New answers tagged

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There is a lot of confusion in the literature regarding the so-called active and passive interpretation of transformations when it comes to scalar fields. However, this terminology and the corresponding dichotomy has its origins in the applications of linear algebra (e.g. computer vision) where it is more relevant and the concepts are more clear. The ...


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you can start with the position vector of the ball bearing: $$\vec{R}_B=\begin{bmatrix} x+f_x(s) \\ f_y(s) \\ \end{bmatrix}\tag 1$$ where x is the car position and $f_x(s)\,,f_y(s)$ are arbitrary function that described the position of the ball bearing relative to the car. and the position vector of the car $$\vec{R}_C=\begin{bmatrix} x \\ 0 \\ \...


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I do not know whether I should analyse the ball bearing in its equilibrium (i.e. A=constant,x′=constant) to obtain some expressions for $y$ I strongly believe that's what the question means. To be more precise, I would frame the question as follows: "The equilibrium position of the ball bearing relative to the ramp is used as a measure of the ...


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Charge is Lorentz invariant, meaning it is the same in all frames of reference. This means that four current is a four vector. This is because, for example, the time-like component is charge density, $\rho =\frac{dq}{dV}$. Because length only contracts in the direction of relative motion, volume only decreases by a Lorentz factor, the same as length. If ...


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If by point particle, you mean really a point, that is something with zero length, well it's not possible because orange has to move with the speed of light in that case, which is not possible to begin with. I am not sure what do you mean by traveling infinitely far though. At some point, the front of needle will arrive at the skin of orange, assuming needle'...


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It depends on context: In the context of affine spaces (such as, Newtonian mechanics or SR), positions are strictly speaking not vectors, but position differences/displacements are vectors, i.e. positions measured relative to some chosen origin/fiducial point are vectors. In the context of manifolds (such as, GR or Newton-Cartan theory), positions are ...


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No, positions are not really vectors in a vector space. They are points in an affine space https://en.m.wikipedia.org/wiki/Affine_space . An affine space comes with a vector space though and you can use the vector space to parametrize the affine space nicely. This, is of course, the classical (and in fact, special relativistic I.e. no general relativity) ...


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Historically, vectors were introduced in geometry and physics (typically in mechanics) Geometry means three dimensional space A Euclidean vector, is thus an entity endowed with a magnitude (the length of the line segment (A, B)) and a direction (the direction from A to B). In physics, Euclidean vectors are used to represent physical quantities that ...


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Setting the partial derivative of $\theta$ with respect to both x and y equal to zero yields the following two equations: $$\frac{5}{4}x+\frac{3}{4}y-100=0$$ $$\frac{5}{4}y+\frac{3}{4}x-100=0$$The solution to these two linear algebraic equations for x and y is $$x=y=50$$The 2nd partial derivatives with respect to both x and y, evaluated at this point, are ...


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That depends on the molecule and its symmetries. For example, a molecule of an ideal monoatomic gas cannot rotate because it is spherically symmetric. A typical diatomic gas molecule, like $N_2$ could rotate about two axes because it is axisymmetric so it cannot rotate about its axis of symmetry. A more complicated gas molecule, like water vapor, can rotate ...


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You need to be careful with words here. First, many of the things that 'happen' in relativity only 'happen' for certain observers. It might be less confusing to think of them as appearing to happen. While your spaceship is not accelerating it is at rest as far as you are concerned. If you put something outside, it stays there. If your spaceship is ...


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I don't agree with your first equation because $\vec{v}_{P/M}$ isn't in the same coordinate frame as $\vec{v}_{O'/F}$, but rotated somewhat. First we consider the general case, where P is not coincident with O' and is not fixed on the body. Position Kinematics $$ \vec{r}_{P/F} = \vec{r}_{O'/F} + \mathrm{R}\, \vec{r}_{O'P/M} \tag{1}$$ Velocity Kinematics ...


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An impulse acting at point A gives the 2m mass its velocity, and the center of mass a parallel velocity of (2/3)V. Since the impulse is not acting in line with the center of mass, it also provide an impulsive torque which causes the system to rotate around the center of mass. Working in the center of mass system is valid and gives a tension which is the ...


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I have come to realise that no direct relationship such as $\omega_x = \frac{d \theta_x}{dt}$ emerges out of the definition for angular velocity $\overrightarrow {\omega} =\widehat{n} \frac{d \theta}{dt}$. If you see the first picture, you will notice that ${\omega_x}$ is 0, since $\overrightarrow {\omega}$ points in $\widehat{k}$ direction. However, in ...


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It is not possible to reverse the structure of the atom. Electrons have a very low mass and (relatively) protons are quite massive. Because of the uncertainty relationship between position and momentum the more tightly confined a particle is the more certain its position and the more uncertain its momentum. As the uncertainty in momentum increases so does ...


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The thing that i find very confusing in that quote … is that Callahan is apparently saying that by virtue of being in a non-inertial frame (while still being in a Minkowski space) spacetime is automatically curved. That is not what he is saying. After considering “disagreement” between “radar grid” and “ruler–clock grid” and noticing that similar ...


