# Tag Info

2

A very simple motivation for writing $\ddot{\bf x}(t) = {\bf F}({\bf x}, \dot{\bf x}, t)$, which might shed some light, is the following. We are given ${\bf x}(t)$ and $\dot{\bf x}(t)$ and we desire to calculate ${\bf x}(t + \delta t)$ and $\dot{\bf x}(t + \delta t)$. Now, ${\bf x}(t + \delta t) = {\bf x}(t) + \dot{\bf x}(t) \delta t$, which we may ...

2

The trajectories are uniquely determined means that the theorem of existence and uniqueness applies (so, the differential equation has to be sufficiently regular). Newton's principle states more: the system is fully determined by the position and the speed, that is, by $2n$ constants, where $n$ is the dimension of the space. As you have $n$ equations (one ...

0

What I understand from the qualitative statement is that of all possible laws that acceleration could "obey," it actually obeys a 2nd ODE! It is irrelevant, whether or not, other functions also obey the same law.

1

I'll answer my own question. Since $\boldsymbol{\omega }=\omega _{z}\hat{\boldsymbol{k}}$ the angular momentum reduces to $\boldsymbol{L}_{O}=-I_{xz}\omega _{z}\hat{\mathbf{i}}-I_{yz}\omega _{z}\boldsymbol{\hat{j}}+I_{zz}\omega_{z}\boldsymbol{\hat{k}}$. We can split the rod in three pieces, calculate moment(product of inertia for each body and sum up. ...

1

As an alternate answer using parallel axis theorem, note that the inertia tensor of a rod pointing in the $y$-direction rotating about its center is $$\mathbf{I}_\text{rod,y}=\left( \begin{array}{ccc} \frac{b^3 \rho }{12} & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & \frac{b^3 \rho }{12} \\ \end{array} \right)$$ and similar for the $x$ and ...

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 ...

0

Another interesting analogy, pointed out by Prof. Graeme Milton, can be seen when you examine Maxwell's equations in media at a fixed frequency $\omega$: $$\boldsymbol{\nabla} \times \mathbf{E} = i\omega\boldsymbol{\mu}(\mathbf{x})\cdot\mathbf{H}(\mathbf{x}) ~;~~ \boldsymbol{\nabla} \times \mathbf{H} = ... 0 You have to remember to express U as$$ U = \frac12 (\sigma_{11} \epsilon_{11} + \sigma_{22} \epsilon_{22} + \sigma_{33} \epsilon_{33} + \sigma_{23} \epsilon_{23} + \sigma_{32} \epsilon_{32} + \sigma_{13} \epsilon_{13} + \sigma_{31} \epsilon_{31}+ \sigma_{12} \epsilon_{12}+\sigma_{12} \epsilon_{12}) $$to get the correct expressions for \sigma_{ij}. ... 3 Let's assume a one litre 1000 W electric kettle, filled with 0.5 kilograms of water at 20^o C: It takes 4.2 joules to warm one gram of water one degree Celsius. So, to warm the 500 grams of water 80 degrees from 20 to 100 takes 168,000 joules. The kettle will supply 1000 joules per second, so it'll take 168 seconds for the kettle to ... 0 Before the water boils, the rate that the water becomes water vapor is slow compared to after it is boiling. The pressure inside the kettle does not increase significantly until after the water is boiling. Only after the water is boiling, is there enough pressure inside the kettle to send a stream of air through the whistle. 3 Your mistake is your definition of the angular velocity$$ \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 ...

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 ...

2

Sure. In the limit $\hbar \to 0$ the Feynman path integral reduces to classical mechanics. You can also take the WKB approximation, which when $\hbar \to 0$ gives classical mechanics in the form of the Hamilton-Jacobi equation.

1

No it's not correct, or at least it's headed in the wrong direction. Properly, you define acceleration, in either 2 or 3 dimensions as fits your situation, by means of vector coefficients. You can use Cartesian (x and y) systems or you can use Polar (radius and angle) systems, and report/calculate the acceleration components for each vector component.) In ...

