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| bio | website | motls.blogspot.com |
|---|---|---|
| location | Czech Republic | |
| age | 39 | |
| visits | member for | 2 years, 5 months |
| seen | 9 hours ago | |
| stats | profile views | 25,130 |
Hi, I am a string theorist and a publicist.
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9h |
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How to derive an oscillation period formula? If you are, it's because the integral sign $\int$ is the opposite of the differentiation sign $d$ so they cancel, at least in this simple physicist's notation. $dt$ is the small interval of time by which the clocks moved during the interval $dt$, sorry for this bizarre near-tautology. If you want to know how much the clocks move after a longer time that is composed of many infinitesimal pieces $dt$, you must sum all these $dt$ terms, and this summation of infinitesimal pieces is denoted $\int$ rather than $\sum$. Note that both symbols $\int,\sum$ originate from "S" or "Sigma" for "sum". |
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14h |
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IR divergence and renormalization scale in dimensional regularization Dear user, mathematical identities like A.5 are meant to be trusted for all values of the dimension or $\nu_i$ or other parameters dictating the dimensional analysis. If there are additional powers of momenta in the numerator or the denominator, the "critical dimension" where the integral starts to be UV-divergent or IR-divergent shifts in the appropriate way, too. In the dimensions where the integral is covergent (or its part), you may actually calculate it. You may analytically continue the result to any $d$. A.5 is an example - and such results tend to have simple poles at integer $d$. |
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14h |
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In Newtonian pressure, what type of function is force? Well, perhaps except for integrating the 2D integral over 1 variable first, and then integrate these 1-dimensional integrals over another coordinate. This may be done but the definition of coordinates to cover the area $A$ is not unique so the function of one variable $F=F(A)$ won't be unique, either. |
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14h |
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In Newtonian pressure, what type of function is force? Pleasure, @DamienIgoe. And Saurabh, no, we can't really write it as an ordinary integral because the integral for the force, $ F = \int dA\,p$, is a two-dimensional integral, e.g. $\int dx\int dy\, p$ when the area is a rectangle in the $xy$-plane. On the other hand, to interpret $dF/dA$ as a derivative that may be inverted to get an integral expression for $F$ as an integral of $p$, we need to interpret the summation as a one-dimensional integral, over "area". There's no natural way to convert one-dimensional integrals to two-dimensional ones or vice versa. |
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2d |
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Why proper acceleration is $du/dt$ and not $du/d\tau$? That's the point. The reading on an accelerometer is invariant - but the derivative of the velocity with respect to coordinate time (in a general inertial system) is not. |
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Jun 13 |
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Ascertaining a mathematical equality to derive a partition function Dear al-Hwarizmi, the Fourier transform of your function that is zero almost everywhere yet finite is zero. To get a nontrivial function in the L2 or Fourier sense, you would have to have delta-functions located at the integer values of $x$, like $\sum_k k\delta(x-k)$, not just a finite value. |
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Jun 13 |
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Symmetry transformation in AdS space You wrote a geometry with the Euclidean $(+++)$ signature which surely can't have that symmetry group. |
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Jun 13 |
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LINAC: why are they being accelerated? If you mean why the particles feel the need that they have to obey some laws and accelerate, i.e. what's the mechanism, well, there is an RF source (radio frequencies) which bring an electric force in the direction of the acceleration at the place where the particle is located. The electric force implies acceleration $ma=qE$. If you mean Why do people accelerate them, what's the goal, well, they need high-energy particles because the energy is needed to to create interesting processes and new particles. |
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Jun 13 |
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Is time quantized? Is there a fundamental time unit that cannot be divided? Mikhail: that's simply not true. The assumption of a discrete time in this sense is a very bold conjecture that has many consequences - independently of other aspects of the theories with which it may co-exist - and one may easily empirically demonstrate that it's wrong. |
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Jun 13 |
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Decomposition of deformation into bend, stretch and twist? In other words, one may say whether a linear transformation is a rotation or not. But one can't say whether a linear transformation is a "pure deformation" because no deformations are "purer" than others. The condition doesn't exist. At most, you may define a "pure deformation" as anything that isn't a rotation-and-translation but almost every deformation would be "pure" by this criterion, it's a negative, not a positive, condition. Do you understand that my answer is No? It sounds like you don't like it and only want to hear the answer yes but this answer is wrong. |
