# Tag Info

0

Yes you would actualy see the car with its speed added with your speed.

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Your understanding is correct. And switching between different perspectives (what physicists would call different inertial frames of reference) like that is a very useful tool in physics, because it turns out that the laws of physics have the same form no matter which inertial frame of reference a problem is described in. For more information, see ...

2

Rough impressions can be misleading. The other car really is moving 200 km/h from your point of view. One thing to keep in mind is that you tend to perceive motion more readily when it is closer to you. This is at least partly due to the fact that you really only can see angular speed across your field of view (like degrees per second). To convert this into ...

3

Suppose the Earth wasn't rotating, but instead you impart a small sideways velocity to the pendulum bob as you release it. What you would have is a conical pendulum that traces out an ellipse instead of a straight line. Now start the Earth rotating. The point of the Foucault's pendulum is that the rotation of the Earth doesn't affect the motion of the ...

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Nothing happens to the initial angular momentum. It is simply irrelevant. Let's imagine that the pendulum is placed on a non-rotating body (or at the earth's equator). When the string is cut, the pendulum falls straight to the center. The path it traces is a straight line. In the non-equatorial case, the pendulum structure has some rotation. At the ...

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You do have to define the velocity with respect to something. The idea (I think) behind defining wrt fixed stars is that the Earth is in motion wrt the stars, too. Barring friction, the moment the rope is burned, the pendulum is only under the influence of gravity and the restraining force of the suspension rope -- analogous to swinging a weight around ...

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Suppose you have a constant angle $\theta$ slope in the original frame (we suppose that the transitions from horizontal movements to the slope is quasi-instantaneous). Call $x$ and $x'$ the horizontal displacements in the original and moving frame. Call $T$ the total time for going to $z=h$ to $z=0$. Then you have $x'= x- v_0 T$ With $x= h \cot \theta$, ...

1

I've heard that inertial frames are frames in which Newton's laws hold. The modern view of Newton's first law is that it defines the concept of an inertial frame. It also, at least conceptually, provides a mechanism for testing whether a frame of reference is an inertial frame. Suppose you know that no forces act on some particle. If that particle ...

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1)Definition: An inertial frame of reference is a frame of reference where Newton's first law applies (uniform motion if without external force). Now if we have other frame of references that are moving relative to this inertial frame with uniform relative velocities, then all the others are also called inertial frame of references. 2)Transformation between ...

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I believe the difference comes from the fact that forces can do different amounts of work in different reference frames. In particular, the normal force by the ramp does no work in the "lab" frame, but does do work in the moving frame (since there is a component of velocity that is now parallel to the normal force). I don't think you accounted for this work ...

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Physicist created momentum as the property of the system that is conserved if on the system act only internal forces. A force is defined internal in the system if, the force itself and its reaction are applied on the system. From the previous hypotesis you can demostrate that momentum is defined as $$\vec{p} = \sum m_i\vec{\upsilon}_i$$ The last example ...

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An elastic collision means that the over all kinetic energy of the entire system before and after the collision is the same. So the ball can bounce off the wall, and the wall can recoil in such a manner that you have an elastic collision.

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Momentum is conserved in magnitude and direction. So in order to analyze any situation of momentum conservation, you should always start with $$\sum \mathbf p_{i}=\sum\mathbf p_f$$ where the subscripts denote the initial and final momenta. As to the ball & wall, you are correct that momentum is not conserved if you are only looking at the ball. If you ...

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It is the momentum of the entire system that is conserved. The fundamental reason for this is that the laws of physics are the same everywhere in space. This argument for momentum conservation is called Noether's Theorem. So where did you go wrong in your original example? Well you assumed that the wall was completely rigid. In reality that isn't actually ...

1

Linear momentum will be conserved when the Lagrangian that describes your system is unchanged by translations in space. This is a consequence of Noether's theorem, and it's as close as you're going to get to a fundamental explanation for the conservation of momentum. In general a Lagrangian written in the coordinate system of an accelerating observer is not ...

0

Okay, let's say you have a friend staying at home measuring the time and observing you with a telescope - the whole situation will be described from his point of view. Namely, we will use his measure of distance $x$ and the time $t$ he measures with the stopwatch in his hand. Special relativity tells us that he has a means of calculating your "proper ...

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My apologies that I can't follow your question enough to answer everything, in particular I can't see how you project your images. If you extended them in what look like straight lines to you, you'll get radial lines expanding from your position, but it seemed like you wanted a Cartesian grid. However I do want to answer the part about strain and its ...

