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Charged particles can't have Majorana masses of any type because they would violate the charge conservation law. The Majorana mass is really a term that is converting a particle into its antiparticle. It implies that the particle must be considered "physically indistinguishable" from its antiparticle. The Majorana mass term violates the lepton number or its ...


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You do not need a unit for force when measuring inertial mass in Newtonian Mechanics. The only things you really need are the Newton's second law and the concepts of inertial frame and acceleration. The way you shall proceed is the following. Take a collection $\{m_i\}$ of (unknown) masses and a spring. Use the spring horizontally to accelerate the masses ...


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Let's look at: $$F_g=Gm_{g_1}m_{g_2}/r^2$$ For an object with mass $m_{g_1}$, on the surface of the earth, then: $$F_g=Gm_{g_1}M_E/R_E^2$$ Where $M_E$ is the mass of the Earth and $R_E$ the radius of the Earth. You can now verify that: $$GM_E/R_E^2=g=9.81\:\mathrm{m/s^2}$$ So we could have written the second expression as: $$F_g=m_{g_1}g$$ An object ...


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One indication whether the 'inner' balloons are filled with helium or something heavier is whether the 'outer' balloon still floats in air or not. Whether it floats or not depends on whether its overall density is lower than air's density or not. Its overall density $d$ is simply given by: $$d=\frac{\Sigma m}{V}$$ Where $\Sigma m$ is the sum of all the ...


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The answer is already as you have given it: $m_{em}=\frac{4}{3}E_{em}/c^2$ The electromagnetic mass depends on the shape you assume for the charged object. In the case above it is assumed the object is a charged, hollow sphere. In general the electromagnetic mass for a charged object producing electric and magnetic fields $E$ and $B$ is: ...


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The left-handed neutrino is a 2-spinor field $\eta_A$, $A=0,1$, and the Majorana mass term is a bilinear, $\Delta L = \pm 2$ term without the complex conjugation in each term, $$ m\cdot \eta_A \eta_B \cdot \epsilon^{AB} + \text{Hermitian conjugate}$$ Note that this Majorana term is the only bilinear term without derivatives that one may construct from a ...


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All three are "correct", and all three refer to mass-energy equivalence discovered by Einstein. Equations (2) and (3) are algebraically identical, and are generalizations of (1). Equation (1) only takes into account an object's rest mass, whereas equation (2) also takes into account the momentum $p$ of the object, and (3) takes into account velocity $v$. ...


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When you add salt to water, sodium chloride dissociates into sodium and chlorine ions. These charged particles alter the intermolecular forces between water molecules increasing the boiling point. The temperature needs to be increased about one half degree Celsius for every 58 grams of dissolved salt per kilogram of water.


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The method with measuring the instantaneous weight while swinging the arm with a fixed angular velocity sounds rather impractical. You might try to combine data from multiple sources. I assumed that you are talking about the bit of limb from elbow up to and including the hand, with the hand stretched, but it turns out that you wish to exclude the hand. The ...


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You have drawn a Feynman diagram. Feynman diagrams are iconic shorthand for integrals over the variables of the problem. The calculation gives the probability for the reaction to happen, in this case the decay of a neutron . The observables are the four vectors of the initial (neutron) and final particles. The integral is over the variables . Here is a ...


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It can be stated this way: In this particular diagram, the W boson is in a state named off-shell i.e. we say that this boson is virtual. Virtual particles are allowed to have any mass value. They can't although violate charge conservation at the vertex. This 80 GeV mass of the W boson, is for a real W boson, which is on the on-shell state. The real ...


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Phonons follow a wave equation, which is at least in first approximation simply a standard wave equation, the only difference to relativistic particles is that the speed of the waves is not c but the speed of sound $c_s$. But this does not change the mathematics of the equation, so that in general there may be phonons which follow a massless wave equations ...


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Phonons are indeed massless, as you can see from their dispersion relation or from the fact that they are Goldstone bosons. The phonon dispersion relation that you wrote down tells us that we can excite a phonon mode, with some finite momentum, using an arbitrarily small amount of energy, hence they have no rest mass (in condensed matter language, they are ...


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It would be more correct to state Newton's second law as following: $$\vec F = \frac{d \vec p}{dt}$$ This holds in relativistic mechanics too: $$\vec F = \frac{d \vec p}{dt} = \frac{d (\gamma m \vec v)}{dt} = \gamma^3 m \vec a_{//}+\gamma m \vec a_{\perp}$$ Where $\vec a_{//}$ is the component of the acceleration which is parallel to $\vec v$ and $\vec ...


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Skipping the normal caveats about the obsolescence of relativistic mass, $F = ma$ still doesn't work. The more proper expression for force is the rate of change of momentum. So, \begin{array}{ll} F &= \frac{dp}{dt} \\ &= \frac{d(\gamma m_0v)}{dt} \\ &= m_0\left(\gamma\frac{dv}{dt} + v\frac{d\gamma}{dt} \right)\\ &= m_0\left(\gamma a + ...


