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When we talk about a (non-spinning) black hole we normally mean the spacetime geometry called the Schwarzschild metric. This is one of the simpler solutions to the Einstein equations, but it's important to understand that the solution is based on a number of assumptions and as a result it is only an approximation to a real black hole. The assumption that is ...

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There is the statement in the video, "particles vibrate", and vibrations lead to the concept of frequency. The confusion comes because in fist quantization, the solutions of the Schroedinger and Dirac equations, the wavefunctions have a sinusoidal dependence, which lead to a probability density distribution for the particles, and the de Broglie wavelength ...

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The gauge symmetry group associated to the SM is $SU\left(3\right)_{c}\times SU\left(2\right)_{L}\times U_{Y}\left(1\right)$. Then we can not build the lagrangian of the SM with terms of the form $m\bar{\psi}\psi$ because they are not gauge invariant. A term of this kind mix the right and left handed parts, which transforms differently. In order to give mass ...

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The time dilation factor with respect to an observer at infinity is $$\sqrt{1-\frac{\text{2 G M}}{\text{c}^2\text{ r}}}$$ so if we plug in G=1, c=1, r=10 and M=+1 we get the clocks running slower by a factor of 0.8944 if they are in a distance of 10GM/c² from the center of the positive mass. If we change the sign of M to M=-1 we get a time dilation factor ...

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The arrow of time is believed to be related to the fact that the universe started in a state of low entropy and is evolving towards a state of larger entropy. The effect of negative mass will not change this. The reason is that any model of negative mass will leave the initial state of the universe as as state of low entropy. A rather uniform distribution of ...

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OK, let's accept your conceit. You ran after $\nu_1$, the lightest neutrino, for the sake of argument, and jumped on it. You ran real fast, as your Lorentz factor γ is several millions, (In natural units $\hbar=c=1$, in the lab frame, $E=p+m_1^2/(2p)+...$, not yours). You watch what is happening around you. You do not oscillate to anybody: You are a non-...

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The simplest answer is that the Earth is even MORE massive. Don't let Earth's lack of stature relative to the Jovian systems fool you. The gas giants are truly massive but are not nor do they have physical land and water at their surface. This may seem laughable on its face as what could be more massive than a gaseous body in a liquid state? Nay, veerily....

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Materials that seem homogeneous often have internal strains, or voids, or even inclusions. Under stress, rather than uniform deformation (bending), those flaws may undergo brittle fracture, or stretch excessively, or become chemically active. A cosmic ray can create internal damage, a particle decay track. So, after some kinds of handling (bending, ...

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Actually, the explanation as to why rotation of a mass affects the metric in principle is simple. Rotation means there is angular momentum, and angular momentum contributes to the energy-momentum-stress tensor in general relativity. If this was a nonrelativistic rotation we would say that the rotation carries kinetic energy. The rotation contributes as a ...

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The bottom line is that space and mass do interact with one another. Otherwise, they would not tell one another "how to curve" and "how to move". Therefore a moving (or rotating) mass and space will interact slightly differently. The interaction will drag with rotating mass. It may or may not be detectable depending upon the mass & speed of rotation and ...

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For an indivisible particle, probably the simplest operational definition of "mass" in your context is $\sqrt{E^2 - (p c)^2}/c^2$, where $E$ is its energy and $p$ its momentum. Equivalently, the mass of an indivisible particle at rest is $E/c^2$ in that frame. A lump of gold and a proton are not indivisible particles, because their constituent parts can ...

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Suppose you start with your (stationary) 1kg block of gold. If you raise its temperature you have to add energy to it, and that means it's different after you've raised its temperature. For example you could shine an infrared lamp on it, in which case you've added the energy from the IR lamp. The mass changes because if you add an energy $E$ the mass goes up ...

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The main issue in your question is that it assumes that motion is absolute. Which it is not. As far as I know only photons are considered to have no rest-mass. In common words when it doesn't move it 'disappears'. Wrong conclusion. It just always move. There is no reference frame where a photon does not move at speed $c$. Electrons and quarks ...

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Not surprisingly, physicists have looked for variations in the speed of light as a function of frequency in vacuum. The state of the art in 1972 can be found in Z. Bay and J. A. White, 'Frequency Dependence of the Speed of Light in Space', Phys. Rev. D 5(4) 796-799 (1972). Using data from pulsar emissions (radio, visible, x-ray) and other sources (see paper),...

