# Tag Info

1

Assume plane waves. The tangential boundary conditions show that the transverse electric and magnetic field vectors must stay in the same direction on transmission or reflexion from the interfaces, assumed aligned with the wavefronts. Since we know the direction of the waves, let's say the $\vec{E}$ fields are all in the $\hat{X}$ direction, the magnetic ...

0

Can matter waves interfere with light? Light - matter interactions are possible, namely, light - atom interactions when you use laser light for atom diffraction: Atom Interferometry For this, I'm pretty sure not, in the same way light does not diffract with sound waves as they operate with different mechanisms. Light does diffract with sound ...

0

Sort of a silly answer, but a bunch of protons, electrons, and neutrons bound together (i.e. an atom or molecule) can certainly diffract through a double slit. This has been taken to ridiculous extremes in recent years: witness the double-slit diffraction of C284H190F320N4S12, a molecule with 810 atoms and over 5000 electrons. But if you're thinking ...

1

Protons and electrons both obey the de Broglie hypothesis: wavelength = Planck constant / momentum. But protons do not act within the atom the same as electrons - they move in a tighter radius at higher speed. They have to be accelerated to reveal their wave nature, and as the momentum of a proton would be much greater than that of an electron at typical ...

1

Let one end of a very long string is being oscillated transversely so as to generate a sinusoidal wave traveling out along the string. In order to set up a wave on a stretched string, the driving force at the end of the string provides energy. This energy is not retained at the source; it flows along the string at the wave speed. The string transports ...

1

No. Energy conservation always applies. The elastic potential energy will be maximum at a wavetop, since here the rope is stretched the most, $U=½kx^2$. The transverse velocity and thus the kinetic energy is zero at this point $K=½mv^2$ since this part of the rope stops and starts moving back again. $$E_{before}=E_{after} \implies K_1+U_1=K_2+U_2$$ Energy ...

1

If the light was emitted after the recombination, it can't have traveled over 13.7 billion years. The EGS-zs8-1 galaxy is located 13.1 billion light years away , which is close to the maximum for a plausible picture. More details on the wiki http://en.wikipedia.org/wiki/EGS-zs8-1 . If the light is not absorbed by an obstacle, it will probably travel to the ...

2

The relationship is true by dint of the trigonometric identity $\cos(u+v)+cos(u-v)=2\,\cos u\,\cos v$. A compelling experiment is to listen to two tones a few hertz apart (the LHS of the identity) and hear the throbbing beats (the AM wave on the RHS). A pitch fork and a guitar / violin string (the latter readily tunable) is a good way to do this; Given the ...

0

As a general rule, two "degenerate" oscillators (i.e. two oscillators with the same frequency) will no longer have the same frequency if they interact with each other, however weakly; the degeneracy is usually broken. So if you have two identical resonators with a fundamental of 11 Hz, and you bring them together, the combined system will have resonant ...

1

For macroscopic systems some physical means must be present for energy to flow between potential and kinetic states, or magnetic and electrical fields. The flow must also not be subject to energy loss. Apples do not have the properties that allow such a flow of energy. Apples do not 'ring' when they are subjected to an impulse of energy. Almost all the ...

1

The phase of a wave varies by $2\pi$ over the distance of one wavelength, $\lambda$. So suppose you have some distance $d$, then the number of wavelengths in this distance is $d/\lambda$, and therefore the phase shift over the distance is: $$\varphi = 2\pi\frac{d}{\lambda}$$ But if the wave is passing through a material with refractive index $n$ this ...

2

$E = mc^2$ is a special case of the full equation $$E^2 = m^2 c^4 + p^2 c^2$$ It only becomes $E = mc^2$ when applied to objects at rest, i.e. with zero momentum. A wave that is transferring energy through space is not at rest, has a nonzero momentum, and is thus not subject to $E = mc^2$. So no, wave propagation does not violate the equation.

-3

It is what you can't see that has a mass and so, $E = mc^2$ still stands. From a different theory of Physics where space is made of invisible particles which are compressed and so generating gravity, energy. I can't explain more here. An ebook is under way for a complete explanation. This theory of mine is the most simple of all and is capable of ...

1

You are confusing in this one sentence two frames of reference, of modeling observations. It is said that waves only transfer energy and not matter, This was first said (modelled mathematically) of waves that are travelling on a medium : water, air , solids. The mathematics is described by second order differential equations, wave equations , which ...

2

You're not wrong, exactly. When a wave pulse travels from one place to another, it does transfer energy, and therefore mass. However, there's a useful distinction between the mass that is transferred by the wave, and the mass of the medium the wave travels through (e.g. the slinky for a slinky wave, or air for a sound wave, or water for a water wave). It ...

0

Since a propagating wave does not involve conversion of mass into energy or vice versa, it cannot violate E=mc2. EDIT: The, e.g., changes in spring potential energy involved in propagation of a wave along a stretched rope do translate into (very) small changes in mass that also propagate with the wave.

