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The mass is given by the Casimir operator $\hat{P^2}$, the square of the momentum, generated by translation invariance. This will give you the actual mass of the particle, as opposed to the bare mass that you will find in the theory, as the actual mass can be affected by the particle's interactions.


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Taking the Hermitian conjugate reverses the order of the $\psi$'s. You have $$ L_M^\dagger = \left( \bar{\psi}\psi\right)^\dagger = \left( \psi^\dagger \gamma^0\psi\right)^\dagger = \psi^\dagger{\gamma^0}^\dagger \psi = \psi^\dagger\gamma^0\psi = \bar{\psi}\psi = L_M \ , $$ where we use that $\gamma^0$ is Hermitian.


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It depends on the scale. Some scales (called balance scales) balance your weight against the weight of a known mass. Your weight is a force, calculated as mass x acceleration. The weight of the known mass is calculated the same way. Generally, there's some kind of adjustable leverage for the known mass. The scale is calculated by measured weight of known ...


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If two black holes hit head on at 99% the speed of light, the result would be one black hole sitting stationary in the center of mass frame with roughly twice the mass and some fraction of the energy expended as gravitational radiation.


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Yes. Gravity bends spacetime and anything occupying it. Edit: https://en.wikipedia.org/wiki/Gravitational_lens


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It sounds like you are trying to reproduce the work of McGaugh (2008, http://arxiv.org/abs/0804.1314 ). In section 3, they adapt a model (the Tuorla-Heidelberg model), based on observations, for the baryonic mass distribution in the Galaxy. The model is based on a synthesis of a number of observational studies. Stacy McGaugh has provided a series of tables ...


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We integrate $x dm$ to say "for every piece of mass $dm$, sum up the amount of mass times the coordinate $x$ of that mass". It makes sense to integrate over $m$, because every piece of mass has an $x$-coordinate. Likewise, in kinematics, it makes sense to integrate over $t$ because for every $t$ we have a velocity $v(t)$. However, calculus doesn't care ...


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You can find an excellent description of what a topological insulator is in this brief presentation from the Yazdani Group at Princeton: Topological Insulators. To answer your question on the meaning of negative effective mass: The effective mass is actually determined by the behavior of the energy levels $E({\bf k})$ as functions of the crystal wave ...


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The violation of gauge invariance by this term is the "only" reason why it's never written down – as long as we define the word "only" to include all other reasons that may be shown to be "physically equivalent" to gauge symmetry. Gauge symmetry is extremely important and its violation would make a similar theory inconsistent, especially at the quantum ...


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When you are given acceleration, density, area and time, you can indeed find an expression for mass in terms of these. Here is how you go about it: Make a table of the units that occur in each, and their exponents: L M T a 1 -2 D -3 1 A 2 t 1 As you can see, you need to use D (density) as the only one that contains mass. But ...


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regarding the first half: Density is Mass/Volume (not the other way around), so you need a Volume.. You could achieve that by a sufficient power of the area or find yourself a length-scale from acceleration and time regarding the second half: that's the whole point of dimensional analysis, a mass-like quantity is a mass-like quantity and must come from ...


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It's clear, higher you go, the Kinetic Energy increases. Let's see how : $v =\sqrt{2ax}$ where $x$ here is the $h$ and $a$ here is $g$. so we can write it like that: $v =\sqrt{2gh}$ And we have the Kinetic energy formula $K=\frac{1}{2}mv^2$ so, as you see, increasing $h$ causes a bigger $v$ that a bigger $v$ effects on the Kinetic Energy you will ...


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When you jump from any given height, the force pulling you down is gravity with $F=mg$. This makes you accelerate to faster speeds as you fall farther, obviously. When you hit the ground, you do not experience the same acceleration. Otherwise, it would take just as long to stop falling as it took to get up to that speed. Hitting the ground imparts a much ...


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You're looking at the net force acting on him at the instant he begins falling. And you are correct that, in the absence of air resistance, the net force that is acting on the man the instant after he begins falling in the same, given by $F_g=mg$. However, there's an old saying that says, "it's not the fall that kills you; it's the sudden stop at the end." ...


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Recall that force is equivalent to, $$F=\frac{dp}{dt} \sim \frac{\Delta p}{\Delta t}$$ Where $p$ is the momentum, and $t$ is time. Momentum is given by, $$p=m \cdot v$$ Where $m$ is mass, and $v$ is velocity. When you hit the ground falling from 2 meters. The change in velocity is very small, and the change in time is small. When you hit the ground ...


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Choosing your system carefully and drawing free-body diagrams is crucial! It is often easiest to start with defining your system as one big block. You can do this because the engine and wagons are all connected by rigid steel locks that cannot compress or stretch, meaning that if one car moves, the others have to move along at the exact same rate. If we ...


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Well, you can think of this as a car pulling a wagon with $a = 0.2 \text{ m/s}^2$. Naturally, the car's motor needs to generate enough force to accelerate both the car itself and the wagon it is pulling. This is true of the situation in this question. The steam engine needs to generate enough force to accelerate both itself and the two wagons at $a = 0.2 ...


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You can actually (in principle) do an experiment for one of those. If you had a huge, dense, thin sheet of matter in empty space you could cut a person sized hole through it then attach a cylinder to the hole and put a flat bottom on the far away end. So on one side it looks like a flat tower coming up from a big plane and on the other side it looks like ...


