New answers tagged

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The Higgs field gives mass to all fermions and three weak gauge bosons (and itself) in the Standard Model. The masses of the fermions are proportional to their Yukawa couplings. The Yukawa couplings of the neutrinos were long neglected due to their small size, and sometimes assumed to be zero. Although the evidence is consistent with nonzero neutrino Yukawa ...


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The Higgs Field gives mass to :- Everything marked in purple i.e. the quarks and anti-quarks Everything marked in green i.e. leptons and anti-leptons, which includes electrons, positrons, tau-particles, neutrinos etc. The yellow marked particle i.e. Higgs Boson, an excitation of the Higgs Field The orange-red marked particles, except for 2 (photon [$\gamma$]...


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The Higgs field $\phi$ undergoes spontaneous symmetry breaking (from a complex doublet to a real scalar field, whose quantum is the Higgs boson) in a process named the Higgs mechanism. This has two consequences: the gauge bosons $W^\pm$ and $Z^0$ acquire a mass term, which they couldn’t have had a priori without breaking gauge invariance. The mass depends ...


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In the Standard Model, the Higgs field also gives mass to the six quarks (up, down, strange, charm, top, bottom) and the three charged leptons (electron, muon, tau) through Yukawa couplings. Some related mechanism may give neutrinos a small mass. (They’re massless in the Standard Model, but we know this is wrong.) Finally, one can argue that the Higgs field ...


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The Higgs field is coupled to the fermions (quarks and charged leptons) in the standard model via Yukawa couplings. As a result of the Higgs mechanism, the Higgs field then gives mass to these fermions, in addition to the weak bosons.


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No. You seem to be mixing up the concepts of average position and the center of mass. It is not as if the electron 'fills up' space all around the center of the atom. If that were the case, one perhaps could make an argument that the 'center of mass' of this is at the nucleus. However, this is emphatically NOT the case: the electron is a point particle with ...


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I've just discovered the answer from an old Physics Forums post Define: $$\psi_+=\psi_L+\psi_R$$ $$\psi_-=\psi_L-\psi_R$$ $$f=e\phi_E$$ Redefine: $$\chi_+=e^{-ift}\psi_+$$ $$\chi_-=e^{-ift}\psi_-$$ The electrostatic potential drops out to leave the standard equations for the electron/positron rest mass energy $m_e$: $$i\frac{\partial\chi_+}{\partial t}=m_e\...


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If the relationship between inertial mass and gravitational mass were not the same for all fundamental particles then one consequence would be that the acceleration due to gravity would depend on what an object was made of. So a hammer and a feather really would fall at different speeds in a vacuum. In fact, it is only a hypothesis that antimatter particles ...


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Correct. Inertial mass is a separate concept than gravitational mass/charge, and does not have to equal it, just as in electromagnetism the electric charge of an item does not equal the inertial mass. Einstein's theory takes as a postulate* that the gravitational 'charge' of an object is equal to the inertial mass and of the same units; this is called the ...


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The gravitational mass of a piece of matter [or more generally the (stress)-energy-(momentum) tensor, but let's stick to mass] is what causes the curvature of spacetime, i.e. the gravitational field, which makes all non-interacting masses move on a geodesic curve. An object falling free in such a spacetime doesn't "resist" to the free fall. Which ...


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There is a slight mistake in the way you are defining the energy. In general, the energy of a relativistic particle is $E^2 = p^2c^2 + m_0^2c^4$. Therefore, for a massless particle--like the photon--the energy becomes $E= pc$, not $E = mc^2$! So in reality, what Plank's relation is telling us is that momentum is quantized for light, not the mass. Edit: By ...


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You seem to only be considering the electromagnetic force, while ignoring the strong nuclear force, which is far stronger (if it wasn't, then nuclei would simply fly apart due to Coulomb repulsion), and also attractive. It is true that pushing two positive nuclei together increases the total mass of the two-nucleon system for relatively large separations. ...


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I'm talking about the weight while the elevator is being pulled up steadily. If the elevator is travelling at constant speed in a constant gravitational field, then its weight will stay the same while it's travelling. However, as it moves upward, it is getting further from the centre of the earth. Hence the earth's gravitational attraction is reducing as it ...


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While the elevator is accelerating upward, which means that its speed upward is increasing, the shaft, elevator, motors, pulleys, etc., will weigh more. If the elevator falls with no resistance from the mechanism, the whole assembly will weigh less; but if the elevator's fall is slowed by a braking system the whole assembly will weigh more.


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I don't see how the collision could be elastic and symmetrical without the two particles having the same mass. You're right, but this isn't a mistake in the argument. The authors are assuming a specific situation and using that to derive general constraints. If you change the assumptions, you would get a different, more complicated setup, which would not be ...


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Weight is the force the earth exerts on the body. Force of gravity is the force per unit mass exert by earth or gravitational field strength.


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Because unfortunately it gained the label "relativistic mass", which gave it a sort of unconscious legitimacy. I propose we consider calling it instead "directional mass". This is IMO much less likely to be taken seriously as a concept. I intend to do it myself from now on, and see how it goes.


