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34

Here is a free body diagram of the balls: … and one of the water volume: The four balance equations are $$ \begin{align} B_1 - T_1 - m_1 g & =0 \\ B_2 + T_2 - m_2 g & = 0 \\ F_1 + T_1 - B_1 - M g & = 0 \\ F_2 - B_2 - M g & = 0 \end{align} $$ where $\color{magenta}{B_1}$,$\color{magenta}{B_2}$ are the buoyancy forces, ...


19

The weight on the left bowl would be the weight of the water plus vase plus ping-pong ball (plus thread, ignored). The weight on the right bowl would be the weight of the water plus vase plus the buoyancy of the steel ball (plus the buoyancy of the submerged thread, ignored). That buoyancy is the weight of an equivalent volume of water. Since the ping-pong ...


17

A Thought Experiment We can arrive at an intuitive explanation without any special knowledge of physics. The strategy is to re-create the setup as closely as possible while keeping the two sides in balance. Imagine that you start with two identical beakers, filled with the same amount of water, no balls. Placed on the scale, they balance. On the left, ...


11

In non-relativistic quantum mechanics the mass can, in principle, be considered an observable and thus described by a self-adjoint operator. In this sense a quantum physical system may have several different values of the mass and a value is fixed as soon as one performs a measurement of the mass observable, exactly as it happens for the momentum for ...


8

Mass-squared is a Hermitian linear operator, it's a Casimir operator $\hat{C}_{1}=\hat{P}_{0}\hat{P}_{0}-\hat{P}_{i}\hat{P}_{i}$ for the Poincare group. It's Hermitian because the translation generators $\hat{P}_{\mu}$ are Hermitian. It commutes with all the generators of the Poincare group and so it's eigenvalues (mass-squared) are constant on each ...


7

$E = mc^2$ is only true for particles at rest (momentum $p = 0$). The full formula is $$E = \sqrt{m^2c^4 + p^2c^2}$$. In these formulas, $m$ is the rest mass (or just "mass"). Photons don't have mass, so for them the formula becomes $E = pc = \frac{hc}{\lambda} = h\nu$ where $\nu$ is the frequency. Therefore, the energy of a photon is proportional to its ...


6

Of course, mass is an observable, although in simple models it is constant. This is already the case classically. One cannot determine the path of as rocket that burns fuel (which forms a large fraction of its mass) without taking into account that the mass is variable. The same holds in quantum mechanics, whenever the mass is not fixed by the modeling ...


5

I know that everyone recognizes $E = mc^2$ as the emblem of relativity, but the expressions is either incomplete or only applies to massive particles (depending on how you understand the $m$ that appears there). The full and complete expression is $$ (mc^2)^2 = E^2 - (pc)^2 \,,$$ where $m$ should be understood to be the invariant rest mass. This allows you ...


5

Yes, massive particles such as W-bosons, Z-bosons, quarks, and leptons couple to the Higgs field via the cubic (Yukawa) interaction, so they may also exchange the virtual Higgs. Yes, because the virtual particle is massive, one gets the Yukawa potential that includes the exponential dumping with distance. This "Higgs force" is much less fundamental and ...


5

To properly understand what is going on you need to understand general relativity. Massless particles, like photons, travel on null geodesics and mass bends spacetime so the null geodesics are not straight lines. The problem is that neither you nor your teacher understand general relativity so this isn't a very convincing argument. But here is an argument to ...


4

I will assume you are talking about the center of mass. If there's no external forces, the center of mass would conserve it's momentum. So, it would stay in constant speed, whatever what that speed is, with respect to whatever inertial frame of reference. This happens because Newton's third law. In the summation of all forces, the internal forces will ...


4

The masses can't repel each other because gravity is mediated by a spin 2 field, and for spin 2 the force between charges of equal signs is attractive. See the question Why is gravitation force always attractive? for an explanation of this. But it's impossible to say why the force can't be zero. Experiment shows that masses do attract each other, and ...


3

Gravity may be treated as a quantum field theory. In this kind of theory, interactions are represented by field correlations, more known as "virtual particles", "virtual gravitons" in the case of gravity. The fact that two charges (more precisely, in the case of the gravitation, $2$ positive energy densities) attract each other is due to the sign ...


3

The underlying level of nature is quantum mechanical and obeys special relativity laws, not Newtonian mechanics. Mass is a conserved quantity in Newtonian mechanics, but not in the underlying quantum mechanical framework. The classical framework emerges as the variables become large enough where h_bar, which characterizes the quantum mechanical level, can ...


3

Let's assume a typical fermionic mass-term (interacting leptons and quarks are spin 1/2-particles): $$ \tag 1 \bar{\Psi}\Psi = \bar{\Psi}\left(\frac{1 + \gamma_{5}}{2} + \frac{1 - \gamma_{5}}{2}\right)\Psi = \left| \bar{\Psi}\left( 1 \pm \gamma_{5} \right) = \left( (1 \mp \gamma_{5})\Psi\right)^{\dagger}\gamma_{0} \right| = $$ $$ =\bar{\Psi}_{L}\Psi_{R} + ...


