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I am posting this answer - based purely upon old comments from other users - to get this question out of the 'Unanswered' section of our site, following David Z's suggestion from a meta post. Note that this answer is a CW, and I am therefore not trying to gain reputation with this answer. As has been shown in this paper, pointed out by Matthew in the ...


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What you read is correct. I am not sure if those were the exact words of your teacher but according to the general theory of relativity, sun doesn't "attract" the photon (or any other body for that matter). In fact gravity is not even a real force. Let me briefly state what the theory of relativity has to say about gravity without going into the complicated ...


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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 ...


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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 ...


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The correct general equation is $E^2=m^2c^4+p^2c^2$, since $m=0$, $E=pc$. A photon's momentum is $p=\hbar k$, where k is the wavenumber ($k=\frac{2\pi}{\lambda}$). This would be consistent with $E=h\nu$ where $\nu$ is the frequency.


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This is because in deep space everything is being attracted equally in all directions since the universe is homogeneous. Hence, all of these tiny gravitational forces acting on the objects in your ship cancel out. Therefore you are left with a net force equal to zero on these objects in your spacecraft. On reflection, forget deep space for the time being ...


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Your question is actually quite a complicated one as there are lots of different factors at play. However if you're asking why the objects inside the rocket don't attract each other then the main reason is that gravity is actually a very weak force. If a book is floating a metre away from me then the gravitational acceleration of the book due to my mass ...


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Your answer of gravity being weak is exactly correct. The forces involved can be calculated with the following equation: $$ F = \frac{GMm}{d^2} $$ A fueled Saturn V was about $3.0 \times 10^6 kg$. Most of it doesn't reach space but we'll be conservative and say your spaceship is that massive. If you had a 1kg book that could somehow be only 1 meter away ...


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The rocket is in free fall along with the book. The nearest gravitating bodies are very far away, so whatever meager acceleration they cause will be almost exactly the same on the rocket and on the book. Suppose you instead went on a very close flyby of a neutron star. Now the book will fall rapidly, and away from the rocket's center of mass. The only way ...


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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 ...


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when low mass object hits high mass object it is reflected gaining opposite velocity almost the same as initial velocity. If I jump onto the wall why my body is not reflected? I know that collision is not fully elastic but it should be at least similar. Human body is not elastic: it cannot be deformed/ compressed in any way and then return to ...


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Changing shape could still be an elastic deformation (of a rigid body). So obviously there are also plastic deformations involved, when jumping against a wall.


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For some height $h$ of the liquid, the pressure of the water on the bottom of the beaker is $P = h \rho g$ where $\rho$ is the density of the water*. Since $h$ is the same for both beakers, $P$ is the same. Net force on the bottom of the left hand beaker is sum of the water pressure $PA$, minus tension of string pulling on the bottom: $$F_{left} = P A - ...


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A partial answer: "I do not see a coupling between the MSSM fields and the messenger fields" There are gauge interactions between observable and messenger fields. Let us discuss with fig $(1)$ page $11$ of this reference. You have not direct interactions between observable sector gauginos ($\lambda$) and s-fermions ($\tilde f$), and the superfield ...


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The real reason is in following. Let's assume Majorana field: $$ \Psi_{M} = \Psi_{L} + \hat{C}\bar{\Psi}^{T}_{L}, \quad \hat{C} = i\gamma_{2}\gamma_{0}, \quad \Psi_{L} = \begin{pmatrix} \psi_{L} \\ 0 \end{pmatrix}. $$ By using this notation it's not hard to see that kinetic term is equal to $$ \bar{\Psi}_{M}\gamma^{\mu}\partial_{\mu}\Psi_{M} = ...


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The short asnwer to your question is that the overall factor $\frac{1}{2}$ from the Lagrangian of a Majorana field (in the 4-component notation) $$\mathcal{L}=\frac{1}{2}(\bar{\psi}i\gamma^{\mu}\partial_{\mu}\psi -m\bar{\psi}\psi)$$ compared to the general Dirac Lagrangian is usual for self-conjugate fields and it is introduced to ensure a consistent ...


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There is geometrical significance. You are sooo close. You are in Euclidean space, but you should be in hyperbolic space. As @fqq points out, you have stumbled upon rapidity, a parameter in hyperbolic geometry that is the analog of angle in Euclidean geometry. In Euclidean geometry an angle (in radians) is a parameter that measures the Euclidean length ...


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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 ...


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If the force applied to each ball is the same then it will provide the same impuls to both, then using J = M(v-u) it can be seen that the lighter ball will reach a higher maximum speed. Following this, the ball's are now experiencing a drag force, depending on the assumptions made we can either assume that the drag force is constant, or that it is a ...


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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 ...


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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 ...


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The harmonic mean involves averaging the reciprocals, so it is dominated by the smaller of the two values. The reduced mass has the same requirement: The orbit of a small body around a large body is almost the same as the orbit of a small body about a fixed point. This also "explains" the factor of 2: in the limiting case m1>>m2, the reduced mass = m2, not ...


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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 ...


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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. ...


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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, ...


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Hm... I believe the scale will remain in a level position, without any tipping to either side. How I see this is this: First, I've distinguished two individual bodies of mass: the first being a steel ball attached to a string, further attached to a stand. Since this combined weight culminates onto the stand itself, and not on the scale, the only interaction ...


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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 ...


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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, ...


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In Newton's Law of Universal Gravitation, the mass of both objects must be entered, but photon has no mass, why should a massless photon be affected by gravity in by Newton's equations? What am I missing? You are forgetting that Newton thought that light is made by particles which at the time were called corpuscles see here, and you surely know that ...


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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 $m/m = 0/0$ ...


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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 ...


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I'd like to answer the first question "Is light affect by gravity? Why?" because I remembered an adorable thought experiment. In the spirit of the equivalence principle consider an observer in a closed box. As we all know, the observer inside that box would be unable to tell (neglecting tidal effects) the difference between a uniformly accelerated box ...


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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 ...


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1) The bending of light rays is a general relativistic effect, not one due to Newton's law of gravity. 2) It's probably better to think about these things from a field perspective -- a distribution of mass-energy moves along, and it creates a gravitational field. Then, when things enter that field, they interact with it, and this changes their motion. ...


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$E$ = $mc^2$ 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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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 ...


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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 ...


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Short answer: It is a combination of the lever/pulley principle plus the transformation of chemical energy into mechanical energy which makes the muscle/weight system work like this. Longer Answer: The arm joints are small levers or pulleys, meaning the amount of force can be multiplied by the relative lenghts of arm joints. The chemical/thermal energy ...


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Not a cosmologist but there some interstellar objects known as neutron stars that are extremely dense. Some neutron stars can have masses of 500,000 times that of earth with a diameter of about 25km. In the core of these stars, it is hypothesized that "quark liquid" exist which is when quarks get pushed into each other with no spacing between them. In such a ...


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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

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 ...


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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 ...


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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 ...


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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 ...


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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 $$ ...


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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 ...


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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 ...


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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 ...


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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} + ...


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Your equation is considering the effective mass of electrons. The holes are lack of electrons. To talk about them, we effectively invert the energy axis, i.e. if we compute electron and hole energies with respect to valence band ceiling $E_V=0$, we have: $$E_e=-E_h.$$ Then it's straightforward to see that $m^*_e=-m^*_h$ in the same valley.



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