New answers tagged

3

If a body is moving, this doesn't mean that a net force certainly must be exerted on it. It can move without any net force (First Law of Newton). You might say "How that body has started its motion without any net fore?" The answer is: "Equations of motion are moment equations, i.e. they are stated for moments not for a time interval ($F(t_0)=ma(t_0)$). So ...


2

You are correct in your definition of force. A car, not accelerating, has zero net force associated with it. However, if the car were to hit something--let's say it's me standing in the middle of the street--it would exert a net force on me, and by Newton's Third Law experience a net force equal in magnitude and opposite in direction. So how can an object ...


5

You are correct. The car has no net force on its environment, and the environment has no net force on the car. This is true of any object traveling with a constant velocity. This is even true in the vertical direction. There is a force of gravity pulling down on the car, and there is a force caused by the road pushing up on the car. If the car is not ...


-1

If the object, when at rest has a mass of $m_0$, then if its moving at a speed $v$, the relativistic mass will be given by $m=\gamma m_0$, where $\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$. Therefore, the way the mass changes depend on the way the speed of the object changes. If the object is experiencing a constant linear force, $f$ it means that its rate ...


0

The is no absolute rest frame as far as we know (and we know quite a bit about it). So there is no such thing as completely still in spacetime. However, that's boring and I'm in the mood for a debate, so I'm going to phrase this answer in the form of a debate where I will be arguing on your behalf. Me: The is no absolute rest frame, so this question makes ...


0

Your mass density must have units kg/m${}^3$. The simplest way to achieve this is to write the density $$\rho({\bf r}) = \frac{M}{V}\left(\frac{r}{r_0}\right)^2$$ where $r_0$ is a constant having units of distance, and $M$ is the total mass, and $V$ is the volume of the cylinder. We choose $r_0$ by integrating to get the total mass $$ M = \int \rho({\bf r}...


1

What you want is $\rho(\textbf{r})=Cr^2$, where $C$ is a constant. Find $C$ by integrating this function over cylinder volume, and equating the result to total mass $m$.


0

The energy momentum equation for a particle of mass $m$, energy $E$ and momentum $p$ is: $$ E^2=p^2c^2+m^2c^4 $$ so if $m=0$ then $E = pc$. For a particle of mass $m$ moving with velocity $v$ we have the standard relations $E=\gamma mc^2$ and $p=\gamma mv$. The above equations imply $v=𝑐$. However, as that implies $\gamma=\infty$ which implies $𝐸=\infty$ ...


0

There is quite a bit of ambiguity in the question(s), so let me start by substituting electro-magnetic (EM) wave for "light." Then, the "universal speed limit," is the speed at which EM waves propagate in "space." The reason I use space (not vacuum), is because it is the characteristics of space ($u_o, \epsilon_o$) that determine the speed of propagation of ...


0

Let's assume that the wheels aren't able to turn. That means the frictional coefficient is kinetic in nature because the car is skidding across the surface...assuming you already overcame static friction. The frictional force is as follows: $$f=u_{k}*m*g$$ You apply a force that opposes and overcomes the kinetic frictional force that is resisting your ...


2

The mass of an object in a physics problem doesn't matter to that objects behavior when all the forces that act on it are fractions of the objects weight so that the acceleration has the from $$\vec{a} = \frac{\vec{F}}{m} = \frac{\vec{k}m}{m} = \vec{k} \,,$$ for some vector $k$. This is true of idealized projectiles and idealized objects siding down ramps ...


0

Drop a piece of paper and it glides sideways as well as flips. So aerodynamics (and hence the shape) affect the way things fall. Specifically aerodynamic forces have a center of pressure, which when ahead of the center of mass the body would rotate and flip, but if behind it will swing and stabilize at this orientation. This is the reason arrows, darts and ...


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 ...


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 _{...


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/


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 ...


0

In any theory with chiral fermions where Left and Right fermions don't have the same charges such fermions cannot have an explicit mass therm and can acquire mass only with spontaneous symmetry breaking. In the standard Model all fermions except Right handed neutrinos are charged under the SM gauge group and all L/R counterparts live in different ...


-1

The least you may require of something to be able to call it a particle is that it can be established where it is when, that it interacts, however weakly, with the objects in its environment. Since according to relativity theory a massless particle must move at the speed of light and at that speed there passes no time at all, the particle -its state- is ...


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 ...


1

This definition isn't obsolete, by definition, because we are still using it. There are, however, efforts underway to redefine the kilogram. One proposed definition is based on the watt balance, basically defining the kilogram in terms of the electrical units volt and ampere. Another proposed definition, called the Avogadro Project, is in terms of the mass ...


-2

I think inertia doesn't depend on speed, it depends on rate of change in speed, i.e. acceleration. The higher you accelerate the more will be the inertia. It can be understood by taking an example of a motorcycle, in which lower gear gives more traction than the higher one. The higher the acceleration you want the more traction is required due to inertia. ...


