92

Quantum mechanical particles have well-defined masses, but they do not have well-defined sizes (radius, volume, etc) in the classical sense. There are multiple ways you could assign a length scale to a particle, but if you think of them as little balls with a well-defined size and shape, then you're making a mistake. de Broglie Wavelength: Particles which ...


77

Fairy Physics It is entirely possible to construct a theory of the universe which states: "All effects are caused by fairies. Each effect has its own fairy, and every fairy is unique. When two fairies produce the same outcome, that is just a happy coincidence." Unfortunately, it is basically impossible to disprove this theory. Also, this theory ...


37

Gravitons do not mediate the gravitational force and you cannot detect gravitons flashing to and fro between objects interacting gravitationally. Since you cannot detect the gravitons you cannot use said gravitons to find out whether acceleration is inertial or gravitational. It is often said that forces are due to the exchange of virtual particles, for ...


25

The question confuses the electron "cloud", which is really the probability for where an electron may be found, with the size of an electron. The electron is sizeless, which can either be taken to mean that it has zero size, or that size in the quantum domain is a meaningless quantity, depending on how one chooses to think of it. An electron has ...


22

We know that the spectral lines in the spectrum of a binary star shift one way and then the other and this is correlated with its position in its orbit around its companion. Clearly, the constituents of the star do not change with each orbit so the shifts in spectral lines must be due to the Doppler shift. Occam’s razor then suggests that we apply the same ...


21

Ultracold neutrons have velocities of a few to tens of meters per second and can be transported from source to experiment on ballistic trajectories (governed by gravity). (See the PF2 experiment at Institut Laue-Langevin ) In terms of high energy experiments, say a positron beam scattering from atomic electrons, the gravitational effect would not only be ...


20

They only differ in their masses. How they interact with other particles, their charge, their lepton number, everything else is the same. In fact they appear in the SM lagrangian just as three identical replica of the basic fermionic family $$\mathcal{L} = \sum_{i=1}^3\bar{\psi}^i(i\not D-m_i)\psi^i$$ where $i$ is the flavour number. One could ask why only $...


18

Davide gives the theoretical side. Experimentally, in detecting them, there is a large difference. Electrons have been detected long ago: cathode ray tubes and then the Millikan oil drop experiment established the existence of electrons. Muons were seen in cosmic rays and it took accelerator experiments to detect them in the lab Taus, because of their ...


17

At the moment mass is one of the axiomatically defined quantities in the MKS (meter, kilogram, second) system. As with axioms in mathematics, other units can and have been defined and then the (MKS) units become derivative. The theoretical models of physics use mathematics with its axioms, and in addition impose additional axioms and axiomatic statements to ...


17

Yes, it's perfectly possible, but extremely unlikely. It can happen through direct resonances, as in $p \overline p \to J/\psi \to e^+e^-$, or through virtual particles, as in $p \overline p \to \gamma^* \to e^+e^-$. (Timescales for the intermediate state are much shorter than microseconds.) It doesn't happen very often because the proton and antiproton can ...


16

You say: a zillion gluons and quarks and anti-quarks self annihilating and popping into existence and while this is a very common way to describe the interior of a hadron like a proton it is actually rather misleading. Nothing is popping into existence then disappearing again. But explaining what is actually happening is a little involved. Our current best ...


15

Yes, mass has a geometrical explanation. The mass of a system is the magnitude or “length” of its energy-momentum four-vector $p=(E,p_x,p_y,p_z)$, using the Minkowski metric $\text{diag}(1,-1,-1,-1)$ of four-dimensional spacetime. This is the standard interpretation of $$m^2=p\cdot p=E^2-p_x^2-p_y^2-p_z^2$$ in units where $c=1$.


15

Good question. Some preliminary remarks. The map "one particle" $\leftrightarrow$ "one field" holds, at best, in the weakly coupled regime, where fields are (by construction, cf. ref.1) interpolating fields for one-particle states. In a strongly-coupled theory, a single field may (and usually does) create many different particles, and ...


15

The electron is defined in the Standard Model as a elementary particle, pointlike, with no size or spatial extension. Protons and neutrons that make up the nucleus are on the other hand composite particles as defined in the Standard Model, and they do have spatial extension. Of course, these are made up of quarks, antiquarks, and gluons, (contrary to popular ...


