# Tag Info

24

First of all, "loopholes" is no disrespect. It's standard nomenclature. Given a law, a "loophole" is a way to circumvent it. Bell's inequalities, in their mathematical formulation, are laws that prevent superdeterminism, so if we believe it should exist, we have to find loopholes in the assumptions. It might be that the loophole is so big that the whole law ...

13

You say "ridiculous sounding ideas often end up becoming standards science" - but take into account that "ridiculous sounding ideas" much more often, by a large factor, end up becoming no science at all. I agree that "that form of idealogical bullying should never exist" - except that it's not ideological, but just a practical matter of deciding which ...

12

Yes, you can. The moment of inertia of the lever will be different in each of the two situations. Let us assume that the lever is massless and the weight is a point mass. In the first situation, $I = mr^2 = (3)(1)^2 = 3 \text{ N} \cdot \text{m}$. In the second situation, $I = mr^2 = (1)(3)^2 = 9 \text{ N} \cdot \text{m}$. Because the moment of inertia is ...

9

Just for fun let me suggest another rather impractical way to tell the difference. The diagram shows the far side of the lever. It has a length $L$ that you don't know and there is mass $m$ on the end that you don't know. The torque is equal to $Fd = FL\cos\theta$, and the force is the gravitational force $F = GMm/r^2$, where $M$ is the mass of the Earth ...

8

$-i ħ \nabla$ is the momentum operator. You have to apply it to a wave function to get the actual momentum. Consider the plane wave solution to the Schrödinger equation: $\Psi = e^{i \mathbf{k} \cdot \mathbf{r} - \omega t}$. Applying the momentum operator gives $-i ħ \mathbf{k} \Psi$. You can see the eigenvalue has units of momentum. (If you can't see ...

7

Perhaps the simplest and most intuitive approach is to regularize the hard wall potential $$V_0(x)~=~\left\{ \begin{array}{rcl} 0 &\text{for}& x<0 \cr\cr \infty &\text{for}& x>0\end{array}\right.$$ as $$\lim_{\varepsilon \to 0^+} V_{\varepsilon}(x) ~=~V_0(x).$$ For instance, one could choose the regularized potential as $$... 7 In the language of differential forms in spacetime, the field strength 2-form F = E\wedge\mathrm{d}\sigma + B gives Gauss's law for magnetism and Faraday induction:$$\mathrm{d}F = 0\text{.}$$Meanwhile, the electromagnetic excitation 2-form H = -\mathcal{H}\wedge\mathrm{d}\sigma + \mathcal{D} provides a natural formulation of Gauss's law and ... 5 You're almost good, you just needed to use the chain rule, which you did perhaps without knowing it. It is clearer if you write it this way perhaps:$$ \frac{d}{dt} m r(t)^2 \dot{\phi}(t) = m\frac{d}{dr}(r(t)^2)\frac{dr}{dt}\dot{\phi}(t)+mr^2\ddot{\phi} = 2mr\dot{r}\dot{\phi}+ mr^2\ddot{\phi}$$5 The moment of inertia is a rank 2 tensor not a scalar. You'll commonly see it written as a scalar, but this is because by choosing your axes to line up with the principal axes of the object the matrix representing the moment of inertia can be diagonalised:$$ {\bf I} = \left( \begin{matrix} I_{00} & 0 & 0 \\ 0 & I_{11} & 0 \\ 0 & 0 ...

4

I) At the classical level, there is no convexity condition. If an action functional $S$ yields a stationary action principle, so will the negative action $-S$. (Under sign changes, a convex function turns in concave function and vice versa.) Or one could imagine a theory, which is convex in one section and concave in other sector. II) On the Lagrangian side ...

4

Refs. 1 and 2 define a canonical transformation (CT) $$\tag{1} (q^i,p_i)~\longrightarrow~ (Q^i,P_i)$$ [together with choices of Hamiltonian $H(q,p,t)$ and Kamiltonian $K(Q,P,t)$] as satisfying $$\tag{2} (p_i\mathrm{d}q^i-H\mathrm{d}t)-(P_i\mathrm{d}Q^i -K\mathrm{d}t) ~=~\mathrm{d}F$$ for some generating function $F$. On the other hand, Wikipedia (March ...

4

The central extensions are classified by the second cohomology group: http://en.wikipedia.org/wiki/Group_extension . If this group is trivial then each central extension is semidirect (and hence in some sense trivial). In particular, this is the case for the Poincare group but not for the Galilei group. However, if you want to take a nonrelativistic limit ...

4

Pressure is a measure of the force applied to a given area. Blades are sharp because they have a small cross sectional area, allowing you to create very high pressure whilst applying only a modest force. This force generates so much of shear stress on the object getting cut that it crushes through the molecular bonds in that object. Cutting through something ...

4

The real issue here is obscured by inane terminology and bad thinking. The real issue is, what sort of "locally deterministic theory" could reproduce the experimentally successful predictions of quantum mechanics? Locality here means, not just that causal influences have to pass through space (rather than acting instantly at a distance), but that they ...

4

When you look at this video you can see that the lower leg seems to maintain a constant velocity. This is probably partially due to the higher total mass of the leg compared to the ball. The ball leaves the foot at a higher velocity. This due to the deformation of the foot and the ball. This deformation is caused by acceleration (initial velocity ...

