# Tag Info

3

The short answer is that what you are proposing would be an extraordinarily challenging task! Simulating a single (non-hydrogen) atom accurately in time requires a huge amount of computational power which scales as roughly (simulation resolution)^(3*number of particles)! In computational biology, nearly all simulations are Newtonian based to avoid this. I ...

1

Strangeness isn't conserved->Weak decay. As for parity conservation, weak decay CAN change parity, it doesn't have to. Basically, knowing that something IS conserved doesn't give you any information, but knowing that it isn't conserved may give you some. (e.g., electrical charge being conserved means nothing, but it being not conserved means this decay is ...

0

You have exactly the right idea. The decay products are only useful in as much as we can compare a previous isotopic ratio to the current (measured) ratio. How this is done differs from dating method to dating method. As an example, C14 dating originally came about with the idea that atmospheric and therefore living plants have a near-constant ratio that ...

0

One needs to account for lead that was formed as lead, rather than by radioactive decay. Wikipedia has a discussion. Lead-204 is not formed by decay, so if you know the primordial distribution of the isotopes, you can compute the primordial amount of each of the others. The rest comes from decay.

0

What you propose is called nuclear democracy, and was very popular in the days before the standard model emerged. See also the discussion in http://www.physicsoverflow.org/22971 , where you can read about (among others) my present view of it.

0

I would say that $D_\mu '\psi' = U(D_\mu \psi)$ is a requirement, rather than an assumption: When $U$ is a global transformation (i.e., when $\lambda$ is independent of $x$), we have trivially $\partial_\mu \psi' = U(\partial_\mu \psi)$, and so $(\partial_\mu \psi)^\dagger (\partial_\mu \psi) = (\partial_\mu \psi')^\dagger (\partial_\mu \psi')$. We define ...

1

If a particle either "is" or "isn't", then its count is either 1 or 0. Even in quantum mechanics it's not possible that half a particle exists. It is possible to detect it with 50% probability, but if you set about counting all the particles one at a time, you necessarily end up with an integer answer.

0

A wavefunction is a function that takes a configuration of all the particles in the universe (if just one particle is in a different place it is a different configuration) and assigns a complex number to that configuration. If you take the square of the length of that complex number you get something people like to call a probability-density. An X-density ...

3

There are two factors at play here. The Lorentz force which causes the paths to bend with a radius proportional to the particles velocity and with a sense that dependent on both the particles charge and the direction of the particles velocity. In high energy (compared to $m_e$ events) such as the one pictured, the particles are nearly co-linear at the ...

-4

The electron has three well known properties, its electric charge, its magnetic dipole moment and its intrinsic spin. All three are constant quantities. And to prevent contradition about the reality of this intrinsic spin, it was shown in the Einstein-de-Haas experiment, that this spin really has to do with a rotation of the electron. It has to be stated, ...

9

But what we never seem to see is why the electron and positron move the way that they do. Saying "they move like they do because of the force on them" doesn't explain anything at all. It's a non-answer. The equation of motion for charge particle (electron,positron) in magnetic field is $$m\frac{d}{dt}\left(\frac{\mathbf ... 4 As mentioned in the OP quarks vastly simplify the theory of hadrons, like atoms did chemistry, and despite confinement Rutherford-like experiments were performed for them too, by Friedman, Kendall and Taylor who received the Nobel prize for it in 1990: "unexpectedly large numbers of electrons being scattered at large angles provided clear evidence for the ... 0 Using c=1, take the familiar equation for energy E^2 = m^2 + |\vec{p}|^2 and substitute in the magnitude of momentum |\vec{p}| in terms of transverse momentum p_T and pseudorapidity \eta |\vec{p}| = p_T\cosh\eta then E = \sqrt{m^2 + p_T^2\cosh^2\eta} 3 In classical electrodynamics, charged particles radiate electromagnetic waves when accelerated. An electron in a circular orbit has radial acceleration. Think about where the energy of the electromagnetic waves could possibly come from, and you'll have your answer. 2 To answer your questions: Yes, fibreoptics transfer light. Maybe. I'll discuss that now Fibreoptics are strands of glass, they're CRAP at going around corners, I mean seriously crap, communications fibre is VERY THIN. Even then it can't go around bends well, they test it at every stage during laying. However with communications stuff the path matters ... 1 Very hard to tell without knowing how the spectrum was produced (type and size of the detector, resolution, anticompton, ...). Anyway 180 counts do not seem so many. The single escape peak is normally weaker and the double escape even more, especially if you are just above the pair production threshold. Sounds reasonable that you may not have significant ... 1 The mathematics of general relativity is clear and unambiguous. The trouble comes when you try and describe what is going on it non-mathematical terms, because there is no precise way to do it. Kip Thorne is attempting to talk about black holes in non-mathematical terms, and he is adopting a different perspective from (probably) most of us. That doesn't mean ... 1 According to wikipedia, Vanadium-48 decays via \beta^+ (positron emission) to Titanium-48, which is a stable isotope. The emission of neutrons for Vanadium-48 isn't allowed becuase it doesn't conserve energy: Vanadium-48 has a mass of 47.9522537 u, and Vanadium-45 plus 3 neutrons have a total mass of 44.965776 u +3·1.00866491600 u = 47.991770 u. For a ... 4 It should be clarified that the Higgs boson does not carry mass. The correct statement is that the Higgs field (not boson) is giving mass to some (not all) particles. In fact most of your mass is not given by the Higgs field. Most of the mass of atomic nucleus (protons and neutrons) is due to the binding energy of strong interaction. The Higgs field is ... 3 To get an understanding on quantum field theory issues, you have to understand the difference between virtual particles and real particles. Virtual particles, in contrast to real particles, are a mathematical construct inspired by the Feynman diagrams used to describe interactions. These diagrams start with real particles, i.e. particles that have the mass ... 0 What you're describing is Brownian motion. If there is a force only on the particle, then you should simulate Brownian motion with an additional vector in the direction of the particle force. Obviously the intensity will determine how quickly a particle will move through the fluid. Rather if the velocity is being applied to all particles, then all ... 3 You should work out the minimum energy state of your system (classically) to find the vacuum expectation value. I assume you're working with the standard \phi^4-Lagrangian$$\mathcal L=\frac{1}{2}(\partial \phi)^2-\frac{1}{2}m^2\phi^2-\frac{\lambda}{4}\phi^4 $$which corresponds to the Hamiltonian$$\mathcal ...

