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A sophisticated, yet easy way to see that this the answer must be "No." is to recall that velocity is relative — that there is no absolute notion of velocity. You said the matter was moving and the antimatter still, but that point of view (AKA frame of reference) is not privileged in any way. An observer at rest with respect to the matter has just as much ...

13

The particle-antiparticle annihilation is on a per-particle basis. One electron annihilates on positron. One up quark annihilates one anti-up quark. One down quark annihilates one anti-down quark. Moving at relativistic speeds doesn't change the number of particles. For that matter, you could annihilate an electron with an anti-muon, since an electron and a ...

11

Each particle only annihilates its exact antiparticle. Electrons annihilate positrons. A blue up quark annihilates an anti-blue anti-up quark. A muon annihilates an anti-muon. The thing about anti-matter is that it postulates an exact opposite of every particular particle type (except for things like photons that are their own antiparticles). It's about ...

7

$\vec{F} = m\vec{a}$ means that an object with a force $\vec{F}$ exerted upon it accelerates by an amount $\vec{a}$, not that an object accelerating with $\vec{a}$ exerts a force $\vec{F}$ on something else. Typically, the force exerted by an object has nothing to do with Newton's second law, but is given by other laws (like Coulomb's law in electrostatics). ...

6

Your question asks why the "current quark masses" [see http://pdg.lbl.gov/2011/download/rpp-2010-booklet.pdf at page 21] of the quarks that make up a proton don't add up to the mass of the proton. The problem is that, for the light quarks, the "current quark masses" are very different from the "constituent quark masses" [see wikipedia]. "Constituent quark ...

3

The lower mass limits of neutrinos is not 0eV. They have to have a mass, since we can observe neutrino oscillations. This is something the standard model did not get right. Looking at PDG (2014) the boundaries seem to be 0 < m < 2ev.

2

The main force keeping the electrons around the nucleus is the electromagnetic one. Electrons do interact gravitationally and weakly but those are very much weaker forces. In principle if the masses were exactly inverted, it would just change the definition of positive and negative charge, which is arbitrary. Generally changes in mass affect the orbitals. ...

2

From "what is a kettlebell" website: So just what is a kettlebell? A kettlebell is a cast iron ball with a handle attached to the top of it (picture a cannonball with a handle on the top). This design makes kettlebells different from training with dumbbells because the weight of a kettlebell is not distributed evenly, thus creating the need to counter ...

2

If there is no resistance then there will be no net torque applied about the center of mass. So any initial rotational speed will remain. The rotation center is going to be the center of mass. The effect of gravity will be to accelerate the center of mass, and it will have no effect on the rotational motion of the body. See this accepted answer for a ...

2

Gravitational force is really weak compared to the other fundamental forces, so it's very difficult to measure the gravitational constant. This is how Cavendish did it without knowing the Earth's mass: He put two lead balls on either end of a long bar. He hung the bar at its center from a long twisted wire with known torque. Then, he placed two really ...

2

I would maintain that we most often explicitly measure mass by comparing the pull of the local (unknown) gravity on the mass to be measured to the pull of the same gravity on a known, reference mass or masses. If you look at the measuring devices in stores, you sometimes see the slogan, "Honest weight; no springs!" They are proudly claiming that they are ...

2

$$\vec{X}_{cm} = \frac{1}{M} \sum_i m_i \vec{x}_i$$ where $m_i$ is your mass of particle $i$, $\vec{x}_i$ is the position of particle i, and $M$ is your total mass.

2

$E=hc/\lambda$ only works for photons, and $E = mc^2$ only works for stationary objects. For moving objects, we use the energy-momentum equation.

2

The equivalence principle tells us that energy and mass are really just two sides of the same coin, and are related by $E = m c^2$. Rearranging, we get that $m = E/c^2$, so instead of asking where all that mass comes from, let's ask where all that energy comes from. In the case of the proton, there are some quarks and gluons that make it up, and those ...

