# Tag Info

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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). ...

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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 ...

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Yes, $B$ does rotate when seen from a static frame of coordinates outside the disk: As to velocities and accelerations, see the article in Wikipedia. It says, $$\vec {v_s} = \vec {v_r} + \vec {\Omega} \times \vec r,$$ where $v_s$ is the velocity in the static frame and $v_r$ in the rotating. If you apply this formula for both points $A$ and $B$, their ...

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If I understand your question correctly you are saying that: $$v = r\omega$$ and therefore: \begin{align} v_A &= r\omega \\ v_B &= \tfrac{1}{2}r\omega \\ v_A &= 2v_B \end{align} but how can $A$ and $B$ have different velocities when they are both attached to the disk so the separation between is fixed? The answer is that $A$ and ...

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No, for several reasons. First, the idea of time "slowing down" is a little bit of a misnomer. If you were traveling at relativistic speeds, you would not perceive the passage of time any differently than you do right now. It's only when you compare your clocks to an observer in another reference frame (let's me, sitting in my living room, at rest with ...

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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 ...

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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 ...

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Let us see a similar example: two people on skates going with some velocity towards each other both a bit on left off their common center, and in the moment of the closest approach, they just catch each other by right arms and they start to rotate. In fact they have (as one system) the same angular momentum all the time. When you have a projectile that ...

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Suppose we play a racing game. I scatter a little bit of dust around space, then you come by me in your spaceship at some speed $v$. Let's start with $v = c/2$, just so we're not contentious. Right as you pass, I fire a really bright laser pulse in the direction you're going. You're racing the laser light. The dust means that you see reflections of it, so ...

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Short answer : the same light goes at the same speed (c) relative to any observer. There is no grid. This is counter-intuitive : if you're standing in a bus traveling at 30 mph and you walk at 3 mph towards the driver, you walk at 30 + 3 = 33 mph relative to the road. But that doesn't work for light : if the bus travels at c/2 and you shine a flashlight ...

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Before the ball was thrown, it was already travelling at the speed of the train. When you are throwing it, you simply increase its speed by how fast you are throwing it. Keep in mind this does not work close to the speed of light.

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Now, after the hammer is released, the thrower still has her same angular momentum (and has to slow herself down), but the hammer no longer seems to have any. Even though the hammer isn't rotating around the axis, it still has the same angular momentum it had at release with respect to the original axis. So the formula $$L = mvd$$ is correct both for ...

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Instead of talking about simultaneity you can talk about the measured time between dropping the object and falling. Indeed if an astronaut drops a feather from his left hand and a hammer from his right on the Moon, an observer in motion relative to the astronaut will not consider the astronaut to drop the objects simultaneously, nor agree that they hit the ...

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The first to do something equivalent to that were Pound and Rebka who first measured the gravitational redshift in 1959. I'm not aware of anyone who actually used a Michelson interferometer in a upright orientation.

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It's relative to all inertial reference frames--in special relativity the coordinates of one inertial frame are related to the coordinates of another by the Lorentz transformation, and this transformation has the property that anything with a coordinate speed (change in coordinate position divided by change in coordinate time) of c in one inertial frame will ...

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Its not possible to stop time but using relativity it can be thought of to be slowed down . Nothing can be faster than the speed of light so its not possible . Even when we near it , energy tends to become infinity .

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An example of a law that is not invariant: $F = -\mu mv$. That is, some kind of universal friction slows down all moving objects. This requires a point of reference they are slowing down compared to, so it is not invariant. Any law that can be written in the form of a tensor is invariant, but this law cannot be written in that form. Not unless you have some ...

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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 - ...

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You say: I found out that the speed of light is NOT invariant in an accelerated reference frame but things are more complicated than this. The local speed of light measured by an observer is always equal to $c$, and this remains the case whether the observer is stationary, moving, accelerating or anything else you might think of. So if your Michelson ...

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Why do airplanes experience negligible Coriolis force while bullets experience the Coriolis force in long range shooting? You are confusing the force with the consequence of the force. Consider a powered parafoil whose total mass is a mere 100 kg (motor+parafoil+pilot) and is moving at a mere 25 km/h and a 50 caliber bullet whose mass is 50 grams and is ...

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Just to make things simple, suppose you are standing at the north pole, and you shoot a bullet south at some speed, aiming for a target 1 km away. In the time it takes the bullet to get there, the target has moved east a certain distance, because the target travels in a complete circle around the north pole in 24 hours. From the viewpoint of the shooter, who ...

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Center of mass before the disintegration is in the initial particle. This means that the center of mass moves with velocity $\vec V$ in lab frame. Thus, to switch from center of mass frame to lab frame you just use the Galilean transformation $$\vec v=\vec v_0+\vec V.$$ As for conservation of mass, it is certainly implied because the system is considered ...

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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 ...

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The columns of a 3×3 rotation matrix contain the coordinates of the local xyz axes (expressed in world coordinates). With Euler angles (321) you apply the elementary rotations $R_Z$, $R_Y$, $R_X$ in sequence to form the local → world rotation matrix. That is $$E = R_Z(\varphi) R_Y(\psi) R_X(\theta)$$ This is interpreted as each rotation occuring about ...

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Airplanes always maneuver with respect to the surrounding air. Something which confuses beginner pilots is the following question: imagine you have a wind from north to south, and you point your airplane to the west. Where is the air pressure higher? a. The left side of the plane b. The right side of the plane The answer is: neither. The pressure is the ...

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Let's say you do the calculation $C_{final}=R_xR_yR_zC_{o}$, where the $C$s are coordinates. The first rotation is about the $z$ axis of $C_{o}$ and will produce a new coordinate system in which the new $z$ is the same as the old, but the $x$ and $y$ axes are different. The next rotation will be about the new $y$ axis and will produce a newer, new $x$ and a ...

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Writing it the way you did, you don't try to compensate the force, but to set it equal. This means you account for the fact that the centripetal force is gravity and nothing more and they do point in the same direction for a circular orbit. Assume a body in circular motion around the origin of a coordinate system at distance $R$. One can express this motion ...

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If it is what you are saying then remember that work done here is neither negative nor positive however "work" doesn't exist here. We define Work as the product of the component of Net force acting on the body and the distance through which Force is exerted. You can say that the Force of gravity and Normal force are cancelling effects of each other since ...

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Angular momentum is conserved in this example! As you already stated, the angular momentum of the thrower doesn't change after the hammer is released. Consider the hammer being in rotation around the origin of our coordinate system for $t < 0$: $$\vec{r}(t) = r_0 \ \ (cos(\omega t), sin(\omega t), 0)^T$$. Its momentum is therefore given by:  ...

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