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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 ...


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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).


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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 ...


0

At a vertex of the diagram the usual conservation rules apply. Lepton number, charge and so on...


-4

Why are electrons alike but photons not? Because it takes a given amount of an energy to make an electron: 511keV. That's the energy of an electron at rest. A fast-moving electron comprises more energy than an electron just sitting there in front of you, but if you were to stop it by removing the kinetic energy, its rest-energy is 511keV, and its mass ...


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 ...


0

Let me begin with David Miller's illustration of the Higgs mechanism. Miller depicts the Higgs field as a throng of journalists and politicians at a cocktail party. A famous politician, Margaret Thatcher (the only bad thing about this analogy), enters the room, playing the role of a particle. As she passes, the crowd gather around her, resisting her ...


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 ...


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

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 ...


0

I would add to that list the masses and mixing angles of the three known neutrinos, which are just as arbitrary as the others within the confines of the standard model. This adds seven parameters to the 19 listed in the Wikipedia article https://en.wikipedia.org/wiki/Standard_Model#Construction_of_the_Standard_Model_Lagrangian


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A hydrogen atom (I.e. a bound electron-proton pair) can travel a distance, during which time the particles must necessarily have the same average velocity; if the electron is oscillating about the proton then its instantaneous velocity will cross over from less than the proton's to greater than the proton's at times, such that there are instants where the ...


0

It depends on the nature of the particles. If we are bosons there is no limitation at all (example is a photon). If we are talking about two fermions (electrons, protons...), then we need to make sure that they do not occupy the same state. Now, what does define the state of a particle? In classical physics the answer would be velocity and position. In ...


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

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 ...


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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


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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 ...


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

We don't currently have an underlying theory for these parameters, we obtain their values experimentally, not theoretically. No, but one day I think the Standard Model will be enhanced to derive some of these paramaters from first principles. I've spoken to a medical doctor called Andrew Worsley who has some interesting "quantum harmonics" ideas about ...


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 ...


0

I would expect a theory of open strings, with Chan-Paton charges at the ends, to be interpreted very much like a theory of preons. It could be claimed that the points at the end of the string are only "mathematical preons".


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

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

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 ...


0

We don't know if the electron is an infinitely small object. It may have size and if it does maybe that will make you feel better about the fact that it creates a magnetic field due to its intrinsic spin.


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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

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 ...


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 ...


1

Well there is no reason of why it would not . But really what is the purpose of putting a turbine inside a particle accelerator.That beats the purpose of the accelerator to work without much resistance.If it is to generate energy or something, I doubt it's a viable solution.hydrogen as a gas is very hard to compress so I don't think you can make a solar ...


0

Let's say there are two twins. One on the ground and one moving through the atmosphere at 1,000,000 m/s which is a few thousand times greater than 1,000 km/h. The moving twin does this for 100 years, or about 3 billion seconds, then returns to the ground to compare clocks. The moving twin will have a clock that ticked more slowly by around 16,600 seconds. ...


0

Here is the most prominent example I know: The Russian astronaut Sergej Awdejew was in orbit for a total of 748 days (traveling at approximately 17000 miles per hour). Therefore he has time travelled a whopping 0,02 seconds into the future!


0

The differential cross section: $$ \frac{\text{d}\sigma(p\bar p \to \gamma\cdots)}{\text{d} p_T^\gamma} $$ is proportional to a probability density function (pdf) for the transverse momentum of the photon $p_T^\gamma$. In fact, if a photon is produced in the above process, the pdf for its transverse momentum would be $$ p(p_T^\gamma) = ...


0

The notion of anti-particle emerges as soon as special relativity is taken into account. For a relativistic particle of mass $m$, its energy and momentum satisfy $E^2 = p^2 + m^2$. To illustrate where the anti-particles are coming from, let's take the simplest "wave function equation" (that's an abuse of language since it's an equation for fields), i.e. the ...


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 ...


0

Decaying exponentials are the result of there being a fixed probability of any specific member of a population being removed. Once removed, the population decreases and so the rate of decay also decreases. The rate of decay is therefore associated with a particular time scale related to the change in population. The same model works if the population is ...


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 ...


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 ...


0

Your question touches upon two different issues. If all the matter in the universe is precisely evenly distributed it can never collapse into denser structures because every particle in the universe feels a net zero force. However this is not physically reasonable. At any temperature above absolute zero there will be some random thermal motion, and this ...


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.


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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 ...



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