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Can all fundamental forces be repulsive? No. If the electric force can be attractive (with opposite charges) or repulsive (same charges), and the magnetic force acts like this too, can all forces be repulsive in some cases? No. For example, could gravity and the strong force actually repel certain things? The strong force can exhibit ...

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Whether forces can be repulsive or not depends on the spin of their mediating field. A scalar (spin-0) force is universally attractive, as is a spin-2 force, while a spin-1 is attractive for different charges and repulsive for like charges. So the electromagnetic, the weak and the strong force can be repulsive, while gravity cannot.

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All are dimensionless constants. With $\frac{e^2}{\hbar c} \approx \frac{1}{137.036}$ Similarly there are constants for weak and colour charge. These basically are the probability over time of a particle emitting a photon, W (or Z)-boson or gluon respectively. The weak constant is of the same order as the electromagnetic constant. The colour constant ...

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I imagine that the simplest experiment you could do to show the non-specialist would be measuring the pion charge ratios near the pion-production threshold on the first few light targets ($^1\mathrm{H}$, $^2\mathrm{H}$, $^3\mathrm{He}$ $^4\mathrm{He}$). That is we're looking at the reactions \begin{align} e^- + {}^Z\!A &\longrightarrow e^- + ... 2 You're question is interesting because it is connected to the notion of elementary particle. As mentioned by anna v, the elementary particles (fermions) of the standard model have very specific properties under the symmetry of the standard model (SU(2)_L\times U(1)_Y \times SU(3)_c): they lie in the fundamental representation of the group, which in ... 11 How can the unstable particles of the standard model be considered particles in their own right if they immediately decay into stable particles? Here I will only consider elementary, non composite particles. All the hadronic resonances are composite particles of quark antiquark combinations as well as the neutron . The standard model of particle ... 4 Inmediately is not really true, there is some proportionality. Sorry I am not answering directly about "the standard model", this is quarks and leptons. But they will fit the general pattern, you will see. Let me first consider all the "particles" listed in the particle data group file. Most of the particles decaying via photons have a half-live about ... 2 How can the unstable particles of the standard model be considered particles in their own right if they immediately decay into stable particles? Nobody has an issue calling the electron a particle. Ditto for a neutron. It's stable in a nucleus, and the fact that a free neutron decays in circa 15 minutes doesn't much matter. It's similar for a muon, ... 4 I think the most direct answer to this would be the fact that a heavier particle can decay into many different lighter particles for different reactions. The abundance of occurence of these relations are const. Again the same heavy particle can be created in multiple types of collisions of various different lighter particles. Thus we cannot say that the ... 24 Take for example an electron and a muon. The muon is unstable because it decays into an electron and two neutrinos in about 2\mus. But a muon is not in some sense an excited electron. Both particles are excitations in a quantum field and they are both as fundamental as each other. The electron is stable only because there is no combination of lighter ... 2 Your question seems quite general, but perhaps you're confused about what "decay" is. When we say something "decays" we don't always mean that it's somehow "breaking up" into it's constituent parts. In fact, we hardly ever do. The heavier particles aren't really "transient interplay of the stable forms", unless I misunderstand, and that isn't something that ... 0 Looking at the literature, and please tell me if I am wrong, it seems that the yukawa couplings fail to unify. This seems to counter the intuition that all the particles in a multiplet should have the same mass, but surely it can be argued that the mass of the multiplet is zero until the higgs mechanism is activated. (Still, comments are welcome about this; ... 0 If we continue the running of quark masses with energy (due to renormalization), what are the mass values we get for the six quarks at Planck energy? Is the sequence of mass values the same at Planck energy or do some quarks "catch" up with others? Here is the definition of Planck energy: Note the 10^19GeV. The electroweak symmetry breaking is at ... 0 I think that your idea of mass is a little wrong. The quark mass is given in a renormalization scheme, if you change it you would have different masses for quarks. But for example the pair production is a physical process, in fact, if you do a pair production with all the radiative corrections you will find the same energy with whatever renormalization you ... 1 Like spin, isospin cannot change in isolation: an intermediate particle is required. For spin that particle is most frequently the photon: if you want to reorient a particle in a magnetic field, the correct model to use is absorption or emission of a virtual photon from the magnet. The most important isospin-changing particle in the nuclear force is the ... 2 This question is experimentally accessible, despite the feebleness of the weak interaction, because the strong and electromagnetic interactions are symmetric under parity transformations and the weak interaction is not. The contribution to the binding energy is small enough that it's not a good way to think of things. Better is to continue the process of ... 1 If one considers the baryon octet the lambda 0 is a singlet in isospin and the sigma zero is in a triplet. The explanation of the mass difference The neutral sigma can decay to the lambda without violating conservation of strangeness, so it proceeds rapidly by the electromagnetic interaction. The sigma-zero and lambda-zero have the same quark ... 2 From this blog post of mine, one should be inclined to think that for large nuclei it can at least contribute to the spin-orbit force, and then to the correction to N=50 and N=82 nuclear shells. As QGR noted above, the exponential suppression of the potential does not need to be all the history. Note that usually the total potential in a nucleus is ... 2 My own impression is that the ball got stuck in someone' roof, but I am partial because some of my numerology did coincide with the relationship (m_u,m_d,m_s) \propto (0, 2 - \sqrt 3, 2+\sqrt 3) proposed by Harari Haut Weyers (1978) (presented by Harari here) when trying to find some first-principled calculation of Cabibbo angle. My understanding is that ... 6 This admittedly, isn't much of an answer, as I'm merely repeating information from the Particle Data Group page about the up-quark, which I consider up-to-date. Their current combination is that m_u = 2.3^{+0.7}_{-0.5}\,\text{MeV}, but they warn that The u-, d-, and s-quark masses are estimates of so-called "current-quark masses," in a ... 4 No, leptons are not coloured. The reason the Pauli principle requires colour is that there are particles made of 3 quarks that would otherwise all be in the same state, so we have to have this other quantum number, as not to violate Pauli (poor Pauli, people always trying to violate him). Even if there was some state where some leptons were all packed ... 3 Particle and anti-particle are described by the same field. Let's look at the Dirac field: $$\psi\left(x\right)=\int\frac{d^{3}p}{\left(2\pi\right)^{3}}\frac{1}{\sqrt{2E_{p}}}\sum_{s}\left(a_{\vec{p}}^{s}u^{s}\left(p\right)e^{-ip\cdot x}+b_{\vec{p}}^{s\dagger}v^{s}\left(p\right)e^{ip\cdot x}\right)$$ where a_{\vec{p}}^{s} ... 7 Both particles and antiparticles arise from the same quantum field. Particles (and antiparticles) are obtained from the Fourier mode expansion of the free quantum field - for a scalar, it is \phi(\vec x) = \int \frac{\mathrm{d}^3 p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}\left(a(\vec p)\mathrm{e}^{\mathrm{i}\vec x\cdot\vec p} + b(\vec ...

