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0

Hard for me to say what you're asking. If you have a particle of mass $M$ in three dimensions such that its positions is described by coordinates $\vec x(t)=(x(t),y(t),z(t))$, then the velocity vector $\vec v=\frac{d\vec x}{dt}=(\dot x(t),\dot y(t),\dot z(t))$ The kinetic energy is then defined as, $$T=\frac{1}{2}M\vec v\cdot \vec v=\frac{1}{2}M\bigg(\dot ...


0

The kinetic energy is $T = \frac{1}{2} m (\frac{d \vec{r}}{d t})^2$ $$\vec{r} = x \vec{i} + y \vec{j} + z \vec{k}$$ The first exprssion is right.


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Always include units Always. No exception. Ever. If your Nobel prize winning professor writes down equations like this without units, you look him in the eye and tell him he's not doing physics. Your analysis suffers from the fatal flaw that you are comparing quantities with different units. $gh/c^2$ is dimensionless. It is the fractional energy change in ...


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In case anyone is interested I'll post here what I learnt. If anyone has any corrections or comments please let me know. The thing that I was missing was that the metric makes the kinetic energy well defined, i.e. independent of coordinates. Working in $\mathbb{R}^2$ with standard coordinates $(x^1,x^2)$ we have the induced coordinates $(x^1,x^2,v^1,v^2)$ ...


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In the case of symmetric double well potential (the Hamiltonian is even under parity) tunneling happens between two states localized at the two minima provided the barrier is finite. Those two states are superposition of the ground and 1st excited states of the Hamiltonian thus they are not eigenstates of the Hamiltonian. If we tune the barrier height as a ...


1

The force that each of the pistons can drive with is given by $$F=\frac14 \pi d^2 P$$ How much lift that produces depends on the exact geometry of the table and in general will be a function of height. In essence if the table moves $x_1$ when the pistons move $x_2$ the force will be $$F_{table}=2F_{piston}\frac{x_2}{x_1}$$


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Until an expert in this field gives a rigorous answer to your question, I will give you my view. If I understand correctly your questions, you are on the one hand asking about other types of existing exchange interactions between virtual particles and real particles, and on the other hand you claim that using the usual virtual photon picture for the ...


1

It seems like a very arbitrary question. It would be much easier if you could just elaborate on one given "specific" and well defined scenario, that you wish to solve, then one can help. From which then you would analogously try to solve other similar scenarios. From what you've asked thus far, your starting point should be Conservation of momentum. Since ...


0

Noether's theorem in Hamiltonian mechanics is saying the same thing as Noether's theorem in the Lagrangian setting, under the Legendre transform. A Hamiltonian system is a triple $(M,\omega, H)$ where $(M,\omega)$ is a symplectic manifold and $H$ is the Hamiltonian. You define a continuous symmetry in the Hamiltonian setting to be a vector field $V$ that ...


3

"but how can you have a "basis" that changes at every point?" This is really the root of your problem. Mathematically (and physically) speaking, such a basis works fine. You just have a (hopefully temporary) conceptual problem. Maybe try thinking of it this way: How does $\hat{x}$ know to point "to the right" in cartesian coordinates, at an arbitrary ...


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It is actually clear that $T$ being NOT homogeneous quadratic in the velocities of the generalized coordinates does not imply that energy is not conserved. E.g. one can consider a boost transformation for a free particle in 1D and then $T=\frac{1}{2}m\dot{x}^2$ becomes $T=\frac{1}{2}m(\dot{x}-c)^2$. Clearly, $E$ is conserved. Unfortunately, I wasn't ...


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Symmetry of potential alone is not enough to guarantee a symmetric orbit. The initial and boundary conditions must also be symmetric. Also, for non-linear systems, numerical treatment has to be sufficiently accurate.


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The behavior of the rocks and yourself (which I'll refer to in the calculations as a general "human body") in collisions has to do largely with conservation of momentum and conservation of kinetic energy. In basic physics, there are two types of collisions that we can consider to model this situation: a perfectly elastic collision and a perfectly inelastic ...