0

Well, if I'm not mistaken, it's pretty straightforward. Let $p(r, \theta, t)$ be separated in two functions with variables of time $T$ and spatial variables $\Theta$ (I'm not using $R$, cause it's already defined): $$p(r,\theta,t)=\Theta(r,\theta)T(t)$$ then: $$T = e^{i\omega t}$$ $$\Theta = i\frac{Q\rho c k}{4\pi R}e^{-ikr}$$ $T$ is given ...

1

Perhaps I could share some idea for further research. If we could make actual and correct pressure measurements in the cochlea to reveal wether the non-stationary Bernoulli effect is a good description of the actual physics-of-how-the-cochlea-isolates-frequencies-along-its-length? I would consider: I would propose to use a pitot tube, with sensor in the ...

3

Every function of the form: $$\psi(x, t) = A\cos[k(x\pm ct) + \phi]$$ is a solution of the wave equation $$\frac{\partial^2\psi}{\partial x^2} = \frac{1}{c^2}\frac{\partial^2\psi}{\partial t^2}$$ The equation is linear, and this means that the general solution is in fact, any linear combination of the possible solutions, something that we can express as ...

1

This is a plane wave. Its an idealization of a wave that doesn't exist. Waves that actually exist are spherical waves. A sinusoidal train of spherical wave are represented as following: $$\psi(r, t) = \frac{A}{r}\cos(kr - \omega t)$$ However, the solution of a general wave for spherical coordinates can be expressed as: \psi(r, t) = \frac{1}{r}F(r - vt) ... 0 You're correct that a wave is a disturbance that propagates at a certain speed, but the disturbance is not a single change. Whatever is being disturbed (the local pressure, the static electric field) will be restored to its "normal" status. If there is a series of waves (a wave train) the medium will be disturbed, restored, disturbed, restored. The ... 3 Classically (since rob has done a thorough job on the quantum picture), the amplitude of a light wave is not related to any physical extent. It is not the size of the wave in space, it is the strength of the fields (electric and magnetic). We often draw wavy lines, but if you look closely the transverse axes will be label differently for, say, waves on a ... 2 If you twisted my arm and forced me to assign an amplitude to a single photon, I'd do it this way: The energy density of a classical electromagnetic field is \begin{align} U &= \frac12 \left( \epsilon_0 E^2 + \frac1{\mu_0} B^2 \right) \\ &= \epsilon_0 E^2 &\text{(only for light in a vacuum)} \end{align} where E,B are the amplitudes of the ... 1 Electric and magnetic fields themselves are totally uncharged. They are always described as totally uncharged things. They can either be described as two uncharged fields (when treated in the more traditional formulation) or as aspects of a unified electromagnetic field. In both descriptions the field(s) interact with charged things without being charged ... 2 Note that e^{jx} - e^{-jx} = 2j \sin(x) So what you have written is not an electromagnetic wave at all. It is an electric field with a fixed direction and an amplitude that varies sinusoidally along the z-axis. Of course if you multiply this by e^{j\omega t}, then you do have a wave. Given the wording I suspect you are meant to assume this (though I ... 5 The cochlea has a complex physical structure, with multiple membranes and fluid-filled chambers. Therefore to explain the separation of frequencies along the basilar membrane of the cochlea is complex to. Sure, there are a lot of very general descriptions (even the answer of theblackcat) and a lot never go into the actual physics of the system. This ... 1 A very simple way is to look at the nodes of the standing wave. As the name suggests, the nodes are not supposed to move as time goes on. Suppose x_0 is one of the nodes at a particular time instant t_0, then we have \psi(x_0,t_0)=0. If position and time can be separated, then we have f(x_0)g(t_0)=0, which gives us two cases: f(x_0)=0, then we ... 4 This depends on the so called dispersion relation of the wave. Even then the answer is complicated, as you can define several velocities (the most commonly used ones are phase, group and signal velocity). I assume you know the relation c = \lambda \cdot f. This is written more conveniently (for working with complex exponentials) as \omega = c k in ... 1 Finally found out, what I was getting so confused about, here's the answer (credit to Dr Sebastian Steinlechner) with a relevant diagram. The incoming light is assumed to be unpolarised. We can, however, describe it as a combination of two orthogonal polarisations: one is polarised in the plane of incidence (the arrows in the picture), and the other is ... 0 The skin depth becomes smaller as the conductivity increases because the Ohmic dissipation of the wave energy increases. The current density is given by \vec{J} =\sigma \vec{E} and the work done per unit volume by the fields in moving charges in the conductor is \vec{J}\cdot \vec{E} = \sigma E^2. The attenuation is in the direction of propagation of ... 1 In general, no you cannot. If you're told that the three source signals are all sinusoidal (for example), then Fourier analysis will give you the answer. But if, e.g., the three source signals are each a combination of various waveforms such as sawtooth or square, then there's no way to separate them unambiguously. I would like to warn you that there's no ... 2 I think that your teacher (?) asked you about thermal de Broglie wavelength, where\lambda_T \propto\frac{1}{\sqrt{T}}.$$You get this expression when you express the momentum in \lambda=h/p in in terms of kinetic energy and the kinetic energy itself in terms of the energy due to temperature. (The derivation is also in the wikipedia article...) Indeed, ... 1 One answer is: it's the expression that is found to be conserved and can be identified as energy. Another answer: consider a harmonic oscillator. It obeys Hooke's Law F=-kx. Find the potential energy at max extension by calculating the work needed to get there starting at the equilibrium position$$\Delta U = -W = -\int_0^{X_{max}} F\,\mathrm{d}x = ...

