# Tag Info

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Your What happens in high-energy physics experiments aside partially contradicts your final question But if we restrict the discussion to radioactive decay, fusion in stars, cosmic rays, is everything a lepton, baryon, or a photon? Radioctive decay has as end products photons, leptons and baryons. Fusion and cosmic rays are the realm of ...

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But if we restrict the discussion to radioactive decay, fusion in stars, cosmic rays, is everything a lepton, baryon, or a photon? No. Baryons are not elementary particle, but are composed of 3 quarks bound by gluons. There is a class of particles called mesons, which consist of a quark and antiquark. The pion is an example of a meson. Pions are ...

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Although the New Scientist requires readers to subscribe in order to read the entire article, this is very much the answer I was looking for. An agreement of 2% of the proton mass with QCD-based calculations is indeed impressive, and a big improvement over previous attempts (and from which I was actually sourcing some of my much less popular responses). ...

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This question is similar to "where is the electromagnetic field?" And the answer is: the electromagnetic field is everywhere; it exists at every point in space-time, but it simply happens that its average value is zero (or close to zero) at points far away from charges, currents, and waves. The Higgs field, like the electromagnetic field, is a quantum ...

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Where there is mass, there is Higgs field. It's everywhere in space.

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You start from the (special) relativistic expression for energy: $$E=\sqrt{m^2c^4+p^2c^2}$$ Now, if the first term is much larger than the second ($mc^2\gg kT$ or $v\ll c$), we should take this first term out of the square root and taylor expand the (then in standard form) rest, discarding all but the leading term: ...

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I believe that this result only holds for $v\ll c$. Having assumed this, lets start: $$Total Energy = Rest Energy + Kinetic Energy$$ Keep in mind that since $v\ll c$, $c^2$ is the energy to mass conversion ratio and $K.E.=\frac{p^2}{2m}$ $$E=Mc^2+\frac{p^2}{2m}$$ There!

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Sure! Take a uranium-238 atom, strip off 91 electrons. Look up the Lamb shift and binding energy. Zowie! Totally ionizing a radioactive isotope that decays only by electron capture indefinitely stabilizes it against decay. http://en.wikipedia.org/wiki/Rydberg_atom Rydberg atoms on the flip side.

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It is small wonder there is no definitive answer to this question here. But I can pass along the best answer I got. I asked the same question, in a slightly different format, to a blog "Of Particular Significance" that is popular with Physics Stack Exchange. My question was whether the famous Goldstone Mexican Hat potential had a value of 245 GeV at the ...

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You have a photon that comes in and splits into $e^-e^+$. One of the electrons escapes, the other has a short segment, where it sheds another photon. The second electron now escapes and the photon is absorbed by the nucleus. The center electron segment and the second photon can be off the mass shell.

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As previous answers have correctly noted gamma matrices do not forma a basis of $M(4,\mathbb{C})$. Nevertheless you can construct one from them in the following way 1 the identity matrix $\mathbb{1}$ 4 matrices $\gamma^\mu$ 6 matrices $\sigma^{\mu\nu}=\gamma^{[\mu}\gamma^{\nu]}$ 4 matrices $\sigma^{\mu\nu\rho}=\gamma^{[\mu}\gamma^{\nu}\gamma^{\rho]}$ 1 ...

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A proton has a positive charge so by charge conservation it is not possible to reduce a proton to uncharged radiation particles such as photons (assuming that is what you mean by "pure-energy") Because of gauge invariance charge conservation is likely to hold good in all future physics, but we can't be totally sure of that. It is possible that some charged ...

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In a similar way that atoms are mostly empty space the same is true for nuclei and nucleons. The nucleus is composed of nucleons and is held together by the spill over strong forces that hold together the quarks within the nucleons ( neutrons and protons are made up of quarks and the energetic interactions between them). The reason a table top is solid is ...