2

$p^2 = m^2$ is the definition (up to a minus sign) of the mass of a momentum eigenstate. He derived that the same quantity (the expectation value of $[P^2,D]$ w.r.t. $\ket{p}$) equals $0$ and $2\mathrm{i}m^2$, so $m^2 = 0$. The $s$ is the scale parameter of the scale transformation induced by $D$, and it is any real number, so, starting from a given state ...

2

First, as has been said in the comments, this has nothing to do with General Relativity per se and can be perfectly explained within Newtonian gravity. The answer is yes, depending on what you mean by weight, since, after all, the building will pull you to the side. Weight is a force and forces are vectors; in this case, your weight will be longer and ...

2

Although an object that moves with constant velocity has no acceleration, it has kinetic energy and it has momentum. Acceleration is not a conserved quantity. It is not passed from one object to another. Momentum and energy, however, are conserved quantities that pass from one object to another. If a moving object hits a target, kinetic energy will be ...

1

2 points: first, $F=ma$ describes the acceleration of an object due to the sum of all forces acting on the object. If these forces are in different directions, they may partly or fully cancel each other out. In the case where the object is not accelerating (so it's moving with constant speed in a constant direction, or it's not moving at all), the sum of ...

1

Assuming we can treat the air in the room as an ideal gas, it will obey the ideal gas equation of state: $$PV = nRT \tag{1}$$ where $n$ is the number of moles of the gas. The question tells us that the pressure is constant, and obviously the volume of the room is constant, so the only things that can vary are $T$ and $n$. The question tells us that $T$ ...

1

The three quarks you talk about are usually called the valence quarks of the proton, and their contribution to the mass of the proton is not it. In particle accelerators, when we hit protons with high energy beams, we discover that protons are made of a cluster of smaller constituents (like quarks and gluons, which constantly are created and destroyed in ...

1

$E=E_1 E_2$ (I am aware that when you decompose energy into two multipliers the units will be different but in purely mathematical sense there should be a way of doing it) Unfortunately, the fact that the unit $\rm J$ is not equal to the unit $\rm J^2$ is more than an inconvenience - it is a fatal flaw in your mathematics, and everything after this ...

1

The observable effects of the gluon field are in the nuclear binding energy curve. The spill over gluon field is what gives the strong nuclear force the attraction that binds protons and neutrons into nuclei. We use this in our every day life through the electricity provided by fission reactions, and shall be using it in the future through fusion reactors. ...

1

Every particle needs to have energy to be a particle (if it had none it wouldn't even exist). Since energy is equivalent to mass and therefore gravitates I would say YES, all particles that have a speed less than the speed of light must also have mass. Because the speed of the particle is less than the speed of light an observer could travel with the same ...

1

First, we will look at the energy of a free relativistic particle of (rest) mass $m$ moving with velocity $v$: $$E = \frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}$$ where $E=mc^2$ when $v=0$. We now consider a few cases: $m\ne0$: In this case, $E\rightarrow\infty$ as $v\rightarrow c$. Therefore, a massive particle that at any point of time is moving at less than ...

1

You dont know the mass of earth, but you know the force earth apply to you. This is F=mg, where g is 9.8 m/seg^2 Buut this is equal to the gravitation force, F=GmM/r^2. G is the gravitation constant, M the mass of earth, m your mass and r the radio of earth. Cavendish was who messure the earth mass. In that time, Cavendish knew the radio of earth and G (he ...

1

A proton is described as a combination of three valence quarks, each with a bayrion number of 1/3 and a charge that adds up to the +1 of the proton, two up and one down. That is a primary constraint from data. Now in QCD, the theory we have developed to describe the strong interactions of quarks, it comes about that overall, in the constraining "bag" ...

1

When considering relativistic speeds, the notion of "particle & anti-particle" somewhat blurs. The correct treatment of a relativistic free electron for example is given by the Dirac Equation, which relates Dirac Spinors. A spinor is something like a 4-vector, describing the wave function of our electron. In it's rest frame, two of the spinor's ...

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