Hot answers tagged

91

What you cannot see by drawing the picture is the velocity of the individual points of the string. Even if the string is flat at the moment of "cancellation", the string is still moving in that instant. It doesn't stop moving just because it looked flat for one instant. Your "extra" or "hidden" energy here is plain old kinetic energy. Mathematically, the ...


63

The work you need to do (to insert the log) against the pressure of the fluid at that depth is equal to the work done by the fluid to get the log up to the height you desire. If you consider a log of volume $V$ and a tank of depth $h$, the pressure at that depth would be $\rho gh$, where $\rho$ is the density of the fluid, and $g$ the acceleration due to ...


48

ACuriousMind's excellent description was missing a picture. Here it is: This clearly shows that for the wave moving to the right, the front is moving up and the rear is moving down. For the opposite wave traveling to the left, the front (now on the left) is moving down and the rear is moving up. Summing them, you get a straight line with significant ...


39

Not if the laws of physics (particularly the laws of gravity) are as we understand them. In general relativity, there are a set of equations, called the Einstein field equations, that relate the curvature of space (roughly speaking, how much gravity there is) to how energy and momentum are distributed in space and time. To be consistent, these equations ...


31

Do black holes violate the first law of thermodynamics? No. See Wikipedia re the first law of thermodynamics: "The first law of thermodynamics is a version of the law of conservation of energy, adapted for thermodynamic systems. The law of conservation of energy states that the total energy of an isolated system is constant; energy can be transformed from ...


31

Can we create an amount of energy at a point in space and destroy an equal amount of energy at another point in space, with both the processes occurring simultaneously? This will not violate energy conservation, as the total energy in the universe is constant. So is it possible? If you learn about Einstein's Special Relativity, you'll discover that the ...


25

Just to complement the other excellent answers, here's an animation showing what two wave pulses with opposite amplitude passing through each other actually look like: You can clearly see that, at the instant when the string is momentarily flat, it's not stationary but rather moving quite rapidly, and thus will not stay flat for long. (Obviously, the ...


24

The short answer to your question is that the statements that "virtual particles need not conserve energy" and "intermediate components of Feynman diagrams need not be on the mass shell" are equivalent statements, but from two different historical perspectives. The concept of a virtual particle was introduced into physics in the mid-1920s while the ...


19

The device is apparently working as a heat pump, for which I give a brief theoretical analysis here. In the example given, the $P_h=69{\rm pW}$ light output comprises the $W=30{\rm pW}$ input by the researchers together with $P_c=39{\rm pW}$ of heat that was formerly in the chip. We can model the process by ideal heat pumping as follows. Heat drawn from ...


17

I'm neither an expert on QFT, nor do I have a very deep knowledge of how the ideas developed - so this is at best a partial answer. I always thought that your first guess is what they actually meant: A virtual particle is an "off-shell"-particle, which means that it does not obey the usual energy-momentum equation. Now people tend to interpret this as the ...


15

Conservation of energy/mass is the result of a symmetry called time shift symmetry, and if this symmetry is broken energy/mass will no longer be conserved. It is far from obvious that time shift symmetry would be preserved if closed timelike curves were possible, so you can't use conservation of energy as an argument that time travel is impossible.


11

Very little of the energy from a rocket engine ever goes to the kinetic energy of the rocket. The only way you get perfect conversion to KE of the rocket is when the propellant is directed in the opposite direction of motion and when the ejection velocity is exactly equal to the speed of the rocket. In that case, the propellant winds up containing 0 kinetic ...


11

What does this small change means in form of Rotational Kinetic Energy? There's a problem with your calculation: You assumed a constant value for the Earth's moment of inertia. The Moon and Sun raise tides on the Earth itself. These Earth tides result in subtle changes in the Earth's moment of inertia. The signature of these tides can easily be seen in the ...


9

The energy of an element of a traveling wave is not constant. Halliday-Resnick-Krane is right. For a string of density $\mu$ and tension $T$ the kinetic energy of an element $dx$ is $$dK=\frac 12\mu dx\left(\frac{\partial \xi}{\partial t}\right)^2.$$ For the potential energy we have $$dU=Tdl,$$ where $dl$ is the stretched amount of the string. A small ...


8

Let's say that the rocket is traveling in the $y$-direction at some velocity $v$, which may or may not be non-zero. A force - in this case, thrust provided by the engine - is applied perpendicular to the direction of motion, in the $x$-direction. This force produces an acceleration, which causes the rocket to move. Therefore, the force is not applied at a 90 ...


