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48

Here is a quick & simple answer until professionals arrive. On the Standard Model, it is zero. This $< 1.10^{-18} \frac{\mathrm{eV}}{c^2}$ is an experimental upper limit (i.e. if it has a rest mass, because of physics beyond the Standard Model, it must be smaller as this value). This value is very small, compare to the estimated rest mass of the ...


47

It's the second one: the reason the speed $299792458\ \mathrm{m/s} = c$ is special is because it's the universal speed limit. Light always travels at the speed $c$, whatever that limit may be. The reason there is a "universal speed limit" at all has to do with the structure of spacetime. Even in a universe without light, that speed limit would still be ...


15

We can't measure to infinite precision; so even if a particle had in fact zero mass we couldn't experimentally measure it to the infinite precision needed to justify this; which is why certain amount of judgement is called for, and that judgement is made in the context of a theoretical framework. The second point to make is that all particles with zero rest ...


11

Massless particles with spin do not have a "$S_z = 0$" state because they actually do not have spin like massive particles do. They have helicity, which is the value of the projection of the spin operator onto the momentum operator. The reason for this is the representation theory of the group of spacetime symmetry, the Poincaré group. To understand this, ...


10

There are indeed massless particles. As of 2015 there were two known massless particles (both gauge bosons): the photon (carrier of electromagnetism) and the gluon (carrier of the strong force). It should be noted, however, that gluons are never observed as free particles, since they are confined within hadrons. Gravitons (if discovered) would be another ...


8

Even if nothing propagated at the speed $c$, it would still be a universal speed limit, and we could still measure it. In fact, it's not impossible that light has a (very tiny) mass in reality. If it does, that wouldn't change anything about special relativity. It would make teaching it even more of a nightmare than it already is, because we'd have to deal ...


6

Above all, speed of light is the speed of propagation of fields through space. While light may be slowed down when crossing matter, fields (electromagnetic fields, gravity) are always propagated at c. One of the consequences is the "speed limit for causality" mentioned by DavidZ and the speed limit for transmission of information.


5

The gravitational mass, $m_g$, gives you the strength of the gravitational interaction while the inertial mass, $m_i$, represents the inertia of the body. The first one is the mass appearing in the Universal Gravitation Law while the second one is the mass appearing in the Newton's second law. The equality between these masses is an empirical fact noticed ...


5

The time dilation caused by the entire mass of the Earth was immeasurable until the invention of the atomic clock. The time dilation caused by a $1$ kg mass remains immeasurable today, and the time dilation caused by an atom of $^{12}$C weighing roughly $2 \times 10^{-26}$ kg is never going to be measurable. That's why we don't measure mass in time dilation ...


5

It is not possible to fairly explain the origin of masses – and most other things – without "any mathematics" at all. Charged fermions get masses through their cubic "Yukawa" interactions with the Higgs $$ L_{Yuk} = y\cdot h \psi_L \psi_R $$ When the Higgs field is $h=v$ in the vacuum, the simple part of this cubic term generates the quadratic term $$ m \...


4

In Quantum Field Theory the one particle states are defined as the states of an irreducible unitary representation of the Poincare Group. If this was not true, there would be states of a reducible representation that would not be connected by a Poincare transformation. These states are rather different particles. The Casimirs If we have an irreducible ...


4

In general, yes you need to know the orbital inclination angle $i$ in order to fully solve the orbit. The radial velocity amplitude $K$ is just modified to $K \sin i$ (where $i=0$ is a face-on orbit). Combining this with the orbital period and Keplerian orbits gives you the "mass function" $$ \frac{M_1^3 \sin^3 i}{\left(M_1 + M_2\right)^2} = \frac{K_{2}^3 \...


4

It is an interesting idea and in principle you could indeed measure mass through the time dilation caused by the curvature of space-time. Experiments to measure the time dilation caused by large masses (the Earth) have also already been performed. But the focus of such a measurement is not the determination of the mass, but simply the confirmation of the ...


4

You need to research the mechanics of drag and the Drag Co-efficient (start with this wiki page). A simple model (where drag is a Ram Pressure) holds the drag to be proportional to the square of the speed. This can be justified on simple momentum conservation grounds, in the case of pure ram pressure. The drag co-efficient is an empirically-found "fudge ...


4

The kilogram is currently defined, as you well note, as exactly the mass of the International Prototype Kilogram, and by definition the mass of the IPK is $1\:\mathrm{kg}$ with zero error. This definition took over from the previous one (the mass of $1\:\mathrm{dm}^3$ of water at $4°\mathrm C$ and sea-level pressure) because the previous one was hard to ...


