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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 previous LHC run which discovered the Higgs. The energies are doubling now for Run II). The main reason for this is as you mentioned, the energy involved. In any frame, we have the following invariant quantity, $s = (p_1 + ... 6 Now, after the hammer is released, the thrower still has her same angular momentum (also 62.8), but the hammer no longer seems to have any. A body does not have angular momentum wrt to a point C only when it is circling around it, you know that planets have elliptical orbits and do have L If a body H has linear momentum p it has also and angular ... 5 Many modern particle accelerators do accelerate both particles towards each other. LEP accelerated electrons and positrons in opposite directions in the same chamber, and the Tevatron did the same for protons and antiprotons. The LHC is a proton-proton collider, and so it has two stacked rings that accelerate protons in different directions. For the BaBar ... 5 Yes,$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 ... 5 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 canA$and$B$have different velocities when they are both attached to the disk so the separation between is fixed? The answer is that$A$and ... 4 No, for several reasons. First, the idea of time "slowing down" is a little bit of a misnomer. If you were traveling at relativistic speeds, you would not perceive the passage of time any differently than you do right now. It's only when you compare your clocks to an observer in another reference frame (let's me, sitting in my living room, at rest with ... 4 You say: I found out that the speed of light is NOT invariant in an accelerated reference frame but things are more complicated than this. The local speed of light measured by an observer is always equal to$c$, and this remains the case whether the observer is stationary, moving, accelerating or anything else you might think of. So if your Michelson ... 3 Let's define the direction of motion of the Solar system as the direction of motion relative to the cosmic microwave background i.e. the direction defined by the dipole anisotropy. Suppose the Solar system is moving in the plane of the ecliptic as shown in (a): In that case no particular part of the Earth is in the forefront of the velocity. Because the ... 3 Suppose we play a racing game. I scatter a little bit of dust around space, then you come by me in your spaceship at some speed$v$. Let's start with$v = c/2$, just so we're not contentious. Right as you pass, I fire a really bright laser pulse in the direction you're going. You're racing the laser light. The dust means that you see reflections of it, so ... 2 Short answer : the same light goes at the same speed (c) relative to any observer. There is no grid. This is counter-intuitive : if you're standing in a bus traveling at 30 mph and you walk at 3 mph towards the driver, you walk at 30 + 3 = 33 mph relative to the road. But that doesn't work for light : if the bus travels at c/2 and you shine a flashlight ... 2 Before the ball was thrown, it was already travelling at the speed of the train. When you are throwing it, you simply increase its speed by how fast you are throwing it. Keep in mind this does not work close to the speed of light. 2 It's relative to all inertial reference frames--in special relativity the coordinates of one inertial frame are related to the coordinates of another by the Lorentz transformation, and this transformation has the property that anything with a coordinate speed (change in coordinate position divided by change in coordinate time) of c in one inertial frame will ... 2 An example of a law that is not invariant:$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 ... 2 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 ... 2 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 ... 2 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. 2 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 ... 2 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 . 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 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 ... 2 In principle in Newtonian mechanics the rest frame can be any of the bodies in a gravitational complex. The geocentric system is one possible rest frame and a one to one mathematical transformation exists going from a heliocentric to a geocentric system. It is when one introduces the concept gravitation, a theory that explains the orbits, that the ... 2 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 ... 1 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 ... 1 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 ... 1 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 ... 1 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 ... 1 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: $$... 1 The columns of a 3×3 rotation matrix contain the coordinates of the local xyz axes (expressed in world coordinates). With Euler angles (321) you apply the elementary rotations R_Z, R_Y, R_X in sequence to form the local → world rotation matrix. That is$$ E = R_Z(\varphi) R_Y(\psi) R_X(\theta)$\$ This is interpreted as each rotation occuring about ...

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In a non-accelerated reference frame, a point or a particle cannot be motionless wrt. light, for the reason that busukxuan mentions: All observers will measure the speed of light to the same value, c. But in an accelerated frame, such as an expanding universe, it is possible. For instance, galaxies that are farther away than about 14 billion lightyears, ...

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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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