I didn't understand what is point 1 and point 2...
Anyhow I found the formula for the integrated 3 body phase space:
\Phi= \frac{1}{\pi^3 2^7 M^2} \int_{s_2}^{s_3}{\frac{d s}{s} \sqrt{(s-s_1)(s-s_2)(s_3-s)(s_4-s)}},
Where M is the mass of the initial state and s_1=(m_1-m_2)^2, \quad...
There is only the differential cross section... I need the total cross section i.e. the integral of your formula. In the case of 3 massive body I think I can't integrate it. I did only the numerical integration...
Hi,
Do you know if there is an explicit formula for the integrated 3 body relativistic phase space of 3 particle with the same mass? I.e. M->3m
Or an approximate one?
Thank you!
In QCD you have an approximate chiral symmetry because 3 of the quarks are light (the mass of u,d,s is less than the typical energy[tex] \Lambda_{qcd}[\tex]). So you have an SU(3)xSU(3) symmetry. The SU(2)xSU(2) is "more" exact because u and d are lighter.
This isn't true for the others quarks...
Not always...
My problem is about Majorana's fermions:
Take the scattering \nu_{\tau}+\bar{\nu}_{\tau}\rightarrow \nu_e+\bar{\nu}_e and the interaction {\cal{L}}=g \sum Z_{\mu}\bar{\psi}_{\nu_l}\gamma^{\mu}(1-\gamma_5)\psi_{\nu_l}.
The \nu are Majorana's fermions (i.e. d_r=b_r ) with...
Yes... For example Z_{\mu} \bar{\psi}_{\nu_l} \gamma^{\mu}(1-\gamma_5)\psi_{\nu_l} . You may take \nu \rightarrow \nu +Z with the first neutrino left-handed and the second right-handed.
The amplitude is {\cal{M}}_{fi} \approx \bar{u}' \gamma^{\mu}(1-\gamma_5)u \epsilon_{\mu}
with...
Hi...
Consider a neutrino with a Dirac mass m_\nu and the weak interaction
{\cal{L}}=\frac{g}{2 \sqrt{2}} \sum_l[{W_{\mu}^+ \cdot \bar{\psi}_{\nu_l} \gamma^{\mu}(1-\gamma_5)\psi_l + W_{\mu}^- \cdot \bar{\psi}_{l} \gamma^{\mu}(1-\gamma_5)\psi_{\nu_l} }\right{]} + \frac{g}{4...
I don't understand... My problem is to express a generic element of the group g=exp[i\xi_ax_a+i\theta_i t_i] as the product of 2 element of the form
g_1=exp[i\xi_ax_a] \ g_2=exp[i\theta_i t_i].
Thank you
Hi,,,
It's better to use always the rest mass and the equation m^2=E^2-|\vec{p}|^2 (the norm of the four vector momenta).
In your case the 4-vectors are P_e=(E_e,\sqrt{(E_e^2-m^2)}\hat{z}) and P_p=(E_p,-\sqrt{(E_p^2-m^2)}\hat{z}).
P_X=P_p+P_e=(E_e+E_p,(E_e-E_p)\hat{z}) where I have...
Hi...
A group G is proken to a subgroup H. Let t_{\alpha} the generator of G and
t_i the generator of H. The t_i form a subalgebra. Take the x_a to be the other indipendent generator of G.
Why any finite element of G may be expressed in the form g=exp[i\xi_ax_a]exp[i\theta_i t_i] even if...
Hi...
The orbital angular momentum L=r \wedge p is the generator of the rotation:
U \Psi(r) = e^{i L \cdot \omega}(r) . For an infinitesimal rotation \omega
U \Psi(r) \approx (1+ iL\cdot \omega)=(1+\omega_i \epsilon_{ijk}x_j \partial_k)\Psi
\approx \Psi(r + \omega \wedge r ) .
A generic...
Hi...
If you change the direction of an electron (for example with a magnetic field) it generates radiation and so its kinetic energy changed (it slows down).