Recent content by elmerx25

Hello: I have had a sabbatical year without physics. Now I try to continue with my study.

How can I send this post to the homework/coursework/textbook area? Thanks.

Ok. Thanks.

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 +...

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...

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...

I need help with this integral: charge distribution Homework Statement 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...

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...

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...

Hello Vela. Thank you for your answer. 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...

Homework Statement 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...

Hello: A spinless electron can interact with A^\mu only via its charge; the coupling is proportional to (p_{f} + p_{i})^{\mu}. An electron with spin, on the other hand, can also interact with the magnetic field via its magnetic moment. This coupling involves the factor i\sigma^{\mu\nu}(p_{f}...