New answers tagged

1

By definition the critical energy is a fixed value, you mixed up the notations. As you mentioned, the critical energy is the energy value where the ionization term and bremsstrahlung term are equal. So (based on the 1st and 2nd formulas in the question) the following equation holds: $$C_iZ\ln{(E_c)}/M=C_bZ^2E_c/M^2$$ where $C_i$ and $C_b$ are the constant ...


0

If $\frac{\partial}{\partial t}B=0$, then $\nabla \times E=0 \rightarrow E=-\nabla \phi$. That is the general form an electro-static potential. And of course $B=\nabla \times A$. You can plug these equations into the expression for the Lorentz force to find the generalized dynamics of a point charge in a static field; however I believe your question is about ...


0

To have a clear understanding of what's going on requires a lot of knowledge. The basic idea is that using experimental, as well as theoretical data, one can constrain various SM parameters coming from different experiments and get out a best fit region in the $(\bar\rho,\bar\eta)$ plane for the unitarity triangle. Clearly being every measurement accompanied ...


1

My question is why is this process defined with the second electron being emitted from the atom instead of just excited to a higher energy state sometimes. The atom is a unit tied up quantum mechanically . To observe transformations of an atom, there must be an interaction that can be measured. An emitted photon can be measured. An emitted electron can also ...


-1

If I understand your thoughts correctly, and you had a typo when you said "Could the 6 neutrons (you wanted to say electrons) and 6 protons cancel each other", you expected a carbon atom to have the same number of protons and neutrons. If you know that there are carbon 13 (which has 7 neutrons) and carbon 14 (which has 8 neutrons), you may think ...


-2

Can Neutrons just be the cancelling out of Electrons and Protons charges, forming a neutral charge inside the nucleus and not an actual particle. What constitutes a unique particle might be the issue here. It is basically true that a neutron is an electron and a proton bound together by nuclear force (by contrast, a hydrogen atom is a proton and electron ...


0

Let me first note that Auger process is due to Coulomb interaction between electrons, so it may be beneficial to think of it in terms of the Fermi golden rule: $$ w_{i_1 i_2\rightarrow f_1 f_2}=\frac{2\pi}{\hbar}|\langle i_1, i_2 | V|f_1, f_2\rangle|^2\delta(\epsilon_{i_1} + \epsilon_{i_2} - \epsilon_{f_1} - \epsilon_{f_2}) $$ One needs to be careful when ...


6

or is the theory we have have no correct and there is no need for further debate. You are a hundred years too late to be able to play with the models of nuclear physics. Physics reasearch at present has progressed to the level that has shown that protons and neutrons, not only are the two versions of the"same" particle called collectively a ...


3

You seem to have a misunderstanding. As already pointed out in the comments, a black hole does not have infinite mass. In fact, the "size" of the black hole (to be precise, the radius of the event horizon since the actual singularity is believed to be point-like), is directly related to the black hole's mass (and only to the mass, since all the ...


2

Usually each laser is specified by emitted maximum power and light wavelength. Using energy-power relation $$E=P\,t$$ and the fact that laser energy is the sum of energy of all photons emitted : $$ E=n~h\nu $$, one can calculate total number of photons $n$, given laser power $P$, wavelength $\lambda$ and time window $t$ : $$ n = \frac {P\lambda}{hc} ~t $$ ...


6

Simplest way? The $\Delta^{++}\sim uuu$ has to be a color singlet. It has spin 3/2, so it is flavor and spin symmetric. But fermion quarks need to be in a fully antisymmetrized state. Can you make an SU(4) singlet out of three antisymmetrized copies of an SU(4) representation, the way you can for SU(3)? (No.) (By now, you simply experimentally check R in $e^+...


0

Yes. For reference, please see the derivation of the MB distribution by Landau and Lifshitz. Their derivation is very general. It depends only on the Gibbs distribution and the usual classical assumption about energy being the sum of kinetic energy and potential energy.


