New answers tagged

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"Could we derive the fact that a falling body has constant acceleration from Galilean relativity and only?" No. However, there's a related fact which is true, which I will explain soon. First, let's define what "Galilean relativity" is. It refers to the fact that the laws of physics are the same for people moving at different velocities. ...


3

A falling body does not have constant acceleration. A body falling toward a planet has an acceleration that is inversely proportional to the square of its distance from the center of the planet, so the acceleration constantly increases as it falls. When a body falls only for a short distance, the distance from the center of the planet changes very little, ...


0

As niels nielsen said, cosmology predicts no such effect. But there's another reason it's unlikely that a universal physical effect was responsible for the mass drift: the drift was only detected by weighing the standard kilogram (IPK) against copies of itself. Many of the copies were made at the same time as the IPK, of the same alloy, so any change in ...


0

No. The expansion of space is actually a pretty tiny effect. It is $73$ km/sec/megaparsec. That is, two points $1$ megaparsec apart are separating by $73$ km/sec. Or, two points $1$ parsec = $3.26$ light years = $3.08 \cdot 10^{13}$ km apart are separating by $73$ mm/sec = $0.16$ mph. Or, two points $1$ km apart are separating by $2.37 \cdot 10^{-12}$ mm/sec ...


1

No it would not, for several reasons. First, the expansion of the universe is manifest only on huge distance scales i.e., the typical distance between galaxies. If you scale it down to human distance scales, it is utterly undetectable. Secondly, it progressively separates objects apart in space but it does not create more atoms inside those objects as it ...


1

$\overline{\Psi} = \Psi^\dagger \gamma^0$ is the definition of $\overline{\Psi}$; this is important because $\Psi^\dagger \Psi$ is not a Lorentz scalar. A change of basis proceeds as usual from elementary linear algebra. Given a COB matrix $N$, spinors transform via $\Psi \mapsto N\Psi$ and matrices via $M \mapsto N M N^{-1}$.


-2

Dark mass is the 'stuff' that fills 'empty' space that is displaced by ordinary matter. Dark mass displaced by the quarks the Earth consists of is the physical manifestation of curved spacetime. The dark mass displaced by a galaxy 'displaces back', causing the stars to orbit the galactic center at the rate in which they do.


1

A consequence of the Heisenberg uncertainty principle is that nothing can ever standstill or not oscillate. Everything has to oscillate. I do not think this is a correct deduction. If one measure momentum there is a limit to the accuracy of measuring position , is all that the HUP says. It is an envelope where measurerements of momentum and position at the ...


0

No. You can't do that because torque and force/weight are different things. Different units. One cannot equal the other anymore than 2 apples equals 3 oranges. The main thing that you need to get clear in your mind is that when lifting something or pushing something horizontally, as soon as you provide enough force to counteract gravity (when lifting) and ...


2

(a) Yes, provided that no other forces act on the mass. (b) "Is the magnitude of the force on a mass due to a spring equal to one value |𝐹|=𝑘𝑥 ? Yes, the magnitude of the force at a particular instant of time is $kx$ in which $x$ is the extension of the spring at that instant. (c) "Or does the spring push with less force as it is released?" ...


2

From The Evolution of Physics: ... all energy resists change of motion; all energy behaves like matter; a piece of iron weighs more when red-hot than when cool ... In answer to your main question, energy from the chemical bonds is converted to energy in the form of heat, so there won't be a change in the mass of the products of combustion and the result ...


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Depends on what your sack manages to capture This was a thing that finally helped kill the phlogiston theory of fire (that burning something means releasing the phlogiston enclosed in it): most things got lighter by burning them, but some got heavier! This could only be explained by having some materials contain negative-mass phlogiston, which pretty much ...


2

Greetings from Mikhail Lomonosov from 18th century. In chemistry (and burning is pretty much a chemistry) mass is considered a conserved property. (It is a conserved property outside of the chemistry as well, but chemists have the luxury not to deal with the heat energy used or produced, because it has a negligible mass in chemical reactions.) If you have a ...


5

Other answers have focused on the combustion products mass, and the mass of oxygen. I want to cover the other part not in those answers. I'm also keeping it very simple, so.bear in mind this isn't how a Chemistry or Physics graduate would describe it. I'm ignoring truly tiny effects that only get noticed at degree level and higher, and being a bit ...