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The key is the coriolis force. The coriolis force is $F_c = -2m\Omega \times v$. Here $\Omega$ is the rotation of the frame of reference and $v$ is the linear speed of the satellite. If you do the calculations, left as an exercies for the reader, you'll get the missing force. In case 2 the coriolis force is 0, because the velocity $v$ has to be used in ...

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I'm afraid your statement that magnetic field at a point can have only on(e) "constant value" is not true, and you will have to learn a bit about special relativity to understand why. There is a classic thought experiment, which is to imagine the electromagnetic fields due to a charged particle at rest in what we'll call the stationary frame. Clearly, the ...

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Answer to first question: At the instant being considered, the space and body axes are identical, so at that moment the matrix $a$ that relates the two sets of axes is simply the identity matrix. $dG'$ is a vector, so $a_{ji}dG_j' = dG_i'$ is simply equivalent to the statement that with $I$ the identity matrix and $V$ an arbitrary vector, $IV=V$. Answer ...

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Planes of simultaneity in special relativity don't really mean much of anything. The real physical structure of spacetime is in the light cones. The takeaway from "relativity of simultaneity" is not that there are "different time orderings for different observers", but rather that there is no meaningful time ordering for spacelike separated events. They ...

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What you're asking about is the existence of surfaces of simultaneity. In SR, surfaces of simultaneity can be defined by measurement procedures such as Einstein synchronization, and they turn out to depend on one's frame of reference. In GR it gets a lot tougher to do this. We don't even have global frames of reference. It turns out that what you need in ...

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"Most probably in reality there are some extremely complex laws and equations which makes this question more complicated." Not really. The equations are rather straightforward. Let's measure velocities in units of the speed of light and let's denote the velocity of $B$ as observed by $A$ as $v_{BA}$, the velocity of $A$ as observed by $C$ (the ...

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It sounds like you're interested in when a spacetime admits a Cauchy surface. The answer is that every spacetime that is globally hyperbolic has this property. This was proved by Geroch in 1970 (article here, see Section 5). This includes most of the textbook relativistic spacetimes --- Schwarzschild, Kerr, FLRW, and many others. But there are some ...

1

That light has a fixed velocity in vacuum comes from observations . In order to fit the data Lorenz transformation were imposed on the rigorous mathematical model for electromagnetism, Maxwell's equations. It was the result of attempts by Lorentz and others to explain how the speed of light was observed to be independent of the reference frame, and to ...

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In your frame of reference, it does indeed look as though the difference in speed between A and B is greater than $c$. But the question is - does A think that B is moving away at that speed? And the answer is "no". There is a thing called the Lorentz transformation which describes how the observed speed of an object is a function of the speed of the ...

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"How do we know that clocks slow down relative to each other?" Experimentally. This has been observed many times in the lab. The same answer is true for ANYTHING in physics and science in general. We only know that it is true, because we have experimental evidence for it.

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No, since by the principle of relativity: A body in constant velocity motion cannot determine whether it is in motion in a certain direction or whether everything else is in motion in the other direction. No physical experiment can determine this hence for all purposes a body in motion will simply claim that it is at rest while the other body is in motion. ...

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But magnetic field at that point can have only on "CONSTANT VALUE". Not true. If a particle is acted on by some combination of electrical and magnetic forces in one frame of reference, then in another frame of reference, it will be a different combination of electrical and magnetic forces. It's possible to have a force that's purely electrical in one ...

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If you set your reference frame to be fixed to the swing (let's call the origin the pivot point of the swing), you are now dealing with a rotating reference frame. In a rotating reference frame, all objects observe a centrifugal force that pulls them outwards from the origin. For any stationary object in your rotating reference frame (the seat on the ...

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Your reasoning should be improved. In an inertial system there are no centrifugal force. You may in that system speak about a centripetal force, which is not a real force but an expression for how much net force is needed for the object to perform a circular motion. You probably mean that the expression for the centripetal force should be equal to the ...

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The crucial point here is, that a reference frame where photons are at rest simply cannot be defined in special relativity: There is no Lorentz transformation which transforms you from a given inertial system into a reference frame or a space-time where some photon is at rest. This, however, means that concepts such as travelled distance and proper time do ...

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Photons are not considered as observers which would be able to observe their proper time. But if you would do so, their hypothetical proper time would be zero. You obtain this result by multiplying the time observed by any real observer with reciprocal gamma (which is 0 for v=c), see http://en.wikipedia.org/wiki/Time_dilation. However, you may not forget ...

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Why do i have the sense that this can have a meaningful physical interpretation (even if velocities are both $c$)? First lets assume that both velocities are $c$ or $1$ in the question's notation. Then the relativistic velocity addition formula simply gives: $1_A \oplus 1_B = 1_B \oplus 1_A = 1_{tot}$, correct? Yes, does the fact that this is mathematicaly ...

