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In the quote cited you could imagine that point particles move in a straight line at a steady velocity and don't rotate when far from massive objects. So to a small accelerating frame, they look like a non rotating point particle moving at a constant velocity would look to an accelerating frame. But maybe point particles near a massive object have similar ...


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Newtons 2nd law only holds in an inertial frame. If you use the center of mass of the rolling object as your frame of reference then the frame will be accelerating and F=MA won't hold. However if you fix the frame of reference to a relatively stationary point, say the surface of the earth, it will. Even in this case the frame is not truly inertial, but ...


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Hint for (a) : Use the Figure above to prove that : \begin{equation} \tan\left(\phi^{\prime}-\phi\right)=\dfrac{w\sin\phi}{c-w\cos\phi}=\dfrac{w\sin\theta}{c+w\cos\theta}\;=\;-\tan\left(\theta^{\prime}-\theta\right) \tag{a-01} \end{equation} This explains why I find the last equality instead of the first as I post in one of my comments : we refer to ...


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I have to say I found Newton's laws very uneasy to understand when I was in high school. I had a lot of questions similar to the OP's. I remember I asked my high school teacher exactly the same question in the OP's post. I was also confused by the meaning of mass, whether the second law is a law or a definition, etc.. My high school teacher couldn't answer ...


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a) Assume that light has to travel 1 m to reach the origin of the reference frame I, so it'll take $t=1/c $ seconds to reach. Now, using trigonometry (the sine rule in the first equation), we get: $$ \frac{\sin{\Delta\theta}}{\omega t} = \frac{\sin{\theta}}{ct} $$ $$ct\cos{\Delta\theta} = 1-{\omega t}\cos{\theta}$$ Dividing the first equation with the ...


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From those we can get $f(x(t),t) = x + Ct + D\,$ after integration. How can I interpret this $f(x(t),t)$? The constants of integration $D$ and $C$ are, respectively, the displacement between the origins of the two frames at time $t=0$ and the constant velocity with which the origin of the primed frame moves with respect to the origin of the unprimed ...


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The above answers do not take into account the vector nature of $\omega$. I venture to provide an answer that hopefully should satisfy the questioner. We Consider a point at colatitude $\lambda$ on the Earth's surface in the Northern hemisphere. We draw a local coordinate system at this point such that the x-axis points south, the y-axis points east, and the ...


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Actually, it's NOT true that in SR the speed of light in vacuum is the same for all observers, regardless of the motion of the light source. This is true only for inertial observers. The same applies for GR, in which the generalization is a "freely falling frame" (a local inertial frame without effects of gravity). A good reference: Speed of Light


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Correct; in general the speed of light is constant only as measured by local inertial observers. As an extreme example, consider a photon emitted from a galaxy far, far away, in our direction. Although it moves away from the galaxy in the direction of the Milky Way, the expansion of space makes it increase its distance from us. Eventually, however, it will ...


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This might or might not be responsive to the question you intended to ask: Suppose you've got a meter stick. Over a period of time, I apply identical forces to the front and back ends of that meter stick, causing them to accelerate in the same direction. Therefore the entire stick, being a rigid body, accelerates in that direction. After a while, the ...


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Despite their superficial similarity Lorentz contraction and time dilation are different things and this is why there isn't a distance version of the twin paradox. To see the difference you need to understand that a clock is a form of odometer. Suppose you start at the origin and travel 100 metres, then the odometer you carry will show the total distance ...


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It is assumed in the box car example that the observer O on the ground is stationary (meaning the box car is moving past her at a speed $v$), and the observer in the box car is traveling with the boxcar's speed of $v$, and would therefore believe herself to be stationary and the observer outside the box car to be moving backwards with speed $v$ (i.e. ...


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Rotation is absolute. And in any non-rotating reference frame, the earth and the sun (ignoring all the other bodies in the solar system, and the rest of the universe) both revolve around their common center of mass. Since the mass of the sun is so much greater than the mass of the earth, it is pretty close to saying that the sun is stationary and the earth ...


