104

There are three parts to the phenomenon, two real and one illusory. While you are lowering the bat, its relative velocity to the approaching ball increases that little bit. The ball bounces off it that bit harder, gaining twice that extra velocity relative to the floor. Repeat for several bounces and the difference might become noticeable. This is one real ...


69

Guy Inchbald mentions the slight force of the paddle on the ball and that velocity relative to the separation increases. For the latter, there's a further issue that every time the ball bounces, it makes a noise, and that noise becomes more frequent as the distance shortens, which increases the perception of speed. Also, I believe there is a third phenomenon:...


6

The question "what happens if a conservation law is observed to be broken?" is not a hypothetical one $-$ it is a historical one. The clearest example of this is the conservation of parity, a quantum-mechanical quantity which is related to the invariance of physical laws under spatial inversion. (There is a deep relation, called Noether's theorem, ...


6

Maxwell's equations place a constraint on the current, namely that it be conserved. To see this, take the divergence of Ampere's law for $$0 = \mu_0 \nabla \cdot \mathbf{J} + \mu_0 \epsilon_0 \nabla \cdot \frac{\partial \mathbf{E}}{\partial t}$$ which is equivalent to $$\nabla \cdot \mathbf{J} = - \epsilon_0 \frac{\partial}{\partial t} (\nabla \cdot \mathbf{...


6

A solution of the free one-dimensional Schroedinger equation: $$ i\hbar \frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2}\,\,\,\quad \text{(1)} $$ is: $$\psi = A e^{i(kx -\omega t)} \quad\quad\quad \text{(2)} $$ where $\omega$ fulfills the condition $\hbar \omega = \frac{(\hbar k)^2}{2m}$. If tentatively one ...


5

The continuity relation holds for solutions of the Schrodinger equation. $A\cos (\omega t - k x)$ is not a solution.


5

It should have said that ''kinetic energy'' is not always conserved. When you throw some sticky dough at a wall the kinetic energy is converted completely to heat and sound energy and the dough sticks to the wall. Momentum is conserved though. When you throw the dough the dough went one way and the entire Earth went the other. When the dough stuck to the ...


4

Momentum should not be confused with energy. In some cases momentum is conserved but energy is definitely not. This is indeed a confusing way of putting things. The first sentence is clear and unequivocal. But the latter sentence should really read: Momentum is always conserved but kinetic energy is not always conserved($^+$). The latter part of this ...


4

Because Ping-Pong balls are elastic And in elastic collisions, kinetic energy is conserved. When your ball is allowed to bounce freely, it reaches a speed of zero at the top of its arc. At that point, all of its kinetic energy has been converted to gravitational potential energy. The opposite is true at the ground, where all its gravitational energy has ...


4

For each continuity equation $$ \sum_{\mu = 0}^n\frac{\partial J^{\mu}_a}{\partial x^{\mu}}~=~0, \tag{A}$$ one can define a conserved quantity $$ Q_a(t)~:=~\int_V \! d^nx~ J^0_a(\vec{x}, t). \tag{B}$$ OP's example: Let $J^{\mu}=J^{\mu}_1 +i J^{\mu}_2$ be a complex current, and introduce a complex space coordinate $z=x^1+ix^2$, i.e. $n=2$. OP's continuity ...


4

is conservation of energy not equivalent to conservation of momentum? The conservation of energy is not equivalent to the conservation of momentum. Per Noether’s theorem momentum is conserved whenever the Lagrangian is symmetric under spatial translations and energy is conserved whenever the Lagrangian is symmetric under time translations. Since it is ...


3

Noether's theorem is the thing to pay attention to here. This theorem basically says that if there is a symmetry in the underlying physics, there is also a conserved quantity. Traditionally the best-known symmetries are: Time invariance, i.e. the laws of physics do not change with time. If this is the case, then energy is conserved. Translational invariance,...


3

Contrary to what science fiction may tell us, quantum teleportation does not involve the physical teleportation of matter. It only teleports the state of matter. In that sense, it teleports the information that is associated with a particle and not the particle itself. So, electric charge remains where it is. In response to the comment: the spin is a degree ...


3

In the above types questions why we don't include the( initial and final both) velocities of bullet while finding the initial kinetic energy of the system in (COE) conservation of energy In the example you gave, kinetic energy is not conserved. During the completely inelastic collision some kinetic energy is spent on permanently deforming the pendulum and ...


2

You are right to question this. Technically the system consisting of the coconut and the firecracker form a closed system$^*$, so that the entire momentum of this system is then conserved (assuming nothing flies into the air of course). If you just considered the coconut as the system, then the force from the explosion is external (since the explosion is not ...


2

For what it's worth, one may show that a 2nd-order ODE $$\ddot{x}~=~f(x,\dot{x})$$ without explicit time-dependence, or equivalently, a pair of 1st-order ODEs of the form $$ \dot{x}~=~v, \qquad \dot{v}~=~f(x,v),$$ always has a local Hamiltonian formulation, cf. e.g. this Phys.SE post. The Hamiltonian is a conserved quantity.


