# Charge distribution with exponential form. Halzem and Martin. exercise 8.4

1. May 27, 2012

### elmerx25

1. The problem statement, all variables and given/known data

In the exercise 8.4 from Quarks and Leptons. An Introductory Course in Modern Particle Physics - F.Halzem,A.Martin we can see:

if the charge distribution $\rho(r)$ has an exponential form $e^{-mr}$, then:

$$F(q) \propto (1 - \frac{q^2}{m^2})^{-2}$$

where F(q) is:

$$F(q) = \int\rho(x) e^{iq.x} d^3x$$

3. The attempt at a solution

The book says that first we integrate the angular part and obtain:

$$F(q) = 2\pi \int\rho(r) (\frac{e^{iqr}-e^{-iqr}}{iqr}) r^2 dr$$

Please, can anyone say me how can I obtain $(\frac{e^{iqr}-e^{-iqr}}{iqr})$

2. May 28, 2012

### vela

Staff Emeritus
As the problem says, you integrate the angular part. Show us your work at doing that.

3. May 29, 2012

### elmerx25

In spherical coordinates, we have

$$F(q) = \int\rho(r) e ^{i q r}r^2 dr \int_{0}^{\pi} sen \theta d\theta \int_{0}^{2\pi} d\phi = 2 * 2\pi \int\rho(r) e ^{i q r} r^2 dr$$

But then, from where come:

$$(\frac {e ^{i q r}- e ^{-i q r}}{iqr} )$$

I think that it comes from a property of Fourier transform. It looks like a $sen (iqr)$ but from where?

And moreover, I am not sure if $e^{iqx}$ in spherical coordinates were $e^{iqr}$

Thank you.

4. May 29, 2012

### vela

Staff Emeritus
$\vec{q}\cdot\vec{x}$ isn't equal to qr. Remember the dot product depends on the angle between the two vectors, not only on their magnitudes.

5. May 29, 2012

### elmerx25

Ok. Then I must write:

$$F(q) = \int\rho(r)\ e ^{i \ q \ r \cos \alpha}r^2 dr \int_{0}^{\pi} sen \theta \ d \theta \int_{0}^{2\pi} d\phi = 2 * 2\pi \int\rho(r) e ^{i \ q \ r \ cos \alpha} r^2 dr$$

And $\alpha$ is the angle between q an r. Sorry, but now I don´t kown how to continue. I can´t integrate by parts... I don´t know which way to take.

6. May 29, 2012

### vela

Staff Emeritus
When evaluating the integral, orient the coordinate system so that q lies along the z-axis.

7. May 29, 2012

### elmerx25

I think that I have it already. I must make a little change:

$$F(q) = \int _{0}^{2\pi} d\phi \int_{0}^{\pi} e^{i \ q \ r \ cos \theta} sen \theta \ d\theta \int \rho (r) \ r^2 \ dr$$

$\theta$ is the angle between z-axe and r, and q is along z-axe, so that $\theta$ is the angle between q and r.

$$\frac {d}{d\theta} e^{i \ q \ r \ cos \theta} = -i \ q \ r \ sen \theta \ e^{i \ q \ r \ cos \theta}$$

so that:

$$\int_{0}^{\pi} e^{i \ q \ r \ cos \theta} sen \theta \ d\theta = [\frac{e^{i \ q \ r \ cos \theta}}{- \ i \ q \ r}]_{0}^{\pi} = [\frac{e^{i \ q \ r \ cos \theta}}{i \ q \ r}]^{0}_{\pi} = \frac{e^{i \ q \ r} - \ e^{- \ i \ q \ r }}{i \ q \ r}$$

And finally we have:

$$F(q) = \int _{0}^{2\pi} d\phi \int_{0}^{\pi} e^{i \ q \ r \ cos \theta} sen \theta \ d\theta \int \rho (r) \ r^2 \ dr = 2\pi \int \rho (r) \ (\frac{e^{i \ q \ r} - \ e^{- \ i \ q \ r }}{i \ q \ r}) \ r^2 \ dr$$

It´s incredible. I didn´t think never that I would be able to solve this problem. I have an exam of "Elementary particles" in the University in a few weeks. I work and I can´t visit the classes. I don´t know anybody that can help me becuase I don´t know the other students. When I am not able to solve a problem, then nobody can help me and it is frustrating.

I am very grateful, vela. Thank you very much.

