Yes, of course I meant that, sorry.
I still don't understand. So if my projectile has low momentum and the spatial resolution is low, the particle doesn't look point-like but like a disc and I have to fold the potential with this disk? If I increase the momentum, the particle will approach...
Dear PF,
although I've gone through many particle phyics lectures and textbooks, I still have problems with wrap my mind around the whole scattering theory and cross section topics.
1. Is there a deep reason why cross sections for charged, point-like particles decrease with the center-of-mass...
Thanks but I solved question 4 now. The energy nonconservation is just due to an approximation of a square root, to be able to compute the answer but it has nothing to do with the parton branching as such. Quigg helped me here.
I think it is not a specific DIS question. The energy nonconservation should imho be a trick but that's exactly my problem, why this trick is allowed. The problem is that I had to ask the prof so much things that I couldn't ask everything and his notes are not selfexplaining.
He called the...
Hi @ all,
during tmy exam preparation I stumbled upon some facts that I couldn't explain to myself.
1. Strong coupling constant g_s is universal -> colour charge is quantised
2. The MS_bar quark mass \bar{m} = m_0 + \delta \bar{m} can be given as a function of the quark pole mass m =...
H_D = \vec{\alpha} \cdot \vec{p} c + \beta mc^2 is the Dirac Hamiltonian. The original Dirac equation looks like the Schrödinger equation but it contains the Dirac Hamiltonian instead of the Schrödinger Hamiltonian:
H_D \Psi(\vec{x},t) = i \hbar \partial_t \Psi(\vec{x},t)
Using a special...
The Hamiltonian is not a vector but an operator on a Hilbert space which we can write as a matrix.
Commutators like [\vec{x}, H] = ... are shortcuts for [x_i, H] = ..., \quad i \in {1,2,3}.
Note that x_i is also an operator on a Hilbert space. Thus a commutator [A,B] = C is also a...
I was tought that if you expand the square root in E = (p^2 + m^2)^{1/2} and rewrite the operators in position space, then we have one time differentiation on the left but a polynomial of space differentiations on the right hence this equation cannot be Lorentz covariant.
One should also point out that the charged weak current interaction (W-boson) just couples to particles in the left-handed chirality eigenstate whereas the neutral weak current interaction (Z-boson) couples to both chirality eigenstates but in general with different strength, depending on the...
If I'm right, your point is that an external field interaction can alter the chirality of a particle. On the other hand, you say that the spin gets flipped. But since an electron is massive, it can be in both helicity states so if its spin flips, its helicity gets changed but the chirality does...