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0

In a conservative field (no dissipation of energy) whatever way follows the small sphere, i.e. rolling along another sphere, or other, the total energy (kinetic + potential + rotational) is conserved.


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You say "Now let a particle move under a central force field where the total energy is constant. Now I want to measure the energy of the particle so I have to apply Hamiltonian and I get one of eigenvalues." By applying the Hamiltonian to the state of the particle you can predict its energy, i.e. on the paper, not measure. We measure in the lab, with ...


1

The equation you wrote $$ H|\psi\rangle=E|\psi\rangle $$ is the time-independent Schrödinger equation for an energy eigenstate. I.e., the state you are considering is already an eigenstate of the Hamiltonian with energy $E$. Therefore, as mentioned in the other answer, its time evolution is a simple phase factor, and you will always measure $E$ if you keep ...


2

If the particle is in an eigenstate of the Hamiltonian, you will get the same energy eigenvalue every time. We know that energy is conserved because the Hamiltonian obviously commutes with itself. The only time it is not conserved is if the Hamiltonian depends explicitly on time.


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Once you measure an eigenvalue, you collapse into a particular eigenstate. Let's call that state $| \alpha \rangle$, such that $\hat{H} | \alpha \rangle = E_\alpha | \alpha \rangle$. Now we use the time-dependent Schrodinger equation: $$ i \hbar \frac{\mathrm{d}}{\mathrm{d} t} | \alpha (t) \rangle = H | \alpha \rangle \\ = E_\alpha | \alpha \rangle \\ ...


1

When they say "Do not ignore electric force", they mean that there is both a magnetic and an electric force on the electron/positron, and you should not forget the electric force. In other words, you are asked to compute, for the $\vec v_+$, $q_+$ of the positron, the effect on the electron of its $\vec E$ and $\vec B$ field. Fortunately, 5 keV (kinetic ...


2

Quoting Sean Carrol's article linked by Симон Тыран, which makes the case for energy not being conserved: Having said all that, it would be irresponsible of me not to mention that plenty of experts in cosmology or GR would not put it in these terms. We all agree on the science; there are just divergent views on what words to attach to the science. In ...


1

Whether energy is or isn't conserved in an expanding universe is a somewhat vexed issue. On the one hand you have an experienced physicist claiming that energy is conserved, and on the other hand you have an experienced physicist claiming that energy is not conserved. The problem is that accounting for energy in general relativity is a complicated business. ...


1

Long story short: conservation of energy only holds locally where you can assume a static spacetime. On large scales the expansion of the universe gets relevant, so energy is said not to be conserved universally since the amount of dark energy per volume stays the same while the volume increases, see Sean Carroll's article, from which I quote: The famous ...


0

Negative energy can refer to several different phenomena: By definition, the potential energy between attracting bodies is negative, so that together with their kinetic energy, the total energy is zero. This is nothing magic, just a convention. In quantum theory, there are several forms of energy which are negative, e.g. the Casimir effect and virtual ...


2

Dark energy is not negative energy. It causes a repulsion because of its unusual equation of state, which causes it to behave as if it has a negative pressure. There is some discussion of this in the answers to Have negative pressures any physical meaning? and possibly also 'Negative pressure' counteracting gravity?. When general relativists talk ...


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I am replying to this because you seem to be a student, and not so clear on the statements. I have read that during fission and fusion processes, there is some kind of equilibrium between the single nucleus and the disintegration products, so they are constantly being converted into each other. " I have heard" is not enough, you should give a quote or ...


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The law of conservation will always hold. The Fist Law of Thermodynamics: The increase in internal energy of a closed system is equal to the difference of the heat supplied to the system and the work done by it: ΔU = Q - W Law of Conservation of Energy: The total amount of energy in any isolated system remains constant, and cannot be created or destroyed, ...


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I can answer the question partially. Energy conservation can be broken in time intervals less than plank time http://en.m.wikipedia.org/wiki/Planck_time (approximately 10^-44s). In time intervals longer than that, energy must be conserved.


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Set the initial total energy (kinetic plus potential at 1m) equal to the final total energy (which will be entirely potential energy at height 2m) There will only be one velocity, the initial velocity. There will be two heights.


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What observations would be needed to falsify the law of conservation of energy? As far as I understand there is no set of observations imaginable from which to draw such a conclusion. The only related conclusion, or measurement, which could be drawn from a suitable set of experimental observations would be: that a particular experimental region ...


1

Law of conservation of energy states that the energy can neither be created nor destroyed but can be transformed from one form to another. Let us now prove that the above law holds good in the case of a freely falling body. Let a body of mass 'm' placed at a height 'h' above the ground, start falling down from rest. In this case we have to show that the ...


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Yes. When you apply a net force to a mass (please note the word "net"), the object become accelerated . This acceleration means the body changes velocity, and a change in velocity means there is a change in the energy, because of the energy formula: $E=\frac{1}{2}mV^2$ The case of the circle is particularly interesting, because there is a force applied ...