Last edited: May 29, 2012
8. Jun 23, 2012

### elmerx25

I need help with this integral: charge distribution

1. The problem statement, all variables and given/known data

In the exercise 8.4 from Quarks and Leptons. An Introductory Course in Modern Particle Physics - F.Halzem,A.Martin we can see:

if the charge distribution $\rho(r)$ has an exponential form $e^{-mr}$, then:

$$F(q) \propto (1 - \frac{q^2}{m^2})^{-2}$$

where F(q) is:

$$F(q) = \int\rho(x) e^{iq.x} d^3x$$

2. Relevant equations

In spherical coordinates, we have

$$F(q) = \int _{0}^{2\pi} d\phi \int_{0}^{\pi} e^{i \ q \ r \ cos \theta} sen \theta \ d\theta \int \rho (r) \ r^2 \ dr = 2\pi \int \rho (r) \ (\frac{e^{i \ q \ r} - \ e^{- \ i \ q \ r }}{i \ q \ r}) \ r^2 \ dr$$

3. The attempt at a solution

I am not able to solve this integral:

$$F(q) = 2\pi \int e^{-mr} \ (\frac{e^{i \ q \ r} - \ e^{- \ i \ q \ r }}{i \ q \ r}) \ r^2 \ dr$$

I have tried:

$$F(q) = \frac{2\pi}{iq} \int (e^{Ar} - e^{-Br}) r \ dr = \frac{2\pi}{iq} ( \int (e^{Ar} \ r \ dr - \int e^{-Br} r \ dr)$$

with $A = iq-m$ and $B = iq+m$

Finally I have this solution :

$$F(q) = \frac{2\pi}{iq} ( [...] + \frac{1}{(1-\frac{q^2}{m^2})^2 m^4 + 4q^2m^2} (2(m^2-q^2) - B^2 e^{Ar} - A^2 e^{-Br}))$$

with [...] igual to the initial conditions when I integrate by parts. I suppose that it is possible to choose initials conditions so that [...] -> 0. But I think that there are too many coefficients. Why do I obtain $4q^2m^2$ ? Is it a mistake or is the result good?

Thank you very much.

9. Jun 23, 2012

### elmerx25

Re: I need help with this integral: charge distribution

I made an error with a sign. The result is:

$$F(q) = \frac{2\pi}{iq} ( [...] + \frac{1}{(1-\frac{q^2}{m^2})^2 m^4 + 4q^2m^2} (4i q m + A^2 e^{-Br} - B^2 e^{Ar}))$$

And if [...] -> 0, then:

$$F(q) = 2\pi \ (\frac{1}{(1-\frac{q^2}{m^2})^2 m^4 + 4q^2m^2} (4m + \frac{A^2 e^{-Br} - B^2 e^{Ar}}{iq}))$$

But, what about $4q^2m^2$?

10. Jun 23, 2012

### vela

Staff Emeritus
It's kind of hard to follow what you did because you left out a lot of steps. Show how you integrated
$$\int_0^\infty re^{-r(m\pm iq)}\,dr$$ If you're familiar with Laplace transforms, you could also combine the exponentials to get
$$\int_0^\infty r\sin(qr)e^{-mr}\,dr$$ which is the Laplace transform of $r\sin qr$.

11. Jun 23, 2012

### elmerx25

I thought also this integral that you say:
$$\int_0^\infty r\sin(qr)e^{-mr} dr$$
but, how can I integrate it?, by parts? When I integrate by parts I use:
$$\int u \ dv = u \ v - \int v \ du$$
In your integral, who is $u$ and who is $v$?

Now I show you the steps in the calculation of my integral:

$$F(q) = \frac{2\pi}{iq} \int (e^{Ar} - e^{-Br}) r \ dr = \frac{2\pi}{iq} ( \int (e^{Ar} \ r \ dr - \int e^{-Br} r \ dr)$$

with $A = iq-m$ and $B = iq+m$

$$F(q) = \frac{2\pi}{iq} \left\{ ( \left[ r´ \frac{e^{Ar´}}{A} \right]_0 ^r - \int \frac{e^{Ar}}{A} dr ) - (\left[ r´ \frac{e^{-Br´}}{-B} \right]_0 ^r - \int \frac{e^{-Br}}{-B} dr) \right\}$$