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Certainly late for your homework problem set, but for future readers: I think it best to go to the source, Ernest Rutherford, "The Scattering of $\alpha$ and $\beta$ Particles by Matter and the Structure of the Atom", London, Edinburgh and Dublin Philosophical Magazine and Journal of Science, Volume 21, Issue 125, pages 669-688 (1911). This is the paper in ...


1

The answer is quite simple. You can't see it because you've forgotten that you make an unphysical simplification when considering "tied to an immovable wall" type situations. What's unphysical about the situation is quite simple : there's no such thing as an immovable wall. Let's say the rocket is tied to the earth. To say where the lost chemical potential ...


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The rocket motor generates thrust by acelerating the exhaust gases that it emits. The force on the rocket is equal to the change in momentum of the exhaust gases per second, i.e. the exhaust velocity times the exhaust mass per second. For the rocket travelling in space the energy generated from burning the fuel goes partly into the kinetic energy of the ...


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If you know the rotational kinetic energy then cut the power and time how long it takes for the motor to come to a stop. Divide the kinetic energy by this time and you have the power dissipation as friction. The difference between this and the input power is then the power dissipated though non-frictional mechanisms. Whether this is related to efficiency is ...


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I think there are 3 factors : 1- gravity, 2- the upward movement of water by wave propagation (and this is the on that works in opposition to gravity) & 3- the displacement of water towards shore. The latter is caused by the fact that the front-face of the water moves slower than the hind-face which causes the shape of the wave to curl.Now, how does the ...


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I) Well, the existence of the involution $$\{ I_i, I_j \}_{PB} ~=~ 0,\qquad i,j\in\{1, \ldots, n\}.\tag{1}$$ is already generically true locally for a Hamiltonian system, cf. e.g. this Phys.SE post. So it is essentially only a global/topological requirement in the definition of (Liouville) integrability. II) Another possible answer (in light of OP's ...


2

It is a Taylor expansion. You might be a little freaked out because they are treating $L$ as a function of $v^2$, but that doesn't matter, consider $$ L(x + \delta) \sim L(x) + \frac{\partial L}{\partial x} \delta + O(\delta^2) $$ but take $x=v^2$ and $\delta = 2 \boldsymbol{v} \cdot \boldsymbol{\epsilon} + \epsilon^2 $, $$ \begin{align*} L(v'^2) &= ...


0

I think the wave nature of water is both transverse and longitudinal. When we drop a stone in water that we can see there the transverse waves but then we can heard sound also, so, water waves are transverse as well as longitudinal waves!


1

Two equal mass bodies orbiting around their centre of mass is exactly how Newtonian gravity works. Your analogy of the runners is actually pretty good. Think about it this way - the two bodies are always attracted toward each other, so they accelerate a bit in that direction. Notice that "toward the other body" necessarily means "toward the centre of mass". ...


4

Yes, that can happen. It is somewhat realized in positronium, a bound electronic state where an electron and a positron revolve around each other. Both have the same mass, so they could (classically) have the same spherical orbit. With Newton, you have an attractive force for two equal bodies of mass $m$ of $$ F = G \frac{m^2}{d^2}. $$ The centripetal ...


0

Suppose that instead, one wrote \begin{align} P = a\sqrt{p} - b\sqrt{q} \end{align} such that $a$ and $b$ had dimensions engineered to make the dimensions match in both terms, then there clearly would be no issue. Now imagine that the values of $a$ and $b$ in a given system of units (say SI units) are $1$; then we obtain the canonical transformation ...


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Calling all finite-element model experts :-) . I can only offer one small tidbit: for wind instruments, aside from the octave hole, which exists primarily to facilitate exciting the higher frequency notes, the tone is primarily defined by the distance from the mouthpiece to the first open hole. As a long-time clarinetist, I'm fully aware that the pitch ...


2

Before we begin, note first of all, that there exist various definitions of a canonical transformation (CT) in the literature, cf. e.g. this Phys.SE post. For instance, OP's last equation (v1) is called an extended canonical transformation (ECT) in Ref. 1. OP is essentially asking (v1): If we have a transformation $$\tag{A} (q,p)~\longrightarrow~ ...