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 ...


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}\...


1

Let's try to make the answer as simple as possible. Static weight is written as $w=mg$.(Notice Newtons 2nd law looks similar $F=ma$.) This is $$\mathrm{mass} \times \mathrm{one\ unit\ of\ gravity}$$. On earth, $g=9.8\frac{m}{s^2}$. On a smaller planet, $g$ is less. On a larger planet, $g$ is more. It is important to understand that despite acceleration in ...


1

The quantity $T^{\mu\nu} T_{\mu\nu}$ appears in the TOSEC (trace of square energy condition). The quantity can become negative, for instance in the stress energy tensor of the Unruh vacuum in the Schwarzschild metric, hence I would not recommend taking its square root.


1

Short answer is we don't know, there may be right handed neutrinos and left handed antineutrinos that either don't interact through the weak force or do so extremely less than neutrinos. Sterile neutrinos a theoretical candidate for these particles.


0

Well, I'm no expert, but I think part of the problem is that the definition of a galaxy is somewhat shaky. This article (which references this paper) has the definition, "A galaxy is a gravitationally bound collection of stars whose properties cannot be explained by a combination of baryons and Newton’s laws of gravity." So I suppose that we can't ...


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/...


0

Consider it in condensed matter physics framework: we all know that around Dirac point, an electron can be considered as massless; however, if we turn on some kind of interactions, say spin orbital interaction, there would be no more linear dispersion, which give the particle somehow a mass.


0

The mass of a quantum particle is derived from the degree to which it is "linked" to the Higgs Field (a job done by the exchange of Higgs Bosons). For instance, a photon has no rest mass as it has no amount of link to the field. In order to "assign" mass to a particle, you would have to increase how much linkage there is between it and the Higgs Field ...


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 \...


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 ...


-10

Articles published in Science and Nature say the speed of light is not constant: http://science.sciencemag.org/content/347/6224/857 "Spatially structured photons that travel in free space slower than the speed of light" Science 20 Feb 2015: Vol. 347, Issue 6224, pp. 857-860 http://www.nature.com/nature/journal/v406/n6793/full/406277a0.html Nature 406, ...


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.


0

The numerical value of $c$ does not have any fundamental significance. Rather it is the number we get based on the experimental fact (according to the number & unit system employed) . If some alien civilization ended with some different value of $c$ compared to us. Even that is not a problem. They will reach the conclusion that this is upper bound 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 ...


0

The equation is built up from considering considering the mass $M$ moving with velocity $v$ splitting into $M+\delta M$ moving at ${\bf v}~+~\delta\bf v$ and a smaller piece with mass $\delta M$ moving with velocity ${\bf v}-\bf V$, for $\bf V$ the velocity of hot gases or plume from the rocket. The diagram below illustrates this Conservation of momentum ...


1

I think both of the pre-existing answer violate (or at least play with the idea of violating) general relativity. General relativity is in a nuthell that gravity and accerlation cannot be distiguished. Just from that principle, we can solve this entire problem! I think the situation is quite simple afterall. The photon will bounce up and down, but ...


0

The original form of the equivalence principle(strong form) is: In a region of gravitational field, for every spacetime point there exists a non-inertial coordinate system where the effect of gravitational field can be nullified in the sense that the laws of nature(laws of special relativity) take the same form as in unaccelerated Cartesian coordinate ...


0

To measure anything, first you need a reference. For example, suppose you want to measure length in terms of meter. There is a particular length which is accepted by everyone as a meter. You say something is so many meters long by calculating how many meter-lengths fit into it, i.e. you take a ratio of length to be measured to standard-meter and express the ...


-1

They are, as Einstein pointed out, equivalent. So why distinguish between the two? Well the only real difference is that they are measured differently. To measure inertial mass, we exert a given force to something with an unknown mass. To measure gravitational mass we compare the force of gravity from an object with an unknown mass to the gravitational force ...


0

Gravity causes acceleration, but acceleration can happen from a lot of other things as well, for example, on electromagnetic effects. In most cases, the acceleration depends on some charge-like quantity. For example, a body with a mass of 1kg and with a charge of 1C will accelerate faster in the same electric field, as a body with 2kg of mass and with the ...


1

This is one of those cases where you need to distinguish mass and weight. Mass (equivalently energy, by the famous $E=mc^2$) is defined in two ways. There's inertial mass, which relates the acceleration of a body to the force applied ($F=ma$), and there's the gravitational mass, which scales the gravitational force exerted by a body ($F=GMm/r^2$). The ...


1

The problem here is that the intuition about fields is different than for particles. And light and particles are pretty similar in that they behave like both or either one on occasion. Regarding the light as a stream of pretty localized photons, I would say the following: The light gives the flashlight a bit of recoil when it leaves. Say the flashlight lies ...


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 ...


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 ...


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 ...


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 ...


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 ...


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, ...



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