# Tag Info

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If the 4-momentum were invariant then it would be a scalar. 4-vectors are defined by the way their components mix when we change coordinates. In particular when we apply a lorentz transformation to our coordinates the inverse transformation is applied to the vector. As a simple example consider what happens to the energy when we boost. If we start in the ...

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Conservation and invariance are fundamentally different things. Conservation means "doesn't change with respect to time". While invariance means "doesn't change with respect to Lorentz transformations". Components of four-momentum transform like vector components and are thus NOT invariant under Lorentz Transformations. But that doesn't prevent them from ...

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I would guess that given that the book was designed to teach amateur astronomers that most of the book refers to measurements made with reference to earth. Now, you can measure the component of velocity along the distance line between the star and the earth by using the doppler effect and the adsorption lines for particular elements(measuring the adsorption ...

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It's pretty natural to think that a star can have velocity - there's no reason a star shouldn't be able to move. The first thing you need to know is "velocity relative to what?" Stars in our galaxy are all in some kind of orbit around the galaxy, so you can talk about velocity in galactic coordinates. Binary stars orbit each other, so you can talk about ...

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There are three mistakes that prevented you from arriving at the correct lagrangian. (1) The correct form for a CM lagrangian should be $L = T_{total} - V_{total}$ instead of $L = T_{total} + V_{total}$ I think this is just a typo since later on you did use the correct lagrangian. (2)It is not valid to assume that $T = T_{space} + T_{rot}$ since energy is ...

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If the velocity of the aether wind is a sizeable fraction of c, the apparent velocity of c will depend strongly and obviously on the direction in which the measurement is taken. Since this is not true, the aether wind velocity must be quite small, which requires a sensitive instrument to detect the effects. It was exactly this range of possible wind speeds ...

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The speed of the earth in its orbit about the sun is about 30 km/s. Michelson assumed that the speed of the earth through the rest frame of the ether was of this order.

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So, special relativity says that every frame is as good as any other frame, and there is no absolute frame of reference. All good. And special relativity is experimentally falsified and general relativity uses the word general to cover the case when inertial frames are only local and not global, such as in the universe we live in. Suppose there is ...

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Consider this. The twin paradox, but with a twist. The twin that accelerates is the one that is younger on return. Statement one It doesnt matter which direction the accelerating twin goes. Ie if she leaves from the equator and goes due north from the earth and returns, it doesnt make any difference if she had gone due south and returned. Statement two ...

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Earth is moving around sun in an orbit with mean radius 1AU. Time of one revolution is 1 year . Thus, its speed is $$v \approx 2 \pi \dfrac{1AU}{1year} = 30km/sec$$ . Sun moves s=around galactic center at a speed of $v'=220km/sec$. Thus, when you are standing still on Earth,you can have a velocity of $v'+v$ with respect to center of Milky way galaxy. This ...

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The robots apply force on ball and calculate the acceleration and discover that Newton's laws are valid; hence both the merry-go-rounds are inertial frame of references. You have leapt from the ball appearing to follow Newton's first law to the frame being inertial. This is not valid. While finding that just one force-free object does not obey Newton's ...

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Yes , I think so. Come to think of it- In earth , we can apply Newton's laws for objects in inertial frames . But Earth itself is accelerating around the sun.

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I did not read this entire post in detail; apologies for that (tl;dr). I will say that you are thinking the right way, but you don't have all the facts right. Consider the situation when stationary observer looks at the orbiting MGR with his telescope. You said that the orbiting observer sees that Newton's laws are obeyed by the orbiting observer. THat ...

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You'll see the object at first accelerate towards the hole (under gravity) and then slow more and more as it approaches the event horizon. It will asympotically freeze in place at the event horizon and then gradually shift redder and redder until it disappears. This is assuming that the black hole is big enough that the acceleration is similar across the ...

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One approach that uses the matrices you've already derived is to set $L_{bh}=L_{bw}L_{wh}$ and then solve for $\gamma$.

