# Tag Info

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 ...

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 ...

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 ...

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 ...

0

For example, when we meassure Higgs boson mass to be 125 GeV, do we think about renormalized or pole mass? Pole mass is the physical mass and independent of any renormalization scheme we use to subtract any infinite parts of the loop corrections. It is what we observe. Should the mass of the Higgs change if it is produced at higher energies? So ...

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 ...

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

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 ...

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 ...

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). ...

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. ...

0

Mass is a scalar. Weight is a vector. 100 grams indicates the amount of matter in an object, and is a measure of the object's inertia, its resistance to acceleration. At relativistic speeds, or when dealing with sub-atomic particles, mass also is a measure of the amount of energy in an object. 1 Newton is a measure of the force exerted on an object by a ...

0

The mass is an intrinsic property of an object, weight is the force with which that object pushes the ground (depends not only on the object but on the gravitational force in it's vicinity as well). On the moon the mass of that object will stay the same, but it's new weight will be 1/6 of it's weight on earth because moon's gravity is much weaker.

0

Well, the difference is that weight $W$ is a force and mass $m$ is... a mass. $$W=mg$$ The difference is the coefficient $g$, roughly equal to 10.

0

The answer to the main question is no. One gram of matter (electrons), will annihilate exactly one gram of antimatter (positrons), regardless of the speed with which they approach each other. If they collide with a speed other than zero, the energy due to the motion will also be added to the energy produced by the annihilation.

0

You could use cylindrical hollow capacitor and electrical bridge circuit and measure changes to the capacitance when the metal balls passing through it. As the ball pass through the capacitor, its capacitance will change due to the change in the dielectric properties inside the capacitor. The changes to the capacitance will be a function of the size of the ...

-1

I believe this question is not worded correctly only because the submitter is not that well versed in scientific terminology. To put it briefly - neither sound waves nor light waves have mass, that is because they are representations of moving particles, they themselves are not entities and do not exist beyond the scope of human-facilitating terminology. As ...

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 ...

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" ...

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 ...

33

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 ...

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 ...

0

I think if we consider a much bigger mass then the force becomes relativistically: $$F=-\frac{GMm}{r^2}+\frac{4G^2M^2m}{r^3c^2}$$

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 ...

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.

0

This is really just a comment on Dargscisyhp's post - please don't upvote this. The equation Dargscisyhp has given you is a vector equation i.e. the $\vec{x}_i$s are vectors. You do the calculation by adding vectors. If you're uncomfortable with this then you can find the $(x_c, y_c, z_c)$ coordinates for the centre of mass by using three separate ...

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

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 ...

0

We almost always determine mass by measuring weight. Weight is the force on an object exterted by a gravitational field, and is proportional to the mass. On the Earth's surface W = m*g. We can convert weight to mass if our measuring scale is calibrated, usually with an object of known mass. This would work for any planet. Even though the weight would change ...

0

You measure mass by observing it's acceleration response to force (i.e by applying Newton's second law). Now, because it is impractical to accurately measure straight-line accelerations over a wide range, we actually use periodic motions and measure frequency. Mass-on-a-spring harmonic oscillator. $\omega = \sqrt{\frac{k}{m}}$ with known spring constant. ...

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 ...

0

Brionius dealt with the value of G. Cavendish's experiments also confirmed the product of masses term. The inverse-square portion of Newton's Theory of Universal Gravitation was immediately accepted, since it straightforwardly produces Keppler's Laws. Further thought shows that, in order to produce stable orbits, the exponent must be exactly 2 - no more, no ...

0

This is explained in the Wikipedia article on the Cavendish Experiment. Cavendish measured the gravitational attraction between two known masses using a torsion balance apparatus, which allowed him to calculate $G$ (to a surprising degree of accuracy given that it was done in the 1790s). That measurement did not involve the mass of the Earth. Once $G$ was ...

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 ...

0

It does seem unlikely. Any non-zero mass density in the region inside the boundary will shift the center of mass away from the boundary, so my guess is no.

0

An experimentalist's answer to: What is the difference between the Higgs Boson particle and an electron moving through the Higgs field? Our experiments found a large number of resonances and particles which fitted beautifully into SU(3) representations, separated by their quantum numbers and occupying unique niches in the representations. The ...

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