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At least on current physics theories, there is nothing "solid" in the sense you mean. In most models of physics elementary particles are zero dimensional, which means, their size is a "point", whatever that means. The nucleus, in particular, is pretty "empty". It consist of soup of quarks and gluons, that is mostly energy; and both quarks and gluons have no ...

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The simple answer to the main question is, yes. There are two ways to annihilate matter without using anti-matter. One is called fission, and the other is called fusion. Although only some of the matter is converted into energy in either of these processes, the efficiency of the "annihilation" is not in the main question. If 100% annihilation is required, ...

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Also, it's worth noting that the quantum mechanical model places the upper limit of the radius of an electron at 2.82e-22, which means that the electrical potential energy that two electrons would be at when they collide (given by k*q1*q2 / r) is in the order of magnitude of 50,000,000 MeV. The mass of an electron, in MeV according to e=mc^2, is closer to ...

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They have 2 main differences. The first of them is very straightforward: They have different spins: As you pointed out, both are bound states of 2 spin 1/2 particles, therefore you can find the possible spins of such a bound state using the usual rules of angular momentum addition in Quantum Mechanics. 1/2 + 1/2 can give you either 3 spin 1 states (the ...

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A $\rho$ meson is the spin-1 (angular momentum, not isospin) excitation of the $\pi$ meson. We have, $$\rho = \begin{cases} \rho ^+ \quad u \bar{d} \\ \rho ^0 \quad \frac{ u \bar{u} - d \bar{d} }{ \sqrt{ 2}} \\ \rho ^- \quad d \bar{ u } \end{cases}$$ where each $\rho$ meson has a different isospin. However they are all spin $... 2 There are two different kinds of symmetry breaking involved in your question. The first would be spontaneous symmetry breaking. In this case we are dealing with a theory that is invariant under a certain symmetry group, but its vacuum is not. The breaking of the symmetry corresponds to a specific choice of the vacuum, the freedom of choosing a vacuum results ... 0 Your question is riddled with ^'s in equations making it hard for me to understand the body of your question. If I understand your question "why is there no weak isospin vacuum angle in analogy with the one in QCD?," then I can answer it easily: Suppose we write that CP-odd term in the Lagrangian. Then, to remove it, all you need to do is to look for a ... 0 As mentioned in the comments,$ C $and$ M are matrices in different spaces. Explicitly showing the flavor space matrices, in the lepton sector perhaps, we have, \begin{align} \overline{\psi} _L M ^\dagger \left( \psi _L \right) ^c & = \overline{\psi} _L M ^\dagger C\bar{ \psi _L} ^T \\ & = \left( \begin{array}{ccc}\overline{ \psi _{e, L} } ... 3 To complement V. Moretti's excellent answer, it's worth emphasizing that the dimension of the four-by-four complex matrices\mathbb C^{4\times 4}$, when seen as a vector space over$\mathbb C$, is$4\!\times\!4=16$. As such, a set of four matrices (i.e. vectors in$\mathbb C^{4\times 4}$) can never be a basis for it. It's also worth saying that the general ... 5 No they do not, due to dimensional reasons, but they are generators of the algebra. That is,$I$and the products of$\gamma^a$(products of one, two, three and four matrices) form such a basis. NOTE ADDED. As Emilio Pisanty correctly remarked (also making some further interesting comments)$GL(4, \mathbb C)$is not a linear space so questions about bases ... 4 The form of the propagator is correct. The expressions from your Wikipedia link are complicated because they show the propagator for the massive theory, where Susskind's argument fails because the propagator can involve any function of the dimension zero combination$m^2 |x-y|^2$. The "simple" massless result is recovered in the$m\to0$limit; for example ... 1 I recently did this lab and the answer I got to this question was: The x-rays ionize the air molecules or gas in between the capacitors. The free electrons then fly off and attach them selves to the oil drops. Alternatively, you could use an electron beam (same effect). 1 According to this link, the x-rays ionize the gas molecules in the apparatus, not the oil drops directly. So yes, you are right, the air is positively ionized. The newly freed electrons from the gas then adhere to the oil droplets (probably due to an induced dipole moment), producing negatively charged oil. 0 If gravity is a fundamental force, then the Higgs mechanism is also. This is is true whether they are related or not. The Higgs mechanism is certainly the source of the inertial mass that inspired Newton to quantify what a force is, and how it behaves. Gravity, like the Higgs mechanism, can add mass/energy to matter in bulk (like constant acceleration), ... 