3

This is one of those situations where you could argue on and on about definitions so let's answer it in all three meanings of the word weight that people might have. Does your mass $m$ increase when you inhale helium? When you fill your lungs with helium, you transfer $\approx1g$ of matter, which is several times less, but roughly of the same order of ...


3

Are you thinking of something like neutronium? This is the (hypothetical) matter formed when you compress the electrons into the protons to make neutrons, then pack the neutrons tightly together. If so, then the density is $4 \times 10^{17}$ kg/m$^3$. However you should note that even neutronium isn't pure matter, because neutrons are made up from quarks ...


3

Suppose classical "pure matter" as you describe it existed and suppose a spherical volume $V$ of $1\,\text{m}^3$ of this stuff has mass $M$. Since it exists of pure matter only, one expects a uniform mass density $\rho$ and $M$ is just $\rho V$. So you'd have to define the mass density of "pure matter" to answer your question. Say you make it 1 Planck mass ...


3

Newton's law does predict the bending of light. However it predicts a value that is a factor of two smaller than actually observed. The Newtonian equation for gravity produces a force: $$ F = \frac{GMm}{r^2} $$ so the acceleration of the smaller mass, $m$, is: $$ a = \frac{F}{m} = \frac{GM}{r^2}\frac{m}{m} $$ If the particle is massless then $m/m = 0/0$ ...


3

I'm amazed that this is so confounding to some. This is too long to be a comment, so I'm making it an answer. The TL;DR version: The answers that say the scale will tilt down to the right are correct. The beaker full of water with the steel ball suspended from above is heavier than is the beaker that contains the ping pong ball anchored from below. ...


3

Notation $W^{-}, W^{+}$ may confuse in a sense that it may seem that here are two different particles which aren't connected by charge conjugation. But of course, $W^{+}$ is only $(W^{-})^{\dagger}$, so it is an antiparticle to $W^{-}$. So term $( W^{-} \cdot W^{+} )$ is simple $|W|^{2}$ (which is standard for the mass-term), and, of course, both of particle ...


3

I agree to many points mentioned in the previous post but the answer is: "Yes, the weight(mass) would change." My reasons are simple. You state that an unbreakable container is being used, so that the products of the explosion will be contained inside. However unbreakable this container may be it will not be able to contain all of the radiation produced in ...


2

Well I got this badly wrong, and grovellingly apologise to those I traduced. It seemed easy: the water in both is the same weight, so I thought that removing it would make no difference to the balance. This was wrong: removing the water from the right hand beaker does have an effect, the presence of the suspended ball does add extra weight to it, so the ...


2

The question is what do we need the matter content of the universe for. As I understand it, in the usual case we want to find the conserved quantity associated with a certain conserved current gained by the projection of the energy-momentum tensor into a Killing vector, as for example in the paper by Abott and Deser. The requirement of asymptotical ...


2

Briefly, the formula $E=mc^2$ applies only particles at rest in an inertial frame of reference. Since there is no rest frame for a photon, no inertial reference frame in which a photon is at rest, one cannot apply the formula $E=mc^2$ to a photon. In more detail, the four-momentum of a particle has components $(\frac{E}{c}, \vec p)$ The four-momentum for ...


2

And so photons have mass No - photons don't have mass - they have momentum. And energy. But just because energy is equivalent to mass, doesn't mean they have mass. And they can only travel at the speed of light. A photon cannot travel at any other speed - so you cannot apply the Lorentz transformation to it. The Lorentz transformation applies to "rest ...


2

Elementary particles are conveniently divided into the fermions and the gauge bosons. The fermions are what we think of as matter, e.g. protons (i.e. quarks) and electrons, while the gauge bosons provide the forces that act between the particles of matter. Fermions get their mass from an interaction with the Higgs field called a Yukawa coupling, and the ...


1

John Rennie's answer is good, but I'll try to explain intuitively why the symmetry breaking leaves some symmetry unbroken. Start with a sphere. You can rotate a sphere in three independent ways—around the x axis, around the y axis, and around the z axis, if you like. All of these are symmetries of the sphere, i.e., they leave the sphere unchanged. These ...


1

From the free body diagram you must have $$ \begin{align} m_1 \ddot{x}_1 &= F_1 - F_2 \\ m_2 \ddot{x}_2 &= F_2 - F_3 \\ m_2 \ddot{x}_3 &= F_3 \end {align} $$ with the spring forces defined as $$ \begin{align} F_1 & = -k_1 x_1 \\ F_2 & = -k_2 (x_2-x_1) \\ F_3 & = -k_3 (x_3-x_1) \end{align} $$ The above is combined as $$ ...


1

$E$ = $mc^2$ A much better expression is $E^2 = (mc^2)^2 + (pc)^2$, where $m$ is the "mass" (also known as "intrinsic mass", also known "rest mass", but most physicists nowadays just use "mass") of the particle and $p$ is the particle's momentum. This reduces to $E=mc^2$ in the special case of a particle with zero momentum, but it also reduces to $E=pc$ ...



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