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Rest mass means the mass which would appear if a paricle were at rest. Do not confuse between particles and photons. These particles are metarialistic particles behaving as energy in some circumstances. Being matter they possess rest mass. While photon is a bundle of energy behaving sometimes as a particle and hence can not possess a rest mass or more ...

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The term rest mass is a poor one because it implies it's the mass measured in the rest frame. But photons have no rest frame, and indeed any particle subject to some form of confinement has a $\Delta p\gt 0$ so its rest frame is somewhat poorly defined. The modern term is invariant mass, which is simply the mass in the equation for the total energy: $$E^2 ... 3 In a vacuum all frequencies and amplitudes of light travel at the same speed of c = 299 792 458 m/s. Frequency is equivalent to colour. Amplitude relates to intensity. When light travels in material mediums (air, water, glass, etc) it travels at a slower speed v < c which depends on frequency. The ratio of c/v is what we measure as the refractive ... 0 There is a clear mistake in what he says : the mass does not come from the interaction with the higgs particles in it self. In essence, for the mass due to the Higgs mechanism, the Lagrangian contains originally massless particles. This is due to the requirement of gauge symmetry. Mass terms are not gauge invariant. The higgs mechanism is a dynamical ... 0 If you have a proton (mass=m_p) and a neutron (mass=m_n) and allow them to join via the strong nuclear force the mass of the ^2_1H nucleus m_{np} is less than the sum of the masses of the individual particles (m_{np} \lt m_p + m_n). During this joining together the proton-neutron system loses potential energy and that energy is called the binding ... -2 Mass certainly will change. In fact we have two connected bodies - brick and earth. The Hamiltonian (full energy operator) of a connected particle system is defined as: H = H(brick) + H(earth) + V, where V is energy of interaction (potential energy in our case). Since potential energy will increase, mass inevitably increases due to the theory of ... 0 No, mass does not increase when you lift an object. Potential energy increases, but that's because of the \Delta h in U_{\mathrm{grav}} = mg\Delta h, not the m. Mass is, to put it a bit imprecisely, the amount of matter in an object. When you lift something, there doesn't become more of it. (Note that I'm ignoring relativistic effects.) 0 In a few words, as Luke Pritchett writes, the Higgs mechanism provides us description of particles mass without breaking of the unitarity, i.e., breaking gauge symmetry explicitly. It is interesting fact that even if You start from electroweak theory in the broken phase and don't know about Higgs boson, W/Z-boson and existense of hidden gauge invariance (i.... 2 Because "spontaneous symmetry breaking" does not actually break any symmetries. This is a pretty important principle that is not always adequately taught. In spontaneous symmetry breaking the symmetry in question is always a full symmetry of the theory. The difference between a spontaneously broken symmetry and an unbroken symmetry is just in how the ... 1 A time dependent coupling would mean that this coupling is in fact a field in its own right. This field would correspond to some new particle that would need to be very heavy, otherwise it would have been detected in experiments directly or indirectly by modifying the way the known standard model particles interact with each other . But if the particle ... 2 No. To make a long story short, if the Higgs field changed its coupling to particles with time then particles in the distant past would have different masses. This would mean atomic spectra of distant galaxies would has differences from spectra now here on Earth. No such change is observed. 0 True weight actually the product of mass and gravitational acceleration which is equal to mg where the apparent weight is the sum of net forces ( when you standing in elevator and elevator is moving either upwards or downwards, either high speed or low speed then you feel your weight heavier or lighter this is the apparent weight that u feels which is equal ... 2 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 ... 1 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 ... 0 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$$An object ... 2 One indication whether the 'inner' balloons are filled with helium or something heavier is whether the 'outer' balloon still floats in air or not. Whether it floats or not depends on whether its overall density is lower than air's density or not. Its overall density d is simply given by:$$d=\frac{\Sigma m}{V} Where $\Sigma m$ is the sum of all the ...

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For the case of a charged hollow sphere the relationship is: $m_{em}=\frac{4}{3}m_0$ because $m_{em}=\frac{4}{3}E_{em}/c^2$ 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 ...

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