4

Sure there was ! Just like we deduce the laws of Newton from relativity. There is a famous theorem in quantum mechanics named Ehrenfest's theorem, which states that quantum mechanical expectation-values follow classical laws. So after averaging out the quantum-behaviour you just get classical mechanics. For the correspondence with the classical mechanics ...

4

It seems to me that you're looking for the Boltzmann transport equation: $$\frac{\partial f}{\partial t}+\frac{\mathbf p}{m}\cdot\nabla f+\mathbf F\cdot\frac{\partial f}{\partial\mathbf p}=Q+\left(\frac{df}{dt}\right)_{\rm coll}$$ with $f$ the distribution in phase-space, $\mathbf p$ the particle momentum, $Q$ some source term, and the RHS an interaction ...

3

Your friends are correct. If there is no force in the left-right direction, then linear momentum will be conserved in that direction. Because the new composite object has more mass than the original object, it will have a lower speed to the right. What about energy? Kinetic energy is not conserved in this case, because the collision is inelastic. The ...

3

There are 3 actions of the Galilean group on the free particle: On the configuration space, on the phase space and on the quantum state space (wave functions). The Galilean Lie algebra is faithfully realized on the configuration space by means of vector fields, but its lifted action on Poisson algebra of functions on the phase apace and on the wave functions ...

3

OP considers an equations of motion of the form $$\tag{1}\dot{\bf x}~=~{\bf B}({\bf x}),$$ where the vector field ${\bf B}$ is of the form$^1$ $$\tag{2} {\bf B}~=~{\bf \nabla}\times {\bf A}.$$ In other words, ${\bf B}$ is divergence-free $$\tag{3} {\bf \nabla}\cdot {\bf B}~=~0.$$ Eq.(3) is locally eqivalent to eq. (2), cf. Poincare's Lemma. Let ...

3

There is one formula relating the speeds of any two "platforms" (say $P$ and $Q$) between each other: $$V_{P}[ Q ] = V_{Q}[ P ].$$ And there's of course the well known symbol for "speed of light (in vacuum)", as determined of light signals exchanged by members of any one platform between each other: $c$. The speed of any one platform ($Q$) as determined ...

3

Why is superdeterminism generally regarded as a joke? My personal (somewhat facetious) answer to that would be because people lack imagination. I don't think of superdeterminism in terms of conspiracies, but rather retrocausality, and do not find it ridiculous if phrased this way. Basically, I don't believe there's really something like a physically ...

3

Your mistake is your definition of the angular velocity $$\omega=rv \tag{not correct}$$ is incorrect. We know that $\omega$ has units of $1/s$, but your assertion gives it units of $m^2/s$. The correct definition is $$\omega=\frac vr \tag{correct}$$ which gives the correct units. Using this: $$\left[L\right] = [m]\left[r^2\right]\cdot\left[v\cdot ... 3 Let's assume a one litre 1000 W electric kettle, filled with 0.5 kilograms of water at 20^o C: It takes 4.2 joules to warm one gram of water one degree Celsius. So, to warm the 500 grams of water 80 degrees from 20 to 100 takes 168,000 joules. The kettle will supply 1000 joules per second, so it'll take 168 seconds for the kettle to ... 2 I) On one hand -i\hbar{\bf\nabla} is the Schrödinger (position) representation of the the canonical/conjugate momentum operator \hat{\bf p} in order to satisfy the CCR$$\tag{1} [x^i, p_j]~=~\hbar {\bf 1}~\delta^i_j. $$II) On the other hand, m\hat{\bf v} is the kinetic/mechanical momentum operator. (Let us for simplicity imagine a non-relativistic ... 2 For the r^2 you will use the power rule of differentiation:$$\frac{d}{dt}mr^{2}\dot{\phi} =2mr\dot{r}\dot{\phi}+mr^{2}\ddot{\phi}$$with the "2" on the first term of the right hand side being the only difference between what I wrote and what you wrote. If you are still confused, think of it this way. Instead of writing r^2 write is as (r)^2 to ... 2 The stable point of equilibrium is at x=0,x=\pm a. This becomes obvious when you realize that for these kinds of problems with linear friction, you can actually ignore the friction term when computing the equilibrium state of the system. Why? If the system is at equilibrium, then it is both at rest (\dot{x}=0) and has no net force acting on it (F=0). ... 2 Use Lagrangian mechanics method to answer this problem because it is easier than Newtonian mechanics (IMHO). Let T be the kinetic energy, V be the potential energy then the Lagrangian L is given by$$ L=T-V $$and the Lagrangian equation is$$ \frac{d}{dt}\left(\frac{dL}{d\dot{q}_k}\right)-\frac{\partial L}{\partial q_k}=0, $$where it is assumed that ... 2 The state from 1\text{ atm} to 2\text{ atm} is normally called decompression or contraction. An equation you can use going from one state to the next is:$$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2} Where $P$ is pressure, $V$ is volume and $T$ is temperature. Now if you want to calculate the force you have to know the surface area of what you are ...

2

(NOTE: This is not a definitive answer and is largely guesswork, but it might be useful for further answerers.) At first I suspected it was some sort of weak spring or magnetic device which applied an opposing torque to the arm, so as to make the center reading unique. Searching Google, I found this explanation of the theory of a triple-beam balance offered ...

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