1

No. Frequency is defined as 2π*θ/t where theta is the angle rotated for a time t. You maybe tempted to equate frequency to angular velocity. But it is not so. Angular velocity = dθ/dt. Angular frequency= 2*pi*(Integral of x over time interval t)/t

1

Unfortunately I cannot comment due to insufficient reputation, so here a comment on the question. There are three cases: $\frac{1}{2}mv_A^2>2mgR$ In this case the pearl has a velocity $v>0$ in the top point and will continue its movement. $\frac{1}{2}mv_A^2<2mgR$ In this case the pearl won't reach the top and will oscillate around point $A$. ...

0

It had a tangential velocity and its weight was enough to provide the centripetal force for motion.

2

You are correct. When the forcing frequency is equal to that of the natural frequency then the external force is constantly doing positive work on the system, which results in resonance. When the forcing frequency is not equal to the natural frequency, then there are times when the force is contributing positive work (adding kinetic energy), and times when ...

0

As noted within my other post at Time dilation in special relativity, here I have basically the same situation yet the box becomes a 300,000 km long spaceship. The basics are explained. However, the only absolute measure that exists within it, is that all objects are constantly traveling at the speed of light within the 4 dimensional environment known as ...

1

Your answer (1) is the correct one. It is actually quite simple if you think in terms of conservation of energy. What you have described is a simplified version of a two body problem. Note that strictly speaking, both the doughnut $(D)$ and the ball $(B)$ will move towards each other. But without outside influence, their combined center of mass should be ...

1

Consider two masses M and m in circular motion with same velocity,v. Both has acceleration v^2/R. The forces acting on the two masses are different. Force will become more on the greater mass. But acceleration of both are same. Because, if you put M and m in the following relation, you get same v^2/R. $$(mv^2/R)/m=v^2/R$$ since we know $$F/m=a$$ where ...

1

You can think of acceleration from a purely mathematical context - it's the rate of change in velocity. (If you're familiar with calculus, you can say that the acceleration is the derivative of velocity.) Because of this, you don't need any mechanics to determine acceleration, so the mass is irrelevant. More concretely, a mathematician could calculate the ...

1

The definition of acceleration is rate of change of velocity. If you know velocity as a function of time, then you know acceleration. No information concerning mass is required.

1

Mass and acceleration are two independent variables. You need to consider them together to arrive at a force with the direct relationship F=ma. In other words, if it's mass doubles, so does the force and the acceleration is the same. Gravity works exactly this way. If you are considering a fixed force from something other than gravity (a force of constant ...

1

I) First some terminology. Consider a symplectic manifold $(M;\omega)$. In a local chart $U\subseteq \mathbb{R}^{2n}$, the symplectic two-form reads $$\tag{1} \omega~=~\frac{1}{2} \omega_{ij}~\mathrm{d}x^i \wedge \mathrm{d}x^j ,$$ and the corresponding Poisson bi-vector $$\tag{2} \pi~=~\frac{1}{2} \pi^{ij}~\partial_i \wedge \partial_i,$$ where $$... 0 Notice that the movement of light must always be treated as a local phenomenon, which in your thought experiment means the light is travelling only within the moving box. It is not travelling from the moving box to a stationary observer. Therefore, throughout the whole experiment the box is stationary for the light and no wall is moving toward or from it. ... 1 I do not know how to deal with non-canonical interpretation of mathematical symbols in integrations, but in this case the proof is straightforward, first changing to t the integration variable x and next using \frac{df}{dt} f = \frac{1}{2}\frac{df^2}{dt}. The position x is parametrized by time t, so:$$W= \int_{x_1}^{x_2} F dx ...

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 ...

0

I think the direction of the answers are a bit off. With conversation of energy, you can get some how estimate the "initial rotational speed" of the hurricane balls. It must less than or equation to the amount of energy of the guy blow out plus the starting rotational energy by hand. However, the key question is why the rotation could keep at the speed for ...

0

Newton's law of universal gravitation holds for point like masses. For spatial masses you would have to integrate of infinitely small parts of that mass. However it does turns out that the result of this for a sphere of constant density at a given radius does yield that simple formula, see Gauss's law. To get a better understanding of this, without having ...

