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Here is the proof not very rigourous. Just to give a slight Idea. Conservation of Energy momentum four vector $(E/c, p_x, p_y, p_z)$ in STR is given by $$p_x^2 + p_y^2 + p_z^2-\frac{E^2}{c^2}=const$$ you can see the proof of this on Wikipedia done using $Action Principle$ Now we Know that $E= \hbar w$ and $p =\hbar k$, and substituting these $$(k_x^2 + ... 0 Actually the phase is exactly \vec{p}\cdot \vec{x} = \eta(\vec{p}, \vec{x}) = \eta_{\alpha\beta}p^\alpha x^\beta = -\omega t+k_xx+k_yy+k_zz where \vec{x} and \vec{p} are 4-vectors in Minkowski space:$$ \vec{x} = \pmatrix{t \\ x \\ y \\ z},\ \vec{p} = \pmatrix{\omega \\ k_x \\ k_y\\ k_z}$$But because a dot product of two vectors is Lorentz Invariant ... 0 Yes, You can! The de Broglie equations relate the wavelength λ to the momentum p, and frequency f to the total energy E of a free particle:$$\mathbf{p}=\hbar\mathbf{k}E=\hbar \omega $$Using two formulas from special relativity, one for the relativistic mass-energy and one for the relativistic momentum$$E=\gamma m_0c^2\mathbf{p}=\gamma m_0\...

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The wave equation is $$\frac{\partial^2 H}{\partial t^2} = \omega^2 \frac{\partial^2 H}{\partial x^2},$$ and you are assuming $H(x,t) = A(x) B(t)$. If you plug this in, $$\frac{\partial^2 A(x) B(t)}{\partial t^2} = \omega^2 \frac{\partial^2 A(x) B(t)}{\partial x^2}\\ A(x) \frac{\partial^2 B(t)}{\partial t^2} = \omega^2 B(t) \frac{\partial^2 A(x) }{\partial ... 2 The following may be useful. The usual form of a plane wave will be something that involves a vector dot product between the \vec{k}-vector and the position vector, call it \vec{x}. Something of the form \vec{k} \cdot \vec{x} will appear in the argument of the sinusoid. I hope this helps. 0 You don’t need a magnetic field analysis; it would be redundant. You don’t need to worry about the boundary conditions. That’s how you derive the Fresnel reflection/transmission coefficients for an interface. Simply use those coefficients (unless your instructor intends you to derive them from scratch here, which seems unlikely), and keep track of the phase ... 2 Frequency is the rate of change of phase with respect to time. As an operator, it's:$$ \hat\omega = -i\frac{\partial}{\partial t} $$so that:$$ \hat\omega B(t) = -i\frac{\partial}{\partial t}e^{i\omega t} = -i^2\omega e^{i\omega t} = \omega B(t) $$is the definition of "constant frequency". 0 A small charged particle will tend to oscillate with the electric field as an EM wave moves by. However, as the particle moves, it is also subject to a magnetic force (in the direction that the wave is moving). This explains how the wave can exert pressure on a surface. 2 I've already accepted Philip Wood's answer as the answer to my question, but I wanted to post the diagram of Huygens' principle and refraction that helped me understand more of the "how/why" that answers my question. I'm new to Stack Exchange, please let me know if this is bad form. I find the image to be self-evident, but there is a description ... 2 In general, n=0,1,2,.... But if n=0, the field is zero, because the field is proportional to \sin(n\pi y/a). So the smallest value of n for which the field can propagate is n=1. In general, k_x=0 just means that the field is constant in the x direction. A way think about this is that k_x=2\pi/\lambda_x, where \lambda_x is the wavelength in ... 3 As you know, the peaks and troughs of water waves are examples of wavefronts. Wavefronts are lines or surfaces along which particles are oscillating in phase, so asking why a wavefront stays in phase would indeed invite the response 'by definition'. But it would not be illogical to ask why a straight or plane wavefront stays this shape when the wave passes ... 2 The diagram you posted shows the fringe width being measured from the center of one light band to the center of the next light band; or from the center of one dark band to the center of the next dark band. Thus the total fringe width is equal to the 1/2 the width of one light band, plus the width of a dark band, plus a second 1/2 the width of one light band. ... 1 It doesn't make sense to critique statements about wave speed when they don't even specify what speeds they are talking about. In n-spherical waves there are at least three speeds, all of which are generally different: phase speed, group speed and leading edge speed. The latter is obviously always c_0, the speed of wave in the medium. The former two are ... 1 Equate the function to half its peak value and solve. For eample, if b\gg a, then the envelope of the overall function is defined by the sinc function, which has a maximum of 1, so it reaches the FWHM points at$$ {\rm sinc}^2 \left( \frac{ax}{2} \right) = \frac{1}{2}$$This must be solved numerically. The solution is ax/2 \simeq \pm 1.392. Thus the FWHM ... 0 You would do just as the name implies. For example, the envelop has a height of about 1. So half max is 0.5. The width of the envelop at 0.5 appears to be 0.25 - -0.25 which equals 0.5. You could also use the function to determine where the maximum occured and how wide it is at half that max. Since you've plotted the data you should have all the info that ... 1 Just a hand-waving argument, but string stiffness refers to the string's resistance to bending or curvature. Curvature is the second derivative of displacement, \psi, and therefore comes into the equation as d2{d2\psi/dx2}/dx2 instead of d2\psi/dx2. 