New answers tagged

5

If you pay careful attention your notice that cosmic ray passing the earth just now. You are moving at 99.999%$c$ relative that particle. In fact the whole planet is doing so. From which we conclude that you don't even need a lot of money to make people go close to $c$, you just need the appropriate frame of reference. Now, I know perfectly well that that ...


1

There is only one real fundamental speed limit, $c$. Historically, we first discovered it in the context of light, so we call it the speed of light. In reality, it has a little bit deeper significance--it turns out all massless particles travel at $c$ and only at $c$, at least in vacuum. A massive particle can theoretically travel at any speed less than ...


2

As far as we know the classical (i.e. non-quantum) laws of gravity apply at all length scales. There are theoretical reasons to suppose that the classical description fails at scales approaching a Planck length, but this is far, far smaller than the size of a neutron. So inside a neutron we would expect the classical laws of gravity to apply, and in ...


-1

If we did have decay of massless particles then there'd be no reason a photon wouldn't decay from an energy of $h\nu$ into two photons of energy $h\nu/2$. Eventually all our photons would be red-shifted into oblivion and we'd have no light left in the universe.


0

One way to establish that $r$ in Schwarzschild coordinates is equivalent to the spherical radial coordinate is its asymptotic behaviour, namely for $r\to\infty$ the metric tends to Minkowski, and the weak field approximation yields Newton's gravitational potential. One way to establish that the parameter $M$ is indeed the mass of the spacetime is to ...


0

Well, in the NJL model you get rid of the gluon. They are considered to be frozen in the low energy limit where you are working because the mass is higher than the energy. Thus you are only working with quarks, and you consider interaction between quarks via effective coupling constants. But in the standard model, gluons (seem to) acquire an effective mass ...


2

Yes the human body has a gravitational field, and yes it's large enough to be measured experimentally (see the Cavendish experiment).


3

See moment of inertia is analogous to mass. Moment of inertia can be thought of as a physical "property" of the object similar to that of mass. And as we know that mass does not depend on any force or gravitational field or any other external effect, so does moment of inertia. Hope this answers your question.


1

If you measure a force (weight) for a given acceleration (gravity) in order to determine the mass of an object and you haven't started measuring then the mass is undefined. As soon as you apply an acceleration $a>0$ and you measure corresponding force $F>0$ you can determine the mass. Equations are useful only when they can be used to measure things, ...


0

It is well understood that if gravity is a 'long range force' its quanta's mass is zero. That is the graviton. Excuse the term in quotation marks, its the easiest way to say what is known. It really means the gravitational field, at infinity in aan asymptotic flat spacetime goes like 1/r But the fact is that the graviton mass could indeed be greater than 0, ...


1

In addition to the already given answears this also might be of interest: When hammer and feather are dropped simultaneously they arrive at the same time, when dropped independently the hammer attracts the planet more than the feather, so you are right, the total time until impact is then smaller for the hammer. If you pick up the hammer and let it fall to ...


1

Many of them are calculated. All the ones that have $u$ or $Da$ as the unit are in atomic mass units, referenced above in the table. They just count up the atoms and add. For bacteria, yeast, and the like it will vary from one specimen to another. Not much precision is quoted and it is likely they use the volume (easy to measure with a photograph) and ...


0

What is more fundamental: Geometry and Topology or physical matter? None of the above. Like CuriousOne said, geometry and topology are mathematical disciplines rather than real objective things. And whilst people talk about electrons and quarks as "fundamental" particles, they aren't really fundamental because after annihilation they aren't there any ...


1

Well when there is no answer available, I do not think it hurts to try to come up with one. As far as existence is concerned, Geometry and Topology can exist without physical matter, but not vice versa. In fact, there is so much empty space without matter, but no matter without being in space. Empty space has some Geometry and Topology. However, to ...


17

The most precise measurement of the mass of an electron was reported by Sturm et al in Nature 506, 467–470 (27 February 2014), quoting a relative precision of $3\times 10^{-11}$, meaning they determined the mass to better than $3\times 10^{-41}~\rm{kg}$. If that is not the best, at least it gives you an upper bound... Note that if you could weigh such a ...


2

Any time you have an equal and opposite force acting on two bodies, the effect is proportional to $\frac{1}{m}$ for each one. The combined effect is proportional to $$\text{(effect)} = \left( \frac{1}{m_1} + \frac{1}{m_2} \right) \text{(action)}$$ When inverted to find which action has a desired effect we get the reduced mass $$ \text{(action)} = \left( ...


1

When you state that the two masses are 'kept at rest' I assume you mean that the pulley is prevented from turning. The load on the pulley is then (M+m)g. [If the pulley is not prevented from turning, then to keep the masses at rest we would have to lift the heavier mass M until it became, effectively, the same weight as the smaller mass. The total load ...


0

The question is given as a practice problem in Poisson's book "Relativists Toolkit" (p. 159, problem #7) along with some helpful hints. The tensor $\gamma_{ab}$ (which is not a tensor at all) must be defined in the following way: Let $h_{ab}$ be the induced 3-metric on the spacelike hypersurface $\Sigma$, and let on its boundary, $\partial \Sigma$, be ...