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It all depends on where you have set your coordinate system. If it is on the earths surface then yes but if you set it say on the sun then no.


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The kinetic energy of a particle is dependent on the reference frame, so if a particle is at rest in a particular reference frame, the KE is zero. If someone moving with respect to the particle calculates the KE based on the reference frame in whichthey are at rest they will say $K \neq 0$. If you are using $$U_g = mgh$$ for gravitational potential energy, ...


2

Yes, absolutely. Potential energy is not a measurable physical quantity. What can be measured are differences in potential energy. So, if you compare the potential energies of a given mass at the Earth's surface and $100 \, \mathrm{m}$ above the surface, you cannot choose their difference, because that is governed by the laws of physics. However, you can ...


1

No. In a uniform circular orbit the orbiting body maintains constant energy while a constant force, only changing in direction, operates on that body. Kinetic energy changes when a net force is applied in the direction that an object is moving. It will reduce if the net force opposes velocity, or increase if net force supports velocity.


2

Where did the energy you spent go? You gained knowledge about the equation I gave. I'll try a different approach before this question is closed or migrated elsewhere. In the context of your question, the answer must be know (heh... no). Consider the case that you expend all the energy calculating incorrectly. Knowledge is, at least, justified ...


1

For a rotating ball you should use for energy $$E=\frac{1}{2}mv^2+\frac{1}{2}I\omega^2+mgh$$ where $I$ is the moment of inertia which for a sphere is $$I=\frac{2}{5}mr^2$$ where $r$ is the radius of the sphere, and $\omega$ it's his angular velocity which is related to the velocity of the center of mass via the equation $$\vec{r}\times \vec{\omega}=\vec{v}$$ ...


0

You assumed a great deal about your engine. You assumed that it "uses" fuel at 1L/s and that it produces a force of 10kN. That's not realistic, but it is possible for slow moving sleds, basically it is possible if the way it "uses" fuel is to burn some to get 10kN of thrust and then dumps the rest to make sure it will "use" 1L/s. It technically meets the ...


1

The damping introduces a dissipative element to the system, that is, energy is leaving the spring-mass system with time. As such, the energy is not conserved. The maximum energy for the system occurs at its initial configuration; the spring is stretched to some extent $A$ (initial amplitude). The initial energy for the spring is thus purely of a potential ...


0

Would one expect this energy to equal the $1/2 kx_0^2$ that we put into the system initially? Yes, it must. Energy conservation applies between any two points in time. If your point one is the initially stretched out spring before it is let go with energy $U_{el}=\frac{1}{2}kx_0^2$ and your point two is the situation of no energy stored in the spring ...


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When I double the energy input, the energy output is quadrupled. The relation should be proportional but it is quadratic. Let's consider firing a double barrel shotgun. If the gun is massive, it gains a small amount of kinetic energy, if the gun is less massive, it gains more kinetic energy. If both barrels are fired at the same time, both barrels may ...


0

you forgot the acceleration part.. as the volume of the fuel tank increases so the maximum velocity acquired at the moment when fuel runs out is also increased.. as the maximum velocity is achieved all other forces except ( kinetic + potential + acceleration - gravity ) is removed.. then the time it requires to all other forces other than gravity to act on ...


1

You assume that if you put the double amount of fuel in the engine, the velocity after the acceleration doubles as well. This is wrong, since when you double the energy input, the velocity only goes up with the square root: $E_{kin} = \frac12 m v^2 <=> v=\sqrt{\frac{2}{m}E_{kin}}$ Edit: 2) is wrong since the effective force is not constant. It's a ...


1

It appears that the confusion both in your question and in the question linked stem from a confusion of reference frames, leading to an apparent difference in outcomes across multiple tests of the "same" process. The original question regards a cannon that fires a 2kg shell at 1km/s muzzle velocity, i.e. imparts 1MJ energy to the projectile. The question ...


0

As you quoted, the Law of Conservation of Energy states that Isolated Physical systems have their energy conserved. But, nobody says that a non-isolated system cannot have its energy conserved. The mass above the Earth is acted upon by a field which is conservative, the gravitational field. In a conservative field the total energy of a body is conserved, ...


1

I think a good way to approach this question is with a Mach-Zehnder interferometer: The field landing on detector 1 is the interference between two waves, one from the lower path, and one from the upper path. Let's suppose the field in each arm is a collimated beam of coherent light, well-approximated as a plane wave, and the interferometer is well-aligned, ...


-2

When there is a complete destructive interference of two light beams, Maxwell’s equations predict that the energy becomes zero. Let’s have the case of two coherent collinear beams, out of phase 180 degrees, like the case of the antilaser. \begin{align} E_1 = E_m \sin (kx - \omega t);\quad E_2 = E_m \sin (kx - \omega t + p) \\ B_1 = B_m \sin (kx - \omega ...