2

Charge conservation can be stated as a continuity equation $$\frac{\partial \rho}{\partial t}+\nabla\textbf{j}=0 \tag{1}$$ where $\rho$ is charge density (measurable in Coulomb/m$^3$) and $\textbf{j}$ is current density (measurable in Ampere/m$^2$). On the other hand, from the magnetostatic equation $$\nabla \times \textbf{B} = \mu_0\textbf{j} \tag{2}$$ you ...


2

The quantity $C = g_{\alpha\beta}\dot{x}^\alpha K^\beta$ is a scalar invariant and is therefore independent of the coordinate system used. Since coordinate systems exist that are not singular at the event horizon (for example Kruskal Szekeres) you could calculate $C$ at the horizon using these coordinates. The quantity is not affected by a change of ...


2

The first thing you have to think is that it took energy to get 2m high, so it may be useful to calculate that energy. The final energy is given by: $mgh = 0.2 \cdot 10 \cdot 2 J = 4 J$ that energy is coming from the kinetic energy at the start of the motion, but that shange of speed is due to the force applied. Since the change of kinetic energy is equal to ...


2

If you just have the free particle Hamiltonian, with $V = 0$ then you'd expect that the dynamics are the same no matter the position of your system and that is precisely what happens. This corresponds, as you mentioned, to the fact that $\left[H,P\right]=0$. However, when you introduce a potential you break this translational symmetry because now the ...


2

First note that $$\nabla_\alpha(F_{\mu\nu}F^{\mu\nu})=F^{\mu\nu}(\nabla_\alpha F_{\mu\nu})+F_{\mu\nu}(\nabla_\alpha F^{\mu\nu})=F^{\mu\nu}(\nabla_\alpha F_{\mu\nu})+F^{\mu\nu}(\nabla_\alpha F_{\mu\nu})$$ $$=2F^{\mu\nu}(\nabla_\alpha F_{\mu\nu})=-2F^{\mu\nu}(\nabla _\mu F_{\nu\alpha} +\nabla _\nu F_{\alpha\mu}).$$ Using this equation, the anti-symmetry of $F_{...


2

Black hole will absorb the momentum and start to move with a constant speed. A black hole solution can be transformed to any other inertial frame via a global Lorentz transformation, so it can move with any speed through space.


2

The entanglement is the coupling of different quantum systems. The experimenters bring entangled systems to their own journies. And when a quantum system is measured, the other quantum system is collapsed at the same time because they are entangled. While this process, no material is transferred. It is the non-locality of the world which makes this possible ...


2

My volume of Symon's Mechanics address' this question: The conservation laws are in a sense not laws at all, but postulates which we insist must hold in any physical theory. If, for example, for moving charged particles, we find that the total energy, defined as (T + V) [kinetic plus potential], is not constant, we do not abandon the law, but change its ...


2

You already have proved the "only if" part. What your derivation shows is that $$\frac{d}{d t}\left(\frac{1}{2} m|\dot{\mathbf{x}}(t)|^{2}+V(\mathbf{x}(t))\right)$$ equals zero only if $$\mathbf{F} = -\nabla V.$$ In other words, your work proves $$\mathbf{F} = -\nabla V \implies \frac{d}{d t}\left(\frac{1}{2} m|\dot{\mathbf{x}}(t)|^{2}+V(\mathbf{x}(...


1

Maybe it's useful to think this question from the Hamiltonian formalism. The Hamiltonian $\cal{H}$ of a system is, in general, the total energy of the system, expressed with a particular set of variables. For example, for a point particle moving in the presence of a constatn gravitational field, the energy is $$E = \frac{mv^2}{2} + mgx$$ but these variables ...


1

the forces between two balls are always perpendicular to the surface, so in direction of the radius, where they touch. That is the reason you know the direction of the acceleration .


1

In principle Noether's theorem (NT) itself also works for Galley's extended action functional (5) and its non-conservative potential $K$, which are still local in time. Galley's peculiar boundary conditions are not a problem as they are not relevant for NT. As usual it is your job to provide an off-shell quasisymmetry of the action in the first place. The ...


1

The assertion/assumption Physics laws should not be changed under translations Does not imply the statement Hamiltonian must commute with the generator of translations If we have a system that has $[H,p]=0$ then it’s a special system where the coordinates of the origin doesn’t matter. It’s a spatially symmetric system. It has translational invariance. ...


1

After the comment by @ZeroTheHero I realized my answer is wrong. Here is the explanation of what went wrong. I will delete the answer later Conservation of ordinary momentum is implied by homogeneity of space and time, not just space. The most easiest way to see it is by the fact that you can simulate decelerating particle by adding a time dependent term to ...


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