0

In the mathematics of quantum chromodynamics, the gluon is the operator which changes a quark's color. Gluons are therefore not color-neutral. (The easiest mental picture is that "a" gluon has a color and an anti-color, but it takes about a year to explain away all of the approximations in that model.) Color-charged objects are confined and ...


1

Lets make the comments into an answer: How does this relate to the image below? Are the black lines representative of gluons? The black line represent quarks, the wiggly ones gluons. So are gluons what mediate pion exchange, which is in turn what binds protons and neutrons with the strong nuclear force? The strong nuclear force is a spill over force from ...


2

It fits with the long discussion we had on this question : What is the connection between quantum optical photons and particle physics' photons? and the answers and comments therein. People working with quantum optics have a more general view of the term "excitations of the electromagnetic field". In the quantum field theory used in the standard ...


3

The "size" of a particle at rest such as the muon, is its Compton wavelength, $1/m_\mu$ in natural units ($\hbar=1, c=1$, easy to reinstate uniquely). You may then check that for the muon this is $$ \lambda_\mu \approx \frac{1}{m_\mu} \approx 2~fm= 2 \cdot 10^{-15}m, $$ a mere speck, close to nuclear size. You are contrasting this to tenths of ...


0

This answer is similar to Ron Maimon's but maybe it will be helpful. In short, the weak force doesn't violate parity; only the fermions do. There are no Dirac fermions in the Standard Model. There are no chiral projections in the Lagrangian because there's nothing for them to act on. There are no gamma matrices at all, just two-component Weyl spinors and ...


3

This feels like a perfect storm of misconceptions, unleashed by unscrupulous popular science writing. Helicity is Lorentz-variant, so, as you envision, may be reversed by changing your frame. It is either positive or negative; never left-or right handed. Chirality is relativistically invariant, so a left-handed particle is left-handed in any frame, and ...


7

The 2018 Particle Data Group gives a value of $880.2\pm 1.0$ s for free neutron lifetime, as an average of the seven best measurements. As can be seen, the measurements have non-overlapping confidence intervals. As discussed in (Wietfeldt 2014) the different experimental methods do not agree on the value. Wietfeldt gives the formula of the lifetime as $$\...


1

The photon goes at the speed of light in a vacuum or water, The photon is an elementary particle with mass zero. All zero mass elementary particles travel in vacuum with the velocity of classical electromagnentic waves,light , c, because of the Lorenz covariance of the quantum field theory. Light is an emergent phenomenon from a large number of photons, ...


2

The neutron is a spin 1/2 neutral particle. That means the only magnetic property it can have is a dipole moment, and it must be aligned with the spin. It can have an electric dipole moment, but that is parity and time-reversal violating (see: Electric Dipole Moment of the neutron). Since it is neutral, but has a magnetic moment, it must have internal ...


3

I dont know what you mean by the two equations will become one equation. But in general, yes. There is a process called cherenkov radiation where, if a charged particle (ie an electron) moves faster than the speed of light in that material, then you get a flash of light (if you use a classical electric field this flash is equivalent to the sonic boom when ...


2

You may be confused by the phrasing of the statement: particles as excitations of a field. This peculiar way of saying it comes from quantum field theory. A 19th-century physicist --before quantum field theory was established-- would probably say "light is the electromagnetic field moving through the vacuum" or "away from sources". It's ...


2

Photons are excitations of electromagnetic field. These are part of the phenomenon that we call "light", so the phrasing in the book is misleading. Closer to the matter: all particles can be though of as excitations of some field. When the field is electromagnetic, they are called photons. In other words, the statement should be Photons are the ...


0

Historicaly: From theoretical considerations, in 1934 Hideki Yukawa predicted the existence and the approximate mass of the "meson" as the carrier of the nuclear force that holds atomic nuclei together. If there were no nuclear force, all nuclei with two or more protons would fly apart due to electromagnetic repulsion. Yukawa called his carrier ...