4

I want to offer a different perspective from the already existing answers, which all seem to somehow refer to the Standard Model or other specific physical theories to say that mass is not an integral multiple of some fundamental mass unit, hence not discretized. The reason why mass is not like that - and can indeed conceivably have continuous values in a ...


0

Just a note, you can rewrite (3) as $$\vec{F} = \frac{d\vec{P}}{dt} = \frac{dm}{dt}\vec{v} + m\frac{d\vec{v}}{dt}$$ If you have a system where the total mass is not changing, then $\frac{dm}{dt}=0$, which brings us back to the form we're more commonly used to.


0

When the blocks are kept at rest , for the vertical equilibrium the pulley has to exert the force equal to the weight of two blocks. But when the heavier block starts moving down the mechanical advantage of pulley with respect to the heavier block increases and hence the weight on the pulley decreases.


5

Newton's 2nd Law of Motion gives the impressed force as $F=dp/dt$, so a physical theory for a massless particle exerting a force requires that the particle have momentum, $p$. First we will discuss mass, momentum, the force law, and Special Relativity. In Newtonian physics mass is identified in two ways: by it's inertia, or as the quantity of matter. The ...


1

One could not defend a book you are most probably misquoting. I strongly suspect the book says "the mass distribution is not constant", that is M is constant but the distribution and number of constituent $m_i$s may vary, i.e. they may split or agregate, a common feature in astrophysics. You are confusing yourself with symbols and definitions and proofs. ...


4

Let us for simplicity work in units where the speed of light $c=1$ is equal to one, and assume that there is no cosmological constant $\Lambda=0$. A spherically symmetric vacuum solution to the EFE of the form $$\tag{1} ds^2~=~g_{tt}(r)dt^2 + g_{rr}(r)dr^2 +r^2 d\Omega^2,$$ and such that it asymtotically becomes Minkowski space $$\tag{2} ...


6

Conserved quantities in GR In GR, energy (or mass) is typically an ill-defined concept. In flat spacetime, we define energy as the conserved quantity corresponding to time translational symmetry. Extending this to GR is quite tricky mainly because, what one is calling time is already observer dependent (this is of course also true in flat spacetime, but at ...


-4

According to the relation $$\mathrm{Force = Mass \times Acceleration}$$ force is defined as the phenomenon, which creates acceleration in a body with mass. The mass is for the body on which it acts and not the body which acts. Theoretically it is possible for light to exert force since it has energy and momentum. Hope your doubt is clear


32

Yes, photons can. See https://en.wikipedia.org/wiki/Radiation_pressure (and photons are certainly massless). PS In fact, any massless particle has momentum (which has a fixed value since they can only travel at the speed of light) and if it is scattered on a body, it changes its own and the body's momentum, which is what a force does.


5

Is there another way to conclude the Schwarzschild solution has a mass M It's not so much a conclusion as a definition. From Schutz in "A first course in general relativity", section 8.4 "Newtonian gravitational fields", pages 207 - 208: Any small body, for example a planet, that falls freely in the relativistic source's gravitational field ...


1

The net force on the string is not F. You are pulling the string forward with force F but I think you are forgetting that the block is pulling the string backwards with a force that is almost equal to F. If the masses of the string and block are m and M then for the whole system (string plus block) F=(M+m)a. The net force on the string is F '=ma=mF/(M+m). ...


0

You wrote it yourself, $F=ma$, so with zero mass there is zero force needed for a finite acceleration. A finite force gives an infinite acceleration. ... Lol, I just reviewed your answer about the electrostatic field being normal on the surface of a conductor. That's exactly the same, here! With a non-zero force the acceleration would be infinite ($F/m$ ...


2

I assume you are familiar with Wigner's classification in d=4, as you might be implying. The m→0 limit is best appreciated on the Poincaré sphere, but let us skip that here to count particle states. So, reviewing Wigner, for a massive state, we can Lorentz-transform the momentum to the rest frame, (m,0,0,0) so the little group is SO(3) and its vector rep ...


0

Q1: The mass splitting he is referring to is between the components of a vector-like quark (VLQ) doublet, $Q=\begin{pmatrix} U \\D \end{pmatrix}$, i.e. between U and D. There is such a splitting because they have different charges, so they get different loop corrections to the tree- level mass term $M_Q \bar{Q}{Q}$. What Sher is saying is simply that we pass ...


1

The density of cotton will be smaller than the density of stone, look at your equation $\rho = \frac{m}{V}$. You have $m = 1 \, \mathrm{kg}$ of cotton and $1 \, \mathrm{kg}$ stone, but the cotton takes up more volume, so $V_\text{cotton} > V_\text{stone}$. (As well, it's wrong to say $1 \, \mathrm{kg}$ of stone is heavier than $1 \, \mathrm{kg}$ of ...


