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The correct answer here is likely to vary depending on your country and institution as what is expected knowledge will change. The best advice I can give is to have a good understanding of the basic maths. From a UK perspective I would recommend knowing the core A-level Maths pretty well (and preferably further maths too). For US people Wikipedia suggests ...


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Books: (To further your textbook knowledge of Physics at the University level): Fundamentals of Physics Problems: (To further your problem solving abilities): Problems in General Physics, Irodov. Knowledge: (To further your knowledge of Theoretical Physics, this book unlike most other theoretical physics book is not for the layman, and contains advanced ...


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Thanks again for this thread. It prompted me to start my own work to look little further into this puzzling controversy surrounding Fabricant laser patent. It is very far from completed, current version is here (sorry its in Russian) http://www.igfarben.ru/index/valentin_fabrikant_laser_patent Briefly several findings: Fabrikant's laser patent application ...


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Newton formulated calculus, so before his time, it was either geometry or difficult algebra.


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In the situation you gave, it's immediately clear what is meant, and there's no possibility for misinterpretation, so yes, it's perfectly acceptable. (Remember that torque is mathematically defined as a vector for convenience, but the direction of that vector isn't really physical.) The only issue I can see with that is that as you leave the simple ...


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The term "shell" originally derives from the non-relativistic version of the answer by @JamalS. In a non-relativistic theory, a free particle satisfies the following dispersion relation $$ E = \frac{ {\bf p}^2 }{ 2m } $$ For a fixed energy a particle satisfies $$ {\bf p}_x^2 + {\bf p}_y^2 + {\bf p}_z^2 = 2 m E $$ In momentum space, this is precisely the ...


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A particle is said to be on-shell if it satisfies the relativistic dispersion relation, $$E^2 = p^2 +m^2$$ in units wherein $c=\hbar=1$. If you graph it, you obtain a parabolic surface for massive particles, and a cone for massless particles, like a photon. This is known as the mass shell, it is quite literally a shell or surface. The momentum of a real ...


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Lets define some variables first. Lets say that the length of the column of air that you trap in the tube between the water level and your thumb is $h_0$ and it is initially at the same pressure as the surrounding air which we will call $P_0$. Lets also define the cross-sectional area of the tube to be $A$. As you draw the tube out of the water, the water ...


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Indeed, the two effects are very much related! I don't know how your background is, so let me start by defining the four-vector $x^\mu=(t,x,y,z)=(t,\vec{x})$ such that $x^0=t$ and $x_i=x,y,z$ for $i=1,2,3$. (Note that it is convention that greek indices run from $0$ to $3$ (space-time) while latin indices run from $1$ to $3$ (space only). Summation over ...


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Because as far as we understand general relativity, it's not doing "the opposite of what gravity does." Gravity can be locally attractive or repulsive, depending on whether the stress-energy content satisfies or violates the strong energy condition. For ordinary matter, the stress-energy is dominated by the mass, the SEC holds, and its gravity is attractive. ...


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The term "Dark Energy" is just a name that we have given it while we try to determine what exactly it is and what a better name for it would then be. However, calling it an energy is appropriate. Our best model, the cosmological constant, says that dark energy has a constant energy density (that is a constant amount of energy per unit volume) in the ...


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I think you are attaching excessive significance to your order of magnitude estimate. If I say something is $10^5$ to within an order of magnitude I mean that the $\log_{10}$ of that quantity is nearer $5$ than $4$ or $6$. If some quantity has a $\log_{10}$ of about $5.5$ does it really matter whether our order of magnitude estimate is $10^5$ or $10^6$? ...


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According to commentator "bn" at The Chalkboards of Quantum Mechanics Professors (via Wired), this one is mostly about holography, how much information can be contained on a 2-D screen of a certain size. The diamonds are referring to domains of causal dependence. The square brackets are commutator brackets (Canonical commutation relation - Wikipedia), and ...


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The full statement seems to be: $T_{dS}\sim\frac{1}{R}\sim \sqrt\Lambda \implies non-SUSY$ In a de Sitter universe, that the temperature (at the horizon) is inversely proportional to radius (distance to horizon) and proportional to the square root of the cosmological constant implies breaking of super symmetry. See for example Temperature at horizon in de ...


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What you're looking for here is the equilibrium case when the wire has stretched out and has a constant length as you're swinging it. When the wire has a constant length each little piece of it with mass $m$ is executing circular motion with a radius $r$, and the force required to keep that piece of the wire moving in a circle is simply the centripetal force ...


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From "The Transistor, A Semi-Conductor Triode", by J. Bardeen and W. H. Brattain, Phys Rev. 74(2), 230-231 (1948): "The device consists of three electrodes placed on a block of germanium as shown schematically in Fig. 1. Two, called the emitter and collector, are of the point-contact rectifier type and are placed in close proximity (separation ~0.005 to ...


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I depends on the pressure ratio between the chamber and ambient . As long as the ratio is critical, the ambient air will flow through the orifice at the speed of sound (with respect to the state in the chamber). For undercritical pressure ratio the flow can basically be found by the Bernoulli Equation.



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