# Tag Info

2

Instead I found (for the logistical equation and the Lorenz system) that the logarithmic plot of the distance between trajectories is constant at the start This should not happen and I cannot confirm this. Here is the separation of two trajectories I get for the logistic map (averaged over 10000 realisations): And here is the same for the Lorenz ...

3

The behavior is approximately constant near the beginning because at very short timescales, the transformation of the phase space induced by the time evolution is basically a deformation where the distances change by a factor of $O(1)$. The middle phase is when the chaos is actually growing and the distances are growing exponentially. At the end, at very ...

2

There is a well-known isomorphism between the linear space ${\mathcal M}_{m, n}$ of $m\times n$ matrices and typical (vectorial) linear spaces ${\mathcal L}_{m\times n}$ of dimension $\text{dim}({\mathcal L}_{m\times n}) = m\times n$. Everything that is valid in ${\mathcal L}_{m\times n}$ has an equivalent in ${\mathcal M}_{m, n}$ and conversely. For this ...

1

This is more or less an exercise in chasing definitions. The adiabatic invariant $I$ is defined as $$I\equiv \oint p \frac{\mathrm{d}q}{2\pi}\tag{49.7}$$ where the integral is taken over the path for given $E$ and $\lambda$. The external parameter $\lambda(t)$ is a slowly varying function of time $t$ in $\S49$, but is assumed to be a constant in $\S50$. ...

1

Here's a simple diagram of the Atwood machine you describe. For reference: $F_{total} = m_{total} \cdot a$ Let us call $g$ the gravitational acceleration, $m$ the mass of the less massive "block," and $a$ the acceleration of the "duel-block-system." $$F_{total} = m_{total} \cdot a$$ $$a = \frac{F_{total}}{m_{total}} = \frac{F_{m} + F_{3m}}{3m + m} = ... 1 I) There are already several good answers. OP is asking about the momentum of the non-relativistic string in the transversal model, which in textbooks usually is given as$$ {\cal L}_T ~:=~\frac{\rho}{2} \dot{\eta}^2 - \frac{\tau}{2} \eta^{\prime 2}. \tag{1}$$II) Let us fix notation: \rho is the 1D mass density; \tau is the string tension; Y is the ... 0 Point C is sure instant centre of rotation for both gears, otherwise they would get teeth broken if any relative slide to each other. Analogy is a wheel on a road having instant centre of rotation at the bottom point thus velocity of top point is twice more than of the car. As far as I understand, confusion point is that first you calculate VO in respect ... 4 The way I understand the setup: The block is initially at rest at some height above the spring. The spring is initially at rest oriented vertically with one end on the ground and at it's natural length (though if its not a massless spring it would compress somewhat due to its own weight). Then, the block is dropped, it lands on the spring compressing it ... 1 Everything is a function of the angle \theta and its derivatives \dot{\theta} and \ddot{\theta}. From there use the chain rule of differentiation.$$\begin{align} x & = \ell \sin \theta & y & = \ell (1-\cos \theta) \\ \dot{x} & = \ell \dot{\theta} \cos \theta & y & = \ell \dot{\theta} \sin \theta \\ \ddot{x} & = ...

1

As you noticed, if we use Euler-Lagrange equation on $L= \frac 1 2 (\dot x^2 + \dot y^2) -mgy$ we get $$\ddot x=0$$ $$\ddot y = -g$$ Something is clearly missing: gravity is not the only force acting on our mass: we have to take into account the tension of the rod/string. But why doesn't it come out from the equations? The point is that system only has ...

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It would be easier to answer your question clearly with a drawing. In the following, the angle coordinate of the pendulum is the angle it makes with the vertical line. When the pendulum swings right(left), the angle will be positive(negative). With this setting, I get the exact same answer as you by working out the equations of motion. However, there ...