1

All the known forces conserve energy, but they don't necessarily conserve energy in macroscopic modes. For instance friction takes some of the energy of macroscopic motion and coverts it into an increase in temperature (i.e. energy in microscopic modes). Total energy is conserved but energy that is useful at the human scale is not. Feynman is talking about ...


2

There are macroscopic forces that admit no description in terms of a potential, for example, any friction force proportional to the velocity of a moving object as path-dependent integral, and is hence non-conservative. But we know the macroscopic description is not the fundamental description. In terms of the interaction of the constituents of matter, all ...


3

Conservation of energy is still fine the following sense: For any region, the energy at a later time is equal to the energy at the current time, plus the net flow of energy in and out of the region during the interval of time. And the energy in question includes the rest energy of bodies, their kinetic energy, their thermal energy, etc. Anything except ...


0

Energy has the capacity to do work. When you feel "full of energy", chemical energy stored in your muscles can be converted into mechanical energy, to enable you to run, or lift small cars above you head. Different types of energy are stored in different ways. Gravitational energy is inherent in any object which the gravitational forced can move downwards ...


1

The wires don't perform work because the direction of motion of the log is always perpendicular to the wires. If the wires don't stretch, then the motion of the log is circular with each wire as the radius of that circle. Therefore, the motion is perpendicular to the tension force. The formula for work $W = F\Delta x \cos \theta$ is zero for $\theta = 90^o$. ...


0

There is a simple trick to achieve really high (>= 90% )efficiency: Just before the flowing (falling) fluid hits the propeller, make it rotate using guides, i.e., give it some angular speed beside the linear speed in a closed tube. To benefit from the fluids kinetic energy (in this case it is saved in both the rotational kinetic energy and linear speed ...


8

First, the energy expectation value of the superposition state you have written down is $$ \left(\frac{n_1 + n_2}{2} + \frac{1}{2}\right)\hbar\omega $$ and one might naively conclude that therefore the energy of the state lies in between the energy of its constituents. This naive concept doesn't work, though - the "energy" of a state that is not an energy ...


1

If energy is conserved, how so that a measurement of the energy of a state, such as $$ \psi = N (\phi_1 + \phi_2), $$ could result in two different energies? I think your question is easiest to tackle with the consistent histories interpretation of QM (though you'll reach similar conclusions with any other interpretation). We must remember that energy can ...


0

Suppose you prepare a state and you want to determine which state you have created. Of course a single measure tells you nothing unless you know a priori that you have created an eigenstate of the Hamiltonian. So from a practical point of view, a measurement is a repetition of the same experimental operations on a large enough ensemble of identical copies of ...


-1

As you write the state $|\psi\rangle$ it seems that your system is in a superposition of $n_1$ oscillators and $n_2$ oscillators, each one of energy $\hbar \omega$. So, in the state $|\psi\rangle$ the number of oscillators in not fixed, and with it the energy is not fixed, it is not a constant of motion. In this case the question where go the other ...


2

Introductory physics problems often limit the momentum economy to the motion of large particles or fragments (collisions and explosions) for simplicity of calculations. In reality, the momentum transferred to any surrounding gas (air) should ideally be part of the conservation. These introductory problems are constructed so that compression waves and huge ...


2

In the case of an explosion, before the explosion the momentum of the bomb is zero, so according to law of conservation of momentum, the momentum after explosion should also be zero. So, momentum of the bomb before collision = momentum of the bomb after collision. As for sound and light energy, I think that it is the chemical energy of the bomb that is ...


2

I'd say part of the answer must be that whatever dynamic variable you use, like Enstrophy, Vorticity, their potential analogues, etc. those are always 'filtered' fields. Filtered in the sense, that you start with the velocity field $\vec v = \sum u_i \vec e_i$ that has full information over the dynamics and then apply some operators (integration and ...


0

Gravitational potential is defined such that it is $0$ at infinity, and has negative values at all points other than the one at infinity. So, an object technically has the highest gravitational potential at infinity. (an extremely extremely far distance.) The incline can be considered as path from an area of lower gravitational potential to a higher one. ...


0

The sun gets its energy from the nuclear fusion, mainly the fusion of hydrogen atoms to form helium atoms. This energy passes through the vacuum of space via Electromagnetic radiation. Plants collect some of this radiation and use it to perform photosyntheses, which in turn is used to make carbohydrates and such. You eat that plant or some other animal ...


1

When you walk up a hill, pushing a bicycle or not, you increase your potential energy by spending chemical energy. One of the reasons you need to eat is to ingest fuel, so to speak, that allows you to spend energy on doing your daily tasks. For example, your body can metabolize sugar (most notably glucose) by oxidizing it, which frees energy that you then ...



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