1

Feynman diagrams are a shorthand for integrals entering a calculation for the crossection of a specific reaction, in your question the generation of the Higgs. When one writes the Feynman diagrams for the crossection of generating a Higgs in a proton proton interaction one must include all possible diagrams that obey the quantum number consrvations for the ...

2

I found my answer, the Higg's Boson is predicted to have a mean lifetime of 1.56 x 10-22 s.

1

Photons are a kind of quanta so if they are interacting with something, that something is also a kind quanta. A classical charge can have a location and a momentum it can look at the value of a classical field at the place it is and adjust it's momentum at a particular rate based on the direction that electromagnetic field points compared to its own motion. ...

7

If you mean actual photons, the particle gets pushed. Photons have momentum that points along their direction of travel. You may be confused because you might have heard that the electromagnetic force comes from the exchange of photons. So if charges throw photons at each other, it looks like they should only be able to repel, and never attract. The ...

3

For a real answer, each particle would have to be discussed individually and that might get long. Dark matter possibly being Neutrinos has certainly been proposed and in many ways, Neutrino's lack of interaction makes them a good candidate, as they are essentially "dark" - though "invisible" is perhaps a more accurate term and Neutrinos fly through stuff ...

1

There exist particles which are their own antiparticle, even at the elementary particle level. Photons, and gluons are their own antiparticle and thus no need to come in pairs.

1

If you count a $\pi^0$ as a particle, you can create and destroy them individually. This is similar for any particle that is its own antiparticle (like the photon, the $Z^0$ and the Higgs.) You can have an arbitrary number of these.

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The creation of particles has to respect a number of conservation laws. For example charge has to be conserved, so it you create an electron with a charge of -1 you have to create a particle with a charge of +1 to balance it out. Likewise lepton number is conserved (in the Standard Model at least). An electron has a lepton number of +1, so if you create an ...

3

Massless particles can carry and do carry confined charges (gluon in QCD) and they may carry charges under spontaneously broken generators (photon is transforming e.g. under the generators of $SU(2)$ associated with the W-bosons) but they cannot carry charges under unconfined $U(1)$ force like electromagnetism. The reason may be explained in different ways. ...

0

If you define spin angular momentum as $\Sigma^{i}= 1/2 e^{ijk} \sigma_{jk}$ and $\sigma_{jk}= e^{ajk} \sigma_{a}$ then there is some inconsistency about when you place upper versus lower indices but you get: $$\Sigma^{i}=\frac{ 1}{2 }e^{ijk} \sigma_{jk}= \frac{ 1}{2 }e^{ijk} e^{ajk} \sigma_{a} =\frac{ 2\delta^i_a}{2}\sigma_{a} =\sigma_i$$ It really ...

4

ALICE is a heavy ion experiment at CERN. Here is a lead lead collision One of the LHC's first lead-ion collisions, as recorded by the ALICE detector. Thanks to the advances of computing the vertex is determined by the tracks , measured and pointing back to it, even though there are thousands of tracks from each vertex. Certain tolerance assumptions ...