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You are absolutely right in everything you said. The momentum is non zero only if the wave has a longitudinal mode, which is in fact the realistic case. Moreover when this is the case, the wave equation is not that simple. Let me try show this. Longitudinal Mode Let us assume that when in equilibrium the string, of density $\mu$, is along with the $x\equiv ... 9 A fake derivation We can rather easily compute a horizontal velocity for the string fi we assume that the total velocity vector is everywhere normal to the string (this assumption is not always valid, see below). The following picture then illustrates the computation: Take two infinitesimally separated points$x$and$x+\mathrm{d}x$and let the wave ... 0 The work done onto the spring is$dE/dt=F(t) \dot x(t)$. You should not look at the direction of$F$alone, but at the the direction of the motion as well. 0 You can't ride it because it's very thin and so is bound to fall over to one side or the other. A more difficult question is why you can ride a bike in normal circumstances. That has already been discussed on this site but as you would expect the answer is related to the steering mechanism. 2 The direction of the force due to surface tension depends on whether the liquid wets the the body or not. In the first picture liquid does not wet the body, so the force is directed in such a way as to decrease the contact surface area of the body with the liquid. That is, upwards. In the second picture liquid wets the body, so the force is directed in ... 0 The physics in the 1st approximation are worked out here: https://en.wikipedia.org/wiki/String_vibration with the result that frequency depends on the length, tension, and mass density as:$ f = \frac{1}{2L}\sqrt{\frac{T}{\mu}} $The 1st factor is why bass strings are long and treble are not (on a piano). The lower 3 strings on your guitar have a ... 0 A Boltzmann distribution is system dependent--it depends on the energy eigenstates. Moreover, if the system lacks symmetry, then Vx, Vy, and Vz may have very different distributions. However, if you're talking about an ideal gas--then it's the standard Maxwell-Boltzmann distribution. To get just Vx, you can use the equal-parition property, and rewrite a 1-d ... 0 If the Lagrangian is linear in$v^2$, then$\partial_{v^2} L = k$,$k$some constant. So the difference becomes $$\delta S \propto \vec \varepsilon \cdot \int dt \frac{d}{dt} \vec x(t) = \vec \varepsilon \cdot \int d \vec x(t) = \varepsilon \cdot (\vec x_b - \vec x_a)$$ That term does not depend on the path$x(t)$, since the endpoint of the variation are ... 0 You are possibly confused by the fact that your textbook, when it says we need this to be a total differential, means with respect to time, not speed. We know that, if we perform a change of${\cal L}$into${\cal L}+\delta\!{\cal L}$such that$\delta\!{\cal L} = df(\vec x)/dt$, the dynamics is unaffected, where$f$is allowed to be a function of position ... 0 Schaum series Differential Geometry will solve part of your problem. Search "problem book in riemannian geometry" on google and it should bring out something useful. Also see V.I. Arnold's books. 1 The expression you derived seems quite correct to me. I'd say the reason why you don't have an explicite dependence on the angle$\theta$is that if there is no observable nutation (that is, if the top’s angular momentum due to precessional motion is small compared to its spin angular momentum), then the torque due to the earth’s gravitational field is ... 0 In 3 dimensions you can use the cross product to get an appropriate rotation axis. If$\hat{x}$and$\hat{n}$are non-parallel 3-dimensional unit vectors then$\vec{s}=\hat{n}\times \hat{x}$is non-zero. Since$\vec{s}$is orthogonal to both$\hat{n}$and$\hat{x}$there is some rotation about$\vec{s}$that takes$\hat{n}$to$\hat{x}$. You can get the ... 1 Let$\theta>0$denote the angle between the$\hat{\mathbf x}$and$\hat{\mathbf n}$. Notice that $$\hat{\mathbf u} = \frac{\hat{\mathbf x}\times\hat{\mathbf n}}{\sin\theta}$$ Is a unit vector perpendicular to both$\hat{\mathbf x}$and$\hat{\mathbf n}$. The desired rotation is a right-handed rotation around$\hat{\mathbf u}$by the angle$\theta$. ... 0 If you are wondering if$F=\frac{-\partial U}{\partial x}$holds, you may check the equivalent condition that you posted,$\nabla\times\vec{F}=0$. In this case the curl will come out to be zero, so you may then proceed to write the force as the negative gradient of a potential. This appears to work because the del operator is not taking time into account, ... 0 To determine the coefficient of friction, (in an ideal situation, of course) push a known mass of the object with a given force F. Also the frictional force has a numerical value of (approximately) coeff. times normal force. So you can calculate the coeff. this way. There are momentary 'bonds' formed between the surfaces. These bonds arise due to Van der ... 0 Average acceleration is defined the same way as average velocity : Average velocity is change in displacement / change in time. Average acceleration is change in velocity / change in time. Your 1st calculation gives the constant acceleration which would give the same change in velocity in the same time. This is correct. Your 2nd calculation gives the ... 7 Dirac does not intend classical to mean non-quantum mechanics; he intends classical to mean pre-quantum mechanics1. So no, this says nothing about de Broglie–Bohm theory. Dirac opens his paragraph with The necessity for a departure from classical mechanics is clearly shown by experimental results. Dirac is not talking about any theory that one ... 4 Yes, absolutely, Dirac's argument shows that one could never construct a complete theory – which specifies the rules for evolution as well as predictions for the measurement and what happens after the measurement – that would be compatible with the basic facts about the atoms. This no-go theorem applies to Bohmian mechanics because it is just another ... 3 The conditions about (i) differentiability of the functions and (ii) the maximal rank of the corresponding rectangular Jacobian matrix are regularization conditions imposed to simplify the mathematical analysis of the physical problem, in particular to legitimate the possible future use of the inverse function theorem. In the affirmative case, the ... 2 The first equation holds good for average acceleration, but the second is the equation for uniform acceleration. The value obtained using option 1 is correct. In the time interval from 0 to 6 s, the acceleration changes (a constant value from 0 to 3 s and another constant value from 3 to 6 s). Then you cannot apply the uniform acceleration equation as ... 2 I believe it would be best (simplest) to describe this problem in the cylindrical frame of reference: that is a frame with a distance from the axis of rotation$R = \ell_1 + (x+r)\sin\theta$, vertical position (relative to the equilibrium position)$Z = r - (x+r)\cos\theta$, and angular position$\Phi = \omega t$. You then have to compute the tension in ... 0 Light interacts with gravitational wells. Relativity makes things far more complicated than in your model. This will give a taste of why you are not correct: https://en.wikipedia.org/wiki/Gravitational_redshift 5 The generalised Lagrange equations are $$\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial T}{\partial \dot{q}_j} - \frac{\partial T}{\partial q_j}=Q_j \tag{1}$$ where$T$is the kinetic energy of the system and$Q$is the generalised force. This is the most general EoM, and is equivalent to Newton's$F_j=m\ddot x_j$. Now, if the generalised force can be ... 0 The basic fact for telling this is that central forces are conservative in nature. This means the work done by a conservative force between any two points is independent of the path, but depends only on the initial and final points only. In such a case, the work done by a conservative force in traversing from point$a$to point$bis given by: ... 1 If the force is conservative then it is the gradient of the potential: $$\mathbf F = -\nabla \mathbf V$$ If we write this out in component form (in polar coordinates) we get: \begin{align} F_r &= \frac{d\mathbf V}{dr} \\ F_\theta &= \frac{d\mathbf V}{d\theta} \\ F_\phi &= \frac{d\mathbf V}{d\phi} \end{align} Since we are told that the ... 1 Energy is a secondary concept, at least in Newtownian mechanics, so let's start with the fundamentals. There exists a force field. A gravitational one, say. An object in this field feels a force. This force "wants" to make the object accelerate. From forces, define work asforce \times distance$and now we can give precise mathematical meaning to the ... 11 This is really a statistical effect, as pretty much all of thermodynamics. You have two free hydrogen atoms. They tend to move around the space they have, and when conditions are favourable (there's enough energy, the atoms come "close enough" together), they might interact - chemically or otherwise. Now, "enough energy" is the important bit here. When a ... 9 I'm going to take a slightly different approach and say it's because we defined energy to make it so. In other words, systems "try" to find the lowest energy state because energy is a concept humans invented in order to describe what we observe. This is the reason that for any given set of constraints, you might need a different "energy" to describe the ... 0 According to sec. 4 in Calculus of Variations, Gelfand, Dover 1991, a theorem due to Bernstein concerns existence and uniqueness of solutions to the equation$y'' = F(x,y,y')$: If the functions$F$,$\partial_y F$and$\partial_{y'} F$are continuous at every finite point$(x,y)$for any finite$y'$, and if a constant$k>0$and functions$\alpha\equiv ...