4 Some of the sound waves do get reflected on their way back, which reduces the intensity of the returning echoes. But enough energy is returned to allow imaging to be done nonetheless. 1 There is a table here that shows a 2% change in refarctive index from 0^\circC to to 100^\circC at 600 nm. Is your setup sufficent to detect a 2% change in n? 0 I assume a wave equation (\Box +m^2) f = 0. This is simply the wave version of Einstein's famous relation E^2 = m^2c^4 + p^2c^2. Thus v \in [0,c) for m\neq0 and v=c for m=0 for any positive number of space dimensions. 1 This is a good question. In the Fraunhofer limit polarisation does not matter but when non parallel interfere polarisation will degrade contrast. It is caused by the polarisation direction parallel to the plane (TM) of incidence, formed by the two k vectors, as opposed to polarisation perpendicular to the plane (TE). Consider two waves propagating at an ... 3 The angle between the light rays is irrelevant. To calculate the interference pattern we are just calculating the electric field vector at a point in space. We consider some point (x,y,z) in space, and at this point the electric fields of the two rays will be \mathbf E_1(x,y,z) \sin(\omega t + \phi_1) and \mathbf E_2(x,y,z) \sin(\omega t + \phi_2). We ... 0 In the first sentence you quote, the "only" is more important than the actual equation for photon energy. For photons of a given frequency, the energy flux through a given area is therefore proportional to the photon flux. From classical electromagnetism, the energy flux in vacuo is proportional to \overline {E^2}. Hence the textbook's claim. 4 |E|^2 here actually means "square of the electric field", not "square of the energy density." In vacuum, it turns out that the energy density of the electromagnetic field is proportional to |E|^2.^\star Since the energy density of the field should be the number of photons in the field^\dagger times the energy of one photon, we ... 4 We hid the time dependence when we said we were using a phasor representation of the field. A phasor is a complex number A representing a time dependent quantity |A|\cos(\omega t+\angle{A}), where \omega is the previously defined angular frequency at which the instantaneous value of the quantity varies. 1 Calculations with the Uncertainty Principle imply that the wave nature of a macroscopic object is too small for us ever to hope to detect. 4 The time dependent factor, e^{j\omega t} has been omitted, to save clutter. Readers are supposed to supply it for themselves. The omission is usually made only when dealing with waves with the same \omega, or with the same wave at different points z in space. In such cases, e^{j\omega t} can be factored out of derivations and put in again at the end, ... 0 Let us take a band-limited sawtooth wave in an electrical circuit. Say its frequency is 440hz. We know that its next harmonic after fundamental is 1320 Hz. Do we actually have something oscillating in this circuit at frequency 1320 Hz in a sinusoidal waveform? No. The period of the waveform is 1/440\text{s}. What is it then?...just a mathematical concept?... 1 The proof can be found in any elementary text on waves. You can do it as follows: We follow the notation as in the figure. Net force upward is$$F_x(t)=T_2\sin\theta_2-T_1\sin\theta_2$$We want to express F_x(t) in terms of \psi(z,t) and its space derivative \partial \psi/\partial z. Now we make an approximation here, In the small-oscillation ... 5 "Vibration" just refers to oscillatory motion around an equilibrium point. \sin^2(x) certainly satisfies this definition. Also it's worth pointing out that \sin^2(x) actually is a sinusoidal function \begin{equation} \sin^2(x) = \frac{1 - \cos(2x)}{2} \end{equation} 0 According to QED, light consists of photons. How do we know that? Because in any experiment with dim light, We always found that light comes in a discrete packet of energy. The phenomena of light can be understood by introducing what is call probability amplitude. With the help of this probability amplitude, We can find out the probability for some event to ... 2 With a compact wave pulse, Huygens’ Principle still holds, and points along the wave front acts as sources. On the advancing side of the pulse, the outgoing waves from different parts of the front interfere constructively. However, on the receding side, the waves have different phases and interfere destructively. That accounts for how the wave packet ... 1 The Doppler Shift formula f'=f(\frac{v \pm v_{obs}}{v \mp v_{source}}) only works if the wind (or the medium that the sound is moving in) is constant. Therefore, if the wind is moving at a constant speed, change the reference frame so that the wind is stationary. In your case, change the reference frame so that it is moving from the train to the boy at ... 1 How can we see colours at the same time ? I am going to assume your question is about the mechanics of human vision. We can see in colour because there are several different types of light sensitive cells in the retina at the back of the human eye. Colour vision comes from cone cells. There are three types of cone cells - one type reacts most strongly to ... 0 Light is a ray photon particle whose energy is given by$$E=h\nu$$Light corresponds to different colors are different energy photons. why don't these electromagnetic waves of each color cancel or add up to each other. Photon particles can not be added nor they can cancel each other. When the two photons see each other inside the LHC, they sometimes ... 