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The thing you have to get your head around is that there is no 'absolute' velocity. This is because there is no 'absolute' reference frame. A reference frame in special relativity is a coordinate system that is not accelerating. If you are traveling at a constant velocity you are a reference frame. Any measurements you make of the velocity or distance or ...


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Let's just say, for the sake of argument, that the equator has a linear eastern velocity of 1,000 mph (actually 1,039). If you release a ball (presumably fired from a cannon) southward from a latitude of 70° north, your ball has an eastern velocity of about 342 mph (cosine of 70° = 0.342). As your ball travels over 45° north, the ground beneath it will have ...


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The direction of deflection of the ball should be independent where the observer is. Left or right deflection is with respect to an observer facing in the direction the ball is moving. Are you asking about a ball being thrown from the northern to the southern hemisphere? In that case the deflection would change from deflecting right to deflecting left as ...


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This is indeed confusing. The confusion comes from this very peculiar hypothesis: What if the person doesn’t apply a tangential friction force at his feet? It implies there is a radial contact force at the person's feet (I prefer "contact" to "friction", which refers to movement). And, indeed, for the person to move radially inwards, or even to stay ...


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According to both observers, if nothing else happen, the 2 ships are going to collide in exactly 10 seconds. For concreteness, stipulate that both spaceship's clocks will read 0 at the instant of the collision (assuming the bomb does not explode first). Now consider the following question regarding the spaceship without a bomb: when the spaceship's ...


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The vector $\mathbf r$ does change, even though its magnitude is nearly constant. Most importantly, the component of $\mathbf r$ which is perpendicular to the rotation axis is decreasing in this example. This fact leads to the explanation for which you are searching. Another way of seeing this would be to use this definition of angular momentum, which ...


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The following will try to briefly address your specific issues: Questions 1-2: The discussion leading to the paragraph in your link starts in Sec.VII of that ref. (see first 3 paragraphs therein), with the assumption of the postulate of the speed of light. Hence Einstein is endeavoring to show that accepting the light postulate necessarily implies ...


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I think that, rather than trying to pick apart the logic of the thought experiment which may or may not be well-worded (yes, I realise I just criticised Einstein), it is easier to just work one out of your own and see what is going on. Experimental setup To that end, consider two frames of reference moving relative to one another: one is going to be his ...


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In my view the motion of the person, seen in lab frame, would be linear motion, because, at the beginning, the person has a tangential velocity $ωr$ and radial velocity $v$, and he will keep these two forever. But then does it makes sense to talk about conservation of angular momentum? I mean it will surely be conserved in the lab frame but the ...


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After reading your question again I gave a try proving why and how angular momentum is conserved. In the example given by the question one needs to understand how the angular momentum of an object moving in a rotating, non-inertial frame is conserved. I will present in brief, my effort of proving how and why the angular momentum is an integral of motion. The ...


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This is not really a relativistic effect. This is simply a consequence of information delay due to the speed of light being finite. $L/c$ is the time it takes for light to travel the distance $L$. Hence the information transmitted by light over the distance $L$ will be delayed by $L/c$. If the two clocks were exactly synchronized (in the stationary reference ...


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The error is just to consider an average speed $h\omega$. When the particle is at height $z$, its horizontal (relative to the Earth) speed is $v=2z\omega$. The time of of flight is $$t=\sqrt{\frac{2z}{g}}.$$ Differentiating this expression we get the time taken by the particle to move a distance $dz$, $$dt=\frac{dz}{\sqrt{2gz}}.$$ The horizontal distance ...


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See the special relativity holds only in inertial frames of reference and in the absence of gravity. Now consider a train moving with uniform velocity comparable to the speed of light. Even at the speed of light an observer in the train cannot tell whether he is moving or the rest of the universe is moving. He feel the exact feeling as he is at rest. This is ...


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I want an answer which is free of mathematical relations.I want an insight rather than a mathematical relation. Very well. Do you see?


