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48

Amazingly this actually happened to a Russian scientist called Anatoli Bugorski (WARNING: this is pretty gruesome). The beam basically just killed all the tissue it passed through. The symptoms were the relatively mundane ones expected from tissue death. The LHC has a much, much greater energy than the one that struck Bugorski, so it would cause a lot more ...


46

This is a tricky question because it asks about the meaning of words. People use the word "particle" to refer to various, not always well defined, notions in physics. In the end, I think the simplest and more correct single way to categorize the terms is to interpret "particle" as "excitation of a field". For example, if someone says "There are two ...


33

A charged particle will create charge separation (ionization) along its path. This will cause harmful chemical reactions to occur in the body, including DNA damage. The effects of these chemical reactions depend on their amount. The body can heal from a low amount on its own, while a high amount will cause radiation sickness and probably death. This can be ...


20

The energy is borrowed from the Heisenberg Uncertainty Principle to create virtual particles and has to be paid back in a very short time. $\Delta{t} \geq \frac{\hbar}{2\Delta{E}}$ This is why virtual particles live for very short times (i.e pop in and out of existence). We cannot manipulate this energy.


18

Taking your question literally, you can see a single barium ion: The TRI┬ÁP group has achieved capturing a single barium ion in a Paul trap. The images show Coulomb crystals formed by a decreasing number of laser-cooled ions as detected with an EMCCD camera. This forms an important step towards the planned experiments on single radium ions to measure ...


14

Whether you can extract energy from this or not (and I strongly suspect not) the Casimir effect is a consequence of vacuum fluctuations. Essentially when two metallic plates are very close to each other, the wavelengths of virtual particles that can be created between the plates is restricted and hence there are fewer particles between the plates and ...


13

Nothing happens obviously, when one high energy particle penetrates flesh as cosmic rays continuously impinge on us and some have the energies of the LHC. The cosmic rays reaching us are mainly muons and the damage they do is with electromagnetic scatters/ionisations in their path. The mean energy of muons reaching sea level is about 4 GeV. Muons, being ...


13

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 ...


12

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 ...


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 ...


12

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 ...


11

Gravitons are the particles you get from quantizing General Relativity. Since we don't know yet how to correctly quantize GR (or whether trying to quantize it is actually the right way to go forward; for all we know it might just be an effective theory where the more fundamental theory has to be quantized instead), we cannot know for sure whether the ...


11

I wonder whether graviton is indeed hypothetical or does its existence directly follow from modern physics? At the moment we believe that at the micro level the underlying framework of nature is quantum mechanical and from that level the classical mechanics, thermodynamics and electromagnetism emerge. "Believe" means physicists have gathered an ...


10

What follows is an answer from an experimental particle physicist, i.e. one who has more knowledge of theoretical physics than the average educated person, but not in a position to teach it :). I can use theoretical results and study data and validate or falsify a theory. I would like to know that if what we conceptualize as a "field" is merely an ...


10

Here is a brief historical ideosyncratic intro to calculus. Calculus of finite differences Consider this problem from a typical IQ test: 2 5 10 17 26 ? What's the next number you expect in the sequence (this is not hard, you should do it). The n-th term in the sequence is given by: $$ n^2 + 1 $$ as you can see by substituting n=1,2,3,4,5, so the next ...


10

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 ...


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 ...


9

There's a subtle distinction here, which is that neutrinos are matter particles (having spin $\hbar/2$ and obeying Pauli's exclusion principle) while photons are force carriers (having spin $\hbar$ and obeying Bose-Einstein statistics). There are three flavors of neutrino and they all have different masses. Therefore at least two of them are massive; ...


8

Gravitons are hypothetical, but they're far less hypothetical than most of the other particles which theorists speculate about (such as axions, magnetic monopoles, strings, sterile neutrinos, and the like). That probably sounds a little strange. Let me explain. We don't have a complete theory of quantum gravity. But we do actually have an extremely ...


8

I don't think you understand QFT. To be fair, I'm no expert myself, but I can certainly point out where you're going wrong here. A particle does not enter the Higgs field. However, the particle field that gets mass from the Higgs field does interact with the Higgs field. What this means is that in the Lagrangian of your model, there exists a term that will ...


8

Does a particle enter/interact with the Higgs Field when created, or at some other time? After reading your question a couple of times as well as your comments, it occurs to me that you're picturing something like this: a massless particle is created, interacts once with the Higgs field to acquire a permanent classical like mass which it then ...


8

The photon couples to all particles with electric charge or magnetic moment. This includes all of the quarks, the charged leptons $e,\mu,\tau$, and their antiparticles. It also includes particles composed of quarks and charged leptons: the proton and neutron (though the neutron only magnetically), the charged mesons, etc. Many electrically neutral mesons, ...


8

The answer is yes. The de Broglie wavelengths of freely propagating particles (i.e. forget the influence of interactions and gravity perturbations, just consider the Universe as a whole) are redshifted by the expansion of the universe. Another way of saying this is that their peculiar momenta with respect to a co-moving local volume decrease as the inverse ...


8

Obviously, the smallest particle that scientists have ever seen directly is a photon. The question is a bit silly because it tries to eliminate any simple device like a photographic plate. But the human eye, its nerves and the visual cortex together are far more complicated.


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 ...


8

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 ...


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

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.


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

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 ...



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