\frac{e^2}{\Omega}\int \frac{1}{r} e^{iq.r.\cos\theta}r^2 \sin\theta d\theta d\phi dr
=\frac{e^2}{\Omega}\int e^{iq.r.\cos\theta}r \sin\theta d\theta d\phi dr
For \phi, I have \int_0^{2\pi} d\phi=2\pi
For \theta, I have \int_0^\pi e^{iq.r.\cos\theta} \sin\theta d\theta
=-\int_0^\pi e^{iq.r.\cos\theta} d(\cos\theta)
=-\frac{1}{iqr} e^{iq.r.\cos\theta}|^{\pi}_{0}
=-\frac{1}{iqr} (e^{-iq.r}-e^{iq.r})
=-\frac{1}{iqr} (-2i \sin(qr))
=\frac{2}{qr} \sin(qr)
Then I have
V(\textbf{q})=\frac{4\pi e^2}{\Omega} \int_0^R \frac{1}{qr}\sin(qr)rdr
=\frac{4\pi e^2}{\Omega} \int_0^R \frac{1}{q}\sin(qr)dr
=\frac{4\pi e^2}{\Omega} \frac{-1}{q^2}\cos(qr)|_0^R
=\frac{4\pi e^2}{\Omega q^2}(1-\cos{qR})
That is my result.