# Tag Info

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Your picture is correct, you have many different energies for electrons with the same quantum number $k$ (the crystal momentum). The reason for this behavior is the following: unlike the free electron case, electrons in a lattice require an additional quantum number to specify the state of an electron. This additional quantum number is the band index $n$. ...

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One can search for "muon spectrometers LHC" and find a number of entries that describe the spectrometers. Look at this dimuon event at CMS Both charges are identified, and the series of detectors used is illustrated behind the event tracks. Identification involves identifying a track in the accurate track detector in the center, with track ...

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As the paper says, the muons are deflected with a magnet into a storage ring, where they are stored for a fraction of a second. The muons are still moving with nearly the speed of light, which extends their average lifetime by a factor of $\gamma$. A small fraction of a second later, the muons have all decayed and the storage ring is ready to receive another ...

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Can spin be used for charged particle identification? Please recall what the conservation rules for each single event studied with the detectors of high energy physics, just energy and momentum. Even though angular momentum is conserved at the interaction point , it is not a value that can be measured per event because all the quantum mechanical effects ...

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In the discovery paper, Chadwick writes that he measured the stopping power of the protons by adding layers of aluminum foil between the paraffin and the ion chamber and watching the proton count decrease. He then looked up this stopping power on a reference curve relating a proton's range in matter to its velocity. (A modern paper would have a more obvious ...

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In my kitchen the background radiation is even around 293K, and yet in the fridge there is 273K and below. The second law of thermodynamics does not say that it is impossible to cool down anything below ambient level, but rather only that this does not happen spontaneously, i.e. without supplying energy that feeds a thermodynamic machine.

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Those ultralow temperature experiments are performed inside extremely well-shielded and insulated test chambers, to isolate them from the environment, and those experiments are performed while powerful refrigeration machines are actively pulling out as much heat as possible from inside the test chamber.

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It is important to specify which multiverse theory one wants to disprove. Some are possible, others are not. Tegmark's "Level I" model, where the universe is spatially infinite and hence contains regions mirroring our world (and other possible worlds) is falsifiable if we ever find evidence that the universe has a finite volume, for example by ...

