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11

Not sure if I can add much to Kyle's comment, but I'll try. Looking closely, he starts with no angular momentum about the vertical axis - the take-off is "straight". Then he moves one arm behind himself and then extends it sideways - generating a torque about the vertical axis. By tucking his other arm in tightly, his body can now rotate. At the end of the ...


10

One of my favorite scientific papers of all times (mainly because it's rather bizarre) explains the basics of what's going on here. That paper is Kane & Scher, "A dynamical explanation of the falling cat phenomenon," International Journal of Solids and Structures 5.7 (1969): 663-666. To get even more mathematical, there's Montgomery, "Gauge theory of the ...


10

Like so many of his colleagues, Michio Kaku has written yet another "layman" book that you can safely dispose of without feeling bad. Modern physics doesn't deal in "electron clouds" any longer. That's an 80 year old paradigm that has outlived its usefulness. Today we are talking about quantum fields. A quantum field (more precisely THE quantum field) is an ...


7

The thing is that not the electron is at all places at one time, but the probability to measure the electron at a certain time isn't localized at a certain point as it is with classical mechanics. The wave function of which the modulus squared gives the probability density of the electron being at a certain place is at first not much more than a mathematical ...


5

The intuitive way to think about this is to consider a gas inside a glass container (that cannot expand). If the gas expands, then what must happen as a result? The gas leaks out of the container. Similarly, if try I put more gas into the container, then the gas compresses. The vector field $\mathbf F$ is what we use to describe the flow of a fluid. The ...


4

There are two and possibly three factors at work here: Factor 1: Nonscalar Inertia Matrix: Angular Momentum and Velocity have Generally Different Directions One which I don't think has been mentioned is that unless a body is rotating about a so-called principal axis, in general the angular momentum and the angular velocity vectors do not point in the same ...


3

Typically, with inductors, we use the complex impedance, $Z=i\omega L$ with current frequency $\omega$ and inductance $L$, for the voltage: $$ V_{ind}=IZ,\quad V_{rms}=I_{rms}|Z| $$ where the left equation is the voltage of the inductor and the right equation the root-mean-square. Surely, however, what goes on inside the inductor doesn't matter, it only is ...


3

Comments to the question (v3): In contrast to QED with fermionic matter, in QED with bosonic matter, the full Noether current ${\cal J}^{\mu}$ (for global gauge transformations) tends to depend explicitly on the gauge potential $A^{\mu}$, see e.g. Refs. 1-2 and this Phys.SE post. The reason for this difference is because the QED Lagrangian for fermionic ...


3

Here's a general overview of how to approach this: Since the only external forces are vertical (gravity pulling the balls down, normal force of the surface holding the balls up), we can use conservation of momentum in the plane. Similarly, there is no external torque rotating things in the plane, so that component of the angular momentum is conserved. And ...


3

In principle there is an effect, but firstly it's tiny and secondly it averages to zero. The mass of the ISS is about 420 tonnes, or about 5000 times the mass of an astronaut. That means if an astronaut pushes themselves off a wall at 1 m/sec the ISS moves in the other direction at about 0.0002 m/sec. But the ISS isn't very large so after only a couple of ...


2

Because of uncertainty, the electron does not exist at any single point, but exists in all possible points around the nucleus. It is all a matter of interpretation. HUP: \begin{align*} \sigma_x\sigma_p &\geq \frac{\hbar}{2} \end{align*} states that you cannot know both position $x$ and momentum $p$ exactly, at an instant. OK, say you want to ...


2

Let's make a concrete example with numbers: Suppose that $v_a = 6m/s$ and $v_b = 0 \rightarrow E_k = 0.5 * 6^2 = 18, p_a = 1 * 6 = 6, v_{cm} = p/M = 2$ . According to the conservation of energy and momentum: Kinetic energy and momentum are conserved only in a perfect elastic collision, if the bodies stick together the collision is inelastic an ...


2

There seems to be some confusion regarding conservation laws vs. invariant quantities. Any (Lorentz) scalar, such as density, pressure, temperature, or charge, will be invariant as reference frames change. So too will any vector or higher other tensorial quantity. That is, even though different observers might assign different numbers to the components of ...


2

To calculate for a situation like this, consider the Law of Conservation of Momentum: Pi = Pf In the case of the billiards: KEi = 1/2 mu1^2 + 1/2 Mu2 (u1, u2 = initial velocity) KEf = 1/2 mv1^2 + 1/2 Mv2^2 (v1, v2 = final velocity) Based on this law, initial kinetic energy and final kinetic energy in an elastic collision are equal. KEi = KEf Hope ...


2

The equation, $v_A-v_B = \frac{4}{5}(u-2u/3)$ is incorrect. The proper equation for the coefficient of restitution is given by, $v_A-v_B$ = $e(u_B-u_A)$, where $u$ and $v$ are velocities along the line of impact. I believe you came across the somewhat incomplete statement that the relative speed of separation after the collision is $e$ times the initial ...


2

"If I am not wrong, the passage says electron (not the parts of electron) can be found at many places at the same time." An electron exists at many points at once, but we may only find it at one. That distinction is the paradox at the heart of quantum mechanics. An experiment that explains this is the double slit experiment. When we fire electrons through ...


2

The following passage has been extracted from the book "Parallel Worlds-Michio Kaku": Because of uncertainty, the electron does not exist at any single point, but exists in all possible points around the nucleus. This electron “cloud” surrounding the nucleus represents the electron being many places at the same time. You say :" If I am not wrong, the ...


1

The electron is not a particle. It's not a wave either. The fact that it seems to behave like one of those under certain conditions is pretty much irrelevant. However, to understand this on a gut level, with the case of an electron in an orbital of an atom, it is very helpful to imagine the electron as a wave. The electron still has just the properties of ...


1

Hint: The equation system is a coupled set of non-linear Schrödinger equations. Here is a trick: The Lagrangian reads$^1$ $$\tag{1} L~=~ i\dot{f}f^{\ast}- i\dot{g}g^{\ast}-a|f|^2-a|g|^2+gf^{\ast}+fg^{\ast} +|f|^2|g|^2 ,$$ where we have assumed that the constant $a\in\mathbb{R}$ is real. Show that the Lagrangian (1) possesses a global $U(1)$ phase symmetry ...


1

Since the book is named "Parallel Worlds", it would not be inappropriate to give the Many World's Interpretion's view on this. A superposition of an electron being at different places should be interpreted as the electron being at all these places in different Worlds. So, there exists a World where the electron is in one position (but note that technically ...


1

A "collision course" is a very fuzzy concept: if you are "barely going to hit" you are on a collision course but don't need a lot of deflection. However, let's assume for a moment a stationary earth, a meteorite of mass $m$ at distance $D$, heading for earth of radius $R$ with velocity $v$. The equations you need are conservation of angular momentum and ...


1

Here is a general figure of an hard spheres collision drawn in the center of mass of the mass $m_2$ before the collision. The black dot is attached to this frame. To solve the problem, you need to observe Conservation of energy: $m_1v_1^2=m_1(v'_1)^2+m_2(v'_2)^2$. Conservation of momentum: $m_1\vec v_1=m_1\vec v'_1+m_2\vec v'_2$ Conservation of torque ...


1

The problem with your solution is that the inelastic collision and assumption that kinetic energy is conserved are mutually exclusive. You can see that in your math when you try to solve for $v_2$. Rewriting equation $(1)$ gives $v_1=\left(m+M\right)v_2/m$ which inserted into $(2)$ yields $$ m \left(\frac{m+M}mv_2\right)^2 = \left(m+M\right)v_2^2.$$ This ...



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