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20

This is a physical rather than a mathematical justification - ignore my answer if that isn't what you wanted! All systems have some thermal motion so they explore the phase space in their immediate vicinity. If there is a nearby point with a free energy lower by some amount $\Delta G$ then the relative probability of finding the system at that point will be ...


11

Here's another way of looking at it. Let M1, M2, M3 be our three masses. In the three body problem we're considering, the whole frame containing M1, M2 and M3 is rotating. You're right to think that if that frame was fixed then the points L4 and L5 would not be stable. After all if you perturb M3 from L4 or L5 then it should just roll down the potential ...


11

Of course it has something to do with the liquid water entering the gas phase just above the cup of tea, but how does that give the bag of tea a directed motion to one side? Nope. The teabag is dangled by a string. Remember that the string is made of wound up threads: Now, the threads stay wound up because they fit well and they have a knack of ...


11

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: \begin{equation} U(\vec{p}_1,\vec{p}_2,\vec{O_1 O_2}) = -\vec{p}_1\cdot \vec{E}_2 = ...


10

The formula you quote does not contain the potential energy, it is valid for a free particle (i.e. a particle which is not affected by external potential). You can link it to classical mechanics by evaluating it for small values of $p$ (more precisely: $ p \ll c$): $$ E = \sqrt{\left(mc^2\right)^2 + p^2 c^2} = c \sqrt{m^2c^2 + p^2} = \cdots $$ $$ \cdots = ...


9

The energy in your equation is for a free rigid body in the absence of a potential. We can see this if we start with a Lagrangian with a scalar function, $\Phi(q)$, and remember $\gamma$ is a function of $\dot{q}$, $$ L=T-V=-\gamma^{-1} (\dot{q}) \, mc^2-\Phi(q) $$ Then if we find the momentum $$ \pi=\frac{\partial L}{\partial ...


8

You're right that if you take Newton's law of gravity as is and apply it to a 2D universe, you'll get an infinite result. So you do need to use a modified theory in two dimensions, or indeed in any number of dimensions other than three. The proper way to do this is using general relativity, and if you apply GR to 2+1D spacetime, you get something that looks ...


8

Gravity is doing that work! If you observe, the domino is in a position of unstable equilibrium. Edit: as pointed out in the comments, this position is of a metastable and not unstable equilibrium. This means that the domino is in a state where it hasn't achieved the minimum possible energy state yet. The energy I'm talking about here is the ...


8

The potential energy only being defined up to a constant does not imply that potential energy differences only depend on differences in position. To see this mathematically, assume that a function $U$ has the property that $U(x_2)-U(x_1) = f(x_2-x_1)$ for some function $f$. Then if we take $x_2 = x+\Delta x$ and $x_1 = x$, and divide both sides by ...


8

Note the title of the link you give : Minimum total potential energy principle bold mine. The only answer to "why" questions about principles in physics is "because the theoretical models dependent on it have been found to describe all known data and can predict new ones". Why questions in physics when they hit postulates and laws, is like asking why ...


7

Actually, your expression for the potential $\Phi(r)$ is incorrect. The expression $\Phi(r) = -\frac{GM(r)}{r}$ is only valid outside the sphere. As an explicit demonstration of its invalidity, note that $$\underset{r\rightarrow0}{\text{lim}}\,\Phi(r)=\underset{r\rightarrow0}{\text{lim}}\,\left[-\frac{G}{r}\int_0^r4\pi r'^2\rho(r')\,dr'\right]=0$$ assuming ...


7

You say: Imagine a book that we lift it with a force that is exactly equal to the force of gravity so the forces cancel out and the book moves with a constant velocity. so I'm guessing your reasoning is that the net force on the book is zero so the amount of work done on the book is zero. And you are absolutely correct - no work is done on the book and ...


7

Let $E$ denote a quantity that does not change over time (from the first principle). Consider a ball with mass $m$ dropped from a height $h$. As the ball drops, its speed changes due to the gravitational acceleration $g$, reaching a final value $v$ at impact. Thus, we can infer that the quantity $E$ depends on these 4 parameters: $$E(m,H,g,V)$$ where $H$ ...


7

While it may be possible to derive a violation of energy conservation due to intersecting equipotentials, there is a much more intuitive and in my opinion a more fundamental reason that equipotentials cannot intersect: Potential is a single-valued function. A good analogy for potential in this case is a map of the ground elevation of the earth; a ...


6

Yes the free body moves outward, but there are two critical things you have to know to interpret this statement correctly. First, this is the effective potential, taking into account gravity and centrifugal force. It has this form because we went into the non-inertial frame co-rotating with the two masses. Mathematically, the potential is $$ ...


6

This effect is called capillarity and is not that straightforward. The contact between water and a solid surface is determined by the chemical bonds. It is macroscopically observed in the contact angle that the water/air surface makes with the solid surface. This angle depends on the strength of the bonds between the solid and the water molecules. You can ...