7

A battery connected to a capacitor is an RC circuit in the limit $R \to 0$ (i.e., there is no resistor and the resistance of the wire is negligible). One might think that the energy loss is zero in this limit, but this is not the case. For an RC circuit with a battery and an initially (i.e., at $t=0$) uncharged capacitor, we have \begin{equation} Q(t) = CV (...


7

Exerting a force and providing energy are quite different things. In particular, to provide energy to a body the force needs to perform work, that is, it needs to move the object in the direction that the force acts in. In the case of the Moon, the movement is circular and perpendicular to the gravitational force, so there is no inwards / outwards motion.* ...


6

Energy and momentum are conserved at every vertex of a Feynman diagram in quantum field theory. No internal lines in a Feynman diagram associated with a virtual particles violate energy-momentum conservation. It is true, however, that virtual particles are off-shell, that is, they do not satisfy the ordinary equations of motion, such as $$E^2=p^2 + m^2.$$ ...


6

I think one simplifying way of thinking about this idea is in terms of causality. That is, here we see a case where two things must be intrinsically linked. The energy being destroyed and the energy being created must have some mediator and they must always at the same time. The problem, however, is that if they must be causally related, there must be some ...


6

Just to expand on Hritik's great answer with a little more physical insight: Energy as a "stuff", friction in this picture. Very often we can define a scalar field (a smoothly-varying set of numbers, one for each point of the space) which describes everything about a pattern of forces. We call this the "potential energy" field. It is useful because a ...


6

When you put the log inside (from the side) you need to place the water from the log position somewhere else. The relevant "somewhere" is at the surface of the tank, so you have to lift the water there (it takes a lot of energy to get a log-equal volume of water to the roof level). The surface is now higher than before (the log-equal volume divided by the ...


6

You are making the mistake of thinking that photons are energy while massive particles are not. Photons are just a particle, albeit a massless one. There are other massless particles, for example gluons, and indeed at energies above the electroweak phase transition all fundamental particles are massless. So the distinction you are making between photons and ...


6

Potential and potential energy are defined for pairs of objects, not individual objects. It's meaningless to say "the potential energy of A". One must say "the potential energy of the system consisting of A and B". There is only one potential and potential energy in your problem. Perhaps the confusion comes from the way potential is introduced in ...


6

If there is friction (air resistance), that friction will extract energy from the spinning fan, thus slowing it to a stop. If you extract energy in any other way, you are also applying friction to the fan, again slowing it to a stop. In other words, yes you can extract energy from the fan, but no more than the rotational energy that the fan possesses: $$E=...


6

Essentially, yes. The derivation of the escape velocity is based only on the energy balance and energy does not depend on the direction. This follows from the property of the gravitational field to be conservative, so work required to move between any 2 points is independent from the path you take. Vague intuition: the vertical speed you estimate is about $...


5

Energy kinematics I have this question, because typically problems that can be solve using conservation of energy or just energy-related principles, can usually be solved sing kinematic equations. Yep. In fact, there are two profound pieces of math, Hamiltonian and Lagrangian dynamics, which say that you can use energies to derive the actual kinematic ...


5

The process of inverse Compton scattering does exactly this. A photon can interact with a relativistic (ie. hot) electron, absorbs some of its energy, and emerges from the interaction with a higher frequency and energy. This has the effect of cooling the electrons. Roughly speaking, the new frequency is related to the original frequency by $\nu \sim \gamma^2 ...


5

As a general rule (regardless of the definition of V) we have: $$\frac{\mathrm{d}}{\mathrm{d}t}(\bf{V}\cdot\bf{V})=\frac{\mathrm{d}\bf{V}}{\mathrm{d}t}\cdot \bf{V}+\bf{V} \cdot \frac{\mathrm{d}\bf{V}}{\mathrm{d}t}$$ $$\frac{\mathrm{d}}{\mathrm{d}t}V^2=2\left(\bf{V}\cdot \frac{\mathrm{d}\bf{V}}{\mathrm{d}t}\right)$$ $$2V\frac{\mathrm{d}V}{\mathrm{d}t}=2\left(...


5

The first method is giving the correct answer. In writing the work done by the force, you are assuming that the force $F$ itself is constant throughout the extension. However, this is not true. While extending the spring in a quasi-static way, the force $F$ must always match exactly the spring force at that time. This is needed so that at the end of the ...



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