$$F(q) = \frac{2\pi}{iq} \left\{ (\frac{re^{Ar}}{A}) - \left[\frac{e^{Ar´}}{A^2} \right]_0 ^r - ((-\frac{re^{-Br}}{B}) - \left[\frac{e^{-Br´}}{B^2} \right]_0 ^r) \right\}$$

$$F(q) = \frac{2\pi}{iq} \left\{ (\frac{re^{Ar}}{A}) - \left[\frac{e^{Ar}-1}{A^2} \right] + (\frac{re^{-Br}}{B}) + \left[\frac{e^{-Br}-1}{B^2} \right] \right\}$$

$$F(q) = \frac{2\pi}{iq} \left\{ r(\frac{e^{Ar}}{A} + \frac{e^{-Br}}{B}) - \frac{e^{Ar}}{A^2} + \frac{e^{-Br}}{B^2} + (\frac{1}{A^2} - \frac{1}{B^2}) \right\}$$

Here I imagine that we can choose the initial conditions so that $r(\frac{e^{Ar}}{A} + \frac{e^{-Br}}{B}) = 0$

$$F(q) = \frac{2\pi}{iq} \left\{ \frac{A^2e^{-Br}-B^2e^{Ar}}{A^2B^2} + \frac{B^2-A^2}{A^2B^2} \right\}$$

Now:

$$A^2 = (iq-m)^2 = m^2 - q^2 - 2i \ qm$$
$$B^2 = (iq+m)^2 = m^2 - q^2 + 2i \ qm$$
$$B^2 - A^2 = 4i \ qm$$
$$A^2B^2 = (m^2 - q^2)^2 + 4q^2m^2$$

$$F(q) = \frac{2\pi}{iq} \left\{ \frac{4i \ qm + A^2e^{-Br}-B^2e^{Ar}}{(m^2 - q^2)^2 + 4q^2m^2} \right\}$$

$$F(q) = \frac{2\pi}{iq} \left\{ \frac{4i \ qm + A^2e^{-Br}-B^2e^{Ar}}{(1 - \frac{q^2}{m^2})^2m^4 + 4q^2m^2} \right\}$$

$$F(q) = 2\pi \left\{ \frac{1}{(1 - \frac{q^2}{m^2})^2m^4 + 4q^2m^2} (4m + \frac{A^2e^{-Br}-B^2e^{Ar}}{iq}) \right\}$$

So, $F(q) \propto \frac{1}{(1 - \frac{q^2}{m^2})^2m^4 + 4q^2m^2}$. But the problem says that $F(q) \propto \frac{1}{(1 - \frac{q^2}{m^2})^2}$

I think that something is wrong in my calculations because the problem don´t says nothing about $m^4$ and $4q^2m^2$.

I thank you vela again for your inestimable help.

12. Jun 23, 2012

### vela

Staff Emeritus
In the denominator, expand $(m^2-|\vec{q}|^2)^2$. Also, note that Halzen and Martin are using the convention $q^2 = -|\vec{q}|^2$.

13. Jun 23, 2012

### vela

Staff Emeritus
I'd actually use $re^{-mr} = \frac{\partial}{\partial m}e^{-mr}$ to get
$$\int_0^\infty r\sin(qr)e^{-mr}\,dr = -\frac{\partial}{\partial m}\int_0^\infty \sin(qr)e^{-mr}\,dr$$

14. Jun 24, 2012

### elmerx25

Ok. Finally I understand. From here:

$$F(q) = \frac{2\pi}{iq} \left\{ \frac{4i \ qm + A^2e^{-Br}-B^2e^{Ar}}{(m^2 - q^2)^2 + 4q^2m^2} \right\}$$

I was working with a vector and finally I transform it into a four-vector using: $q^2 = -|\vec{p}|^2$

$$(m^2 - |\vec{q}|^2)^2 + 4|\vec{q}|^2m^2 = m^4 + |\vec{q}|^4 - 2m^2|\vec{q}|^2 + 4|\vec{q}|^2m^2 = m^4 + |\vec{q}|^4 + 2m^2|\vec{q}|^2 = (m^2 + |\vec{q}|^2)^2 = (m^2 - q^2)^2$$

Oh vela!! , you are really great!!

Thank you very much. When I make my exam next week I´ll keep you in my mind.

15. Jun 24, 2012

### vela

Staff Emeritus
The variable r shouldn't be in your final result you're integrating with respect to it.

16. Jun 24, 2012

Ok. Thanks.