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12

The rubber-sheet analogy is often used to "explain" the basics of GR to beginners, but actually it has nothing to do with real gravity. It acts much more like a scalar field (the up/down freedom degree) - and there were several attempts to build a scalar gravity. But the correct description turned out to be tensorial and purely geometrical. GR has 10 ...


11

Estimating the mass of a "single star" can be a very difficult task, though perhaps your question is too pessimistic. There are a number of suggested relationships linking the mass of a star to its luminosity. These can of course come from stellar evolution models, but they can also be empirically calibrated using stars in resolved binary systems of known ...


11

In quantum mechanics a particle can be treated as a wave and a wave can be treated as a particle. This is the notorious wave particle duality. I won't go into this any further here because it's been discussed to death in lots of previous questions. Search this site for wave particle duality if you're interested in finding out more. Anyhow, assuming I ...


9

Gluons and photons are similar in the sense that they are both massless gauge bosons. They do, however, correspond to different gauge symmetries: photons arise due to $\mathrm{U}(1)$ symmetry, while gluons follow from $\mathrm{SU}(3)$. This leads to a different number of particles: there is only one photon, while there are eight different gluons, ...


9

The source of gravity is not mass, but stress-energy-momentum, so you are correct that the energy converted in this process already has gravity and that that gravity is only rearranged The change in the gravitational field needs time to propagate, though, and this does indeed happen at the speed of light.


8

I) At the perturbative/diagrammatic level of photon self-energy/vacuum-polarization $\Pi^{\mu\nu}$ , the photon masslessness is protected by the Ward identity, which in turn is a consequence of - you guessed it - gauge invariance. For the explanation in the setting of QED, see e.g. Ref. 1. Fig. 1: A one-loop contribution to the photon ...


7

A binary star system, which is quite common, will allow us to determine mass with great accuracy, using Kepler's Third Law of Planetary Motion which is as follows: $$ \frac{T^2}{r^3} = \frac{4 \pi^2}{GM_{sum}} $$ Using the orbital period, $T$, we can determine the acceleration and affect that stars have on one another to determine mass. Once we have the ...


7

Relativistic mass is obsolete. See Why is there a controversy on whether mass increases with speed? . Therefore this is not a question that modern physicists would consider of interest. Furthermore, the usual motivation given for using the relativistic mass convention is that it lets you use Newton's second law without modification, but Newton's second law ...


7

No, you don't need to account for the mass of humans. We're a part of the total mass. Children who grow and gain weight are merely transferring mass from other parts of the Earth to them. The mass of humanity versus the mass of everything else on the Earth is a zero sum game. To see whether the Earth is changing mass, you need to look outside the box (or in ...


6

Yes you need to include the mass of the 7 billion or so humans in the total mass of the Earth, though since their total mass is something like $5 \times 10^{11}$kg and the mass of the Earth is around $6 \times 10^{24}$kg humans make up only about 0.00000000001% of the total mass. We don't make any great contribution to the escape velocity. The point of your ...


6

From the perspective of fundamental quantum field theory, gluons and photons are quite similar. Both of them are gauge bosons, meaning that their existence is required by a mathematical mechanism called local gauge invariance. However, as particles, there isn't any particular connection between them. For instance, there's no reason they both have to be ...


6

We use the term mass, when we mean the mass of a weight, and we use the term weight, when we mean the weight of a mass. :-) The important thing to remember is, that the mass is the same everywhere, while the weight varies with the local gravity. So if you are referring to the constant mass of an object, you use mass expressed in kg. If, however, you mean ...


5

Your position that mass measurement is done by measuring gravitational force is not quite correct. A balance measures the mass of an object by comparing the force of gravity on the mass in question to the force of gravity exerted on a reference mass. In some cases there is also a factor based on the geometry of the scale. The measurement is based on the ...


5

There is approximately 7 billion people on earth. If we take some average of about 75 kg each, that means $$ m_{\rm population}=7\cdot10^9\times75\,{\rm kg}=5.25\cdot10^{11}\,{\rm kg} $$ The mass of the earth is roughly $$ m_{\rm earth}=5.97\cdot10^{24}\,{\rm kg} $$ which is about 13 orders of magnitude bigger than the total population of the world. So no, ...


5

To add to Hindsight's great answer: one of the reasons that the analogy fails is the same reason why Nordström's Scalar Theory of Gravitation fails: Waves on rubber sheets are described by linear wave equations; at least in the small amplitude limit. However, by analogy with Maxwell's equations, waves in gravitation should bear energy. But we are also ...