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the most clarifying information about that that I am aware of is in a physics blog post (2007) by a physicist called Jim Pivarski, titlled The origin of mass?. The theory of electroweak interaction offers a way of unifying the descriptions of electomagnetism and the weak nuclear interaction. This requires a mechanism to account for the fact that while the ...


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Indeed, the properties of inertia must be assumed as they are in order to formulate general relativity at all. It could be, for example, that the attempts at formulating a form of emergent gravity will also lead to formulation of a theory of emergent inertia. It could be, I don't know, that some physicists regard inertia as a phenomenon that is irreducible. ...


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The Einstein equations imply that a suitably localized blob of energy-momentum that is not interacting with other field will follow (ignoring finite size effects) a geodesic with mass $m$, where $m$ is the mass monopole moment of the blob. In this sense, the connection between inertia and the gravitational mass is made. Any interactions will cause the blob ...


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In a nutshell: weight is the application of gravity on mass. Under a lack of gravity (or under micro-gravity conditions), objects still have inertia. Inertia cannot depend on gravity.


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If by same force, you mean same change in momentum, $\Delta\vec p$, then the change in four momentum: $$ \Delta p^{\mu} = (\Delta E/c, \Delta\vec p)$$ is orthogonal to the 4-momentum: $$ \Delta p^{\mu}p_{\mu} = (\Delta E/c, \Delta\vec p)(E/c, \vec p)=\frac{E\Delta E}{c^2}-(\Delta\vec p)\cdot \vec p = 0 $$ which means: $$ \Delta E = \frac{(\Delta\vec p)\cdot \...


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You can attribute properties to things for e.g. colour, smell, a name etc. One such property is the ratio of applied net force $\vec F$ to induced acceleration $\vec a$. Like other properties, you don't expect the ratio to be independent of almost everything: it could depend on the material, the place of doing the experiment, surroundings, the temperature ...


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One interesting aspect is that when you accelerate a body against a gravitational field, for example by starting a rocket, you actually accelerate the gravitationally bound bodies as well! 1 No wonder it's hard — you are dragging Earth behind you! ;-) 1 The "starting rocket" scenario is on closer examination a bit complicated because a substantial,...


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we can easily notice that one body will be easier to move as compared to another This is the error in your reasoning. The low-gravity object will only be easier to move in two situations: Moving it against the gravitational potential (e.g. lifting it) Moving it against friction that is proportional to the gravitational force (e.g. sliding it across the ...


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I remember reading one of Arthur Clarke's books years ago where he pointed out the misconception that massive objects (which would be heavy on Earth) would be easy to move around in the weightlessness of space. They still have mass and therefore inertia.


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Expanding on Harry Johnston's answer, if you had a fairly large, nice round pebble weighing 1kg and held it your hand while standing on Earth, it would exert 9.8N force and feel about the same as an everyday bag of flower. If you threw that stone, you would expend effort and feel a force against your palm as you accelerated the stone. If you then flew up to ...


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Imagine a 10kg curling stone on a flat ice surface on Earth. If we apply 10N of horizontal force, the stone will accelerate at about 1 meter per second per second. On the Earth, a 10kg stone weighs approximately 98N. Now imagine the same 10kg stone on a flat ice surface on the Moon. If we apply 10N of horizontal force in this scenario, the stone will ...


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Physicists distinguish gravitational mass from inertial mass. In practice we find that gravitational mass is equal to inertial mass, but the distinction is important because conceptually they need not be the same. A measurement of weight is, in effect, a measurement of gravitational mass. That is to say, the amount of gravitational force acting on a body as ...


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So does this mean that weight is measure for inertia rather than mass being the unit to measure inertia. No. Inertia is resistance to change in velocity (acceleration, $a$). From Newton's second law $$a=\frac{F_{net}}{m}$$ where $F_{net}$ is the net force acting on the mass $m$.. It's true that a mass $m$ will be harder to accelerate upward in opposition to ...


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As follows from energy conservation, the gravitational time dilation near a Schwarzschild object is equal to the velocity time dilation at the escape speed: $$\dfrac{\tau}{t}=\sqrt{1-\dfrac{r_s}{r}}=\sqrt{1-\dfrac{v^2}{c^2}}$$ $$v=c\sqrt{\dfrac{r_s}{r}}$$ $$v=\sqrt{\dfrac{2GM}{r}}$$ Where $c$ is the speed of light, $G$ is the gravitational constant, $M$ is ...


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Inertial mass in classical mechanics can be measured by the relation between force and acceleration. So it is necessary some time to make the measurement, but it doesn't mean it is a function of time. For the relativistic equation, $E = m$ for the frame where the mass is at rest. Velocity of light is only a constant.


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First of all, it is generally a bad idea to divide vectors. Write the equation as $m \vec{a} = \vec{F}$ or $\vec{a} = \vec{F} / m$. The mass is a scalar number (it has no time component) and describes the amount of matter or inertia. It does not change in the dynamics.


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The $c^2$ term is the conversion constant for going back and forth between mass and energy units. It has nothing at all to do with Newton's $\vec F = m \vec a$ law. In $m = F/a$ as written, the mass term is not changing that is, varying the acceleration does not magically cause the mass to change.