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Jun 13 |
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Decomposition of deformation into bend, stretch and twist? It's exactly the same question, @FeiZhu, and I think I have already answered it. There is no way to define a "pure deformation". For example, if you declare that a shear deformation sliding horizontal planes on others is a "pure deformation", then you should allow the sliding vertical planes to be a "pure deformation" as well - but these two shear deformations differ by a simple rotation. There is no way to say "how much rotation" is included in a general deformation. |
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Jun 13 |
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Reflections in a glass of water A metallic mirror is a conductor so it sets the electric field to zero - because of Ohm's law, $j=\sigma E$, a nonzero $E$ would immediately create currents that would compensate the distribution of charges. With the $\vec E = 0$ boundary conditions at the mirror (and with some related boundary conditions for the magnetic fields), the only allowed waves are those that vanish inside the metal but get reflected on the side of the air. |
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Jun 12 |
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Global part of a local symmetry? I don't believe that @Trimok's explanation is so incomplete or unclear that it would require an example before it's usable. If you need an example after Trimok's explanation, then it's because you misunderstand not only what is a "global part of a local symmetry" but you misunderstand what a local symmetry itself means, too. |
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Jun 12 |
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air molicules repell Equivalently, the molecules are attracted to Earth and their initial velocity only gets them to a certain height. BTW the airfoil theory assumes that the fluid is ideal and has no viscosity. In fact, one also assumes the fluid is incompressible, an approximation. For liquids, the incompressibility follows from the molecules of the liquid that are close to "touch". Gases deserving the name are always compressible but the approximation of incompressibility simply means that the mechanical pressures added by the experiment are negligible relatively to the pre-existing atmospheric pressure. |
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Jun 12 |
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air molicules repell Sorry, I don't believe you have actually recorded any evidence of repulsion between air molecules because in the real world, this repulsion is negligible and the air is close to an ideal gas. The pressure of a gas is due to the collisions of the gas with the walls and each molecule of the gas acts independently (mutual interactions are negligible). Gas is more likely to be found at lower altitudes because the probability that the total energy $E=mgh+mv^2/2$ is $E$ decreases as $\exp(-E/kT)$, it's the Maxwell-Boltzmann distribution, so one gets an exponential decrease with $h$. |
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Jun 12 |
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How exactly are the degrees of freedom seen by a falling into a black hole observer related to the ones seen by a staying outside observer? Dear @Dilaton, that's a very good - and very hard - question that hasn't been fully answered, as the continued stream of new papers attempting to answer the "code" - such as the Maldacena+Susskind paper a few days ago - demonstrates. So while it seems clear that the degrees of freedom inside can't be treated quite as independent ones from those outside - i.e. that classical GR's causality is misleading in principle - the precise dependence hasn't been clarified. |
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Jun 12 |
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Gauge fields — why are they traceless hermitian? Dear Prathyush, thanks. The subtlety is simple but I omitted it because it is a distracting global detail: $U(N)$ is isomorphic to the quotient $(SU(N)\times U(1))/Z_n$ because the $U(1)$ are the transformations that multiply the whole matrix by a phase, including the phase $\exp(2\pi i / N)$ (and its integer powers) which may also be achieved by the $SU(N)$ factor: $Z_N$ is the center of $SU(N)$. So some non-identity elements of $SU(N)$, the center, may be combined with a $U(1)$ phase to produce an identity element of $U(N)$, therefore the quotient is needed. Otherwise no subtleties. |
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Jun 12 |
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How does light behave within a black hole's event horizon? Sorry, this question makes no sense. Which slice are you talking about? There are infinitely many slices - even infinitely many slices that intersect a light ray. |
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Jun 12 |
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Aharonov-Bohm Effect in Torus Dear Shiki, $A$ is equal to $B$ mod $2\pi$ (or "modulo $2\pi$") if the difference between $A$ and $B$, i.e. $A-B$, is an integer multiple of $2\pi$. So if $A,B$ are angles showing a direction in plane, they correspond to the same direction because if you turn around by $2\pi$, you return to the same direction. |
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Jun 11 |
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Thermal physics question for calculating temperature It's $C_1 \Delta T_1+C_2\Delta T_2=0$ because it's the total change of energy (via flowing heat) where $\Delta T_i$ are changes of the temperature (with the right signs) between the final and initial states of both parts and $C_i$ are their heat capacities. |