4

If you care about the inertia you use "mass". When you are considering the force of gravity you use "weight". So when you do calculations about the force in the cables due to acceleration of an elevator car you need to know both it's mass and it's weight... The (calibrated) object you place on a scale is called a "weight" - because that is the property you ...


4

Comment to the question (v4): It seems relevant to mention that there in principle could be a difference between the universal speed limit constant $c$ (which is usually casually referred to as the speed of light in vacuum), and the actual speed of light in vacuum if the photon has a rest mass, at least from an experimental point of view. Of course, no ...


3

If rest mass does not change with v then why is infinite energy required to accelerate an object to the speed of light? The momentum of a material particle, a conserved quantity, is theoretically and experimentally a non-linear function of velocity given by $$\vec p = m \frac{\vec v}{\sqrt{1 - \frac{v^2}{c^2}}}$$ which goes to infinity as $v ...


3

The problem is that while mass is the same everywhere on earth, weight is not - it can vary as much as 0.7% from the North Pole (heavy) to the mountains of Peru (light). This is in part caused by the rotation of the earth, and in part by the fact that it is not (quite) a sphere. When you are interested in "how much" of something there is - say, a bag of ...


3

I'm not sure whether these theoretical ideas are is included in what you have in mind. They are only good (and the first , as far as I know, only in theory) for fundamental particles and not for measuring masses of everyday things, but here goes. The second - inference from cross coupling co-efficient between otherwise dispersionless, massless states - is ...


3

Newton law of gravitation is given by: $$F = G \frac{m_1 m_2}{r^2}=\left(G \frac{m_1}{r^2}\right) m_2$$ The gravitational constant, $G$, the weight of Earth, $m_1$, and the radius are constants, so: $$G \frac {m_1}{r^2}=(6.6742 \times 10^{-11}) \frac{5.9736 \times 10^{24}}{(6.37101 \times 10^6)^2}=9.822$$ Hence, the equation simplifies to $$F =(9.822) ...


3

This is a hypothetical question, since electrons are elementary particles and protons are composite. The solutions of the potential problem would give stable orbitals with smaller average radii. Here is a Bohr model solution for the muonic hydrogen, where the muon is 200 times heavier than the elecron. The energies become KeV instead of eV. To go to the ...


2

Technically the electron and proton are both orbiting the barycenter of the system, both in classical and quantum mechanics, just as in gravitational systems. You find the same dynamics for the system if you assume the proton and electron are moving independently about the barycenter, or if you convert to a one-body problem of a single "particle" with the ...


2

There are two issues at play here. If you lift something upward at constant speed, then the acceleration $\vec a$ is zero. This means that the net force $\vec F_\text{net}$ is zero by Newton's second law ($\vec{F}_\text{net}=m\vec{a}$). As long as something moves with constant velocity, all of the forces add up to zero (i.e., they cancel out). Yes, that is ...


1

The short answer is: protons are much more (1800 times) massive than electrons. That makes them (approximately) the center of mass of the system, that's why electrons are the ones orbiting protons and not vice versa. The term 'orbiting', however, means something essentially quantum. It is the reason of the stability of the atom (electrons don't radiate ...


1

I think that the answer should be searched in surface metal (dielectric)free electrons. When an electron start to move it became heavier or easyer throgh absorbtion or radiation of some Planck energy.The change in speed is : speed is increasing if electron go through band gap by absorbtion.(Fourier Generalised Tf. and antenna radiation) In orbital valent ...


1

In the equation $F_{net}=ma$, normally we would assume that $F_{net}=0$ implies $a=0$ on the right-hand side. However, for a massless object, we can satisfy the equation by having $F_{net}=0$, $m=0$, and $a\ne0$. In reality, of course, the pulley is not massless, so $m$ is small, $a$ is some nonzero number, and $F_{net}$ is small. The above reasoning is the ...


1

In relativity the rest mass is the mass of an object measured from a reference frame in which it is at rest. But this is not the mass involved in acceleration or inertial mass. Inertial mass, or the opposition of the body to the change of movement (directional or in magnitude), will grow with the speed of the body: $$m = \frac{m_o}{\sqrt{1-v^2/c^2}}$$ ...


1

See the wiki page . There is no difference between the two . Weight of a body of mass $M$ is $M.g$ which is equal to gravitational force on the body.


1

To start with, special relativity is necessary when describing elementary particles and hadrons composed by them. Special relativity defines the invariant mass of a particle or an ensemble of particles as the measure of the four momentum vector carried by the particle/ensemble. In the case of one elementary particle, an electron for example, it is ...



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