4

You're both half right. Your teacher is thinking of the invariant mass or rest mass(see footnote [1]), which is thinking in the more modern and much simpler paradigm of naming properties of a particle-like "thing" in a frame of reference at rest relative to that thing and you're thinking of the transformations of the properties to effective values as seen ...


4

No, in our own frame (comoving frame) we are not moving at all. Even though the CMB radiation is defining a preferred frame, to us it looks like the CMB is moving with respect to us - that's why we see a kinematic dipole in the CMB maps: http://scienceblogs.com/startswithabang/2013/06/28/our-great-cosmic-motion/


3

The upper limit you mention reflects the hypothesis that photons in vacuum could have some tiny rest mass. But it seems to be more important that c is the velocity of massless particles such as photons in vacuum. However, there is no real vacuum in the universe: Not only that even in outer space you will always find some interstellar atoms. But also, the ...


3

If you mean the mass of the hadrons, it is because of the energy/mass contributed by the gluons carrying the SNF and virtual quarks and antiquarks, which is high compared the three valance quarks. The valance quarks are the three "permanent" quarks that we think make up the proton and neutron. The Higgs field has a much lower effect on the mass of these ...


3

That light moves with a fixed speed in vacuum, in all reference systems is an experimental fact. Maxwell's equations fit so well all macroscopic electromagnetic data that the speed of light is fixed is not under question. It is inherent in the construction of the classical theory. Light is made up by a zillion of photons. Photons are elementary particles in ...


2

According to quantum electrodynamics, the most accurately verified theory in physics, a photon is a single-particle excitation of the free quantum electromagnetic field. More formally, it is a state of the free electromagnetic field which is an eigenstate of the photon number operator with eigenvalue 1. The single-particle Hilbert space of the photon ...


2

An explicit mass term violates Gauge invariance, because left and right particles belong to different representations 2.At one loop, the lepton mass is given by $m_{1L} = M_{bare} + \Delta M_{1L}(\mu = m_{1L})$ this condition uniquely defines the bare mass. The correction $\Delta M(\mu)$ is anyway proportional to some power of the yukawa, thus is very ...


2

The way to see to this a la Peskin and Schroeder is by staring at the renormalized propagator in equation 7.75, \begin{equation} P _{ \mu \nu } \equiv \frac{ - i g _{ \mu \nu } }{ q ^2 ( 1 - \Pi ( q ^2 ) ) } \end{equation} In $ 4 $ dimensions, $ \Pi ( q ^2 ) $ doesn't contain a pole (as expected) and so, \begin{equation} P _{ \mu \nu } = \frac{ - i g _{...


2

I would call this quantity "first moment of mass" or just "moment of mass". Have a look at this wikipedia article to read about the general concept of moments in physics. As pointed out in the comments to your question, the first moment of mass is closely related to the center of mass (CoM). For a collection of particles with masses $m_i$ and positions $\...


2

The mass of a field is completely well-defined only for free field theories. There it is the mass of the particle whose 1-particle equation is second quantized. For fields of spin $>1/2$, massless is equivalent to the existence of local gauge transformations. Once one adds interactions, the language is kept (based on the noninteracting part) but the ...


2

This is a fairly subjective question. I do not think there is a need for such a constant. Why? First, lets look at... The definition of the mole Taken from Wikipedia: ... is defined as the amount of a chemical substance that contains as many elementary entities, e.g., atoms, molecules, ions, electrons, or photons, as there are atoms in 12 grams of ...


2

In Newtons three laws of motion, its the second law that introduces the concept of mass, here its the linking term between force applied on an object and the motion or acceleration that results. The more 'stuff' there in the object, that is the more mass it has, the harder it is to accelerate it. That is, it has more inertia; so we call this concept of mass, ...


2

Einstein had his "happiest thought" about his equivalence principle one day as he imagined himself in freefall towards the earth. A typical mind wouldve thought "okay, so i feel as though im floating" but Einstein and his fabulous mind thought about it a different way. Einstein connected his freefall with being in motion outside of a gravitational source ...


2

Give it a try. The four momentum of a massive particle is $P~=~(E,~\vec p)$ and the invariant momentum interval is $$ P^2~=~E^2~-~|\vec p|^2c^2. $$ Since E = $\gamma mc^2$ and $\vec p~=~\gamma m\vec v$ then $P^2~=~m^2c^4$ using $\gamma^2~=~1/(1 - v^2/c^2)$. Now suppose that this particle is converted into a photons with zero mass. This means $P~=~(h\nu,~h\nu/...


2

Your want to add a drag term to your force equation. It will be more complicated and involve the geometry of the objects in question. Note also your supposition will not always hold, an adult in a parachute will not fall faster than a child without a parachute. {disclaimer: Not tested empirically!!!!} You will have drag force term like: $$\vec{F}_{drag}\...



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