14

The choice of particles for a collider depends on what needs to be clarified next in the zoo of particle physics and the theory of the Standard model, and also depends on the difficulties introduced by the particular particles used. Here are the current accelerators at CERN. Creating a beam means to accelerate charged particles, whether electrons, protons, ...


13

Photons are quantum mechanical elementary particles in the very successful quantum field theory standard model. If you look at the table, they are on par with electrons, and individual photons are point particles, not waves in any space. What waves is the mathematical complex value wavefunction $Ψ$, whose only measurable prediction is the probability of ...


13

Firstly, pure General Relativity theory doesn't have gravitons, it just has spacetime curvature. Gravitons are a quantum particle, and GR isn't a quantum theory. Hopefully, some kind of Quantum Gravity theory will unite GR & quantum field theory, but we don't have a successful QG theory yet. So we don't know if gravitons even exist, but considering how ...


13

You are describing a resonance in the scattering of protons on antiprotons. This is the measurement of the total crossection as found in the particle data group No resonances are seen, so the answer is that no, there is no such resonance in the region of energy explored up to now, to go into the probability of its having an $e^+ e^-$ decay. In contrast ...


12

Since a meson is composed of two spin 1/2 particles, its total spin must be an integer, which makes it a boson.


12

Neutrinos and antineutrinos are indistinguishable by most of their qualities, butt not all. One of the quantities that distinguish them is exactly the lepton number, which make them interact with other particles in a different way. For example, a neutrino can take part in the reaction $$ n + \nu_e \rightarrow p^+ + e^-$$ but antineutrino can't; on the other ...


11

In the center-of-mass frame of the collision, the total momentum is zero. Therefore, the neutral pion must be produced at rest in this frame, or else the interaction violates momentum conservation. But the neutral pion's mass is lower than the mass of the proton and antiproton, so producing it at rest violates energy conservation. Therefore, the interaction ...


11

The Higgs field $\phi$ undergoes spontaneous symmetry breaking$^\dagger$ (from a complex doublet to a real scalar field, whose quantum is the Higgs boson) in a process named the Higgs mechanism. $^\dagger$: well it's a local/gauge symmetry, not global, so it's not "real" SSB, hence the different name "Higgs mechanism". This has two ...


11

Yes. In fact that was what Rutherford got the Nobel (chemistry!) prize for. He trapped alpha particles from radium decay and shows that they produced a gas which, when excited, gave off light with the same spectral lines as helium. https://web.lemoyne.edu/~giunta/ea/ROYDSann.HTML


11

Could anybody explain to me the errors in my thinking? Your fundamental error is thinking that the lines in a Feynman diagram are actual trajectories. Quantum particles don’t have trajectories. Furthermore, virtual “particles” aren’t real particles; they don’t even obey basic relations between the energy, momentum, and mass of a real particle, such as $E^2-\...


10

Yes; you can derive mass from the representation theory of the Lie group ${\rm Spin}(3,1)$. It's a long story, but I think I can usefully summarize it. If you need some areas further expanded, please say so in the comments. First, a definition. The Lie group ${\rm SO}(3,1)$ describes the world that we live in according to special relativity. It is the ...


10

In the Standard Model, the Higgs field also gives mass to the six quarks (up, down, strange, charm, top, bottom) and the three charged leptons (electron, muon, tau) through Yukawa couplings. Some related mechanism may give neutrinos a small mass. (They’re massless in the Standard Model, but we know this is wrong.) Finally, one can argue that the Higgs field ...


10

Even if there were some valid relaxed sense in which every field in every QFT has an associated particle, the important point is that fields are inputs (used to define the theory mathematically) and particles are outputs (phenomena that we derive from the theory). Particles are transient and not always sharply defined. As examples that challenge the ...


10

Other answers have correctly given some of the reasons. I am merely going to add that ultimately this is a question that bears on a very large number of different observations, on the concept of constructing a model, and on Occam's razor. The observations are the accumulated data from all of astronomy, combined with knowledge of matters here on Earth, such ...


9

The reason that no other Standard Model particles oscillate into each other is that they are defined not to. An "oscillation" is another name for an interaction in which one particle goes in and one particle goes out. In the standard model (after electroweak symmetry breaking) any fermion of a particular type (neutrino, electron, up, down) can ...


9

The mass of the proton has been measured to be $938.27208816(29) MeV/c^2$, the value in parenthesis the error in the measurement. The job of a theoretical model in modeling the proton is to attempt to explain the measurement. what is often described as "binding energy" It is not a good description, as if you imagine an atomic type model just ...


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