0 You are right! You can see the following way: Suppose the following setup : The path difference between the rays reflected from the upped surface and the lower surface is given by$$\Delta x= 2\mu \lambda $$and phase difference given by (with additional \pi phase) :$$\Delta \phi=\frac{2\pi}{\lambda}\Delta x \pm \pi2 Typically we assume that the distance to the screen is much larger than the width of the slits or the separation of the slits. The light that emerges from the two slits at an angle of theta are almost parallel. Thus, theoretically and approximately there is no angle S_1 P S_2. 0 A wave function can be thought to be analogous to a vector in the sense that the set of all functions on a domain forms a vector space. Therefore, each function can be thought of as a vector in the vector space of all continuous functions on that domain. The reason you might have seen these differently is that matrix formulation can be easier to deal with in ... 0 The energy balance is indeed an interesting problem. For a monochromatic plane wave the source is an infinite sheet of sinusoidal current. It is not trivial but is straightforward to calculate the Poynting vector for this arrangement. When you do so you find that energy propagates away from the current sheet with equal power density on both sides of the ... 1 This is an artifact of the so-called conversion process, by which a continuous ribbon of paper hundreds or thousands of feet long and four feet wide on a huge roll is slit to width and then sheared to length and the cut sheets then stacked and wrapped into packages for shipment. Since the initial spooling process by which the master roll is made occurs when ... 0 I notice one tendency. Biggest sheets distortion is in the bottom pages. If you go upwards, you'll see that distortions will become smaller and smaller and paper will be getting more straight and almost plane-like at the top. This seems a little-bit counter-intuitive, because bottom sheets experiences highest load from paper mass above, still they warped ... 0 Let's assume that they were originally perfectly stacked. If you apply some nonuniform external force to the whole stack, some sheets might slightly slide around (which is true in this case, as visible in the photo; note sides of the stack). This can be viewed as a deformation of the whole stack. Due to being stacked, upper sheets will press down on the ... 4 I do not think "energy flow" is a well defined variable for electromagnetic waves. The single photon at a time double slit experiment, shows that the energy from the dark lines has gone to the bright ones Single-photon camera recording of photons from a double slit illuminated by very weak laser light. Left to right: single frame, superposition ... 0 I can not understand the phenomenon but I can describe it to you. I'll take light as a particle (photon particle). Considering the following is your main concern: I was wondering exactly why a ray of light changes direction when moving from one optical medium to another. The discussion will be followed by the Feynmann lectures on QED: Grand Principle: The ... 1 When an electromagnetic wave is reflected from a perfectly conducting mirror,* the electric field {\bf E} is phase shifted by 180^{\circ}. However, there is also the magnetic field {\bf B}, and the magnetic field is not phase shifted. There is a tendency to identify the wave with its electric field, and this is entirely natural, since in most ... 1 For an ideal mirror it is shifted by 180 degrees. For imaging this means a shift of \lambda/2 along the propagation direction. This is not related to the mirroring effect, which is caused by the change of sign of k_\perp. 0 After some bewilderment I perhaps understand now where your question is coming from. At the end of your question you suddenly mention speed! Without specifying the initial distribution of velocities in your long slinky thing we cannot say how the compressed region will progress. With zero initial velocities the compressed region will split into two parts, ... 0 There is a difference between a pinhole (which by def. is circular in shape) from a narrow slit, which isn't. The slit cuts out all electric field vectors except the ones close in alignment with the long direction of the slit. It works similarly to a polarization lens. 3 To address this question I need to get something else out of the way first. Experiments that are designed to obtain a specific kind of interference pattern need to manipulate the light in order to give it one or more particular desired properties. This may give the student the impression that in order for any interference effect to occur the light must be ... 1 Consider a localized volume or "bubble" of an ideal gas (V_1) at a pressure P_1. Something happens to compress that bubble isothermally to a new volume V_2. What would be the new local pressure of the bubble? P_1V_1 = P_2V_2  P_1 = \frac{P_2 V_2}{V_1} $$So$$ \Delta P = P_2-P_1 \Delta P = P_2\left(\frac{V_1 - V_2}{V_1}\right)  ...

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The restoring force here would be the fact that gasses always seeks to equalise pressure. Thus, when a sound wave moves around some particles, the pressure of the surrounding particles will eventually get the system back to rest.

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