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You would have much more mass than 100 kg after the wood was burned. As it turns out, wood is made of cellulose and lignin. Both are cross-linked glucose polymers, so a good approximation of what you would get is given by the chemical reaction of burning glucose: $$\rm C_6H_{12}O_6 + 6O_2 \to 6CO_2 + 6 H_2O$$ This means that 6 oxygen molecules combine with ...


13

Relativistic loss of mass is unmeasurable here, but in principle, you’d lose some tiny fraction of the mass by heat transfer to the surroundings. Whether the smoke would weigh more or less than the wood depends on your definition. Oxygen from the air is combining with carbon in the wood to form carbon dioxide. If this counts as smoke, then the smoke weighs ...


1

All permanent magnets have approximately the same density, the density of iron, because they mostly consist of iron or cobalt. Their maximum magnetization is also about the same, about two tesla (lower for ferrites). The difference between materials of permanent magnets is their coercivity, their ability to keep their magnetization against demagnetizing ...


1

Strictly speaking a scale measures weight and a balance measures mass, but it is common, especially in non-technical English, to refer to a "scale" in either case. Most probably your home device is actually a scale and is somehow measuring the force imparted by whatever you put on the platform due to your local gravity. The function to display the ...


0

The simple answer is no, mass does not affect electromagnetic forces: only the position and movement of charges do. See the Lorentz force $\mathbf{F} = q\left(\mathbf{E} + \mathbf{v}\times\mathbf{B}\right)$. Note that mass is nowhere to be found in this equation. But what produces the $\mathbf{E}$ and $\mathbf{B}$, you may ask? Well, those can be determined ...


1

Your implicit assumption is that if A and B are both uncertain, then A+B is uncertain. This is not true. The ground state of an atom is not a state of a definite kinetic energy or a state of a definite potential energy, but it is a state of good energy. Relativistically, it's a state with a definite mass-energy.


2

In general relativity, which is the consistent picture relativistic gravity, the gravitating source is not "mass", but rather the stress-energy tensor. This is a two dimensional matrix $T_{ab}$, whose components are: $T_{tt} = $ density of matter $T_{ti} = $ ith component of the 3-momentum density of the matter $T_{ii} = $ pressure felt by ...


0

The speed of light is "broken" each time when a particle moves faster than the speed of light in some transparent media with $n\gt 1$. No infinite mass/energy is necessary for that. Remember Cherenkov's radiation for fast charged particles.


1

In simple terms: No amount of finite numbers will ever sum to infinity. Scoop as much mass as you want, magically accelerate it so it doesn't slow you down, carry it on board. You will have a finite spaceship containing finite mass and obtaining finite mass from the ambient environment requiring infinite acceleration. Your mass inflow rate would need to be ...


2

There is no minimal force to overcome inertia. As you said, because $F=ma$ whatever total force $|F|>0$ will cause an acceleration $|a|>0$, even the smallest force will accelerate the object. How big, given a force $F$, is the acceleration going to be, that depends on the inertial mass $m$. So On Mars, because $m$ is a constant, the same force will ...


0

The first line of Wikipedia states that Inertia opposes change in velocity. My teachers also told that Inertia opposes change in state of rest or motion...shouldn't every body oppose the acceleration caused from that force because that force will change its state from rest to motion? Changing from rest to motion is a change in velocity (the velocity changes ...


1

Inertia is an intrinsic property of matter. The term inertia is used in two different senses. In the directional sense, it means "momentum". Think of it as suddenly stopping while traveling by car. You would move forward. Is there any physical force that makes you move forward, no, there isn't. This is why we don't have a special variable for ...


3

We already include inertia in that equation, so there’s no need for another term. The mass, m, is a quantification of the inertia of the body, which is why the acceleration that a given force produces is proportional to the mass (which is to say the inertia) of the body being accelerated. Without inertia, any net force would produce an infinite acceleration. ...


2

This is more of a computer science question, but it's also something that I find physics students get confused by quite often and it doesn't seem to be mentioned as often as it should be, so I'm not voting to close it. Basically, you're using the wrong arctan function to plot your results. If you use the right one, everything checks out. The arctan function ...


1

They have different masses. Since the black hole horizon is order of magnitude $M$ from the center of the black hole (the exact proportionality constant depends on spin, ranging from 1 to 2), and since the "surface gravitational force" at the horizon is proportional to $\frac{M}{r^{2}} = \frac{1}{M}$, this means that the smaller the black hole is, ...


