Hot answers tagged particle-physics
11
If you want to print it on a T-shirt - print the Lagrangian of the Standard Model - then the T-shirt has already been printed and it looks like this:
http://www.flickr.com/photos/joshmt/7503048090/
The physicist on it is John Ellis, the 2nd most cited living particle physicist. Of course, the terms could be elaborated upon, written a bit more ...
8
I'll state one version of the theorem, valid for classical systems. I'll not give the most general framework, as things become messy, but this should still give you an idea of how general the result is.
We need the following ingredients:
Spins: to each vertex of the lattice $\mathbb{Z}^2$, we attach a spin $\phi_x$ taking values in some compact ...
5
In order to understand asymptotic freedom, you need to be aware of the concept of renormalization. Since you want a qualitative description, just think of renormalization a modification of the coupling strengths and masses of particles at high energies. This is roughly like pushing a ball through the water; the harder you push, the more the water sticks ...
5
The first generation of elementary particles are by observation not composite and therefore not seen to decay. They are shown in this table of the standard model of particle physics in column I.
The Standard Model of elementary particles, with the three generations of matter, gauge bosons in the fourth column and the Higgs boson in the fifth.
All ...
4
In an experiment in which particles are collided, a resonance is a large peak in a cross section (rate at which a process occurs) when plotted against the energy of the incoming particles. For example, when LEP collided electrons with positrons, they saw a resonance when the energy of the incoming particles equalled the mass of the $Z$-boson.
Resonances ...
4
As far as I know, nobody has ever done this, at least not at what we currently consider high energy. (Electron-electron collisions happen at low energy all the time, of course.) I doubt that anything interesting would happen, primarily because electrons are mutually repulsive, and they have a low mass. That means two colliding electrons would just bounce ...
3
Let us clear some misunderstandings:
if an electron absorbs a photon to get exited to a higher energy level,
It is not the electron that absorbs the photon to go to a higher energy level. It is the whole atom, which is represented by a potential well with energy levels filled by electrons up to a point. A photon with the correct energy, i.e. an ...
3
This is a plot where the incoming pion beam is varied and the total interaction cross section is measured , showing higher scattering cross section at the resonances, as the beam kinetic energy is varied. The interactions are measured by looking how many pions are left in the beam direction after the beam has passed has passed the target. The difference is ...
3
There are at least two issues:
@Lagerbaer's comment points out that that you have to convert the bin numbers into time intervals. The time for your experiment is already fixed, so if you use more bins, then each bin indicates a shorter time interval. The final bin in your third plot means a time twice as much as the final bin in your second plot, and three ...
2
The problem is that your coordinates aren't well defined at $\theta=0$ and $\phi=\pi/2$. Note in particular that
$$
U|_{(0,\frac{\pi}{2},\gamma)} = \begin{pmatrix}1&0\\0&1\end{pmatrix}
$$
for any value of $\gamma$. A simpler choice is
$$
\tilde{U} =
\begin{pmatrix}
x+iy & z+iw \\
-z+iw & x-iy
\end{pmatrix},
$$
with
$$
x = \sqrt{1 - y^2 - z^2 ...
2
The magnetic quadrupole moment tensor is given by
$$m_{ij}=\left\langle \frac{2}{3}\left(\mathbf{r}\times\mathbf{J}\right)_i r_j \right\rangle,$$
in analogy with the magnetic dipole moment vector
$$m_i=\left\langle \frac{1}{2}\left(\mathbf{r}\times\mathbf{J}\right)_i \right\rangle.$$
The magnetic field at a point $\mathbf{R}$ is then, up to quadrupole ...
2
In the quantum mechanical conceptual framework, the boundary between particles and (excited or ground) states vanishes. A particle is a state, and a state is a particle. (More precisely, particles are eigenstates of the operator of energy - in the low-energy case - or the operator of mass - in the high-energy case.) A physical system with a ladder of energy ...
2
Charged particles can't be used in air. In outer space, they can be deflected by electric or magnetic fields. A collimated beam will also become defocused by ambient fields over a sufficiently long distance -- and the distances are typically very long in outer space. If you make a beam of, say, negatively charged particles, then conservation of charge says ...
2
First, of course there's no perfect mirror. But let's assume there was one.
Next, the question is: Is the bouncing off the mirrors elastic or inelastic. If the photon is absorbed and re-emitted with the same frequency, then the bouncing is elastic and no energy is lost by the photon. It would then go on forever and ever.
But what if it does lose energy ...
2
You can generate Dirac neutrino masses through the Higgs mechanism by introducing right handed neutrinos (in the same way you generate masses for the upper quarks). Since neutrino masses are at the sub-$eV$ scale, this means that the Yukawa couplings have to be unnaturally small, of order $10^{-12}$.
People prefer to keep $\mathcal{O}(1)$ Yukawa's and ...
2
First I should say "we don't know". The fact that an electron has an intrinsic magnetic moment (spin) suggests it probably has some structure though.
See What is the mass density distribution of an electron? for more on this subject.
2
The natural diversity of energy levels at which matter organizes itself tends to isolate related phenomena in interesting and often confusing ways.
Consequently, physics has found that phenomena that seemed inexplicable often began to unravel and start to show revealing internal structure when energy levels are raised high enough to overcome those internal ...
2
The Pion fields are the coordinates of the Stereographic projection:
$\phi_i = \frac{2 \pi_i}{1 + \pi^2} , i = 1, ..., n-1$
Where:
$\pi^2 = \sum_{i=1}^{N-1} \pi_i\pi_i$
And:
$\phi_n = \frac{-1 + \pi^2}{1 + \pi^2} $
As can be seen, this construction solves the constraint equation: $ \sum_{a=1}^{N} \phi_a\phi_a= 1$.
Substituting in the Lagrangian, we ...
1
The fact that electrons are identical particles doesn't mean you can't separate one from another and keep track of which is which. They can be told apart according to their positions, energies, and momenta. Say I stick a $\beta^-$ source a foot away from a Geiger counter and I get a click. I know that the electron I detected was 1 ns old (c=1 ft/ns). This ...
1
Ignoring the fact that this is a reaction with very low probability and the difficulty of detecting such an interaction the answer is : no, nothing happens to the quantum mechanical status of the other gamma.
If by some ingenious method one managed to determine the spins in the pair produced, still the spin of the parent gamma would be unknown, since the ...
1
To me, it seems like there are 3 different concepts being discussed: (1) fine-tuning, (2) wanting unitless constants to be of order unity, and (3) wanting theories to have a simple form. The WP link defines "naturalness" as #2, although I don't think that's universally understood.
A very old example of #3 would be the history of models of the solar system, ...
1
David Zaslavsky has given a solid, relatively model-independent explanation of the empirical bounds on the size of an electron based on particle-physics experiments that probe short distance scales by using collisions at short wavelengths. There is also another way of getting at this question, which has been studied by people who have tried to model quarks ...
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