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11

In many cases our systems are described by linear differential equations, and these have the property that any linear combination of solutions to the differential equation is also a solution to the differential equation. This is useful because usually any arbitrary solution can be Fourier transformed to express it as a sum of plane waves. So if we can find ...


7

The "modes" in this case refer to the standing waves that can exist in a cavity. A very nice diagram / explanation is given at http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/rayj.html To summarize: if you consider a cavity of dimension $L$, the modes that fit inside the cavity have wave numbers $n_1$, $n_2$, $n_3$ such that $$n_1^2 + n_2^2 + n_3^3 = ...


6

It would be physically impossible to be able to "see" light as anything other than a particle (photon). The only time photons, or any other subatomic particle for that matter, can be described as a wave is when we are NOT looking at them.


6

The two solutions are different because they have different boundary conditions. In the first case, the equation is indeed $$ \frac{\partial^2u}{\partial t^2} = c^2 \nabla^2 u = c^2 \left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}\right) u. $$ Here though we specify $u(t,x=x_0)$ to be some value ...


6

What's the mathematical process and physical logic? The Fourier transform of position space ($\vec x$ domain) is wave number space ($\vec k$ domain). This is an unambiguous, well understood mathematical result. By the De Broglie hypothesis, the momentum is $\vec p = \hbar \vec k$. This is physical hypothesis with experimental confirmation. Although ...


6

Sure. Un-polarized light is just a superposition of many polarizations. Even if you are in vacuum you can use some beam splitters in cascade to obtain many rays, change (rotate) the polarization of each one in a different way, and then recombine the beam.


5

Use the light to excite a gas. The re-emission would be un-polarized. That is given you have enough energy in the wave to do so. That's one way I could think of off my head. EDIT: As per DarioP's suggestion, a solid fluorescent material would be nicer, as it is certainly more opaque than a gas.I am not aware of emission characteristics of a fluoroscent ...


5

No; wave interference takes place whenever two waves of any frequency, same, nearly the same or widely different interact. An air molecule next to your ear, for example, can only respond to the sum of all the different sound waves reaching it at any moment. The results are simpler when the two waves are closely related, or some simple multiple of each ...


4

Your idea that a quantum fluctuation created the universe is a misinterpretation of the suggestions that I have heard. Explaining why requires introducing a few ideas, so bear with me while I do this. The object we think of as the universe is made up of two bits: a manifold equipped with a metric = spacetime some matter/energy The manifold and metric ...


4

The answer is no. The reason is that waves are the solution of the wave equation, and the wave equation cannot be derived withouth implicitly assuming Newton's third law. It is intuitive to see why in waves that propagate in a medium. For instance, the bouncing back could be up and down in a transversal wave (for instance, the atoms of a string) or back and ...


4

John has answered this partially, however the fundamental mathematical idea is missing: If we think of a function being member of some vectorspace, then basisvectors exist. This concept will surely be familiar to you from Quantum mechanics. But also from there we remember, that problems were alot easier to handle, if we know the Eigenbasis of the Operators ...


4

The statement: $$ -exp(i(kz-wt)) = exp(i(wt-kz)) $$ is obviously false as you can prove very easily by expanding the equation using Euler's formula. However they are both plane waves. So if you're studying some system where you're expanding solutions as linear combinations of plane waves then they are both valid ways to write down your plane waves.


4

Edited and simplified on behalf of the crowd (useless at this point of other very good answers): Consider a cube of edge length L in which radiation is being reflected and re-reflected off its walls. Standing waves occur for radiation of a wavelength λ only if an integral number of half-wave cycles fit into an interval in the cube. In other words, ...


3

Suppose we have a drum. When you bang on the drum it will vibrate. When you look at any point on the surface of the drum head, you will see it go up and down, similar to if to take a mass hanging from a spring attached to the ceiling and watch it bob up and down. However, there is also an important difference between these two situations. In the situation ...


3

You're perfectly correct. Referring to Classical Electrodynamics by Jackson, we see that the index of refraction $n$ is given by: $$n=\sqrt{\frac{\mu}{\mu_{0}}\frac{\epsilon}{\epsilon_{0}}} = \sqrt{\mu_{r}\epsilon_{r}}.$$ But Jackson notes that for most optical frequencies (and non-meta-material media), $\frac{\mu}{\mu_{0}}=\mu_{r}\approx 1$, which is why ...