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When you want to see if two quantities with uncertainty agree, you should compare their difference with zero. If $\gamma_t-\gamma_m=0$ within the uncertainty, they agree. The difference is the best way to check the degree of (dis)agreement, too. The number of standard deviations, $n_\sigma$, is: $$n_\sigma = \frac{\gamma_t-\gamma_m}{\sigma},$$ where $\... 1 The best approach is to plate a gold film directly on a flat side of a semi-circular polarizer. The next best approach is to have the gold film surface clamped tightly down on the polarizer. The third best approach is to have a liquid film between the polarizer and the gold surface. The index of refraction changes as the light exits from the polarizer to air/... 6 This part-per-million measurement of the muon's anomalous magnetic moment is a part-per-billion measurement of the muons total magnetic moment. Every part-per-billion measurement is hard. This one is, fundamentally, a measurement of a frequency: the frequency at which the muon's spin precesses in a magnetic field. In order to measure a frequency with ... 1 I'm sure someone will give a more detailed answer, but here's a brief one: These effects do occur for the electron, however they are much larger for the muon due to its larger mass. I imagine these sorts of experiments aren't really doable for tau particles due to their extremely short lifetime of$\sim 3\times 10^{-13}$seconds, in contrast to 2 ... 3 You may or may not call it a photograph, as this is not a precise term. But it is direct empirical evidence for a black hole. In astrophysics, there are often pretty images of things, but physics is always more than meets the eye. The image is not so interesting in itself (any capable person with Photoshop is able to create such a photo), but the context and ... 2 Stirring helps heat transfer (i.o.w. cooling) because it decreases the film thickness of the tea boundary layer with the cup and thereby increases the convection heat transfer coefficient$h$. Stirring also ensures the tea is at homogeneous temperature, which maximises the temperature difference between the tea and the environment and this, in accordance ... 6 The spin magnetic moment of a fundamental particle with mass$m$, charge$q$, and spin 1/2 is $$\vec\mu=g\frac{q}{2m}\vec S$$ where$\vec S$is its spin vector. The "$g$-factor" is a dimensionless number. The Dirac equation predicts that it should be exactly 2. However, the complications of quantum field theory cause$g$to differ slightly from 2, ... 1 Surfactants can be added to change the surface tension (even in the ppm region). Polymers as well can affect the viscosity and there is a phenomenon known as polymer surfactant interaction that decreases the concentration required to achieve similar changes in viscosity. Typically both surfactants and polymers decrease the shear viscosity while increasing ... 1 regarding surface tension: Surface tension of water is easily reduced by adding chemicals called surfactants to it. The quantity of surfactant required to get a big change in surface tension is so small that it will not affect the viscosity. Surfactants are the active ingredients in dishwashing detergent liquid, which is the classic method of reducing ... 1 As I recall, surface tension can be changed by an additive, such as a detergent. Viscosity probably requires dilution with another liquid. 2 A Nature paper was published on the same day, which seems to have attracted a lot less press. This presents a recalculation of the muon$g-2$value, using standard model physics and their value is consistent with the new experimental value (Borsanyi et al. 2021). So there's one theoretical explanation of the result! 1 Yes there is a theory that explains the results ... the Standard Model. In other words, the claim is that the Standard Model already is consistent with the experimental data, and the original "prediction" was calculated wrong. Check out the paper published in Nature together with the muon g-2 results, or the writeup at popular level. 5 In the standard model the$α=(g-2)/2$of an elementary particle should be calculable , the calculations as accurate as the higher orders are computed. For the electron the calculations coincide with the experimental value to great accuracy The muon$α=(g-2)/2$has different diagrams dominant so the theoretical value will be different, but it was seen ,... 1 To add onto what has already been said: the "theory" width tells you how far away the mass of a particle can be from its nominal mass. That is an energy spread. And that spread is related to the lifetime. Due to the uncertainty principle, a short lifetime requires a large energy range (or mass range or natural/theory width, whichever term you ... 1 The JouleThomson coefficient is a function of pressure P and temperature T: $$\mu=\mu(P,T)$$What you do is, for different values of the downstream pressure (at a fixed upstream pressure), measure the temperature change. You then calculate the ratio of the temperature change to the pressure change. You then plot the ratio as a function of the downstream ... 3 Why isn't the data producing a constant when I divide the force at x distance by the force at 2x distance? Shouldn't all inverse relationships produce a constant if you divide by the same ratio? Let me address both questions. To answer the first question: because the relationship between force and distance is non-linear. The data you collected via your ... 0 Each of the two magnets contains billions of atomic magnetic dipoles. Their response to the field from the other magnet would be about the same as if each magnet had a sheet of electric current flowing around their outer surface (giving the same dipole moment/unit volume). Each small loop of current (or atom) experiences a force which depends on the ... 2 Comments along the lines of “the electron is perfectly spherical” are a low-jargon shorthand for describing the continued non-observation of any permanent electric dipole moment for the electron; for example. Permanent electric dipole moments probe CP-violating physics. The fact that the universe contains more matter than antimatter, and in particular the ... 8 The basis for deriving the Poisson law is the assumption that the events occur with a constant rate,$\lambda$(although one can generalize to a varying rate). We can then divide the time interval$t$into$N$intervals of size$\Delta t = t/N$, with the probability of an event occurring in each interval being$\lambda \Delta t$, and calculate the ... 3 I think they refer to the decay width of the Higgs boson. By "natural", I think, they mean the ideal width that is predicted by the theory and that would be observed in the plots if everything was measured with an infinite precision. In reality the detectors are not perfect and measure the energy/momentum of particles with some error, which leades ... 