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You would have to apply a force upwards to stop the body. Stopping the gravity would stop the acceleration but not the speed that it already has. A good place to start to check the effects of g forces in a human body is this wiki Changing the mass won't stop the fall either. You cannot make the mass zero, you can cut legs and arms but I think you'll be ...


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Massless particles can carry and do carry confined charges (gluon in QCD) and they may carry charges under spontaneously broken generators (photon is transforming e.g. under the generators of $SU(2)$ associated with the W-bosons) but they cannot carry charges under unconfined $U(1)$ force like electromagnetism. The reason may be explained in different ways. ...


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This new paper addresses exactly this question, albeit with simulations. Here's a partial breakdown of the distribution of matter in the Universe, summarized from the above paper: Dark matter makes up about 26% of the critical energy density budget of the Universe, while "baryonic matter" (which is jargon for "visible matter" and includes all baryons as ...


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I believe the quark masses are obtained from lattice Quantum Chromodynamics (QCD) calculations on a giant computer. In essence, they use the quark masses as free parameters and calculate hadron masses. The adopted masses are those that give the best fit. See this link for details.


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Archimedes: Get a bucket big enough to contain your whole body at the most shrinking position (like a baby in the womb). Take a note for these following weight scale: Strip off all your clothes - fully naked. Weigh your naked body - Record it as $W_n$ Weigh the bucket - Record it as $W_b$ Put the bucket on the weigh scale Fill that bucket with water ...


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There are 4 general contributors to the mass density of our co-moving patch of what may be a larger universe: (1) Visible baryonic matter (including clouds of baryonic matter which may be visible only as shadows blocking galaxies). NASA estimates 4.6% of all matter is baryonic (http://map.gsfc.nasa.gov/universe/uni_matter.html). (2) Dark matter (not ...


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I would say (almost) yes... If you are familiar with QFT, then you probably know that the mass $m$ of a given field, say a scalar $\phi$, is introduced via the square term $\frac{m^2}{2} \phi^2$ in the Lagrangian. This however comes at the expense of some of the symmetry of the theory. In particular, if you naively apply this idea within the Standard Model, ...


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A classical "shell" of charge of radius $R$ will appear to have a field, outside the shell, corresponding to the charge all being at the center of the shell. If the charge on the shell is $q$ then a charge $dq$ being pulled in from infinity will cost an energy $k_e~q~dq / R$; without loss of generality this is the same if $dq$ is spread over an infinitely ...


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The mass of a fundamental particle turns out to be quite an elusive concept, because massless particles act as a source of gravity and they carry momentum. What then is special about mass? Where mass comes in is in explaining the relationship between the total energy of a particle and its momentum. For any particle we have the expression for the total ...


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By your logic, there is no gravitational force on the ball while you are holding it, because it is not accelerating. And if there is no gravitational force you should be able to throw the ball as high as you want, right? The truth, though, is that there are two forces on the ball - one from your hand and one from gravity - which are in balance. Keeping that ...


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The force that you "feel" is not just the current force of gravity but the sum effect of lots of gravitational force. There are a couple equivalent ways to sum up the forces involved. One is called "impulse" and "momentum". This says that when a (constant) force acts on an object for a unit of time, it gets an "impulse" equal to the product of the two. So ...


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Let's say that you have a 1 kg object that you want to throw directly upward and have it reach a height of 1 meter after you let go of it. Using $$2a(y -y_o)=v^2-v_o^2$$ with a constant acceleration of $a=-g=-9.80 m/s^2$ (up is my choice of positive direction) we get that $v_0= \pm 4.43 $ m/s. We choose the positive solution. That means my hand must exert ...


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While your object is in motion its acceleration is $g$, or -9.81 m/sec$^2$ (we'll take the upwards direction to be positive). This constant acceleration is why the velocity decreases from its initial value of $+v$ when you throw it to $-v$ when it lands. So far so good. But I think the force you are talking about is the force required to stop the object ...


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I don't see the paradox you are describing and i have some issues with your text: "it will always fall with an acceleration of 9.8m/s": If there is air resistance, it will reach a terminal velocity at which the object is no longer accelerating. Let's assume no air resistance to make things easy. "If I throw it very high up in the sky, it falls with a ...


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For one constant acceleration $a_{\mathrm{Gravitation}}$ is an approximation, that holds because the radius of the earth is so big compared to the height of your thrown mass. That being said, what you probably "feel" is, that something with greater height gains more momentum on his way back to the ground (as the distance is greater so is the time needed to ...


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The mass of each of the three mass types of neutrinos is unknown (although a lower limit has been placed, as Anna V's answer indicates), chiefly because they exist in three weak classifications as well as the three mass classifications, and the combinations can not be simultaneously isolated and identified with certainty. The combined mass of all three mass ...


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The particle data group has a thorough article on neutrino oscillation experiments. Oscillations are the way to measure mass differences between neutrinos, and in table 13.7 experimental values are given for delta(m^2) of the three neutrino species. These put a lower limit to the mass values. From the definition it is obvious that the limits are of the ...


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Nobdody knows. There are many experiments that try to find this out right now. For example the Karlsruhe Tritium Neutrino (KATRIN) experiment. We only know that they must be very light $\approx$ eV.



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