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The usual way of deriving the equipartition theorem is through manipulation of an ensemble average. Equilibrium ensemble probability densities in phase space depend on the canonical coordinates through the Hamiltonian. The kinetic part of the Hamiltonian of a system of $N$ particles in the 3D space has the complete rotational symmetry, therefore, at the ...


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After going through the text again, I wanted to give an answer more motivated by the exposition in the text. From, p.63, since the laws of nature are invariant under certain space-time transformations, if $A | \phi_n \rangle = a_n | \phi_n$ with $A$ representing an observable, and $\phi_n$ an eigenvector, then after the transformation $A' | \phi_n' \rangle =...


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The $E=\frac{k_B T}{2}$ per degree of freedom is an average value. We do not consider the motion of a particular particle, but the ensemble average. Thus, if we have a single particle in free space, we can not associate a temperature with its motion. We need random motion to associate the kinetic energy per particle with a temperature.


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It's easy to fudge signs in this business, so why don't you just utilize the coordinate space representation of the momentum operator, $$ \mathbf{P} = \int d^3 \mathbf{x} ~| \mathbf{x}\rangle ( - i \hbar \nabla) \langle \mathbf{x}| \implies e^{-i\mathbf{a}\cdot \mathbf{P}/\hbar}= \int d^3 \mathbf{x}' ~~| \mathbf{x}'\rangle e^{- \mathbf{a}\cdot \...


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I will give this a try. Shift to the reference frame of the center of the ball. The point on the ball (call it A) now rotates wrt the center at an angular velocity $\omega_{1} = 2\pi r$ whereas the point O rotates around it at an angular velocity $\omega_{2} = 2\pi R$ in the opposite direction. [Since rotation and revolution of the ball around O are in the ...


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You put considerable effort on merry go rounds (Even when seated) in order to subconsciously show there is no such thing/s. Or you are just wired up to face the rotating system you are in and your brain is taking charge. and $a=0$. You can't fool the Coriolis part is a lot more sneaky. Just look at that term with the angular velcoity vector crossed with ...


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The confusion is not realizing that real forces are real, and fictitious forces are not real. Therefore, they can not "cancel each other out" in real life. Using a rotating frame of reference and introducing fictitious forces is just adding and equal quantities (a force, and a mass $\times$ acceleration) to both sides of the math equations. It has nothing ...


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For a mass moving in circles, like a child in a merry-go-round, there has to be a centripetal force that we (I suppose) have to write. It is a real force, after all, making it move in circles. But, in that case, wouldn't the centripetal term and the centrifugal term cancel each other out every time we have a situation like this? If you are in a frame ...


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To solve it in the ground plane you need to separate out in your mind the rotation about a common centre of mass and the movement of the centre of mass. You might then need to think about the special case of such motion in which one of the masses stops moving from time to time in a particular reference frame. Googling 'cycloid' might also shed some light. ...


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If the force is directly in line with the center of mass of any unrestrained object, it will cause linear acceleration only, if the force is not directly in line with the center of mass, it will cause linear acceleration and rotation. Consider the book on the table, if you push it away from you on the center of an edge of the book (in line with it's com), it ...


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The relationship $\boldsymbol{F} = m\, \boldsymbol{a}_{\rm CM}$ always holds true as long the acceleration is that of the center of mass. The is true because the definition of momentum is $\boldsymbol{p} = m\, \boldsymbol{v}_{\rm CM}$, and force is the rate of change of momentum. When a force is applied away from the center of mass (say point A), not only ...


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In the case of extended bodies $\mathbf F = m \mathbf a$ applies only to the centre of mass. Other points experience net torque which causes them to gain additional angular acceleration hence for such particles $\mathbf F \neq m \mathbf a$


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Almost exactly right. The axle provides a force in the opposite direction that's equal and opposite to your applied force. But it's not a reaction force to your force. It's a separate force. An analogy would be a book lying on a table. The book applies a force to the table. What keeps the table from accelerating? The normal force of the floor on the ...


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In classical mechanics the addition of velocities is based on Galilean transformations whereas in special relativity it’s based on the Lorentz transformations. Hope this helps.


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Start with equation (38) $$M(q)\,\ddot{q}+V(q,\dot{q})\,\dot(q)=B(q)\,\tau-\Lambda^T\,\lambda$$ with : $\dot{q}=S(q)\,\eta$ and $S^T\,\Lambda^T=0$ you can eliminate the constrain forces in (38) and get equation (46) $$\bar{M}(q)\dot{\eta}+\bar{V}(q,\dot{q})=\bar{B}(q)\,\tau$$ to solve this equation, you use equation (15) and (16) and get: $$\eta=J\,\...


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The solution given in the paper is correct. The additional terms are the equivalent inertia due to the rotating wheels and terms stemming from centrifugal forces. As your chosen reference frame is not an inertial reference frame, you have include fictitious forces that you wrongfully did not consider in your derivation. I have only taken a quick glance at ...


