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5

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 ...


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 ...


4

The fundamental difference between an electron's spin and that of a baseball is that the electron is (as far as we know) a point particle. It therefore cannot rotate in the usual sense, where individual parts move relative to the center of mass; we say that its angular momentum is intrinsic. The magnitude $\lvert\vec{S}\rvert^2$ of a particle's intrinsic ...


3

We integrate $x dm$ to say "for every piece of mass $dm$, sum up the amount of mass times the coordinate $x$ of that mass". It makes sense to integrate over $m$, because every piece of mass has an $x$-coordinate. Likewise, in kinematics, it makes sense to integrate over $t$ because for every $t$ we have a velocity $v(t)$. However, calculus doesn't care ...


3

My questions here are how is it possible for us to visualize the position of say 700 particles using just a single point in 2100 dimensional space? Think of it as a map. You can read a map and loom around and you can learn how to relate the common points. So of you have 700 particles there are 2100 scalars. So have a giant 2100 dimensional vector ...


2

The notion of work in physics was first formulated by the French mathematician Gustave Coriolis in Calculation of the Effect of Machines, or Considerations on the Use of Engines and their Evaluation published in 1829. Coriolis defined work as "weight lifted through a height". He was concerned with developing a term that could measure the units of work ...


2

Draw an arrow to represent a vector, with its length representing the vector magnitude. Draw a coordinate system and get the components of the vector. Now draw another coordinate basis, rotated with respect to the first, and get the components with respect to the new basis. The length of the arrow is the same in both systems - i.e length is invariant - ...


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

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

The velocity of the standing wave is the velocity of the incoming and reflecting wave that formed this standing wave. See http://www.physicsclassroom.com/class/waves/Lesson-4/Formation-of-Standing-Waves


1

First you need to define the orientation of the cube relative to the axis you want to measure. Typically a 3×3 rotation matrix $E$ does the job transforming local coordinates along the principal axes to the world coordinates. The use the transformation $E I_{body} E^\intercal$ Example: A single rotation $\theta$ about the world $z$ axis is $$E = ...


1

Liouville's theorem says the accessible volume in phase space does not increase, but it tends to become narrow filaments that "fill up" a much larger volume. If you think of a particle in a reflecting box, you might start it with a known position $\pm 1$ mm in all three axes and a known velocity $\pm 1$ mm/sec in all three axes. This is a phase space ...


1

Your logic is actually correct. The discordance between the conservation of phase-space volume according to the Liouville theorem and the Second Law is known as the Ergodic Problem. Heuristic explanations as the one provided by Ross Millikan, or course graining the dynamics for another example, do not hold under closer formal examination, since the math ...


1

On an intuitive level if $a\ll R$ then nearly all of the deformation will occur close to the surface. Imagine for a bit that R is radius of the earth and you're pushing on some dirt with base ball so there's a circular contact patch with a radius of about 1/2 an inch. Now if earth were half as big would the forces/stresses/strain/contact area be any ...


1

Engineers usually treat thermal expansion as isotropic, which means the expansion occurs with the same magnitude in every direction. This means that an unconstrained object will have a constant strain and zero stress associated with thermal expansion, it's as if object just scaled up. However, as you suspected, materials with an organized structure can be ...


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 ...



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