2 Non-conservation of charge in Majorana terms The Dirac mass term is$m\bar\psi \psi$where one field-factor$\bar\psi$is complex conjugated (aside from other transpositions included in the Dirac conjugation) and the other is not. So one may assign a fermion number$1$to$\psi$which means that$\bar\psi$automatically carries$-1$and in the product, the ... -2 Mass so small that it cannot be observed, measured, tested, is unknown, but believed to exist, is not mass. it is pre-mass. It is alpha. The substance and evidence is confirmed only by our strong, unwavering belief that all things visible are made of things not visible. 1 Now when we operate parity operator, does that mean we are taking any physical entity at x to −x. Or we are just reverting axes of the co-ordinate system? Well, either operation should adhere to the same rules, and you mention the correct term: it depends on whether we see the operation as active or passive. Either view has the same end result: we move ... 1 Let us suppose that that the Standard Model is an effective field theory, valid below a scale$\Lambda$, and that its bare parameters are set at the scale$\Lambda$by a fundamental, UV-complete theory, maybe string theory. The logarithmic corrections to bare fermion masses if$\Lambda\sim M_P$is a few percent of their masses. The quadratic correction to ... 1 Yes it fluctuates but it is a very small fluctuation. Note that unstable particles have a decay rate or width$\Gamma$that is related to its lifetime$\tau$by $$\Gamma=\frac{\hbar}{\tau}$$ when you measure the mass/energy of such particles in experiments you always get a Lorentzian or Breit-Wigner distribution like this from which you can measure the ... 0 It sounds like a ballistic problem with initial conditions: position = (0,Y0), velocity = (1.55,0) - to me. This is considering a coordinate system where the Y axis is the vertical just at the edge of the table and the X axis the ground. Start from Newton's 2nd law for a freefall, project it on both axes, integrate over time (known) with the initial ... 0 Take a simple quantum mechanical potential that describes an atom. The mass of the atom is fixed. Take hydrogen. The electron is in an orbital around the proton which is a probability distribution of its location in time and space: if you measure it, i.e. interact with it, where you may find it. Correspondingly there exists an energy width to the energy ... 0 The antineutrinos do indeed form a doublet. The particle-antiparticle conjugation operator is usually denoted by$\hat{C}$and is defined through: $$\hat{ C}: \psi \rightarrow \psi ^c = C \bar{\psi} ^T$$ where$ C \equiv i \gamma _2 \gamma _0 $. So given a neutrino you can always get its complex conjugate with this operator: ... 0 You're asking about the leptonic sector of the standard model of Glashow, Salam, and Weinberg*, in particular the lightest generation$(\nu_e,e), of which there are three, sometimes denoted: \begin{align} L_e &= \frac{1-\gamma_5}{2}\left(\begin{matrix}\nu_e \\ e\end{matrix}\right) & L_\mu &= \frac{1-\gamma_5}{2}\left(\begin{matrix}\nu_\mu \\ ... 13 The definition of an antiparticle is dependent on having the opposite quantum numbers of the particle so that they can annihilate, i.e. the sum of the conserved quantum numbers are zero. Thus the answer by @mpv is adequate. The implication of your question is then: is baryon number conservation a strict law or an emergent law that may be violated at some ... 2 I just started here so I don't have the rep. to comment and I don't have the time for a full answer, but the black hole idea mentioned in the comments above is a fine answer. See, for example, http://arxiv.org/abs/0908.1803v1 and How would a black hole power plant work? 16 I am assuming that by "energy" you mean photons. So you want to transform protons into photons. It is not possible. It would violate several conservation laws - mainly the charge conservation (protons are positively charged), but also baryon number conservation. The antiparticle is necessary to cancel these quantum charges to make the transition possible. 0 Think dynamically - I suggest that inside the Nucleus protons and Neutrons are continuously 'flipping' Neutrons decaying into protons and protons absorbing the electron and neutrino to become Neutrons again and so forth. It is this mechanism which gives rise to the binding Force. The Neutron obviously plays a key role in the stability of any heavier then ... 0 Benford's law is pretty cool. It states that, for many sets of data, a leading digit of n has a probability ofPr(n) = log_{10}(1+1(n))$Plugging in our n values we find that we can expect low values of n to have a higher probability of being our leading digit. The most (initially) boggling thing is that our$Pr(1) = .301$stays independent of units. If ... 0 Each free particle or field (each component of corresponding field or wave-function) must satisfy the Klein-Gordon equation, because it corresponds to relativistic energy-momentum relation (or, more formally, refers to the Casimir operator$p^{\mu}p_{\mu}$of the Poincare group). But each free particle or field must satisfy relation$W^{\mu}W_{\mu} \Psi = ...

