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6

It's a good question, and one that puzzled me for a while as well. However the answer is very simple. For a massive particle like an electron the total energy is given by: $$ E^2 = p^2c^2 + m^2 c^4 $$ where $p$ is the momentum and $m$ is the rest mass of the electron. Electrons can obviously have any momentum you want, so the total energy can be any value ...


5

Intuitively, the moment of inertia of a single atom is far smaller than a diatomic molecule because the nucleus is at the origin, while in a diatomic molecule the nuclei are half the bond length from the origin. The minimum excitation energy for rotation is then much higher, well above room temperature, so it doesn't contribute, because $E=\frac ...


3

The term "preliminary" does not refer to the runs. It refers to the status of results and figures that have not been properly reviewed and/or approved for publication yet.


3

This won't work, though possibly not for the reason you think. High energy protons will go straight through a turbine blade without transferring any significant amount of momentum to it. The LHC uses a seven metre long block of graphite to catch the proton beam if there's a beam dump. Steel has greater stopping power than carbon, but even so a turbine blade ...


3

Elementary particle physics is an outgrowth of what was high energy physics, historically at the time. X-rays were high energy physics when first discovered, they are part of the tools of solid state physics now. Alpha particles and gamma rays were high energy physics at their time, they are nuclear physics now. Mesons discovered in cosmic rays started ...


2

As an alternative to Anna's nice historical discourse a heuristic that covers modern uses of the phrase would be that energies are "high" when the QCD can be treated as perturbative. That regime sets in considerably above the nucleon mass scale, say 10s of GeV. So LHC physics is in, JLAB physics is out (even with the 12 GeV upgrade).


2

Within the standard model alone, all these parameters are independent, and to those you can add the masses and mixing angles of the neutrinos. Possible additional symmetries beyond the standard model suggest some relations between the gauge couplings, since renormalization group analyses based on these symmetries lead to unification of these couplings at ...


2

No, if you observe which slit they traveled through then there is NOT an interference pattern. The act of observing, or more accurately, the need for the location of the electron to be resolved causes it to take on a definite position and then continue on from that position as a particle. If it is not observed or interacted with in some way that would make ...


2

One thing that you have to remember is that the laws of physics are the same in all reference frames. Lets say there exists some particle $A$ which is stable for $t_o$ seconds when it is at rest. From its own reference frame, its lifetime HAS to be $t_o$ seconds regardless of the velocity with which it travels, because the laws of physics which apply to it ...


2

Just to complete Hritik Narayan's answer: the lifetime of an unstable particle is by definition the average time before it decays in its rest-frame. So whatever the frame used, the lifetime remains the same. Now, as already mentioned, what you measure experimentally does depend on the frame and thus on the velocity of the particle. You do have to correct for ...


2

People are usually more interested in the reverse process of production, that is the annihilation of dark matter particles. This is simply because it may be easier to see the products of annihilation (which might produce photons as a by-product) than to notice a small amount of ordinary matter that has "disappeared" to produce dark matter. And finding ...


1

We have a number m that shows up in a purely mathematical context and then we interpret it as having a physical meaning. The question is: What is the intuition that connects the two? The intuition connecting the two is essentially identical with the reason why Wigner connects the purely physical notion of an elementary particle with the purely ...


1

Strangely enough, when you do a quantum mechanics experiment, you get a result that says something about what you already know. If you place a detector at one (or both) of the slits, you watch individual particles go through the slits, and you get a particle result from your detector (e.g., no interference pattern). If you don't know which slit the ...


1

Your intuition is quite reasonable. We observe (based on cosmic rays) that the universe is basically all matter, no antimatter. At the high energy state of the big bang we would expect an even balance. Cosmological theories need to explain the asymmetry in baryons vs antibaryons


1

Have a look at this Big Bang graph: Annihilation means that when a particle hits an antiparticle, there exists a probability that they would both disappear and the energy turned into other particle antiparticles. It is the quantum numbers that cancel each other. At the elementary particle stage of the Big Bang, the energy is carried by elementary particles ...


1

Do matter and antimatter annihilate or release energy? They typically annihilate and release energy. Check out electron-positron annihilation, and low-energy proton-antiproton annihilation. Image credit CSIRO, see The Big Bang & the Standard Model of the Universe Do matter and antimatter eliminate each other or release their equivalent ...


1

Your question seems to look for a fundamental picture for particles. We have been colliding particles since long time, actually trying to find what is there more fundamental to protons. Since we cannot know in detail what hapens in the short time and space where they occur, what we can do is measure everything that comes out and try to understand from this ...


1

An electron does not spin! Its intrinsic angular momentum (the so called spin), should not be confused with the point-like electron rotating in configuration space (then the gyromagnetic factor would be one which is in a way related to charge spinning in configuration space. Actually the gyromagnetic factor of the electron spin is approximately 2.) A black ...


1

As the comment above says, the word "spin" should not be taken literally, as in the spin of a beachball. The word spin came about as an attempt to physically understand the differing energy levels an electron can have, due to the magnetic field associated with it. The idea behind it goes back to when the electron was discovered experimentally to have a ...


1

If my understanding is correct, an electron is an elementary particle which means that it is just a point in space, ... The electron spin is a special case of the general concept of angular momentum, which is a physical quantity generated by rotations. This is completely analogous to energy being generated by time translations and momentum by spatial ...


1

W-Boson events decaying into two leptons (e.g. electron and electron-neutrino) The convention is of course to say that a leptonic decay of a $W^-$ boson produces a negatively charged lepton (such as an electron, $e^-$) together with an anti-neutrino (of the matching weak state, such as an anti-electron-neutrino, $\overline\nu_e$). the angle $\theta$ ...


1

The general theory of relativity predicts that kinetic energy will contribute to gravitational mass. Here is a paper that explores the gravitational effect of kinetically energetic particles within a system: http://arxiv.org/PS_cache/gr-qc/pdf/9909/9909014v1.pdf. Here is an interesting article by Frank Helle on the production of gravity by relativistic ...


1

A particle with such a wave function is in a momentum eigenstate i.e. all measurements of momentum for the particle will always return the value p. The physically relevant quantity with respect to position is the amplitude modulus squared which by the Born interpretation gives us the probability of finding the particle at a given point. In this case, this ...


1

There's 2 types of particles that quarks can form, Baryons and Mesons. List of Baryons and List of Mesons All of these, except for the Proton are unstable. The Neutron is unstable on it's own but it's stable when bound to Protons in an atomic Nucleus. A Baryon is 3 quarks and a Meson is 2 quarks, so that limits the number of combinations. Description ...



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