3

Your equation is actually a statement of either Faraday's law or Ampère's law and it only holds for plane waves, i.e. waves whose field vectors vary with position $\vec{r}$ and time as a vector of the form $\vec{X}\,\exp\left(i\,(\vec{k}\,\cdot\,\vec{r}-\omega\,t)\right)$, where $\vec{X}$ is a constant. For such a wave, it is not too hard to show that the ...


3

An on-axis pointlike object, infinitely far away will produce rays that focus at the focal point of your converging lens. The sun and the trees are far enough away that "infinity" is not such a bad approximation, but they sun and the trees are neither pointlike, nor on-axis. An off-axis pointlike object, infinitely far away will produce rays that (nearly) ...


3

Technically, $\omega^2/1^2-k^2/c^2=0$ is a degenerate hyperbola if that counts. But I don't think you can derive an equation of the form $\omega^2/a^2 - k^2/b^2 = 1$ for waves propagating in free space. You may however find something of the kind if you consider materials with fancier dispersion relations than $\omega = ck$, like e.g. plasmas.


3

It is not amplification! The purpose of the guitar body is to impedance and mode match between the string and the surrounding air. Intuition When a an object vibrates it pushes on the surrounding air creating pressure waves which we hear as sound. A string vibrating alone without the body of the instrument doesn't make a very loud sound because exchange of ...


2

There are some situations in which the plane wave ansatz is useful. It is a solution to many wave equations. Plane waves are also familiar and we have mathematical techniques to handle them, such as Fourier series. However, in some situations you can quite badly wrong using the planewave ansatz. Notably when the wave interacts with a structure that has ...


2

Note that you haven't actually found the general solution in spherical coordinates... What you have there is a solution known as a spherical wave, which describes a set of spherically symmetric wave fronts that diverges from (or converges towards) the origin $r=0$. However, in general, a wave could also be a function of the angles $\theta$ and $\phi$, which ...


2

The freespace dispersion equation is $\omega^2 = k^2\,c^2$ and this cannot change: this simply follows from considering plane wave components of propagating fields, which all fulfil the Helmholtz equation $$\nabla^2 A_j + \frac{\omega^2}{c^2} A_j = 0\tag{1}$$ which is fulfilled by all Cartesian components of the moncrhomatic EM field vectors and, for a ...


2

The direction of the electric field is called polarization. The direction of the electrical field in the free space lies in the plan perpendicular to the direction of propagation, and if this direction is unique for all the beam, it is said to be linear polarization. So, for linear polarization, the electric field can point in whatever direction that lies in ...


2

No, there is no such convention. The sine and cosine functions can be used interchangably since: $$ \cos(\theta) = \sin(\theta+\tfrac{\pi}{2}) $$ So simply changing the origin for the time or $x$ coordinate can switch the equation describing the wave between sine and cosine. As CuriousOne says in a comment, it's common to describe the wave by a complex ...


2

If we try to polarize the same beam of light in two planes, or if we mix two planar polarized beams, the light will interfere. If the phases of two beams will be identical, then we get 45 degrees polarized light. If the phases of two beams will be different, then we will get so called circular polarized light In other words, any sort of polarized light ...


2

Electric fields created in storm clouds could also be a contributing factor in interfeareing with your radio, a small field created by my bathroom fan constantly interfeares with with my tv signal and it drives me mad.


1

To quantize a classical system, start from the Poisson bracket $$\{x_i, p_j\} = \delta_{ij}.$$ This relation defines $p_i$ as the momentum canonically conjugate to $x_i$ and is equivalent to Hamilton's equations. Quantize by letting $x_i, p_j$ be Hermitian operators on a Hilbert space, with commutator $$[\hat x_i, \hat p_j] = i\delta_{ij} $$ (identity ...


1

Because (as you say) the ODE is linear we have that if $\phi_i$ are all valid solutions then so is $$\sum_ia_i\phi_i$$ for any real $a_i$. You can convince yourself of this by substituting the sum into the ODE and showing it is satisfied assuming the $\phi_i$ are solutions. We use the sin and cosine functions for our decomposition because they happen to ...


1

Usually examples in textbooks, the internet, and others will use two waves of the same frequency for experiments such as the Double Slit experiment, or else to show complete constructive or destructive interference. All EM waves will interfere with one another to varying degrees. Usually if we are considering frequencies in white light the intereference will ...


1

Propagation of electromagnetic waves on Earth is highly wavelength dependent, which is also true for the question whether the atmosphere has anything to do with it. What you are looking at is actually a formula for the geometric attenuation (aka free space or path loss), which has absolutely nothing to do with the atmosphere. As ACuriousMind points out, your ...



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