0

Your text sensibly assumes you have internalized the lesson of the previous page, p 483. For the Ws, couplings to the Higgs trail masses faithfully, scaled by a factor of 2, $$ \frac{ g^2}{4} W^- W^+ (v+h)^2 = \frac{ g^2 v^2}{4} W^- W^+ + \frac{ g^2 v}{2} W^- W^+ h + O(h^2)\\ =m_W^2 W^- W^+ + g_W m_W W^- W^+ h + O(h^2), $$ where I have supplanted $g_W=g$,...


1

They're definitely wrong, because what sets a black hole apart is not its mass, but its density. For example a primordial black hole (if it exists) might have a mass of a large mountain, yet it would still be a black hole. On the other hand if you changed the question to "what happens when a galactic body gets denser" then the relation still falls ...


0

There is one thing that comes to my mind when thinking of the connection between the strong coupling constant $\alpha_{S}$ and the proton structure functions $F_{1}\left( x, Q^{2}\right)$ and $F_{2}\left(x, Q^{2} \right)$. When you studied elastic electron-proton scattering (such as in Mott or Rutherford scattering), you might have come across the term form ...


2

Before going to dangerous irrelevance, it helps to briefly recapitulate what irrelevance under RG means in itself. When thinking of an RG fixed point, the scaling behaviour at low energies/long wavelengths is typically controlled only by a handful of relevant operators which dominate the physics, while all irrelevant terms progressively get smaller and ...


1

The electron is a spin 1/2 particle. This severely limits the possible structure it can have. If we ignore the weak interaction, and only consider its charge, then it talks to the rest of the world through a photon vertex, e.g.: The figure shows a presumably structureless electron on the left (coupling through the Dirac matrices, $\gamma_{\mu}$, and a well ...


3

The models that describe photons used in quantum optics and in particle physics are one and the same: the Standard Model of particle physics (often replaceable with just its quantum electrodynamics component) as encased within the formalism of quantum field theory. Moreover, the definition of photons (more specifically, single-photon states of the field) are ...


-2

What is important to keep in mind is what is considered the wave characterizing a particle in particle physics. The wave is a probability wave, the probability function given by the solution of the quantum mechanical equation corresponding to the given point particle in the table of elementary particles. That photons have a footprint of a particle as seen ...


2

Physicists used accelerators such as the Stanford Linear Collider to study high-energy collisions between electrons and positrons. These experiments found that neither particle has internal constituents at the length scales that the accelerator could probe. They behave as fundamental, non-composite particles, as the Standard Model of particle physics assumes....


0

The answer is some kind of a compromise between the fact that the b-quark has a high mass, and the composition of the CKM matrix which governs weak decays of quarks. As one can quickly notice from the CKM matrix, each quark has the largest probability (order of 1) to evolve into a quark of the same generation (diagonal elements of the matrix): d<->u, s&...


0

This video has the definition also the history and the use of the effect mainly in neutrino experiments aiming at very high energy neutrinos. Very high energy cosmic rays hitting the atmosphere create air showers An air shower is an extensive (many kilometres wide) cascade of ionized particles and electromagnetic radiation produced in the atmosphere when a ...


1

The point is that we define the new coupling parameter e as $$e:=g\sin\theta_W=g'\cos\theta_W$$ in order to avoid carrying around these longer expressions


1

As an analogy, consider a toss of two coins. The outcome of each coin toss can be heads (H) or tails (T): HH HT TH TT Since we don't care about the order of the coins (just as we don't care which $W$ actually decayed to quarks/leptons), we can write think of this as: HH 2*HT TT


0

I would like to add an explanation to @OON's statement that $I_{3}$ is an eigenvalue to $T^{3}$: If we consider at least an $SU(2)_{L}$-doublet of the form $$ T^{3}\psi = T^{3}\begin{pmatrix} \nu_{L} \\ e_{L} \end{pmatrix}= \frac{1}{2}\sigma^{3}\begin{pmatrix} \nu_{L} \\ e_{L} \end{pmatrix} = \frac{1}{2}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\...