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Based on the rewording well-given by @WhatRoughBeast we can consider two events: the astronaut passes by Earth at $t_1 = t_1' = 0$ and $x_1=x_1'=0$, where the unprimed frame is at rest w.r.t. Earth, and the primed frame is at rest w.r.t. the astronaut. the astronaut arrives at the planet: $t_2=D/\beta$ and $x_2=D$ where $D$ is measured in light-time units ...

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But then I'm assuming the time it takes him for an observer is equal to 16 years, am I correct in doing this? I think not since that means the astronomer would be travelling at the speed of light to an observer... Okay, first thing: an astronaut (Greek, star sailor) is different from an astronomer (Greek, star namer). The former is seen at the wheel of ...

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You got it half right, but you got so focused on the correction factor that you forgot to calculate the trip time in your rest frame. In deference to Bill N, let me rephrase the question, hopefully more to his liking. The astronaut is travelling to a star sixteen light years away which is stationary with respect to earth. During this trip he ages fifteen ...

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No, you're not misreading the question. From reference frame here on Earth, light would take 16 years to reach the start 16 light years away. However, this problem takes a look at special relativity. As the astronaut travels at relativistic speed, he experiences time dilation. That means that clocks back on Earth will appear to tick fast as opposed to ...

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It's actually not a contradictory result. Each observer sees the other's clock running more slowly. You're tacitly assuming the existence of a third, "absolute" frame wherein times of other frames can be unambiguously compared, which would be a contradiction because it would require duration measurements to be well ordered. However, one of the fundamental ...

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Yes frame dragging is related to vorticity; consider Kerr spacetime for simplicity in the following. In Kerr space-time one has a time-like Killing field $\xi^{\alpha}$. The congruence of observers $u^{\alpha} = \xi^{\alpha}/|\xi|$ following orbits of $\xi^{\alpha}$ are of course the observers who are at rest with respect to the central mass and hence are ...

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I will use this reference Chapter 3 Time dilation from the classical wave equation from Understanding Relativistic Quantum Field Theory by Hans de Vries 3.1 Signal propagation: The bouncing photon clock The classical wave equation tells us that the propagation is on the light cone, and the propagation speed is c. With this as a starting point we will show ...

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The classical fluid mechanical idea of a vortex is well illustrated by this picture, which shows a vortex created by the passage of an aircraft wing, revealed by colored smoke. In classical fluid dynamics, a region of mostly rotational fluid around an axis line is known as a vortex. This axis line can be curved or straight, depending on the physical ...

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Why don't we consider the centrifugal force acting on the bob of a pendulum while drawing the Free Body Diagram of a pendulum? It's also a sort of circular motion. First off, you meant centripetal rather than centrifugal. Circular motion requires a centripetal force, not a centrifugal force. There are two forces acting on the pendulum bob: ...

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The reason these statements are consistent becomes clear if we quote from the Landau & Rumer book a little more extensively: Ahead of us is a very long railway line with Einstein's train moving along it. At a distance of 864,000,000 kilometers from each other there are two stations. At its speed of 240,000 kilometers per second, Einstein's ...

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Yes. Under any normal operating mode today, the station rotates at the rate of once per orbit so that a particular part of the station always points at the earth. This is almost identical to the way the moon orbits the earth. But this is not something that has to happen. During assembly, there were times when the station did not rotate. So if there were ...

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Once it goes into orbit, it will remain in the same attitude, what I mean is the cupola the astronauts look out of always looks "down," was far as I know. I don't think the ISS has any spin, so it should stay in the same position. The International Space Station orbits about 354 kilometers (220 miles) above the Earth and travels at approximately  ...

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Ultimately, we are just converting between a spherical coordinate system and a Cartesian one. I'll try to stick to the notation of the paper, with one critical change: components of vectors will be denoted with superscripts, while subscripts will be reserved for identifying which vector from a set is being referenced. We have a vector $\mathbf{n}$ (aka ...

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