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As per the comments, I wasn't taking into account the relativistic addition of velocities, which is becomes relevant when designing scenarios with such high velocities. So for a observer in the point specified in my argument, the fastest objects (object #1 million, object #999.999, ...) would appear to have velocities close to light speed, but they would ...


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Take a free particle moving on a plane in polar coordinates $$ \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} r \cos \theta \\ r \sin \theta \end{pmatrix}$$ The velocity is found from the chain rule, with clear separation for radial and tangential components: $$\begin{pmatrix} \dot{x} \\ \dot{y} \end{pmatrix} = \begin{vmatrix} \cos \theta & ...


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Here is one way of looking at it via a velocity-dependent potential.$^1$ The Coriolis potential is $$\tag{1} U_{\rm cor} ~=~ -m({\bf v} \times {\bf \Omega})\cdot{\bf r} ~=~-{\bf v}\cdot ({\bf \Omega}\times{\bf r} ),$$ cf. Ref. 1. The factor $2$ comes from two different terms in the corresponding force formula $$\tag{2} {\bf F}~=~\frac{\mathrm d}{\mathrm ...


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It does indeed have something to do with time dilation. You can use the formula $$s=\frac{v+u}{1+(uv/c^{^{2}})}$$ where s is the speed of one spaceship relative to the other while u and v are their speeds relative to the Earth. I think you will find whatever values of u and v you use s will always be smaller than c.


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Fictitious forces do not exist in inertial frames. Fictitious forces result from force-fitting Newtonian mechanics to non-inertial frames.


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It is possible as soon as one is sure to be very distant from every body (gravitational source) in the universe. This is because all inertial forces behave as gravitational forces. If one is confined to stay in a closed room and observes the motion of bodies therein, he/she cannot decide whether the observed accelerated motion is due to a gravitational field ...


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1) This is one of Einstein's relativity statement. An observer in an inertial frame cannot tell whether he is in motion or at uniform velocity as both corresponds to a non- accelerating situation. You can't "feel" velocity or rest, but acceleration, because an acceleration is accompanied by a force which changes your state of inertia (equilibrium). It is ...


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In a non-inertial frame, since the frame is accelerating the person feels a force. A simple example is this. You just stand on a bathroom scale. While the frame moves, if the scale reading is not changing, then the frame is inertial. Otherwise it's non-inertial. This is how gravity is explained in general relativity. So if the person is aware of the ...


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In a non-inertial frame, observers will see fictitious forces with no reaction pair. For example, in a frame accelerating linearly forward, there appears to be a force acting backwards, and one cannot find the reaction (or source) of this force.


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The $\Omega \times$ terms signify change due to the moving orientation of the item. So the velocity vector when attached to a moving body will cause a $\Omega \times v$ term in the acceleration vector.


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If one takes a vector in an inertial frame say r and the same vector viewed from an observer in a rotating frame say r' (call it dashed frame) and observe the the rate of change with respect to time- Due to rotation of the dashed frame the tip of the vector starts rotating about the axis of rotation and any change in time dt leads to a change dr ...


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Imagine two ficed chares on an axis that can rotate (like merry go round). On the one chair is your friend that holds a ball and on the other it's you. At some point the hole system starts rotating and your friend throws the ball towards you. After he throws the ball you have changes position so the ball doen't come to you. Instead according to you it moves ...


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To add more details to jimjo's answer, I would like to explain the "at most" in my comment Vector, at most, can be covariant. Three-vectors are only covariant under rotations, but if you include boosts then three-vectors transform in a non-covariant way. Therefore, Newton's second Law is non-covariant under the full Lorentz Group. To get a covariant ...


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Newton's second law is covariant, as it does not change its form if we switch to another frame of reference. As already explained by @AccidentailFourierTransform in his comment, Newton's second law is a vector law. This means the quantities in the law are vectors, which have different values in different frames of reference. Only scalars, by definition, do ...



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