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This is a consequence of the second law of thermodynamics, which states that In a closed system with fixed internal energy (i.e. an isolated system), entropy is maximized at equilibrium. It can be shown that this statement is equivalent to the following: In a closed system with fixed entropy, the energy is minimized at equilibrium. Callen in his ...

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It is caused by the fundamental forces, and Tendency of energy to escape Number 2 itself is caused by fundamental forces, so it is the forces. Most questions end up at why/how there are fundamental forces.

2

Consider a smooth 2D sphere and a point of positive mass constrained to move on that without friction. The only force is the reactive force normal to the sphere. The trajectories of the motions of the point are geodesics of the spherical surface. Therefore if you fix the north and south poles as boundary conditions you find infinitely many solutions ...

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The anthropomorphic formulation "tries to" is misleading. Under the effect of ambient noise, matter explores the possible configurations around its current state: e.g., two single hydrogen atoms wiggle around and meet. If they happen to bind, this releases energy which goes away, and we say that the energetic state of this new $H_2$ molecule is lower than ...

1

Quite simply, a viscous flow is a flow where viscosity is important, while an inviscid flow is a flow where viscosity is not important. Gases and liquids alike are considered fluids and any fluid has a viscosity. So a gas bubble surely has a viscosity, albeit relatively low compared to some liquids; liquids are generally more viscous by a factor of 1000. ...

1

In fluid mechanics, a gas (like air) is considered a fluid and operates under the mathematical formulae developed in this field. I believe the very essence of viscosity is displayed in the deformation characteristics of a fluid with regard to shear stresses. For example, the shear stress for a Newtonian fluid with respect to a flat plate is given by Newton's ...

2

From your question, I guess that you don't really visualize what the flow geometry is. It's true that there's one unclear bit in the exercise: I guess it should read "The ﬂow in the boundary layer must match with the constant inviscid bulk velocity $\vec{U} = (U, 0)$". The overall geometry is a flow meeeting a plate which is aligned with it (0 incidence). ...

1

$\mathbf{u\cdot n}=0$ is indeed the answer for an inviscid fluid, and $\mathbf{u = 0}$ for a viscous one.

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In the classical electromagnetic theory the word "photon" has no meaning. Nature as we know it from our experiments is basically "quantum" , i.e. it follows the equations and postulates of quantum mechanics. The equations and laws of classical mechanics can be shown to emerge from the quantum mechanical underlying framework. Classical mechanics is very ...

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