2

If the rocket maintains the same rest mass and also the same thrust in the center of mass system, its terminal velocity can be as close to the speed of light as you wish. It will just take forever. It is interesting to ask when a given velocity is reached. Say the thrust remains the same all the time in a stationary frame. The mass of the rocket will grow ...


0

It depends on what other forces act on the spaceship. If the spaceship starts from rest on the surface of the earth, pointing up, its maximum speed will be 0, no matter how long it thrusts. That's because the force of gravity that pulls the spaceship down to earth is 980 Newtons (read the definition of N in wikipedia), which is greater than the thruster. ...


0

You can indeed balance the horizontal forces at each point, and the sum of vertical components should equal the weight. That does seem to leave you with an over constrained problem (four equations with three unknowns) which will only have a solution when the angles are chosen "just so". If one of the angles was not given you could solve. Pick one and prove ...


0

Edit I was wrong. Look at Floris's answer. If you split it into sections it should be easier than combining it into one. If you have $F_1$ , $F_2$, $F_3$ as the tensions of the strings $$ F_{1x} = F_{2x} $$ $$ F_{2x} = F_{3x} $$ $$ F_{1y} + F_{2y} = 12g $$ $$ F_{2y} + 7g = F_{3y} $$ Then you can use trigonometry and then eliminate. Edit: Eliminate ...


0

Somebody correct me if I'm wrong but I do believe you would have five tensions. There would be the tension in the one rope joining the two ropes holding the masses, the tension in the two ropes from the intersection to the mass, then the tension in the two rope from the top of the diagram to the intersection of all the ropes. The tension in the rope from ...


-1

No, we don't know that for sure about gravity. One object cannot have 2 different masses: The force that makes things roll, and fall, is the same force; it is gravity. This is what my question is about. We don't know for sure that Newton was correct: The fact that something heavier rolls at a higher speed shows us that Newton's theory of gravity is wrong. ...


0

I guess the smallest mass would be the neutrino's, though their mass hasn't technically been determined, but it is thought to be of the order 0.05 eV.


0

Fundamental values are quantised so I believe there is a 'minimum' or 'smallest' mass. If you look at the similarities between fundamental electromagnetic and gravitational interactions then its clear to see the two are very similar, with gravitational interactions being related to Mass and EM related to Charge. There is a minimum quantised Charge (e, or ...


2

In general relativity, the field equation relates the metric (through the associated curvature tensor) to the stress energy tensor $T^{\mu\nu}$. This can be interpreted as a flux of energy and momentum in spacetime (i.e. integrating $T^{\mu\nu}$ over a spacetime hypersurface, like a three dimensional hypersurface of constant time, tells you the rate at which ...


2

Why does a mass attract all the masses around it? Because this is what has been observed . No apples falling up have been observed. Why should't it repel Because no repulsion of masses has been observed up to now. There exist experiments at CERN where the gravitational behavior of antimatter is probed and maybe in the future there will be an ...


-1

In refference to: http://physics.stackexchange.com/a/250821/84895 The reason you are sticking to the floor right now is that the shortest distance between today and tomorrow is through the center of the Earth.


0

BP = 101.325 kPa STP = 101.325 kPa, 0°C NTP = 101.325 kPa, 20°C Molar mass of oxygen = 31.9988 g [1] Molar volume of gaseous oxygen at STP = 22.41 L [5] Molar volume of liquid oxygen at BP = 28.04 cm³ - Molar volume of metallic oxygen = 23.5 cm³ [4] Density of gaseous oxygen at STP = 1,429 g/m³ [2] Density of ...


0

If we work on the theory that over decades we have been able to with the assistance of more unique scopes been able to break down these sub atomic particles then in theory, as we have in discovering quarks, these particles must be made up of even smaller particles. Our perception is only confronted by our current knowledge and capabilities, therefore in say ...


0

If you neglect air resistance the mass of the golf ball does not influence the range with same initial velocity , only with air resistance it does. But if you hit the ball the kinetic energy you transfer gets divided through a smaller mass, so the velocity of the lighter ball should be higher if hit both balls the same way: $$e_{kin}=m\cdot v^2/2 \to ...


0

In this case, it is the momentum that must be considered. Impulse is defined as the change in momentum of an object. The golf ball will always start with 0 momentum. If we assume that all of the clubs momentum is imparted on the ball (unlikely, but simplifies the math), then they both receive the same impulse. The impulse is also equal to force multiplied by ...


3

For Newton's Laws to hold, mass must not vary. Wherever you read otherwise was mistaken. Take a look at this answer. It contains a description of why mass must be constant in Newton's Laws in the context of the rocket equation ... but the analysis applies generally. Newton's Laws are not valid for variable mass systems.


2

The rotation will not necessarily be parallel to the ground. The general motion will be a combination of rotation in a horizontal plane (the conical pendulum) and oscillation in a vertical plane (the simple pendulum). If the support (pivot) is a fixed point, the motion you get depends on the starting conditions. If you launch the mass horizontally at the ...



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