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The reason that the gauge particle must be a spin 1 gauge boson is because there aren't any renormalizable alternatives. To see this consider the Dirac Lagrangian: $$\bar{\psi} i \gamma ^\mu \partial _\mu \psi$$ This term is not gauge invariant under the transformation, $\psi \rightarrow e ^{ i T ^a \theta ^a (x) } \psi$, ...

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There are two possible Feynman graphs for Bhabha scattering at tree level. I have shown them below. Are you asking what will go wrong if these two are modified like shown below. If this is what you are asking then the only thing that we should be concerned about is the conservation of charge at each vertex point. We can ...

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Is an excited atom more likely to emit a photon if there is a similar atom in the ground state nearby ready to absorb it? The probability of decay to a ground state is independent of the proximity of other atoms, except so far as the wavefunction's change due to the different boundary conditions that the proximity imposes on all atoms in a lattice. The ...

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There is no such tree-level interaction in the conventional theory. But then free-propagating neutrinos are not in pure flavor states except by chance anyway, so any pair of neutrino and anti-neutrino1 could participate in a vertex $$\nu_l + \bar\nu_l \to Z^0 \,,$$ which is roughly equivalent to Drell-Yan in the charged lepton sector with a projection into ...

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Electrons around the nucleus are in orbitals, whose shape depends on the angular momentum eigenvalues of the energy levels which the electrons occupy. An orbital is a probability locus of finding the bound electron at that (x,y,z). The five d orbitals in ψ(x, y, z)2 form, with a combination diagram showing how they fit together to fill space around an ...

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An atom does not have a discrete boundary, like the surface of a baseball. There's always some chance that an electron will be found at any distance from the center, although the probability becomes vanishingly small at distances that are not all that far from the atom. Thus the "boundary" is kinda fuzzy, perhaps somewhat like the surface of a tennis ball. ...

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Yes, as you say, there is a built-in time symmetry in the mechanical laws that underlie our universe. At the moment the most accurate statement seems to be CPT symmetry. Under a CPT reversal (particles -> antiparticles, flip space, flip time), mechanics works identically. On a practical level though, even time symmetry alone holds to a good degree. It is of ...

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Assuming that $\tau^-\tau^+$ can form a bound state similarly to positronium($e^-e^+$), all we need is the form of the ground state of positronium, specifically that it is proportional to the reduced mass of the pair: $\mu = \frac{m_1m_2}{m_1+m_2}$. Knowing that positronium's ground state is $(-13.6/2)= -6.8$ eV and that the new reduced mass for the Tau ...

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