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No, because $$F = \frac{dp}{dt}$$ (p is momentum) $$\rightarrow F = \frac{d \ m \times v}{dt}$$ $$\rightarrow F = \frac{dm}{dt} + \frac{dv}{dt}$$ cross-product rule $$\rightarrow F = \frac{dm}{dt} + \frac{dv}{dt}$$ $$\rightarrow F = \frac{d\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}}{dt} + \frac{dv}{dt}$$ which gives the correct relation between Force, mass and ...


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Negative energy or mass is not forbidden in Relativity, but gravity is not a force but geometry, so if you have a negative mass it would repell positive mass as well as negative mass, just like positive mass would attract negative and positive mass all together. If you place a positive and a negative mass near each other the positive mass would attract the ...


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If a black hole travels from point A to point B, long distance at the speed of light you can say it's a wormhole.


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This is a quite subtle problem. You have to be careful about three different situations. A ball can be thrown with velocity (relative to the ground): a) $v_0-v_e$. b) $v(t)-v_e$, where $v(t)$ is the velocity of the car just after the ball is thrown. c) $v(t)-v_e$, where $v(t)$ is the velocity of the car just before the ball is thrown. You actually stated ...


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All of your math is correct, but your result conflicts with our intuition as well as the reality of how rockets work. Here is why: You assume that "The ball will be moving at $v_\mathrm{ball}=v_0-v_e$". This approximation is valid only in the limit where $m$ is much smaller than $M$. (i.e. the limit of a continuous stream of tiny balls, like a normal ...


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In a vacuum the rod will keep its orientation because of the equivalence principle, but with air resistance the heavier part will tilt toward the ground because its mass and therefore force is higher. The force depends not only on the mass of the planet but also on the mass of the falling object, and since air resistance is also a force that acts on the ...


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Depends in the context you are considering. There are two aspects to think about: the theoretical definition, which depends on a model, an image of reality the operational definition: the way you measure it. Rest mass is a concept from Relativity, which is a part of classical physics. Here, we can think of particles as points of mass, with no wave ...


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UPDATED ANSWER : The centre of gravity will always be the same as the centre of mass in a uniform gravitational field (constant in magnitude and direction). This applies for bodies with non-uniform density as well as those with uniform density. The Earth's gravitational field can be considered uniform if the dimensions of the object are much smaller than ...


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You are asking about micro black holes. Some hypotheses involving additional space dimensions predict that micro black holes could be formed at energies as low as the TeV range, which are available in particle accelerators such as the LHC (Large Hadron Collider). Popular concerns have then been raised over end-of-the-world scenarios (see Safety of ...


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The limit c comes from Maxwell's equations for electromagnetic waves in vacuum. This does not know about photons, is pure mathematics given E and B fields and permittivity and permeability of free space. $$c= \frac1{\sqrt{\mu_0\varepsilon_0}}= 2.99792458\times 10^8~\mathrm{ms^{-1}}\;.$$ If photons do have a mass within the experimental limit, Quantum ...


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Komar mass is associated to one asymptotically flat end. So a wormhole has two Komar mass, one for each side, and in principle they can even be different! Indeed the simplest wormhole you can imagine is simply two copies schwarzschild spacetimes glued together at the would-be horizon. As an aside, even time can run differently on the two sides (often ...


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If you pay careful attention your notice that cosmic ray passing the earth just now. You are moving at 99.999%$c$ relative that particle. In fact the whole planet is doing so. From which we conclude that you don't even need a lot of money to make people go close to $c$, you just need the appropriate frame of reference. Now, I know perfectly well that that ...


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There is only one real fundamental speed limit, $c$. Historically, we first discovered it in the context of light, so we call it the speed of light. In reality, it has a little bit deeper significance--it turns out all massless particles travel at $c$ and only at $c$, at least in vacuum. A massive particle can theoretically travel at any speed less than ...


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As far as we know the classical (i.e. non-quantum) laws of gravity apply at all length scales. There are theoretical reasons to suppose that the classical description fails at scales approaching a Planck length, but this is far, far smaller than the size of a neutron. So inside a neutron we would expect the classical laws of gravity to apply, and in ...


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If we did have decay of massless particles then there'd be no reason a photon wouldn't decay from an energy of $h\nu$ into two photons of energy $h\nu/2$. Eventually all our photons would be red-shifted into oblivion and we'd have no light left in the universe.


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One way to establish that $r$ in Schwarzschild coordinates is equivalent to the spherical radial coordinate is its asymptotic behaviour, namely for $r\to\infty$ the metric tends to Minkowski, and the weak field approximation yields Newton's gravitational potential. One way to establish that the parameter $M$ is indeed the mass of the spacetime is to ...


0

Well, in the NJL model you get rid of the gluon. They are considered to be frozen in the low energy limit where you are working because the mass is higher than the energy. Thus you are only working with quarks, and you consider interaction between quarks via effective coupling constants. But in the standard model, gluons (seem to) acquire an effective mass ...


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Yes the human body has a gravitational field, and yes it's large enough to be measured experimentally (see the Cavendish experiment).



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