# Tag Info

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The "simplest" classical explanation I know is the van der Waals interaction described by Keesom between two permanent dipoles. Let us consider two permanent dipoles $\vec{p}_1$ (located at $O_1$) and $\vec{p}_2$ located at $O_2$. Their potential energy of interaction is: U(\vec{p}_1,\vec{p}_2,\vec{O_1 O_2}) = -\vec{p}_1\cdot \vec{E}_2 = ...

3

You are right, there is most likely a typo in the paper. Presumably they mean $$U(r_{ij}; \lambda) = 4\epsilon\left(1-\lambda\right)^{n}\left\{ \frac{1}{\left[\alpha\lambda^{2}+\left(\frac{r_{ij}}{\sigma_{ij}}\right)^{6}\right]^{2}}-\frac{1}{\alpha\lambda^{2}+\left(\frac{r_{ij}}{\sigma_{ij}}\right)^{6}}\right\}.$$ This is what's given in their reference ...

3

The Lagrangian should not only be independent of the direction of $\vec{v}$ but it should also change correctly under a Galilean transformation. For instance, if $K$ and $K'$ are two frames of reference with a relative velocity $\vec{V}$ then the two Lagrangians $L$ and $L'$ should differ only by a total time derivative. If $L$ is a function of fourth power ...

2

Angular momentum is a bivector, $J = x\wedge p$, and since the exterior/wedge product is antisymmetric, you do indeed get $n(n-1)/2$ independent components for any bivector. In general the Hodge dual provides an isomorphism between $k$-vectors and $(n-k)$-vectors. In three dimensions, $\star(x\wedge p)$ is an axial vector which and we call this ...

2

There's are many things wrong in your concepts . Let's attend them one by one , A body without a force acting on it always can never rotate as its velocity is changing at each hence momentum is changing at each instant . But to maintain constant velocity , the force must be such that it never does any work , so as to maintain the constancy of Kinetic ...

2

Let's assume no explicit time dependence and that our Poisson bracket $\{,\}$ - I prefer curly brackets so square ones $[,]$ can be used to denote the commutator of vector fields - is non-singular, ie there's a corresponding symplectic product $\omega$. The time derivative $$\frac{\mathrm d}{\mathrm dt}=\{\,\cdot\,,H\}$$ is actually the Lie derivative ...

1

For a given configuration of blocks, the COM of all the blocks over any block ( or pair of blocks )must not be outside the corner of the block beneath it. For example here: The blue dots are approx. COM of each block-pair Then the effective COM of the first 6 block-pairs must lie in the region above the 7th one (from top). Whenever this condition is ...

1

The pieces that support the most weight have higher friction and are more difficult to remove. The easier it is to remove a piece the less important it is structurally. Each block needs to support the weight of all the blocks above it, and it has to have at least 3 contact points spread apart like a three legged chair. With two contacts points it will create ...

1

For pushing it up, we have to overcome friction(act downwards) as well as the $mg\sin\theta$. So, $$3N=f+mg\sin\theta$$ Now the block is just slipping , so friction is acting upwards, and so does the force applied externally.So, $$N+f=mg\sin\theta$$ Eliminate $N$ and use $f=\mu mg\cos\theta$. Solve for $\mu$ you get your answer.

1

Assuming you want to place the $5$ $kg$ mass at a point such that entire COM coincides with $(1,2,3)$. Then we can assume a $6$ $kg$ mass at $(1,2,3)$ and a $5$ $kg$ point mass lying at $(-1,3,-2)$. And then the new $5$ $kg$ mass should be placed such that $COM$ of these two masses is lying at $(1,2,3)$ . So the new $5$ $kg$ mass should be placed at ...

1

The key inside to OP's question has already been provided by Ikiperu in above comments. Here we just want to show that the problem becomes very simple to study in the corresponding Lagrangian formalism. The Hamiltonian reads $$\tag{1} H(p,q) ~:=~ \frac{p^2}{2m} + \lambda pq + \frac{m\lambda^2}{2}\frac{q^6}{q^4+\alpha^4}.$$ Since there is no explicit time ...