0

If we take the mass term $-m_{e}\bar{e}e$ and apply an SU(2)$_{L}$ gauge transformation $U$, we get: $$U\left( -m_{e}\bar{e}e\right) = -m_{e}U\bar{e}e \overset{(17.16)}{=} -m_{e}U\left( \bar{e}_{R}e_{L} + \bar{e}_{L}e_{R} \right) = -m_{e}\left\{ \left(U\bar{e}_{R}\right)e_{L} + \bar{e}_{R}\left(Ue_{L}\right) + \left(U\bar{e}_{L}\right)e_{R} + \bar{e}_{L}\...


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The limit in the 2020 Particle Data Group Summary is $m_\gamma < 10^{-18}\mathrm{eV} \approx 10^{-27}m_\text{proton}$, based on a 2007 analysis of the magnetohydrodynamics of the solar wind. The PDG reviews are excellent reading on these kinds of questions.


0

The magnetic moment is expressed in terms of the spin angular momentum, which is proportional to mass. Magnetic moment is instead proportional to charge. Hence there is a factor q/m in its expression.


12

The mass that is known to exist within (non-primordial) black holes is much less than that known to exist in stars. Even the biggest supermassive black holes are typically 1-10% of the stellar mass of the galactic bulges they reside in, and more typically 0.5% (e.g. see the M-sigma relation). Only the most massive stars produce "stellar mass" ...


0

Perhaps you heard it a little wrong. There is no such thing as a rigid stick. A stick is made of atoms connected by bonds. Bonds exert electromagnetic forces on neighboring atoms, and those forces cannot travel faster than light. So if you accelerate one end, one atom pulls on the next. The forces between atoms change. The other end does not begin to ...


0

There is a limited notion of rigid motion in relativity. In principle you can rigidly accelerate an extended object, but only via precisely applied forces spread across its whole bulk. If you push or pull it in some other way, it will still accelerate, but the way it accelerates won't meet the definition of rigid motion. It isn't possible to construct an ...


1

Some aspects of matter-energy are captured in geometry Let us first consider what the question is even about. How can matter be identified with curvature? It would mean that one should be able to read off all the properties of matter only from space-time quantities such as the Riemann curvature tensor $R^\mu{}_{\nu\kappa\lambda}$ and it's derivatives. A part ...


3

The equation $G=T$ (let me put the proportionality constant equal to one for simplicity) is extremely beautiful in its conciseness. But this conciseness can also cause various misconceptions (which indeed appeared among the very developers of the theory). Something that is often omitted in cursory accounts of the Einstein equations is that they are an ...


5

General relativity is that theory. The field equation of general relativity says that mass (more precisely four-momentum or stress-energy) equals Ricci curvature, up to a constant that's merely a unit conversion factor like $c$. There's nothing about one side of the equation causing the other. They're just equal, everywhere, always. In present physical ...


3

It is not possible: curvature propagates outside masses, classically gravity propagates from sources. So there is curvature in regions where there is no mass. They are distinct notions though related. The spacetime Riemann curvature is a dynamical notion, it measures the relative acceleration of free falling bodies close to each other. There is also the ...


1

To accurately account for how the object feels when rotated through more than one axis (e.g., a turn of the wrist will feel different than swinging with the arm), you will also need the full inertia tensor. This can be fully specified by three numbers, the moments of inertia about the principal axes, instead of just one number for the moment of inertia you ...


1

It is implicitly explained on the second page of the paper. Interacting relativistic effects should be intended as the spin-orbit effects, while by shell structure effects Pyykko means the effect due to the combination of the contraction of the inner $s$ and $p$ shells and the need for $d$ and $f$ shells to be orthogonal to the core states. The two effects ...


0

assuming all of the other conditions are controlled I would still have doubt of seeing confirmed data on a uniform basis wouldn't state any results as fact= unanswerable


1

Measuring in forward and then reverse order is a good idea. Another good idea is to use a random ordering. For more sophistication, employ a model which includes parameters indicating the size and type of hysteresis, and make the hysteresis itself the focus of study until you have got a good model of it. Once you have got that then you can adjust for the ...


2

So my question is this - does light in fact have gravity or does it just have Gravitomagnetism meaning the ability to pull nearby object in its direction of travel rather than toward the light-beam. Here is a list of the abilities that light has, and after each ability in parenthesis there is the name of the ability. Light has the ability to pull a nearby ...


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