Two identical waves, of amplitude $A$, have energy proportional to $A^2$. The sum of the energies of these two waves being proportional to $2A^2$.  If these two waves superimpose to form a wave of the same form with amplitude $2A$. The energy of this wave will now be proportional to $(2A)^2 = 4A^2$. This suggests an increase in energy by a factor of $2$, which is a contradiction to the conservation of energy.  How is this rectified please?
51952
I have a problem with energy conservation in case of interfering waves.  Imagine two harmonic waves with amplitudes $A$. They both carry energy that is proportional to $A^2$, so the total energy is proportional to $2A^2$. When they interfere, the amplitude raises to $2A$, so energy is now proportional to $4A^2$ and bigger than before.  The equivalent question is what happens to the energy with the superposition of two waves that interfere destructively.  Also, if someone could comment on the statement about this problem in my physics book (Bykow, Butikow, Kondratiew):      The sources of the waves work with increased power during the interference because they feel the wave from the other source.
21316
The intensity of an electromagnetic wave is only related to its amplitude $E^2$ and not its frequency. A photon has the same wavelength as the wave that's carrying it, and its energy is $h  f$.   So if a laser wave is kept at the same amplitude and the wave length is reduced, why does its intensity remain the same even though its photons now carry less energy?  Why are the intensities of electromagnetic waves so different to sound waves (and other waves travelling through a medium) which are related to $f^2  E^2$?
15721
I'm looking for an example of a Hamiltonian $H$, where $H\neq T+V$, but the total energy in the system, $E=T+V$, is still conserved.   While I'm at it, I might as well add that I'd be most interested in an example from a classical field theory. Furthermore, I am looking for a nontrivial example, e.g. where $H$ doesn't just differ from $E$ by some gradient function that integrates to 0 in the action.   In particular, I am trying to get a better understanding of the physical implications of $H$ being conserved when nontrivial variables are used, i.e. physically, what does $H$ this correspond to. Conversely, can one have a system where $E = H$ in one frame, and is conserved, while in another frame $E'\neq H'$, and $H'$ is conserved while $E'$ is not?
105904
If you increase the number of photons leaving a light source you increase the brightness/intensity, so it could be said that intensity is directly proportional to number of photons. However intensity is directly proportional to amplitude squared, so this leads me to wonder that if you increase the amplitude of a light wave, are you just creating more photons rather than affecting each individual photon in some way?
1164
I have the following statement which I don't know how to explain:  Suppose I have 2 identical monochromatic waves (same intensity and phase) shooting into the same receiver. If each wave's intensity is I, based on energy conservation I would expect the 2 waves together will bring a total intensity of 2I. For example if each wave carries 100mW power, I'm expecting a 200mW total power on the receiver side.  However the summation using phasor gives a different result: if we consider that intensity is proportional to the square of wave's amplitude, the square root of 100 gives 10 for the amplitude of 1 wave (for simplicity I use 10 as amplitude which included the constants), adding the other wave of the same phase which gives 20 as amplitude of the new beam, then square it to get intensity which is 400mW instead of 200 in the above example.  If we keep going by summing 4 50mW waves with the same convention, we get (4 x sqrt(50))^2 = 800mW ... which the logic is obviously not correct.  My ultimate goal is to sum up power of beams with different intensities and phases. Phasor addition works great if I have amplitude and phase, but when I try to use intensity to get wave's amplitude I got the above dilemma. Could someone point out where my logic go wrong, and please explain the way to do power summation with intensity and phase known for each wave? Thanks!
97475
Consider the  of Bennett et. al. How does one prove that this protocol would never violate the conservation of energy? At the face of it, it doesn't seem to be something obvious, as the various measurements, communications, rotations etc. don't seem to be able to account for the differences in the Hamiltonians of Alice's and Bob's labs. The state $\psi$ might have an expected energy $E_1$ in Alice's lab and a completely different expected energy $E_2$ in Bob's. Where does the difference come from?
39291
Kinetic energy is simple, right?! from the equation it seems as though if you know the speed and mass you'll know the work. Let's try to prove it. let's assume we know the acceleration and distance for now. $$w=mda$$ So the distance is $$d= 1/2 *at^2$$ Then  $$t = \sqrt{2d/a}$$ So $a*t$, which is the speed, is  $$a*\sqrt{2d/a} = a* \sqrt{2d}/\sqrt{a} = \sqrt{2da} = \sqrt{2w/m}$$ So regardless of the distance or acceleration, if you know the work and the mass, you can calculate the speed which is  $$v=\sqrt{2w/m}$$  What if you know the speed and the mass? You can easily calculate the work. $$v = \sqrt{2w/m}$$ Then  $$v^2 = 2w/m$$  That would mean  $$m*1/2*v^2=W $$ Rearrange it and  $$W= 1/2 *m *v^2$$ would be the equation.  What if you have an initial speed and a new speed and want calculate the amount of work done after an initial speed to get the new speed? You'll just subtract the total work from the initial speeds work.  $$1/2mv^2 - 1/2ms^2$$  where s is  the initial speed and v is the new speed. However if you do it the other way which is  $$d= 1/2at^2 + vt$$ and solve it the answer would be $$1/2ms^2  + 1/2msv$$ Why aren't they equal?