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The strong interaction explains the characteristics and operation of the strong force, one of the four fundamental forces of nature (strong force, weak force, electromagnetism, gravity). The strong force binds together quarks that compose protons, neutrons, and mesons. Gluons are carriers of the strong force. The nuclear force is a residual manifestation ...

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The force responsible for the stability of the atoms and the nucleons is the Strong force. Just like electric charge is a value of how much a particle interact with the electromagnetic field, the color of a particle is a value of how much a particle interact with the Strong field.

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Isospin is an approximate symmetry of nuclear interactions. Roughly, the "x" component of isospin consists of changing u quarks to d quarks and d quarks to u quarks. Because u quarks and d quarks are very similar - have different charges but about the same mass and reasonably precisely the same interactions with gluons and other quarks, you can quickly ...

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A small point in addition to Slereah's solution: it is only CP symmetry that is required to have the same half life for particles and antiparticles. A brief story of fundamental symmetries: for a long time it was thought that the laws of physics behave the same in a mirror situation (parity P), and also that antiparticles are exactly indistinguishable from ...

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Particles and their antiparticles having the same half life is related to the C symmetry (charge symmetry), which roughly states that processes for particles and antiparticles (that is, if you have a system and you apply the C operator on) have identical probabilities. It is not true that all of them do, though, as the weak interaction breaks C symmetry, ...

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Some lecture notes, e.g. http://arxiv.org/pdf/0906.1271v4.pdf p. 149, are more concrete about the "order one" condition by asking that the $J=0$ channel of the diagram $t\bar t \to ZZ$ that emits two $Z$ must be canceled with the diagram that actually aniquilates $t \bar t$ into $H$ and then decays to two $Z$. The first diagram goes as \$\alpha^2 m_t^2 / ...

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