3 First, to be clear, physics, with its five sigma threshold, has a higher standard of "proof" than almost any other discipline. In the social sciences, for example, two sigma is considered credible and three sigma is considered something close to divine truth. This is simply the wisdom of collective experience, although it has a solid basis that is ... 13 The$g$factor describes the magnetic moment of a spinning particle. The$g$factor for a classically spinning particle is equal to 1, but in the "basic" (ie, non-interacting) quantum field theory of spin-1/2 particles you would expect this number to come out to be 2, or in other words you would expect$g-2$to be zero. It turns out that in "... 1 It seems that muon g-2 is finally in the news : https://www.quantamagazine.org/muon-g-2-experiment-at-fermilab-finds-hint-of-new-particles-20210407 -1 Every coincidence test compares to the accidental chance rate. To find chance you use only one detector at a time to read the click rates,$R_1$and$R_2$, for each detector. There is a time window,$\Delta t$, within which you will use to determine both the chance rate and the$\Delta t$within which you will count as a coincidence in your experiment. The ... 0 With any cathode ray tube you are working with a beam of electrons (some with positive ions). My college physics lab had one which operated at a very low voltage to measure, e/m, for electrons. If you have a radioactive source, a cloud chamber lets you see the tracks of individual particles. 0 There are certainly other type of experiments to detect particles. Take for example: https://iopscience.iop.org/article/10.1088/1475-7516/2011/06/035/meta Where dark matter signatures are searched using gamma-ray emission. Particle detection experiments can be classified as "direct" or "indirect". In the direct experiments (such as ... 1 Classical physics is not just a set of independent ideas, it is a set of connected ideas, and consequently almost any process involving mass moving from one system to another will offer evidence of the correctness of the 2nd law as it applies to such systems. Simple examples include collisions, rockets, and moving platforms (e.g. a cart, a boat, a car, a ... 0 Another version of this is where the mass changes under a scenario of no external force. There is a classic problem of rain falling into the bed of truck, where one ignores the friction between the truck and the road. If you have an airtrack, you could do the experiment where you a car slowly slides under its own inertia (F = 0) with a cup attached to the ... 2 The Rocket Equation makes use of this variable mass relationship since rockets change weight as they use propellant. So every rocket is experimental evidence of this relationship. 1 In the Poisson formula the average value$\lambda = s + b$does not have to be small. E.g. if we have many particles passing through a detector and most of them do not contribute to the obtained signal (background or real), the conditions for a Poisson process are satisfied. This is exactly why we use this distribution to model the signal. Here an example: ... 2 Here is another reason for studying high-energy particle collisions, which produce unstable particles not found in ordinary matter. There was a time immediately after the big bang when all the particles present had truly gigantic energies, and were experiencing constant collisions with all their neighbors at those high energies. This meant that within that ... 0 As a matter of principle, it's nearly always easier to make a differential measurement directly than it is to measure two values absolutely and compute some small difference. This is a theme that runs across many experiments, in many disciplines. I am a strong proponent of reading classic papers directly: in general they are written quite clearly, in order ... 1 For Galilean and Special Relativity, the "Law of Inertia" suggests that an affine space is an appropriate model of spacetime. For more details, consult these articles by Andrzej Trautman. From Warsaw U's Physics page for https://en.wikipedia.org/wiki/Andrzej_Trautman , http://trautman.fuw.edu.pl/publications/Books/... 2 The history of physics experimental research shows that the higher the energy of the particles used in interactions the more one learns how to build theoretical models that fit and are predictive of new data. The aim of particle research now is to find a mathematical theory that fits all known forces and is predictive of new situations. The higher energy ... 1 When you do Millikans experiment you don't know n or what two neighbouring groups are. The experiment shows, with enough droplets measured, that you get only multiples of some Number approximately 1.6*10^-19 As to really measure e with some accuracy this is not the best experiment. Once you have established, that its multiples of a number you have a good ... 0 It is good practice to present the steps which you used to calculate the$p$-value. Hence, if you show that the average value of your experiment is$\bar g = 10.1m/s^2$, the standard deviation associated with the average value is$s=Sd[\bar g] = 0.2m/s^2$, most people are able to eyeball the$p$-value of the hypothesis test. Thus, your result can be ... 0 In my opinion (1) is the better technique. It avoids having to know the exact viscosity of the air, questions about the Reynolds number of the flow etc - all the things that make drag such a difficult subject. Further, you get "infinitely long time" to take a good measurement - you can iteratively adjust your voltage until the drops float "... 3 There seems to be some confusion about how collider experiments works, particularly with the question, "Don’t you believe this rate to be too SLOW and that a very fast decay event could be missed". There is no relation between the event rate and the decay rate of an event. I am familiar with DESY, not CERN, so I don't have any numbers (other than ... 1 Sodium is not a blackbody. Rather the spectrum is determined by all the possible transitions between energy levels of it's electrons. Some of those levels are degenerate to varying degrees. If there are a lot of possible transitions with the same energy, the spectrum of the material will have a higher intensity for those energies, or wavelengths, ... 1 These two examples are of different origins. The black body spectrum is the results of two factors: Bloltzmann distribuition$e^{-\beta \hbar \omega}$and the available number of modes between$\nu$and$\nu + d\nu$,$N(\nu)$. The intensity of spectrum of frequency$\nu$, $$I(\nu) = N(\nu) \sum_{j=1}^\infty j e^{-j\beta h\nu} = \frac{N(\nu)}{ e^{\beta h \... 0 Given that the probability of transition is obtained:$$P_{\nu_\alpha \to \nu_\beta}(L, E) = sin^22\vartheta sin^2(1.27 \frac{\Delta m^2[eV^2]L[km]}{E[GeV]})$$On the other hand,$\Delta m\$ is fixed by nature. Therefore, in the experiment, the ratio of length to energy is important to us. And we categorize the experiments accordingly: SBL = Short Base Line, ...

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