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7

For a single particle, yes they're parallel. For a system of particles, $$\sum_i \frac{{\bf r}_i\times \dot{\bf r}_i}{\|{\bf r }_i\|^2}\neq \alpha\sum_i{m_i{\bf r}_i\times \dot{\bf r}_i}$$ (you can come up with a specific counterexample but it should be obvious the two sides don't have to be proportional/collinear -- each vector in the sum is weighted ...


6

No. The Einstein field equations are the equation of motion for the metric (i.e. gravity) in the Einstein-Hilbert action. If you add other dynamical fields to the action, you not only change the stress-energy tensor appearing in the EFE, but you also have to vary the action with respect to the new fields to obtain e.o.m. for them.


4

Multiple classical solutions to Euler-Lagrange equations with pertinent/well-posed boundary conditions (such solutions are sometimes called instantons) are a common phenomenon in physics, cf. e.g. this related Phys.SE post and links therein. In optics, it is well-known that already e.g. two mirrors can create multiple classical paths.


4

Actually, the extra path is not irrelevant. If you put a light bulb at A and a $4\pi$ detector (this means $4\pi$ steradian coverage, i.e. it detects incoming light in any direction) at B, the detector will see light along both paths: direct, and bounced off the mirror, which is exactly the result you got from Fermat's Principle. If you want to exclude the ...


3

The equivariant moment map has several applications. Its meaning is that it provides an encoding of how the Lie group $G$ acts on the phase space, and it gives you a way to find the observables corresponding to the conserved quantities/generators of the symmetry $G$: It also defines the process of symplectic reduction to a reduced phase space. Given that $$ ...


3

Usually in this case, the constrait is assumed to be the surface of the table itself - eg. the block is not allowed to leave the surface of the table anyhow. Or more mathematically speaking, the configuration space is restricted to the submanifold that represents the table's surface. Of course this is an idealization (even if we don't account for the block ...


2

This equation reduces to the Hamiltonian-Jacobi equations in certain specific examples. The Hamiltonian formalism in symplectic geometry, which you've written above, is dependent on the fact that for a symplectic manifold $(M, \omega)$ and a smooth function $H : M \rightarrow \mathbb{R}$, there exists a unique vector field $X_H$ satisfying $$\iota_{X_H} ...


2

Holding the rubber band at a constant stretched position/length causes you to contract your muscles, but the energy you expend (from food) just heats your muscles instead of doing mechanical work on the rubber band.


1

Holding a rubber band stretched is the same as holding a weight above the ground. You aren't adding any energy to the weight, just maintaining its position. However, your muscles' actin and myosin require energy input just to maintain a force. This energy ends up heating the muscles, and is lost.


1

Assuming the hose length is the same, and ignoring any restrictions caused by bends in the hose, you want as great a gravity and pressure difference as possible to pull the liquid through. If you have the hose end at the top of the destination container, with the liquid falling down to the level of the already-moved liquid, then you aren't getting as much ...


1

Initially your ball has some energy ($2J$) and some momentum ($2Ns$). And the wall has some energy ($0J$) and some momentum ($0Ns$). And there is some internal energy, $U=U_0,$ the thing that heat increases. Afterwards the ball has some energy ($0J$) and some momentum ($0Ns$). And the wall has some energy ($0J$) and some momentum ($2Ns$). And there is some ...


1

Timaeus has given the full technical answer: the kinetic energy of the wall itself changes a tiny bit. Since kinetic energy scales as $v^2$, this is totally negligible in the first case (where the wall starts with $v = 0$) but actually significant in the second case (where the wall starts with $v = 2$), and that's where the missing energy goes. Luckily, in ...


1

As a simplification, you can consider that you have a 2D viscous flow between two boundaries that approach each other. Assuming that the flow is symmetrical about the line (with the line along the Y direction), you can simplify this further to "no flow at x=0". What you are left with is a pressure distribution $p(x,t)$ whose integral in $x$ should equal the ...


1

Here is one way to imagine a variation. You have a path $\vec y(t)=(y_1(t), ..., y_n(t)),$ where $y_i(t) = x_i(q_1(t),q_2(t),\dots,q_{n-k}(t),t).$ So the idea is that for every time $t$ you have some $q$s and they (together with the $t$) give you all your $y_i$ hence give you your $\vec y.$ Seems basic but the idea is that given a $t$ you have an $\vec y.$ ...


1

Reading about Computable numbers I wondered if there is any physical experiment that returns non-computable numbers or if there is any physical theory that needs non-computable numbers. Because if that would be the case, we would have prove that the universe is not "simply" a simulation inside a Turing machine. Measurements and experiments result in ...


1

So, the short answer is that you're quite correct: if the dynamics of a system is subject to Liouville's theorem, then phase space volume is conserved, so the entropy associated to a given probability distribution remains constant as it evolves under those dynamics. This is actually just one instance of a much more general puzzle: how do we reconcile the ...


1

To a stationary observer on Earth watching you travel in the same direction in which the Earth rotates, you might be going 500 mph. Galilean relativity says that you are moving 500 mph relative to Earth, and the Earth is moving 500 mph relative to you.However, since the Earth is rotating, that stationary observer is also rotating. So if you're travelling in ...



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