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@freecharly has given an excellent answer! In human words it means that the effective mass is the curvature of the conduction/valence band near its minimum/maximum (respectively for electrons/holes). It is worth adding that electrons and holes in a crystal lattice are not free particles, but excitation of a complex system. Effective mass is just a way of ...


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The mass number of Carbon-14 is $A=14$, while its atomic mass is given by $$m(^{14}C) = A\cdot 1u + \textrm{mass defect} = 14u + \textrm{mass defect} \tag{1}$$ Here $1u \approx 1.66054 \cdot 10^{-27}kg$ is the mass unit defined by the atom $^{12}C$. The interesting part for answering your question is the "mass defect", which is due to the ...


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The mass number (symbol A, from the German word Atomgewicht [atomic weight]),1 also called atomic mass number or nucleon number, is the total number of protons and neutrons (together known as nucleons) in an atomic nucleus. Note: atomic mass number . numbers come in integers, so it is the number of baryons in a nucleus, i.e ignoring that that protons and ...


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When you apply a force to an extended object (with density profile $\rho(\vec r)$), you can treat is as accelerating the center of mass: $$ r_{i,\, cm} = \frac{\int_Vr_i \rho(\vec r)d^3r}{\int_V\rho(\vec r)d^3r}$$ The numerator is the 1st moment of the mass distribution. The denominator, which is the zeroth moment, is just the total mass: $$ m = \int_V\rho(\...


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Center of Mass is a mathematical concept used for ease of analysis because we are familiar with Equations of Motion for a point particle. Center of Gravity is a quite similar concept but very special to gravitational forces acting on a body. Now, Center of Mass is the averaging of the position of masses while Center of Gravity is the averaging the position ...


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As for chemical potential, from the point of view of grand-canonical ensemble, it is a constant, coupled to the conserved number of particles: $$ \mathcal{Z} = \text{Tr} \ e^{-\beta (H - \mu N)} $$ The definition of chemical potential assumes the conservation of $N$. It seems not to be connected with the 'zero rest mass' property. For instance, one may ...


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It might help to think of Wilsonian RG in terms of coarse graining in real/coordinate space. Say we are on a square lattice, and there is a “structure” on the lattice of linear dimensions $L_0 = m_0^{-1}$, which can be written in units of the lattice spacing ($a$) as $L_0 = a \tilde{m}^{-1}$. The lowering of the cutoff in momentum space is equivalent to ...


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You said that if potential energy does not have an equivalent mass, then the entire relation can't be true. Actually, the $E = mc^2$ corresponds to an sort of 'internal energy', something which you may think of as needed to hold the entire object together. When reactions like nuclear fusion or fission occur, some of the internal energy is released. Similarly,...


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Actually, in a thought experiment with a box with perfectly reflective interior walls, the box will have an observable additional mass if there is electromagnetic radiation inside the box. Conceptually, you can understand this as the photons getting a redshift (from general relativity) when they travel upwards and a blueshift when they travel downwards. That ...


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What Einstein proved with his famous formula, is that your first definition is just wrong. Mass isn't a measure of the amount of matter an object consists of. Instead, it is a measure of how much energy that is contained in the object or system. I like your analogy with currencies. Just like you can convert from dollars to euros using some conversion factor, ...


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One of the first steps in developing the kinematics and dynamics of special relativity is to extend three-dimensional momentum to a four dimensional four-vector. This is done by multiplying the four-velocity $(\gamma,\gamma{\bf v})$ by a constant, $m$, with the dimensions of energy (in natural units with c equal to 1). The invariant length squared of the ...


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Here's a hint: If the piston is displaced a distance $\delta$ to the right (where $\delta$ is small) and the adiabatic relation $PV^{\gamma}=P_1V_1^{\gamma}$ applies, by how much does the pressure increase and decrease in each of the compartments? What is the net force on the plunger $\delta F$? What is the "spring constant"?


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Written as $ΔE = (Δm)c^2$ the $ΔE$ is the increase in the kinetic energy of a mass being accelerated, and the $Δm$ is the corresponding increase in the inertial mass (as predicted by special relativity).


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You are fundamentally correct, but are mixing up what's necessary for each. Special relativity is both necessary and sufficient to construct the idea of rest energy, but it is merely necessary for constructing a model of conversion of rest energy to other forms of energy. In order to do that, you need particle physics, and the only way we know how to ...


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In case you have access to the Journal of Applied Physics, this work is a comprehensive resource of III-V material parameters, including a brief discussion of the available references: I. Vurgaftman et al., J. Appl. Phys. 89(11), 5815 (2001) link to article $In_{0.52}Al_{0.48}As$ is discussed on page 5840 with a recommended value around $0.07 m_0$. Edit: I ...


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Summary: Your question is tantamount to asking whether mass-energy equivalence has been extended to mass-energy-information (or mass-energy-entropy). As far as I can see, essentially no one in the physics community (outside of a handful of outliers) accepts such an extension. So the answer to your question seems to be, as far it's known, no. I discuss this ...


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