0

when your ball is far from a donut, there's a potential energy of the gravitational field. when the ball reaches the donut, that potential energy must be converted into something else, it can't disappear. do it must be that it converted into the kinetic energy of movement, i.e. the ball must be moving at this point, hence, it'll pass the center at some ...

1

1) is correct. The wrong reasoning about 2) is that what you have in mind is probably Newtons Law for point masses. When the sphere is close to the doghnut the gravitational force will be more complicated, but still point towards the centre of the doughnut due to the symmetries in the situation. It will however stay finite, because all points of the sphere ...

0

The key point that you are missing is that the speed of light is constant for all inertial frames of reference. If you are going $0.99\ c$ and you are holding a flashlight and you turn it on, the photons emitted from the flashlight will appear to you to be leaving the flashlight at exactly $c$ (the speed of light). The key point of special relativity is ...

0

About derived equations: Keep in mind that all of mechanics effectively builds onto $F = ma$. This means that when you overload derived formulas in unfamiliar situations without verifying that the derivation holds you may be walking into a minefield. For example: consider the formula $s(t) = s_0 + v_0(t) + \frac{1}{2} at^2$. This was derived with a ...

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 ...

2

The quote is taken from just above eq. (1.32) in Ref. 1: [...] If the internal forces are also conservative, then the mutual forces between the $i$th and $j$th particles, ${\bf F}_{ij}$ and ${\bf F}_{ji}$, can be obtained from a potential function $V_{ij}$. To satisfy the strong law of action and reaction, $V_{ij}$ can be a function only of the distance ...

0

The strong law of action and reaction says that the forces that two bodies exert on each other have the same magnitude, opposite direction and act along the line joining the particles. When you want that last bit to be true and you want to write the force on particle $i$ as $-\nabla_i V_{ij}$, then the potential has to be a function of the relative distance. ...

0

What is force? It is the transfer of momentum with respect to time. Momentum is the product of mass of a body, and it relative velocity with respect to something. If we imagine, for once, our world made up of tiny uniform particles, then we can take the mass of one particle as one unit mass. This simplifies the situation, as the momentum, 'mass x velocity' ...

1

They are not the same thing. Suppose you are standing still and your friend is moving at 5 meters per second on a train. Let's say your frame is the $K'$ frame and your friend's frame is the $K$ frame. Now you throw a ball in the direction of the trains motion at eight meters per second. Then the speed you see is $\mathbf{v}'=8\mathrm{m/s} \hat{x}$, where ...

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 ...

0

I think you are prejudiced yourself if you say 'I do think superdeterminism is the most natural approach to how the universe works'. This is not meant as criticism, I just want to say that your mind is shaped by the world you perceive, and that world looks deterministic. Quantum mechanics and its randomness represent a different view on the (microscopic) ...

0

This site derives the principle of least action from Newton's laws. http://www.damtp.cam.ac.uk/user/tong/dynamics/two.pdf

0

Newton's second law, $\mathbf{F}_{net}=\dot{\mathbf{p}}$, is the definition of force. Lagrangian and action are defined to be $T-V$ and $\int L\: \mathrm {d} t$ (and not $\mathrm {d} x$) respectively. You don't derive anything from anything here (however we can talk about how $T$ and $V$ come about).

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 ...

1

Ideas shouldn't be treated as jokes but superdeterminism is false: it is incompatible with the best available theory of knowledge. First, a digression on the theory of knowledge. Any instance of adaptive complexity has to be explained. For example, as Paley noted, if you find a watch and notice that its parts fit together in such a way that they accurately ...

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 ...

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 ...

0

Once you measure a single electron's property, it collapses. It is a quantum randomness. Thus there is no classical analogy. There are two kinds of randomness: classical randomness, unknown because we are lazy. (contribute to entropy) quantum randomness, unknown because the God plays dice. (does not contribute to entropy) Classical randomness ...

0

Not possible, because if you want a very sharp blade, it should be very thin. On the other hand, cut material into two parts need to break the chemical bond of the molecules. However the blade is too thin that even if it cut off the bond(rather than saying get through), the bond will form again.

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