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19

I am wondering whether is it taken as a postulate or a proven phenomenon that c is constant irrespective of observer's speed? Either one. Both. Einstein took it as a postulate in his 1905 paper on special relativity. From it, he proved various things about space and time. The frame-independence of $c$ is also experimentally supported. This is what the ...


5

The statement from the Wikipedia articles is, as written, wrong. The EM field tensor - as a tensor - does change under change of reference frames. It is covariant, but not invariant under the Lorentz group, while the electric and magnetic field are neither, but they are covariant under the rotation group. The electric and magnetic fields are ordinary, ...


4

By using orthogonal optical resonators, laboratory tests concerning verifying the isotropy of c have come a long way. As quoted from http://journals.aps.org/prd/abstract/10.1103/PhysRevD.80.105011 "An analysis of data recorded over the course of one year sets a limit on an anisotropy of the speed of light of $\Delta c/c \sim 10^{-17}.$ This constitutes the ...


3

It's intuitive that while accelerating in a locally constant gravitational field, there is no perception of acceleration, since the body accelerates uniformly. The reason you can't perceive it is not that it's uniform, the reason is that there's nothing to compare with. If there's something to compare with, then you can see the difference. For instance, ...


3

If rest mass does not change with v then why is infinite energy required to accelerate an object to the speed of light? The momentum of a material particle, a conserved quantity, is theoretically and experimentally a non-linear function of velocity given by $$\vec p = m \frac{\vec v}{\sqrt{1 - \frac{v^2}{c^2}}}$$ which goes to infinity as $v ...


3

In principle there is an effect, but firstly it's tiny and secondly it averages to zero. The mass of the ISS is about 420 tonnes, or about 5000 times the mass of an astronaut. That means if an astronaut pushes themselves off a wall at 1 m/sec the ISS moves in the other direction at about 0.0002 m/sec. But the ISS isn't very large so after only a couple of ...


2

Option 4, none of the above. Your option 1 is wrong because points don't rotate. Your option 2 is closer to correct, but ultimately still wrong. You're overly hung up on points (the origin). It might help to get a handle on what "rotation" is. Points don't rotate. Better said, a rotated point is indistinguishable from the original. What about one ...


2

First, to clarify: A body does not "gain mass" upon acceleration, it "gains mass" at high speeds. That is, whether or not the body's velocity is changing is not relevant, only its speed relative to the observer is important. That being said, a body doesn't actually "gain mass" when it moves at a high velocity. The mass of a body is always the same, and is ...


2

If I understand correctly you are asking how observer dependent is electromagnetic radiation. The first thing is that non uniformly accelerated charges are described in a inertial frame by Larmor's formula and Abrahm-Lorentz force which take into account the radiated field and the recoil on the particle. Now in Newtonian mechanics and special relativity ...


2

Technically the electron and proton are both orbiting the barycenter of the system, both in classical and quantum mechanics, just as in gravitational systems. You find the same dynamics for the system if you assume the proton and electron are moving independently about the barycenter, or if you convert to a one-body problem of a single "particle" with the ...


2

Torque is defined as $\vec \tau = \vec r \times \vec F$, where $\vec r$ is the displacement vector from the origin to the point at which the force is applied. This means that torque depends very much on the choice of origin. Then again, the choice of origin also affects the inertia tensor. So long as you get all of the physics correct, you can choose any ...


2

It's just a drawing convention. Rather than "vertical", time is orthogonal to the x-axis. The reason it is not shown vertical is because the paper surface is 2D, and the author uses the vertical axis for drawing the altitude with respect to the ground. Just recall the way you draw the 3D axis. Here, the author uses X (horizontal),Y (vertical) and time ...


2

I think you are considering two different situations: 1) Team Pole passes through the barn at constant velocity, simultaneously (in Barn's frame) grab the pole, and continue on with the same constant velocity. In this case, your second calculation is correct. The pole's length does not change from Team Barn's perspective. The pole remains length $L_0$ ...


2

At any given moment, the pendulum is swinging in a certain plane. The driving force should be within this plane. If the earth weren't rotating, then such a force could never cause the plane of swing to rotate about a vertical axis, since by symmetry there would be no preferred direction for the rotation.