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I found a couple definitions of impulsive force, one of which states The force that two colliding bodies exert on one another acts only for a short time, giving a brief but strong push. This force is called an impulsive force. Frictional forces don't typically satisfy the criterion of strong in magnitude. The idea behind this definition is that a ...


0

An isotropic 2D oscillator, when taken into action-angle variables by doing a canonical transformation of the Hamiltonian, $$ H(q_1, p_1, q_2, p_2 ) = \frac{q_1^2}{2m} + \frac{kq_1^2}{2} + \frac{q_2^2}{2m} + \frac{kq_2^2}{2} $$ will yield two constants of the motion, the actions , and two angles running from 0 to $2\pi$ which generates a torus. I'm not ...


1

First things first: waves that have already reached close vicinity to the beach DO displace water towards the shoreline - just notice how the water moves back and forth at the point it's ankle-deep. This is related to the phenomenon by which they lose their wave form and get a crest. There are many forces acting on a surfer, but two of them are the ...


1

Yes, take the orbits of Neptune and Pluto, for example, which are inclined to each other. Furthermore, it is possible for both satellites to influence each other's orbit and in doing so, to run along chaotic trajectories.


0

No, because $F$, in fact, has a statistical meaning, $F$, or rather $e^{- \frac{F}{kT}}$, is a sum along different possible energies configurations : $Z = e^{- \frac{F}{kT}} = \sum_i e^{- \frac{E_i}{kT}}$ So you cannot compare with one-particle (non statistical) physical quantities (for instance the energy of the particle would have only one value). ...


2

As user ACuriousMind correctly writes: What Goldstein calls the principle of least action $\int p~\mathrm{d}q$ is usually called Maupertuis' principle or the principle of abbreviated action. What Goldstein calls the Hamilton's variational principle is often also called the the principle of least/extremal/stationary action $\int L~\mathrm{d}t$. This is ...


0

This is a good question, but the answer is that the energy equations of thermodynamics and dynamics now cannot correspond to one by one, completely, in that the theoretical structures of the two theories are different. Since the internal energy \begin{align}U=TS+Yx+\sum_j\mu_jN_j-pV.\end{align} Helmholtz free energy ...


6

The more common names for what you are talking about are the abbreviated action $$S_0[q] := \int p \mathrm{d}q$$ versus the action $$ S[q] := \int_{t_1}^{t_2}L(q,\dot q,t)\mathrm{d}t$$ Both are used in different formulations of classical mechanics, and deliver a different "flavor" of solutions. On both one can do variations calculus and obtains the ...


0

As you state in the comments, $$ \frac{dF}{dt}=\frac{\partial F}{\partial q}\dot{q}+\frac{\partial F}{\partial t} $$ So popping this into the Lagrangian, $$ L'=L+\frac{\partial F}{\partial q}\dot{q}+\frac{\partial F}{\partial t} $$ The Hamiltonian $H=p\dot q-L$ implies $$ H'=p'\dot{q}-L'=p\dot q+something\tag{1} $$ where $something$ is for you to work out. ...


0

Thermodynamic free energy is a subset of potential energy. There are at least two types of thermodynamic free energy - Gibbs free energy and Helmholtz free energy. Both describe the potential energy of a system under certain conditions. For Gibbs free energy, the conditions are constant temperature and pressure - useful for biochemists - because the two are ...


2

This is low-Reynolds number particle sedimentation. It turns out that the problem is strikingly difficult, despite the simplicity of the setup and even of the equations (Stokes plus dynamics of pointwise solid particles). Check the webpage of E Guazzelli who's been working a lot on this. However, I believe you can get a fair rendering with simply a ...


0

Let's estimate some of the contributions discussed: Gravity We can compute the power spent gaining altitude. $$ W_\text{grav} = mg \dot h = m g v \sin \theta \sim m g v \theta $$ for small angles, with some typical numbers $$ W_\text{grav} = ( 180 \text{ lbs} ) ( 9.8 \text{ m/s}^2 ) ( 1 \text{ mile} / 10 \text{ minutes}) ( 5 \text{ degrees} ) \sim 200 ...