2

To make this easy, I will assume that all speeds are small compared to the speed of light $c$. The motion is in one dimension, so the trajectory of the space vehicle will be the following: $$x(t) = x_0 + v_0 t + \frac12 a t^2$$ We can easily set $x_0 = 0$ if the put the origin of the coordinate system at the start point where the space vehicle started. If ...

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Your friend is correct. The acceleration of the projectile is determined by the thrust its rocket motor can produce. If the acceleration of the projectile is $2g$ then the thrust of its motor would be $2mg$. But launching the projectile from your space vehicle can't increase it's thrust. You can increase its initial velocity, but once the projectile has ...

0

You are right. Torques are not free vectors, the same way that linear velocity is not a free vector and it is associated with a specific point. When the point moves, the components of torque and velocity change according to the same transformation law \begin{align} \vec{v}_B & = \vec{v}_A + (\vec{r}_{A}-\vec{r}_B) \times \vec{\omega} \\ \vec{\tau}_B ... 1 Torque is defined as the cross product of two vectors: r and F. r points from the pivot point to the point where the force F is applied. Both vectors are not bound and stay the same vectors no matter where you put them. However, if you combine them via mathematical operations, you have to stick to their rules (here: T1=r2*F3-r3*F2 etc...). There are no ... 2 Some key reading if you want to understand this stuff is chapters two to five of the IERS Technical Note 36, the IERS Conventions (2010). It's not just the J2000/FK5 frame (aka the EME2000 frame) that is associated with some epoch date. Every Earth-centered inertial frame has some epoch date. There are two fundamental reasons why this must be the case: ... 0 A simple pendulum would be a good experiment to detect both non-inertial frames - rotating and linearly accelerating: If you know weight of the pendulum in inertial frame, in rotating frame, its weight would decrease because of radially outward centrifugal force acting on it. (If earth stops rotating, our weight would increase!) Think: what would happen if ... 4 That is exactly the point: if the field is in a vacuum with respect to observer A, and observer B accelerates uniformly with respect to A, then B will observe a field state with nonzero particle content. It doesn't really matter whether you talk about different observers on different frames of reference, or of a single observer who 'switches' their frame of ... 2 Yes, I can think of two ways to do this and there may be more. In a rotating frame the acceleration is a function of distance from the pivot. If B has sufficiently precise instruments the variation of acceleration with position will be detectable. However if B is confined to a very small space, or doesn't have precise enough instruments the variation of ... 1 The friend rotating and experiencing the centrifugal force may observe several effects that his linearly accelerating friend doesn't: the acceleration at different points of the box is slightly different i.e. the apparent gravitational field is non-uniform there is the extra Coriolis force acting on objects that are moving relatively to the rotating frame ... 1 Where's the flaw? Here's one: Let γ be the squareroot of 1−v²/c² but, in fact,\gamma_v = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$But, there is a flaw in your reasoning too so correcting the error in the formula for \gamma will not get you to the correct answer. Let's work it out with coordinates to see explicitly what's going on. Let the ... 0 Your question exposes the importance of defining notions in physics unambiguously and universally in terms of "How to measure?". As Einstein put it explicitly (however, referring specificly only to the notion of "simultaneity", and unfortunately only as late as 1917): "We thus require a definition of simultaneity such that this definition supplies us with ... 2 For Special Relativity (SR) i think the Michelson-Morley experiment is compatible and provides a verification of SR principle (some other formulations are also compatible with the experiment). Quantum Field Theory and especially the Dirac prediction and verification of positron is also a verification of SR (and many other expreriments in this context) For ... 0 I don't have the reputation to comment, so I'll comment about David Hammen's (accepted) answer here. His conclusion is correct, but mentions "Those formulae do imply a singularity for the clock that is closest to you" in reference to the forumale he thought you were referring to. He also mentions "In between, you'll get a nice continuous change from faster ... 1 Actually it's not that difficult (but a neat problem), there's only one crucial step in the development that I will show you, but let's start from a bit earlier. Let's first write out the two forces on interest here (in terms of magnitudes, as we know already they're on parallel trajectories): The coulomb repulsion between the charges:$$F_C = ...

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A more particle-based view of zakk and Moonraker's observation that photons experience zero time between emission and absorption is this: The postulated event that a massless particle (say a photon) could emit another massless particle (say another photon) in the retrograde direction — or in any direction for that matter — is always zero. A photon from its ...

0

I want to check that I get it right. [...] [...] we would be associating time with objects and not with the points of space itself. In physics, the word "time" is used in various related but quite different meanings (which I describe in some detail below). Therefore, if you care to "get it right" (which is of course commendable) you should avoid ...

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