2

Fundamentally it is that the $1/N!$ for the classical system only correctly compensates for overcounting of indistinguishable states if the particles are always in different states. For a system of Bosons at low temperature, where it is quite likely that many particles are in the same state, this breaks down. For a very understandable introduction to this ...

1

The states of charmonium are treated as bound states of a charmed quark ($c$) and its anti-quark ($\bar{c}$). Since the binding energy of the $c-\bar{c}$ system is relatively small, compared to the rest energy of the charmed quarks, it is a reasonable starting point to analyze the states using the non-relativistic Schroedinger equation with a potential ...

0

The simple answer to your question (What implications does a quark quartet have on QCD?) is that it has no fundamental implications that I can imagine. It is an open question whether or not tetraquark ($q q \bar{q}\bar{q}$) or pentaquark ($qqqq\bar{q}$) states that live long enough to be classified as particles actually exist -- at present, our ability to ...

1

We usually say that in geometric optics we use the "ray model", which is neither the particle model nor the wave model. Given that particles (subject to no forces...) travel in straight lines and that "light rays" go in straight lines when they are moving through a uniform medium I suppose one could say that the light rays look like particle trajectories. ...

3

It's possible to detect neutrinos in whichever flavor they are oscillating through, so that won't necessarily cause a "dropped packet" problem. The answer is, technically, yes, there is no physical law preventing the use of neutrinos as a communication medium. It has been demonstrated that we can cause the emission and detection of neutrinos. For example, ...

1

Protons and neutrons are extended object, being roughly $1\,\mathrm{fm}$ in radius. Give or take a bit, depending on how you want to define their size. Evidence for this comes in three basic forms: first from scattering experiments in which we bounce other particles off of them (electrons are particularly good for this applications); second from packing ...

0

The elementary particles of the Standard model are pointlike. This has been determined by numerous experiments. All other particles except the ones in the table are composites of some of the ones in the table, and these have dimensions that have been measured using scattering experiments to find their form. Here is an answer on how this is done.

0

I believe the quark masses are obtained from lattice Quantum Chromodynamics (QCD) calculations on a giant computer. In essence, they use the quark masses as free parameters and calculate hadron masses. The adopted masses are those that give the best fit. See this link for details.

0

Any particle can be a virtual particle, even an electron. The incoming electron and positron are real, so they must be on shell. The virtual electron-positron can be off shell, which means different frames might even disagree about whether it is a virtual electron or a virtual positron . But it is not really an electron or a positron. It is a stand in for ...

-3

For myself I created a model where exist two quanta e and p which form electrons and positrons too. The number of e-quanta equals the number of p-quanta. Unfortunately the quanta are so tiny that from the light spectra until now it is not possible to conclude about the energy of such quanta. The only difference between electrons and positrons is, that at the ...

2

You have surely seen slit experiments, where waves which pass through the slit and scatter to produce a pattern on a screen placed behind the slits. In the far-field approximation (also known as Fraunhofer diffraction) this pattern is precisely the Fourier transform of the slits. Perhaps you even remember that waves passing by lines produce the same pattern ...

2

Once one specifies a quantum field theory, typically in the form of a Lagrangian density, one can calculate the probabilities of various outcomes in collisions. A quantum field theory is a theory based on fields that obeys quantum mechanics and special relativity. The so-called Standard Model is perhaps the most famous quantum field theory, and certainly ...

1

This is actually an open question. So far what we are able to state from theoretical considerations are restrictions in terms of the conservation laws that we have observed. These tells you for example that the whole momentum in a reaction is conserved, an the mass-energy, or some quantum numbers. And this already constraints much of what can come out from ...

2

A common method is to do the same thing for when a beam of light is needed and no laser is to be found -- collimate by elimination. That is, have an omnidirectional source and just filter out the particles moving in the wrong direction. This is illustrated in the below Wikimedia image: A source is heated so that atoms fly off in all directions. Small ...

11

In principle, yes. You can reverse any decay process and the corresponding synthesis will be valid - in this case, since $H_0\to\gamma\gamma$ happens, then $\gamma\gamma\to H_0$ will also happen, assuming the kinematics work out. However, the corresponding probability is very small. Out of all the possible things that could happen when two photons cross ...

0

I would say a black hole passing or being sunked in is not reversing gravity. both objects are producing positive gravity. Reversing would be more like totally reversing gravity and all its effects all matter would repel itself. i doubt this would have any effect on the strong force that binds matter together but stars would explode black holes would become ...

0

The only possible way Bosons admit anomalies in flat space (analogous to the case for fermions you have mentioned) is when they do not admit a covariant lagrangian. These are called by a special name, "Chiral Bosons". The usual Bose lagrangian can always be regulated to give an action free of anomalies. See section 8 of this paper. ...

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