1

The physical interpretation is integrability conditions being satisfied on the manifold. From the first equation, if you would take A not depending on 't' explicitly then dA/dt = [A,H]. The Poisson bracket contains in it the dynamics involved in canonically conjugate variables and in classical mechanics, we can measure them simultaneously. Apart from this, ...

1

Let's assume that we are considering the effective acceleration due to gravity felt by an object at rest relative the the Earth's surface at lattitude $\phi$. Since the object is at rest, $\dot{\vec r} = 0$, and the Coriolis term vanishes. With the $z$-axis along the axis of rotation, we can write $$\vec\omega = \omega \hat z$$ On the other hand, the ...

1

The Coriolis term is neglected since it does not add to the uniform acceleration. We can consider only the case of starting motion. The last term is calculated straight out, and then added to the first by the rule of vector sum. To ease that: 1. Consider Cartesian coordinates aligned with the Earth axis of rotation. That will give you a simple way to get ...

1

You don't need to apply Steiner's theorem onto the point mass. The point mass finds itself at a distance (apparently) $R$ of the x-axis. Since the moment of inertia is an extensive value, you can simply add all moments of inertia. There's the moment of inertia of the solid disk with respect to it's diameter. You have to 'Steiner' that away from a distance ...

1

You should decompose Normal force along the weight so that their horizontal component cancel out each other, not the weight along normal forces (as you end up with two weight component along two surfaces which won't cancel each other). That is $F_n sin (20) = {W\over2}$ so $F_n = {W\over2 sin(20)} = 1.4 N$ Also in problems like these you should start by ...

1

Your premise violates Newton's first law of motion: If there is no net force on an object, then its velocity is constant. The object is either at rest (if its velocity is equal to zero), or it moves with constant speed in a single direction. For an object (a body) to be accelerating there must be an external force applied. One of the reasons for ...

1

New answer What you've done here is just dimensional analysis. But you've gone a little too far. In particular, just because two things have the same dimensions doesn't mean that they are equal. If you want to expand $F/A$ a little more, you can choose your favorite from F = \frac{d p}{dt} = \frac{d}{dt}(m\, v) = m\, \frac{d}{dt} v = m\, ...

1

Here are two simple rules about reaction forces. 1) they do no work and 2) their job is to enforce a motion constraint. In the simplest term when an object of mass $m$ is sliding on a horizontal plane with velocity vector $\vec{v}$, there is a motion constraint of the form $y=0$. In Pfafian form this constraint is $\dot{y}=0$ and $\ddot{y}=0$. The ...

1

Newton's third law (1) relates forces that are of the same type, and (2) always involves exactly two objects. Let's call the object resting on the inclined plane A. You proposed linking the following two forces through Newton's third law: earth's gravitational force on A inclined plane's normal force on A These are of different types (gravitational and ...

1

I might have misunderstood your question. If you mean, is there an absolute "motionless"? The answer is No, because all motion is relative. Inertial reference frames are reference frames that are not accelerating - two inertial reference frames move at a constant velocity with respect to each other. Whilst there is no absolute motion, there is absolute ...

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I am not sure if this is what you are up to (it is related to what Xiao-Qi Sun said) to but I'll give it a try too ... At the beginning of Chapter V.2 of his QFT Nutshell, Anthony Zee explains how classical statistical mechanics (characterized by the corresponding partition function involving the Hamilton function) in $d$- dimensional space is related to ...

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Consider a uniform rod of length $L$ pivoting and sliding on a horizontal plane. Kinematics You want to describe the relationship between the coordinates (and their derivatives) $x$ and $\theta$ and the motion of points A, B, and C. Position Kinematics  \begin{matrix} \vec{r}_A = x\,\hat{i} & \hat{i} = (1,0,0)\\ \vec{r}_B = \vec{r}_A + ...

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