0

Your calculations are correct. The light does indeed take longer to reach the other end of the box when they are moving in the same direction, and vice-versa. How could this help, however, to convince the guy in the moving box that he was the one moving, and not me? When he compares the time that the light takes reaching the end of the box, and the time ...


5

Explicitly proving non-integrability of an arbitrary Hamiltonian system is an open problem. For some classes of Hamiltonian systems (e.g systems on a plane) is possible to prove explicitly the non-integrability of the system, using theorems of Poincare, Burns, Ziglin and Yoshida (and generalizations). For example there is a theorem of Poincare: For a ...


1

Second answer, what about the phenomenon of “Quantum Locking”? Right now it is being used to levitate superconductors over magnets, but I am sure you could exploit the phenomenon to transmit torque. Plus, you can put the superconductor on the vacuum side of the seal to keep it cold.


0

Take a look at the torque converters of cars with automatic transmissions. They are rotational couplings where the working fluid is completely sealed from both the source and sink environments, so one of those could be vacuum. One shell of the device rotates inside the other, to the outer one could be stationary and form an integral (welded) part of the wall ...


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One standard way is a rotary fitting using a ferrofluidic seal. These are fairly standard parts in your favorite vacuum components catalog. Often used in semiconductor processing equipment.


4

Consider a non-relativistic massless particle with charge $q$ on a 2D torus $$\tag{1} x ~\sim~ x + L_x , \qquad y ~\sim~ y + L_y, $$ in a constant non-zero magnetic field $B$ along the $z$-axis. Locally, we can choose a magnetic vector potential $$\tag{2} A_x ~=~ \partial_x\Lambda, \qquad A_y ~=~ Bx +\partial_y\Lambda, $$ where $\Lambda(x,y)$ is ...


0

I think the key to the question is considering that $U(x)$ is periodic function therefore one has to consider an appropriate domain for integral otherwise you cannot keep the sign of $E-U(x)$ positive, in general. The fact that the particle is in a bound state implies that: $$ E<0, $$ then we should have $$ E-U(x)\geq 0 \Rightarrow U(x)<0 $$ otherwise ...


4

$U(1)$ Chern-Simons theory with (physical) space a 2-torus is such an example. Its phase space is the gauge equivalence classes of flat connections on the 2-torus. These are specified by the holonomies around two 1-cycles forming a basis of $H_1(T^2)$. This is of course a 2-torus $U(1) \times U(1)$. Because of the form of the Chern-Simons action, these ...


2

In solid state physics, the bulk of a crystal is usually given periodic boundary conditions to avoid the sticky problem of what to do at the termination of the crystal. So the crystal is all bulk, no surface. This turns out to be a very good approximation to the bulk of a real crystal. It also gives the solid the topology of a 3-torus.


1

Differentiating the whole integral with respect to $f$ shows that it is strictly monotone decreasing in $f$ for $x_1<x_2$. Therefore $f=0$ is the only solution. $$\frac{d}{df}\int_{x_1}^{x_2} \frac{\sin x dx}{\sqrt{E - U(x)}}=-\frac{1}{2}\int_{x_1}^{x_2} \frac{\sin^2 x dx}{\sqrt{E - U(x)}^3}<0$$


1

This may be cheating, but I think the problem is easier if you use conservation of energy. If you set the gravitational potential energy reference to the height of m1, then initially you have $$ U_i = k\frac{q^2}{d}. $$ The final energy will have a gravitational potential energy and an electrical potential energy: $$ U_f = mgh + k\frac{q^2}{r} $$ A bit of ...


1

I worked the problem out a ways and it involves quite a bit of tedious algebra. The technique I used was to include the electric force in the Fx and Fy equations by looking at the angle that $ \hat{r}$ (the vector between the two charges) makes between the charges. For example, the equation i came up for Fx is $T_x - \frac{Kq^2}{r^2} \cos{\theta} =0$ where ...



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