2

I'll go with an Equivalence Principle argument. For a model system, consider a test particle in a highly elliptical orbit around a neutron star; the particle will pass through regions of greatly different field strength. But it feels no force as it "falls" around the star. Per the Equivalence Principle, at each point there is a locally inertial ...


2

It is a well-substantiated observed phenomenon. Science deals only with provisional truths, but this hypothesis has undergone (and passed) immense amounts of scrupulous experimentation and mathematical formulation. In a Neo-Lorentzian interpretation, physics works differently in all reference frames except for one single, undetectable, privileged reference ...


2

Take the reference frame as centered in the fixed axis. The $R$ that connects the origin to the centre of the spinning disk forms an angle $\phi$ with the horizontal. Now, inside the disk of radius $r$, the angle of a certain point mass is given by the angle it forms inside the spinning circle, which we'll call $\theta$. Now take as generalised coordinates ...


1

Quantum-mechanicaly and relativisticaly the energy of a given object is never removed completely (Heisenberg uncertainty relations, relativistisc rest mass) Assuming we talk about kinetic energy, kinetic energy is defined with respect to a specific reference frame. This reference frame can be related to another object or not. The kinetic energy can be ...


1

As long as you don't forget that Andromeda galaxy is cca. 2.5million light years away, then there should be no paradox at all. The observers are only seeing different slices of history that took place 2.5million years ago (plus/minus 1day), the decisions has already been made long time ago. It's like when you take a newspaper from a pile, more recent papers ...


1

The interpretation is that two events being simultaneous as measured in frame $S$ doesn't imply that the events are simultaneous in frame $S'$. Which events count as being "simultaneous" depends on the frame of reference. This is known as the relativity of simultaneity. Added clarification due to comment: The coinciding of the origins is an event, call ...


1

One expects the energy stored in the capacitor to transform like the zeroth component of the four-vector $(U,\vec p)$. In its rest frame the field configuration around the capacitor has $$(U,\vec p)_\text{rest}=(U_0,\vec 0),$$ and by the Lorentz transformation the moving observer will see $$(U,\vec p)_\text{moving}=(\gamma U_0, \gamma\vec\beta U_0),$$ where ...


1

$m_2$ will leave with the same magnitude of momentum but opposite direction. Now the assertion is made that in an elastic collision, $m_1$ and $m_2$ have the same speeds leaving the collision as entering it. In other words, the speed of $m_1$ is $v-v_c$ and the speed of $m_2$ is $v_c$ after the collision. In order to simplify things and to ...


1

In relativity the rest mass is the mass of an object measured from a reference frame in which it is at rest. But this is not the mass involved in acceleration or inertial mass. Inertial mass, or the opposition of the body to the change of movement (directional or in magnitude), will grow with the speed of the body: $$m = \frac{m_o}{\sqrt{1-v^2/c^2}}$$ ...


1

The short answer is: protons are much more (1800 times) massive than electrons. That makes them (approximately) the center of mass of the system, that's why electrons are the ones orbiting protons and not vice versa. The term 'orbiting', however, means something essentially quantum. It is the reason of the stability of the atom (electrons don't radiate ...


1

The fact that different observers in relative motion can measure the same light ray to move at a speed of c has to do with the fact that each observer defines the "speed" in terms of distance/time on rulers and clocks at rest relative to themselves. It's crucial to understand that different observers use different rulers and clocks to measure speed, because ...


1

No. My answer is negative, even if I confirm the statements of other answers: "The first thing is almost completely arbitrary, especially in full general relativity. The second thing is an unambiguous result of an experiment."(Jerry Schirmer) "In Einsteinian relativity all observers can still agree on a number of facts, they are just ...


1

In order for a body to move with uniform velocity in a circular path, there must exist some force towards the centre of curvature of the circular path. This is centripetal force. By Newton's Third Law, there must exist a reactive force that is equal in magnitude and opposite in direction. True, although the adjective "reactive" is meaningless. There is ...


1

A particle moving at the speed of of light does not experience time, as it has no rest frame. Furthermore, a particle cannot continuously accelerate and eventually reach the speed of light, since massless particles can only move as fast as light. They either move at the speed of light or do not exist at all.



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