# Tag Info

12

The Dirac equation implies negative energies as well as positive. This is due to energy-momentum relation $E=\pm \sqrt{m^2+p^2 }$. If we replace $E$ and $p$ by operators $E\to i\frac{\partial }{\partial t}$ and $p\to -i\nabla$ we get the Klein-Gordon equation $(\Box+m^2)\phi=0$ for scalar (spinless) fields $\phi$. The problem with this equation is that it ...

11

You basically just need to be careful about the distinction between velocity and speed. In particular, you say that Won't the particles change velocity when exposed to the magnetic field, and therefore change KE? A change in velocity is not necessarily accompanied by a change in speed, and it's the speed that determines the kinetic energy. The ...

9

Massless charged particles can't exist in Nature because they would be easily produced by the colliders, and they haven't been. Such a production would simply arise from the Feynman diagram with an intermediate photon that "splits" into the new charged massless particle and its antiparticle. The cross section of this process would be calculable, and not ...

9

Elementary particles, like photons and electrons, are not more elementary in the sense that there are underlying theories, such as quantum spin model on lattice, from which they can be derived as an effective approximation (see for example arXiv:hep-th/0302201). In particular, the string-net condensation provides a unified origin for gauge interactions and ...

9

You first have to understand what a "white hole" is. It's the time reverse of a black hole. It was rightly pointed out in previous answers that white holes violate the second law of thermodynamics. Now, like anything in thermodynamics, this makes them unlikely but not impossible (unlikely here usually means unlikely even in an astronomical number of ...

8

As you may know, when particles are scattered off a target, what actually gets measured is the differential cross section $\frac{\mathrm{d}\sigma}{\mathrm{d}\Omega}$. This can basically be thought of as being related to the fraction of particles that come out of the collision in a particular direction. It's possible to calculate this quantity using quantum ...

8

No, it's not a problem. The reason is that, in order for expressions like $$\mu=-T\left(\tfrac{\partial S}{\partial N}\right)_{E,V}.$$ to be meaningful, you have to be using the grand canonical ensemble (or a generalisation thereof), in which particles are able to enter and leave the system. Consequently, $N$ stands not for an integer number of particles, ...

8

What exactly is a boson? A boson is a particle whose spin (= intrinsic angular momentum) is an integer number. For example, the photon (the particle that is responsible for the electromagnetic force) is a boson. Contrast this with a fermion, such as the electron, whose spin is a half integer. In everyday terms, the bosons are the microscopic particles ...

8

I think this is mostly a philosophy question. We can say whatever we want. Your question title talks about truth. I hope you won't be disappointed if I tell you that science is not about finding the truth. Rather, science is about contructing models that are useful. I had a friend in university who actually quit chemistry and went to study philosophy when ...

7

There's no problem in writing down a theory that contains massless charged particles. Simple $\mathcal{L} = \partial_{\mu} \phi \partial^{\mu} \phi^*$ for a complex field $\phi$ will do the job. You might run into problems with renormalization but I don't want to get into that here (mostly because there are better people here who can fill in the details if ...

7

One of the questions under investigation in the data being gathered at LHC is the search for compositeness of quarks and leptons. They gave limits for quark compositeness from the data of 2010. So the answer is, it is an open question under investigation, though not popular with the theorists.

7

You'll find Dirac's 1933 Nobel lecture on the Nobelprize.org website. The pdf is quite brief (5 pages long) and speaks on the antiproton at the end (p4). The argument is the following : In any case I think it is probable that negative protons can exist, since as far as the theory is yet definite, there is a complete and perfect symmetry between positive ...

7

From the very basic understanding that they are created out of nothing mutually and collide to annihilate each other seems to indicate this happens due to an attraction. Why? this just means that if two of them are nearby, they can annihilate. Remember that particles are waves, and thus are quite spread out. They don't have to be directed to collide ...

7

My questions are: 1) The word "quantum" is a perfectly good word meaning a wave with minimum amplitude; why are physicists insisting on calling a quantum a particle? The word "quantum" in physics is used as a "definite quantity of a variable" in contrast to "a continuous quantity of that variable". There are quanta of energy, for example the energy ...

6

Errors in particle physics are of two kinds. Statistical, and systematic. Statistical is the usual standard deviation of gausian distributions, sqrt(n)/N for 1 sigma. It is the systematics that take a lot of effort, and often are not taken well into account. Systematic errors come from 1) the background to the signal expected. The background is calculated ...

