# Tag Info

8

Of course you can define such a quantity, but the question is: does it mean anything physically? Contrary to what has been stated in some of the answers/comments, this quantity is not comparable to a "normalized" dipole moment. A dipole is a system of two charges equal in magnitude but opposite in sign. The corresponding dipole moment, which is of great ...

6

The electron is stuck to the atom, which isn't going anywhere. This means that $\langle p \rangle = 0$. The uncertainty $\Delta p$ measures the RMS fluctuation of the momentum: $$\Delta p^2 = \langle p^2 - \langle p \rangle^2 \rangle = \langle p^2 \rangle$$ Since $E = p^2 / 2m$, this means that $$\langle E \rangle = \frac{\Delta p^2}{2m}$$ The ...

6

As weird as it sounds, the answer is "yes." Take, for instance, a satellite in gravitational orbit around some heavy body. It's energy is given by $$H=\frac{p^2}{2m}-\frac{GMm}{r}$$ Clearly, there are solutions to this equation which have $0$ energy (look at a slowly moving particle that's really far away), but those solutions necessarily involve a ...

4

You have already got "practical" answers, so I intend to answer form another point of view. There is a quite famous theorem due to Stone and von Neumann, later improved by Mackay, and finally by Dixmier and Nelson, roughly speaking establishing the following result within the most elementary version. (Another version of the theorem focuses on the unitary ...

3

Mass and charge are not so similar for the charge having "center of charge". The notion of "center of mass" appears in many applications when number of bodies move. In this situation, the movement can be splitted into movement of center of mass and individual movements of bodies relative to the center of mass. This occurs because of dual role of mass: it ...

2

You may use the following notation for hypersurfaces in four dimensions : $d\sigma_\mu = \epsilon_{\mu\alpha\beta\gamma}dx^\alpha dx^\beta dx^\gamma$ For instance $d\sigma_0= d^3x$ The expression of the momentum-energy is then : $P_\nu = \int d\sigma^\mu \Theta_{\mu\nu}$ The same kind of expression could be used with the charge : $Q = \int ... 2 Say that we are handed a 1D quantum mechanical system, which satisfies the canonical commutation relation $$\tag{1} [\hat{Q},\hat{P}]~=~ i\hbar~{\bf 1},$$ and handed some choice of eigenstates$|q\rangle$and$|p\rangle$for every value of$q,p\in\mathbb{R}. The eigenstates satisfy $$\tag{2} \hat{Q} \mid q \rangle ~=~q\mid q \rangle, \qquad ... 2 I) At least three different quantities in physics are customary called an action and denoted with the letter S: The off-shell action S[q;t_i,t_f], The (Dirichlet) on-shell action S(q_f,t_f;q_i,t_i), and Hamilton's principal function S(q,\alpha, t). For their definitions and how they are related, see e.g. this Phys.SE answer. II) OP's ... 2 If we have some coordinates q_i and some momenta p_i, then a generator of a transformation is defined as a function g(q_i, p_i). By definition, this generates the transformation$$q_i \to q_i + \epsilon \frac{\partial g}{\partial p_i}p_i \to p_i + \epsilon \frac{\partial g}{\partial q_i}$$So if we want the generator of translations, we want ... 2 Frankly, until you bring along an accelerometer as well as tracking the bus's speed thru corners (I'm assuming you mean you sway during a turn, not just going down the road), I'm going to remain skeptical of your claim. Centripetal force is centripetal force. Now, if you happen to counteract turning force unconsciously, it may be that you feel more ... 2 It is not immediately obvious, but the block has calculable angular momentum at the point just before impact. the block has velocity v tangential to the disk's center of rotation which is a distance r away., and so has angular velocity \omega=v/r. the block also has calculable moment of inertia around that center, I=mr^2. Then, it is simply ... 2 As I cannot post any comments I have to post this as an answer although the essential points were already given: In classical mechanics energy itself was no meaning. Only energy differences have a physical interpretation. Thus in the classical case energy is only defined up to an arbitrary constant. So any fixed state's energy can be set to zero (but of ... 1 They're not the same thing. They have very different implications. You can imagine Force and thus Momentum as the "push" that will happen to the target, while Kinetic energy is the damage it causes. E(k) is equal the Work the object will perform, let it be penetration, fracture, etc. As soon as the object hits the target, the E(k) applies (i.e. the ... 1 Lionel's answer affords good intuition. For me, another notion that really motivates all this is the notion of symmetry and the proof (in a particular sense) that for every symmetry we can see in nature, there must be a conserved quantity. This idea is embodied in Noether's theorem. At high school it wouldn't be reasonable to expect you to have the ... 1 One way to think of it is "flux", i.e., the amount of stuff flowing or moving through somewhere. If I wanted to quantify the rate at which fish are escaping from a pen, I could say that x\,\mathrm{kg} fish swim past the opening in the pen in 1s, or that the flux is x\, \frac{\mathrm{kg}}{\mathrm{s}}. Faster fish or heavier fish would contribute more to ... 1 A force affects the motion of the center of mass only (call it point C). The rotational motion is defined by the total torque applied on the center of mass. If the applied force (F_x,F_y) is at location (r_x, r_y) relative to the center of mass then$$ \begin{aligned} F_x &= m \ddot{x}_C \\ F_y &= m \ddot{y}_C \\ r_x F_y - r_y F_x ... 1 When you hit anything with a force, it causes an acceleration in the direction of the force. So if you hit the square at the corner like so: It would start moving upwards. However, it also experiences a torque about its center of mass. This torque can be calculated as the force multiplied by the perpendicular distance between the center of mass and the ... 1 The reason why you arrive at different solutions is that the assumptions in the assignment are inconsistent. One way to show this is to show in isothermal flow if ideal gas in straight duct of constant cross-section with no friction the gas has to have the same pressure everywhere; different pressures at the entrance and the exitP_1, P_2\$ are not possible. ...

1

The "center-of-charge" is part of a more general concept that is used quite often in physics: Multipole expansion. The general idea of multipole expansion is the following: If you view a charge (or mass) distribution from a large distance, then most of its internal structure is irrelevant to you. Instead, it suffices to do all calculations based on a few, ...

Only top voted, non community-wiki answers of a minimum length are eligible