1

From the Wikipedia article on fog: Sound typically travels fastest and farthest through solids, then liquids, then gases such as our atmosphere. The distance the water molecules are from each other, and temperature, are the reasons sound is affected during a fog condition. Molecule effect: Though fog is essentially water, the molecules are barely ...

14

According to this link sound (especially high frequency sound) is more attenuated in fog, because it is dispersed by the (billions of) air-water interfaces of all the droplets. This is one reason why a fog horn is a very low sound - low frequencies travel further, especially in fog. For echolocation you want to use high frequencies, and fog is more ...

0

I understand that Vg is different for different frequensies. Vg is defined as dω/dk for a specific frequency ω. So for big phasmatic areas (Δω) Group velocity Δω/Δκ is the average of the speed of all phasmatic componets in the area, so it is close but it differs from any individual Vg within the group.

1

Do the electric and magnetic components of an electromagnetic wave really generate each other? No they don't. Like Andrea said, they're two "aspects" of the same thing. And like you said, it's an electromagnetic wave. See the wiki article for electromagnetic radiation where you can read that "the curl operator on one side of these equations results in ...

6

This plane polarized wave from wikipedia may help Electromagnetic waves can be imagined as a self-propagating transverse oscillating wave of electric and magnetic fields. This 3D animation shows a plane linearly polarized wave propagating from left to right. Note that the electric and magnetic fields in such a wave are in-phase with each other, ...

2

Maxwell's equations in vacuum are: $$\nabla\cdot\mathbf{E} = 0$$ $$\nabla\cdot\mathbf{B} = 0$$ $$\nabla\times\mathbf{E} = -\frac{\partial\mathbf{B}}{\partial t}$$ $$\nabla\times\mathbf{B} = \frac{1}{c^2}\frac{\partial\mathbf{E}}{\partial t}$$ It's the last two of these that give rise to the interpretation that a changing magnetic field generates an electric ...

8

As you say, a changing magnetic field is always associated with a changing electric field, and in fact in relativity they are finally revealed to be the same field. So at this level it cannot be said that the one field generates the other, as they are merely two aspects of the same object. But maybe you still want to look at it from the perspective of ...

3

"a changing magnetic field is not generated by a changing electric field, but instead just happens to always be present perpendicular to a changing electric field due to the laws of electromagnetism." So ... it is due to but not caused by. What is the difference? Short answer: it is not only "a thing" it is a correct thing. This is much more clearly ...

1

For our three compartment hearing sense, from a physics point of view, there is a basilar membrane stimulation, from base to apex, in its pathway in the cochlea, to a place on the basilar membrane. By periodic movement of perilymph, non viscous fluid, backwards and forewards, in the cochlear duct meet the conditions of a potential flow. The basilar ...

-1

I believe this question is not worded correctly only because the submitter is not that well versed in scientific terminology. To put it briefly - neither sound waves nor light waves have mass, that is because they are representations of moving particles, they themselves are not entities and do not exist beyond the scope of human-facilitating terminology. As ...

6

Lets suppose the amplitude of each wave is $A$ and thus intensity will be $I_0 =A^2$. After superposition amplitude of the resultant wave becomes $2A$. but intensity becomes $I=(2A)^2$ Implies $I=4A^2 =4I_0$

4

Actually, the light beam does not follow the shortest path, but rather the faster path. Else the light would not bend but go straight there. This is Fermat’s principle. what point is the photon trying to reach? Good question. This point you are talking about, is in fact your eye. A straw in a glass of water visually bends at the interface. Look at ...

2

For the modeling of surface wave motion there are only two restoring forces to consider: surface tension and gravity. Compared to gravity, surface tension forces are very weak and therefore have a greater influence on the regime of the smaller, capillary waves. Waves in deep water carry away the energy dissipated by shear wind forces - perhaps from a storm ...

1

For any arbitrary collection of such travelling waves will always be a wave envelope that retains the same shape as the collection of waves propagate? No, it will not. For example, a Gaussian wave-packet will spread out in time. Wave packets are used to represent localization of particles in Quantum Mechanics.Group velocity will give the physical velocity of ...

4

It is commonly believed that the speed of sound at high densities is bounded from above by $c/\sqrt{3}$, where $c$ is the speed of light. Calculations of this quantity in many theories, ranging from QCD to systems with scale invariance, have all shown it to either stay below or exactly saturate the bound. See the introduction of this paper for a recent ...

2

You have that $$E=2A_0\cos\left(\frac{\phi}{2}\right)\sin\left(\omega t + \frac{\phi}{2}\right)$$ which is correct. To get the intensity, you then square and time average this: \begin{align} I=\langle E^2\rangle&=4A_0^2\cos^2\left(\frac{\phi}{2}\right)\left\langle\sin^2\left(\omega t + \frac{\phi}{2}\right)\right\rangle\\ ...

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