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14

We do. The LHC accelerates two protons, each with 3.5 TeV of energy, giving a total of 7 TeV in the CoM frame (The energies are from the initial phase of the previous LHC run. Later in the run this was increased to 8 TeV and the combination of the two dataset was what discovered the Higgs boson. The energies are roughly doubling now for Run II, to 13 TeV). ...


5

Many modern particle accelerators do accelerate both particles towards each other. LEP accelerated electrons and positrons in opposite directions in the same chamber, and the Tevatron did the same for protons and antiprotons. The LHC is a proton-proton collider, and so it has two stacked rings that accelerate protons in different directions. For the BaBar ...


4

From scholarpedia: The Unruh effect is a surprising prediction of quantum field theory: From the point of view of an accelerating observer or detector, empty space contains a gas of particles at a temperature proportional to the acceleration. Direct experimental confirmation is difficult because the linear acceleration needed to reach a temperature 1 K ...


4

Assume the vehicle is already there before the space station is built. So it is floating. If it is floating slightly above space station ground, this does not change if the station starts to spin (neglecting its acceleration due to air friction). In the reference frame of the station it will move with the velocity of the outer cylinder. If you get in the ...


3

I think the answer has to be yes, it would be much easier to detect the cosmic neutrino background if you could arrange for your laboratory to be travelling at close to the speed of light in the local, co-moving reference frame defined by the cosmic microwave background. The current energies of neutrinos in the cooled cosmic neutrino background are expected ...


3

Let $M$ be your spacetime, a smooth manifold equipped with (pseudo) Riemannian metric (for example $\mathbb{R}^{(1,3)}$ for special relativity). The set of reference frames is the frame bundle over $M$, usually denoted $FM$. Explicitly a frame at point $p$ in $M$ can be viewed as an ordered orthonormal basis (with respect to the the inner product defined ...


2

We conclude that the lenght perceived by the observer in S is bigger than the lenght perceived by the observer in S. No, that would be an error. Why? The short answer is that the coordinate difference $(x_2 - x_1)$ is not a length in S. A length is the (spatial) coordinate difference at the same coordinate time. To be clear, there are two events - ...


2

There is no contradiction. From your quote: The proper length of an object is the length of the object in the frame in which the object is at rest. and The proper time between two events - such as the event of light being emitted on the vehicle and the event of light being received on the vehicle - is the time between the two events in a frame ...


2

You are right. In Newtonian physics, work depends on reference frame. Force does not. Let's start with a book sitting on a table. The table exerts a normal force on the book, but it does no work because there's no motion. Next, imagine that the table is in an elevator and the elevator is going up at constant speed. The force between the book and table is ...


1

The point of view of a photon is not really a valid reference frame. This can be seen by asking yourself this question: if the speed of light is constant in all reference frames (a basic postulate of special relativity), then does a photon see itself traveling at the speed of light? As can be seen, this is utterly nonsensical.


1

An "east wind" blows from the east; your intuition is correct. As far as why, in the northern hemisphere, the Coriolis force makes a south wind bend into a west wind, and a north wind bend into a east wind: it's conservation of momentum. An observer on the ice at the North Pole doesn't have any east-west momentum — he can't, because "east" isn't a ...


1

There's a trick which is often used in computergraphics to account for rotations + translations in one single matrix multiplication. If $R$ is you rotation matrix and $\vec{t}$ is your translation vector you construct the following rotation-translation-matrix: $$ M = \left( \matrix{. . . \ \ |\ \ . \\ . R . \ \ |\ \ \vec{t} \\ ...


1

The technique would work, but whether you would use it would depend on many many details. Rotating frames of motion come with all sorts of counter intuitive bits, so lets look at it from an inertial frame outside the station. We perceive the station as rotating. The velocity vector of any object on the station is a tangent, so we see the result of every ...


1

If I understand the question right, we suppose we want to prove to someone that the earth orbits the sun. I'm not quite sure that' the case from a scientific point of view. Literally speaking, we can choose any reference frame we like and thus prove a heliocentric system or a or a geocentric. Quoting Einstein:" The struggle, so violent in the early days ...


1

Let's define the direction of motion of the Solar system as the direction of motion relative to the cosmic microwave background i.e. the direction defined by the dipole anisotropy. Suppose the Solar system is moving in the plane of the ecliptic as shown in (a): In that case no particular part of the Earth is in the forefront of the velocity. Because the ...


1

No definite answer to this question, the effect in some sources is accepted and other sources dispute it. From Wikipedia: The hypothetical Unruh effect (or sometimes Fulling–Davies–Unruh effect) is the prediction that an accelerating observer will observe black-body radiation where an inertial observer would observe none. In other words, the ...


1

Yes I'd agree that reference frames are sections in the frame bundle. For physical intuition, consider the notion of frames in classical GR. An observer, at any given point in spacetime, would measure things in an orthogonal basis. That is, to the observer, the time he observes should be orthogonal to the spacial distances he observes, and spacial ...


1

anna v's nice answer didn't go where I expected: the big acceleration at the LHC isn't in the accelerator, it's in the collisions. Let's suppose we have a proton in the LHC that undergoes an elastic, billiard-ball type collision and ends up with its original momentum in the opposite direction: \begin{align} \vec p_\text{initial} &= +7\,\mathrm{TeV}/c ...



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