0

$I_3$ denotes the eigenvalue of the $T_3$ matrix corresponding to the fermionic field in the sum. In the very first equation you write the sum is not over electrons and neutrinos. Instead you have the left doublets on which $T_3$ acts as the Pauli matrix, \begin{equation} l_k= \begin{pmatrix}\nu_{Lk}\\e_{Lk}\end{pmatrix} \end{equation} (where $k$ denotes the ...


1

You have two $W$. For the final state $qql\nu$, one $W$ decays to $qq$, the other to $l\nu$. But the decay to $qq$ could come either from the first W or the second W, thus you have the following possibilities : First possibility -first $W$ decays to $qq$, thus the second $W$ decays to $l\nu$. -second $W$ decays to $l\nu$, thus the first $W$ decays to $qq$. ...


3

You question begins An electromagnetic wave is made of an electric wave and a magnetic wave, so 2 waves. and asks “How many waves for a particle wave?” However, I think it’s important to realize that “there are two waves” is just one model of the electromagnetic field, and it’s not necessarily the best one. For one thing, electric field $\vec E$ and ...


1

Phonons do not have all the ingredients required in the Standard Model (local symmetry, three fermion generations, scalar multiplet, SSB). Is this the reason why phonon are not in the boson list of SM? The table of particles axiomatically assumed for the standard model consists of point elementary particles, i.e.non composite. Phonons by the your ...


3

The answers are good, but I need to explicitly state what is conceptually wrong in your question. An electromagnetic wave is made of an electric wave and a magnetic wave, so 2 waves. The classical electromagnetic wave is one wave, its intensity and direction depending on two variables that are a function of each other through the solutions of Maxwell's ...


1

Actually in electro-magnetic wave there is effectively only one "wave", because the electric and magnetic components are strictly coupled. (Not only in waves, but also in general.) Also photons and electro-magnetic waves are the "exact" same things. The names are only distinguished to emhpasise if we are talking about the particle ...


4

For a particle propagating through space, its wavefunction furnishes the probability of finding it in a given place which is a different number for every different place. This is a single-valued variable. Electromagnetic waves are very different; they propagate through space as the interplay between a linked pair of an electric field and a magnetic field, ...


0

The reason massive neutrinos is considered beyond-the-standard-model (BSM) is because even if you want to add neutrino mass in the most straight forward way -- add a right-handed neutrino current and having it couple to the Higgs, then just make that coupling arbitrarily small to match whatever the neutrino mass is -- it still leaves you with a problem. ...


3

Virtual particles are better regarded as a convenient way of representing terms in the infinite sums of quantum field theory than as particles. Feynman introduced his diagrams to organise the calculations but did not claim the virtual particles were "real" in any useful sense. They are not ontological explanations for things, but often used in ...


1

Look at it this way: It is the present day mathematical model of nature in the microcosm. Particles are quantum mechanical entities on which quantum numbers are absolutely attached , and a four vector $(E,p_x,p_y,p_z)$. If that four vector's length is fixed, it is called the invariant mass of the particle. If the length is variable within a mathematical ...


1

What I have so far understood from my limited knowledge of quantum field theory and standard model is that the unification of strong force with electroweak force is still a conjecture that needs to be verified by experiments. So far there has not been conclusive evidence of unification of two domains. The conjecture is that that at high enough energies the ...


1

One does not talk about dangerously irrelevant fixed point but of a dangerously irrelevant coupling (or operator). The gaussian fixed point of a $\phi^4$ theory (describing phase transitions in dimension $d\geq 4$) has one relevant coupling (corresponding to the mass/correlation length) and an infinity of irrelevant couplings (one marginally so in $d=4$). ...


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