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Time dilation on living creatures (and more in general to any experiment scenario not limited to light or e.m. wave propagation phenomena) is at the moment still only a speculation of the applicability of SR theory and no experiment has been conducted as far as I know to prove it. To note that such a situation, i.e. with different ageing scenarios depending ...


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Let me see if I understand your question correctly: you have an object (say a rod or board) inclined at some angle $\alpha$ in some reference frame $S$, and you would like the find the angle $\alpha^\prime$ in a reference frame $S^\prime$. Since you've mentioned $\alpha^\prime$ as the angle from the rod's frame, I'm using the convention that the rest frame (...


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Short answer: centrifugal force may not be a real force but centripetal force certainly is. Long answer: A weighing machine (which I am assuming is equivalent to a spring balance) is not measuring the force with which the Earth attracts an object. It is actually measuring the reaction force that the ground (or, to be exact, the spring) exerts in the object. ...


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Psuedo-forces aren't reality, they are the need of humanity! You must account for a centrifugal force in a non-inertial reference frame otherwise you can't use $F = ma$ to figure out the dynamics of the problem. Though as you may have noticed Newton's third law still isn't valid in such frames. And you must have observed that Earth is a non-inertial ...


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My friend, however, made a simple answer to this anomaly as to which I am not convinced. He claimed that EEP is applicable only for the observers inside the shuttle and on the planet. That is, a Schwarzschild observer is necessarily not able to use EEP for other noninertial observers from his own viewpoint. Do you think his answer is valid? Yes, your ...


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As Floris pointed out centrifugal force is a pseudo force. To get a deeper insight you may refer this: https://youtu.be/nh26C519wuo


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The equivalence principle gets stated in all sorts of different ways, but it is basically the statement that the four-acceleration is given by the equation: $$ A^\alpha = \frac{\mathrm d^2x^\alpha}{\mathrm d\tau^2} + \Gamma^\alpha{}_{\mu\nu}U^\mu U^\nu $$ The first term on the right hand side is basically the coordinate acceleration, i.e. rate of change of ...


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EEP tells you that effect of gravitation is locally indistinguishable from that of an acceleration, so locally you can forget gravitational field and you can just use special theory of relativity (STR). Locality is important though. For example, you cannot simulate radial field by acceleration. Moreover you can only consider area so small, that even tidal ...


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Regarding the aforementioned Michelson, Gale and Pearson experiment in 1925, this commentary (below) appears on the flat earth society wikisite. The analysis by Dr. Paulo Correa and Alexandra Correa can be found at www.aetherometry.com/publications/direct/AToS/AS3-I.2.pdf. These scientists stated that the Michelson Gale Pearson experiment was inconclusive ...


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Let we just stick to the point of view of the traveler. To him, as he is accelerating, he would continuosly see outside distances contract and outside time slows down. The correct frame to represent the point of view of a uniformly accelerating traveler is called the Rindler frame. This frame does not behave as you suggest. As other things move faster ...


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This is a question that has to wait for the definitive quantization of gravity. And in this case , it would not be definitive if, in the phase space region of overlap with special relativity and classical General Relativity ,it would give different results. At the moment gravitons are invented so that gravitational interactions will be analogous to the ...


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Note that special relativity is all about flat spacetime, so bringing the rest of the universe into these types of problems can lead to a lack of clarity. Regarding your inertial observer who has accelerated to $c-\epsilon$ for some time, the magic of special relativity is that he is still at rest. Locally, his spacetime looks exactly the same as before he ...


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I think I know what you mean. You are asking "how does the traveler account for the fact that she is unable to reach or surpass the speed of light relative to the rest of the Universe, in terms of what she is observing happening in her rest frame?" Qualitatively, you have the right idea regarding that as she accelerates, the "size" of objects in the rest ...


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Though I also learned the energy requirement argument from pop-sci source first, once I studied relativity I've come to feel that it skips past a more fundamental reasoning. Let's define a minimal experiment with which to discuss the problem. We start by assuming a heavily fueled and very efficient spacecraft in free-fall in an otherwise empty universe. ...


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If the small disk did not rotate relative to the large disk then its rotation rate (angular velocity) relative to us would be the same as that of the large disk $\omega_0$. This answer does not depend on the distance $x$ between the centres of the two disks. If the small disk rotates at rate $\omega_1$ relative to the large disk then its rotation rate ...


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If I'm understanding your question correctly, and after having read the comments, I'll just try to put my two cents in. First, you're asking about something a traveler sees from the traveler's point of view. I take that to mean in the traveler's rest frame, because by "point of view" we usually mean this. In the traveler's rest frame, the traveler is always ...


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This is the case of hyperbolic motion. Let me shift the thought experiment to a related case. In a linear accelerator a particle is accelerated by an electric field. You can accumulate kinetic energy as follows: You create an electric potential that extends from the current position of the particle to some distance ahead of it. By the time the particle ...


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