120347
We've all heard of the inverse square law, but apparently that refers to the flux or intensity or number of photons hitting an imaginary surface area. This is not exactly what I want to ask about.  I'm asking about the energy level of an individual photon, or individual ray of light. The light could be any frequency or wavelength.  If I'm not mistaken, the amplitude of the light ray is directly related to its energy.  Consider a ray of light travelling through a vacuum. Does its energy decrease over time according to inverse square law? Maybe inverse linear? Or not at all?  This question comes from . If it's true that the energy doesn't diminish at all, then that would mean a laser beam shot into space (and never hits anything) would never lose any energy and arrive at the other side of the Universe with the same energy it started with. That's kinda hard to believe.  EDIT: It appears "energy" and "amplitude" are not synonyms for light rays. Apparently the concept of waves is tricky for light, since after all, a vacuum has no medium to propagate through in the normal wave sense. So let the question be, "Does an individual photon lose energy solely from the act of traveling through a vacuum?"
108289
Let's say we have an harmonic oscillator (at Temperature $T$) in a superposition of state 1 and 2:  $$\Psi = \frac{\phi_1+\phi_2}{\sqrt{2}}$$ where each $\phi_i$ has energy $E_i \, .$  The probability of finding each the $i$ state would be 50% in this case. However, approaching this problem with statistical mechanics the probability would be proportional to $e^{-E_i/k_BT}$  What is wrong with my approach?
120828
In wave motion of a string both kinetic energy and potential energy are minimum at $y=y_\text{max}$ then why does the string come down again? As everything in nature tries to attain the lowest energy possible, what brings that string element back to its original position?
58219
Is $Q=\Delta U+W$ for when the the work is done from the system while $\Delta U=Q+W$ is for when the work is done by the system?  Will anybody explain this to me, please?   When do we use $Q=\Delta U+W$ and when do we use $\Delta U=Q+W$?
97127
What is the underlying reason that the same pairs of conjugate variables (e.g. energy &amp; time, momentum &amp; position) are related in  (e.g. time symmetry implies energy conservation) and likewise in QM (e.g. $\Delta E \Delta t \ge \hbar$)?
35643
I have read the explanation for this in several textbooks, but I am struggling to understand it via Archimedes' principle. If someone can clarify with a diagram or something so I can understand or a clear equation explanation that would be great.
12801
In the book &quot;Quantum Mechanics and Path Integrals&quot; Feynman &amp; Hibbs state that  the probability $P(b,a)$ to go from point $x_a$ at time $t_a$ to the point $x_b$ at the time $t_b$ is $P(b,a) = \|K(b,a)\|^2$ of an amplitude $K(b,a)$ to go from $a$ to $b$. This amplitude is the sum of contributions $\phi[x(t)]$ from each path. $$ K(b,a) = \sum_{\text{paths from $a$ to $b$}} \phi[x(t)]$$ The contributions of a path has a phase proportional to the action $S$: $$ \phi[x(t)] = \text{const}\ e^{(i/\hbar)S[x(t)]}$$  Why must the contribution of a path be $\sim e^{(i/\hbar)S[x(t)]}$? Can this be somehow derived or explained? Why can't the contribution of a path be something else e.g. $\sim \frac{S}{\hbar}$, $\sim \cos(S/\hbar)$, $\log(S/\hbar)$ or $e^{- (S[x(t)]/\hbar)^2}$ ? Edit: I have to admit that in the first version of this question, I didn't exclude the possibility to derive the contribution of a path directly from Schrödinger's equation. So answers along this line are valid although not so interesting. I think when Feynman developed his formalism his goal was to find a way to quantize systems, which cannot be treated by Schrödinger's equation, because they cannot be described in terms of a Hamiltonian (e.g. the Wheeler-Feynman absorber theory). So I think a good answer would explain Feynman's Ansatz without referring to Schrödinger's equation, because I think Schrödinger's equation can only handle a specific subset of all the systems that can be treated by Feynman's more general principle.
59333
Please consider the following derivation of the wave equation:    Questions:   I am just starting to learn about waves so this might be trivial. In the above, why are we interested in the transverse force which seems to be defined as the component of the tension force (which I am assuming is the force in the direction of the curve of the string) in the $f$ axis direction? Why is the wave equation modeled on the vertical displacement of this line segment under tension force?  Lastly, is this supposed to model a fixed profile wave moving at a constant velocity? Or is the string supposed to change shape as a function of time? What is the geometric interpretation?
27801
Let's say that we have a car travelling at a constant velocity (V) on a flat plane , let's assume that it's applied force or the force applied from its engine ( Fapp ) balances the frictional force ( Ffrict ) , hence the power due to Fapp is equal to Fapp×V , now let's assume that the same car now moves up an incline with the same engine power as previously stated and at a constant velocity ...but this time it will travel at a much lower constant velocity since the force which the engine exerts is much higher now in order to balance Fparalell and Ffrict , but what I don't understand is what caused this reduction in velocity , if the engine applies an equal force to that of the counteracting forces shouldn't that mean that there is no net force hence no change in velocity ,also if p=FV and if we now apply a greater force for the same velocity V why does the power increase , I know the answer can be simply explained using mathematics but my intuition says that power is the rate at which kinetic energy is transferred to a particle , so why would a greater force at the same constant velocity mean that you are transferring a greater amount of KE per unit time ?
12641
Consider 3 waves of frequency 101, 103, 106 hz, and of same intensity. What should be the beat frequencies?     Now I can calc it for 2 waves, and i know how to write the combined equation of the two. But the addition of a third causes lots of problems...the equation is getting cumbersome. Can you please help me here? I dug around online, and some responses were neglecting the third wave as it was too close to one of the waves. If I want to avoid that, is there any other solution?