6

Suppose you treat scattering of a particle in a central potential. This means that the Hamiltonian $H$ commutes with the angular momentum operators $L^2$ and $L_z$. Hence, you can find simultaneous eigenfunctions $\psi_{k,l,m}$. You might know, for example from the solution of the hydrogen atom, that these functions can be expressed in terms of the ...

6

From an experimental point of view, we know one mass less particle, the photon. We cannot describe the photon relativistically by E=mc^2. Its energy is E=h*nu, When it interacts and loses energy, it is the frequency that changes. Thus I would expect, if a massless charged particle could exist on shell, a corresponding energy definition would give it a ...

6

Martinus Veltman Facts and Mysteries in Elementary particle physics He introduces almost all the concepts he talks about, provides interesting biographies and stories, and keeps the material at a level perfect for layman and high school students. And he is a Nobel Laureate.

6

It's just what it means for a particle to be massless. For a general motion of a particle in the vacuum, the rest mass $m_0$ satisfies $$m_0^2 c^4 = E^2-p^2 c^2$$ If the left hand side is zero, it follows that $$E=|p|c$$ which means that the energy-momentum vector is null (light-like). In general, the speed may be calculated from the direction of the ...

6

particles are ordinary quanta of the corresponding quantum fields - without any knots or other topologically nontrivial features. (You have to get used to the wave-particle duality, probabilities, and the uncertainty principle - they're fundamental features of the world around us.) However, this is only true for "weakly coupled particles" that are directly ...

6

First, an assignment of the parities. The parity of fermions is a bit ambiguous because one may always redefine parity by $$P \to P (-1)^{2J}, P(-1)^L, P(-1)^{3B}, P(-1)^{3Q}$$ or one may add the product of several factors of this kind because the second factor is a multiplicatively conserved sign. By this definition, one gets another parity that is still ...

6

They are more elementary in the sense that there is no accepted underlying theory from which they can be derived as an effective approximation. On the other hand, what is elementaty changes with time. At some time, protons and neutrons were considered to be elementary particles, whily they are now considered to be composed of quarks. There are various ...

6

It is not possible that SUSY particles are hiding in keV or MeV range. In particular, there can't be any new charged particles (and similarly new color-charged particles) that would be this light because they would be easily pair-produced and easily detected. The first (February 2012) claims by different authors (the original ones, Rupp and van Beveren, who ...

6

The above equation solves for the average kinetic energy of a gaseous particle at a given temperature. k is known as Boltzman's constant, $k_B = 1.3806503 × 10^{-23}\frac{m^2kg}{s^2K}$ and is equal to the ideal gas constant divided by Avagadro's number, $\frac{R}{N_A}$. So where does the equation come from? The short answer: The equation above is ...

6

In mathematics, there is a complete symmetry between $+i$ and $-i$. Both the imaginary unit and the minus imaginary unit obey $$i^2 = (-i)^2 = -1$$ The exchange of $i$ and $-i$ is known as the ${\mathbb Z}_2$ automorphism group of the complex numbers ${\mathbb C}$. When you introduce the complex numbers for the first time, it's a complete convention ...

6

So there are two different things you might be talking about which often get conflated in popular level discussions of the topic: Real particle/anti-particle pair production in reactions. An example reaction (one among many) would be $p + \gamma \rightarrow p + e^{-} + e^{+}$ (a proton and a gamma ray collide to produce a proton and electron positron ...

6

The first quote that Feynman wrote in the 1960s remained as accurate and crisp as it was 50 years ago. The quantum randomness can't ever be predicted, even in principle, so there are no hidden wheels here (which we would call "hidden variables"). The proof that hidden theories couldn't be right were more or less available during the publication of Feynman's ...

5

I would like to add that radioactive isotopes are a constant background in all natural environments, and more so in stone and concrete buildings. The relevant question is how much more than the natural background is the artificial radiation induced by human activities. In a recent viewpoint in BBC news professor Alison gives relevant numbers. For example, ...

5

It is not so hypothetical situation if you consider very relativistic electrons, for example. The velocity is always $c$ but the energy-momentum changes. Quantum mechanically it means a change in the De Broglie wave frequency. So a less energetic relativistic electron is like a less energetic photon - it has lower frequency. Classically it correspond to ...

Only top voted, non community-wiki answers of a minimum length are eligible