6

Yes, quantum tunnelling in the double well potential can be solved in a Wick-rotated Euclidean formulation $$ S_E[x]~=~\int \! dt_E \left[ \frac{1}{2}\left(\frac{dx}{dt_E}\right)^2 - (-V) \right], $$ see e.g. Ref 1. Here $t_E=it_M$ denotes Euclidean time. The Euclidean action is in turn interpreted as the usual kinetic minus potential term with a potential ...


6

The question about minimizing potential energy and the replies that such questions do not make much sense is a typical conversation between a physicist and a mathematician: Physicist: - Why systems tend to minimize potential energy? Mathematician: - Look around, lots of things follow this principle: potential energy, entropy... Physicist: - OK, I can see ...


6

Schrödinger's Wave Equation is an application of Hamiltonian Mechanics. Unlike Newtonian Mechanics, Hamiltonian Mechanics relies on knowing about the things that contribute to the energy of the system. If you know the things which contribute to the energy of a system, then you can determine things like forces, accelerations, and positions. (All through the ...


5

It's valid in the sense that it does tell you the rest energy of a 200-pound person, but it does not tell you how much energy you could get by splitting all those atoms. As a matter of fact, most of the atoms in a human body are carbon, nitrogen, and oxygen; splitting these atoms takes energy, it doesn't produce it. Your character would need to tap into a ...


5

Think about the work-kinetic energy theorem, which states that the net work done on an object is equal to its change in kinetic energy: $$W_{net}=\Delta\mathrm{KE}.$$ You are right that when lifting an object of mass $m$ by a height $h,$ in a uniform gravitational field, the work you do is $W_{you}=mgh$ (assuming, as you said, that you're applying a force ...


5

Your teacher's explanation is incorrect. A simple counterexample can be constructed to illustrate this by considering what happens when the role of your arm is replaced by that of a rubber band. When a weight is suspended from the ceiling by a rubber band, the band stretches and its polymer chains become more ordered, in exact analogy to your teachers ...


5

It isn't possible to measure potential energy because it has a (global) gauge symmetry. It's like trying to measure the height of a mountain - this could be the height above sea level, the height relative to the deepest sea trench, the height relative to the centre of the earth and so on. Any measurement can only measure the change in potential energy, and ...


5

For forces that change along the way, displacement is not the thing to calculate work with. Let $\gamma : [0,1] \rightarrow \mathbb{R}^3$ be the (closed or open) path that the particle the force is exerted on follows. Then, the work done along that path is $$ W[\gamma,F] = \oint_\gamma \vec{F}(\vec{x})\cdot \mathrm{d}\vec{x}$$ which is a line integral. If ...


5

Well, we can do a simple counter-example. Let $$ \vec{F}(\vec{x}) = F_0 \cdot \varrho(\vec x) $$ where $\varrho$ is the function that rotates vectors by 90° counter-clockwise (in matrix form $(\begin{smallmatrix}0 & -1\\1 & 0\end{smallmatrix})$ if you prefer that). Clearly, for the closed path $$ \vec{\gamma}\colon\quad [0, 2\pi]\ \to\ ...


5

Special relativity doesn't alter the fact that interactions between particles "store energy" in the form of "potential energy," alhtough special relativity does alter the terms you listed, all of which have to do with the energies possessed by particles either by virtue of their motion, or their mass. For example, in special relativity, electromagnetic ...


5

Yup. Inside the (uniform spherical) mass, IIRC $\phi=-\frac{GM}{2R^3}\left(3R^2-r^2\right)$. Or something like that. So, $$\phi=\begin{cases} -\frac{GM}{r}, & r>R \\ -\frac{GM}{2R^3}\left(3R^2-r^2\right), & r<R \end{cases}$$ The laplacian $\nabla^2\phi$ should be $$4\pi G\rho=\nabla^2\phi=\begin{cases} 0, &r>R \\ 4\pi G\rho_0 ...


5

About negative energies: they set no problem: On this context, only energy differences have significance. Negative energy appears because when you've made the integration, you've set one point where you set your energy to 0. In this case, you have chosen that $PE_1 = 0$ for $r = \infty$. If you've set $PE_1 = 1000$ at $r = \infty$, the energy was positive ...


5

Yes, u is indeed the potential energy. And yes, you can calculate the force acting on a particle by calculating the gradient of the potential energy field at the position the particle is in. Computationally you will want to calculate the force on particle 1, by taking the gradient at the position particle 1 is in, of the potential energy field created by ...


5

When you look at the dynamics in the rotating reference frame, there are 4 forces acting on the particle: the two gravitational pulls from the massive bodies, the centrifugal push away from the center of rotation (located between the massive objects) and the Coriolis force. The first three forces depend on the position of the particle, and can be derived ...



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