What DIS reveals is that, as you increase the energy of the probe particle, it starts to scatter off lighter and lighter constituents of the proton. At higher energies it starts to scatter off virtual quarks and anti-quarks that are fleetingly created from the gluons that mediate the strong...
In the case of nuclear isotopes, the reason why they decay is because there is an alternative state of lower energy that can be "reached" by one of the available types of interaction - in this case, a weak decay.
The mass of a composite system of particles is defined as
\sqrt{(\sum E_i)^2 - (\sum \textbf{p}_i)^2}where E_i and \textbf{p}_i are the energy and momentum, respectively, of the ith particle.
It is not true that this gives a lower bound on the sum of the masses of the constituents. Consider...
The concept of force as in classical physics is not so prominent at quantum level as it is only really meaningful when discussing stable or metastable particles that are involved in "elastic" collisions, ie where the particles coming out are the same as those that went in. Obviously a particle...
The Planck energy is not directly relevant to the formation of black holes - the latter is governed primarily by the Schwarzschild radius of the quantity of mass under consideration.
See also http://en.wikipedia.org/wiki/Planck_energy.
Surely, if the electron is moving along the z axis then p_z = p and then
1 - (\frac{p}{(E + m)})^2 = \frac{E^2 + 2Em + m^2 - p^2}{(E + m)^2} = \frac{2m}{(E + m)}
I'm no expert, but this doesn't look invariant to me.
Looking back at this again, a couple of quick follow-ups:
The proton will not necessarily just be transformed into a neutron. At high energies a shower of mesons is likely to emerge; the one remaining baryon may be either a neutron or a proton (eg if one more \pi^- than \pi^+s is produced)...
In Deep Inelastic Scattering you basically 'fire' beams of electrons or neutrinos at targets containing nucleons and observe the scattering angles and outgoing energies of the particles that have interacted. From these observations some of the properties of the particles that scattered them can...
OK. But these sphaleron transitions don't seem to be observed in any experiments I'm aware of. Is this because they would require even higher energy levels than (say) LHC can muster, or is there a bit more to it than that. Unfortunately, this area is some way beyond my own current understanding.
The photon comes from a three-way interaction 'vertex' that allows electrons (or any other electrically charged particles) to emit or absorb them spontaneously.
Electrons have no currently known sub-structure, so it's impossible to say anything about "constituents" of them.
Assuming you mean a process like e- → W- + ve → W- + vμ → μ-, this is prohibited for a free particle by energy/momentum conservation even if the total energy of the incoming electron is sufficient to create a muon. This is similar to why a free electon cannot radiate photons.
Hadrons have a property called the Baryon number which is conserved by all everyday reactions*.
Protons, neutrons and other baryons have Baryon numbers of 1 (with -1 for the corresponding antiparticles).
All mesons have Baryon number 0. Therefore, mesons can never decay into baryons...
But surely the thing that isn't defined until symmetric breaking is the "direction" within the SU(2) "space" along which it breaks. Isn't this analogous to the difference between empty space where there is no magnetic field, when there is no preferred direction, and when a magnetic field is...
The coupling constants are parameters of the theory. Whatever their values, there will exist a value of θ that yields a massless photon.
They are also related to the electromagnetic coupling constant e by the relation
e = g_L\ sin (\theta) = g_Y\ cos (\theta)
So by measuing e and...