# Tag Info

16

This Isn't About Water Microwave heating actually has nothing to do with the moisture content of items. It has everything to do with the amount of electric dipoles (polar molecules) in the item of concern. Water molecules (with many other organic molecules) happen to be electric dipoles. (That is, one side of the molecule has a positive charge and the other ...

13

Microwave heating is largely caused by the changing electric and magnetic fields (i.e. the "microwaves") which are emitted by your microwave oven affecting polar molecules. As the direction of the electric field changes over time, the polar molecules (often, of water) attempt to follow the field by changing their orientation inside the material to line up ...

12

CMB spectrum has near-perfect black-body Planck spectra. Planck distribution has temperature as a parameter. By fitting observed spectrum by Planck function, we can "measure" temperature of CMB. This way we observe 2.7 K temperature today. In the context of photons temperature describes degree of disorder of photons of radiation in thermal equilibrium ...

12

The universe was a thermal plasma prior to the recombination epoch, consisting mostly of photons, protons, electrons, and alpha particles (helium-4 nuclei). There were also a small portion of deuterium, helium-3, and lithium-7 nuclei. All of this primordial stuff was in thermal equilibrium. The temperature of a gas is a result of the random velocities of ...

9

Very simply: just because something is true in the static case doesn't automatically make it true in the dynamic case. To reject a static electrical field from inside a shell, it is sufficient for the charges on the surface on the shell to move however slowly they want in order to arrange themselves so as to cancel the field. This is Gauss's Law. When an ...

7

Your misconception is the application of Gauss's law. There are solutions to Maxwell's equations (including Gauss's law), that permit an electric field inside the shell. If there are no charges inside the shell, all Gauss's law tells you is that the number of electric field lines entering the shell must equal the number coming out. An example of a field ...

7

Next to the very detailed and good qualified answers, here is a simplified alternative: If you had a very dark body very far from any radiating thing, the cosmical background heated it until 2.7K.

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.

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 ...

5

The ${\rm sech}$ pulse is, in Kerr effect nonlinear optical mediums, an Optical Soliton. This means that it is the particular time variation such that the tendency of the pulse to spread out in time owing to linear dispersion is exactly counterbalanced by the nonlinear effect that tends to confine pulses in time. This balance is a stable one in a Kerr ...

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 ...

4

Calculating the path that a light ray takes in a gravitational field is a complicated business, but for the special case of a light ray coming from infinity and escaping to infinity there is a convenient approximate formula for the angle, $\theta$, the light ray is deflected: $$\theta \approx \frac{4GM}{r_0 c^2}$$ Where $M$ is the mass of the ...

4

The electromagnetic field itself contains energy distinct from the energy of charged bodies, the energy in a given volume of empty space can be found by integrating the energy densities $\frac{1}{2}\epsilon E^2$ and $\frac{1}{2} \frac{B^2}{\mu}$ over the region. When the EM fields increase the kinetic energy of charged particles, there is a corresponding ...

3

That is the point of a Faraday cage. Although some of the EM field can penetrate inside the shell, under various circumstances, there is an effect where a conductive shell forms a void in a pervasive EM field. To be clear, though, EM fields do not just 'stop' - they are formed of photons, which are reflected at such an interface. Consider the microwave ...

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

X-rays very much do travel through water. I think your quote may be out of context. For example, being deep in the ocean would protect against X-rays because there is so much water above you. However, concrete or lead are two more common materials which provide more protection against X-rays. Using the link you provided, I generated the following plot of ...

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 key word here is continuity. The continuity boundary conditions for the electromagnetic field vectors sets these phenomenons. The tangential components of $\vec{E}$ and $\vec{H}$ must be continuous across an interface - the only way that they can not be is if there is a surface current flowing (which cannot happen in dielectrics). Likewise, the normal ...

2

Let me answer your question with a question. How would you use $E$ as it is to calculate $f$? Suppose you plug $E = 20\text{ keV}$ into the formula and solve it for $f$. \begin{align} f &= \frac{E}{h} \\ &= \frac{20\text{ keV}}{4.135\times 10^{-21}\text{ MeV s}} \\ &= \frac{20}{4.135\times 10^{-21}}\times ... 2 The polarisation field is given by {\bf P} = \chi_e \epsilon_0 {\bf E} in linear dielectric materials, where \chi_e=\epsilon_r - 1. ie. the polarisation is defined to be in the same direction as the electric field. Therefore the equation {\bf D} = \epsilon_0 \epsilon_r {\bf E} is equivalent to {\bf D} = \epsilon_0 (1+\chi_e) {\bf E} = \epsilon_0 ... 2 Since the emitter is a point, the wave will never get "flatter". It will always look like a dipole pattern (for the non-relativistic case, for the relativistic case the forward and backward lobes become asymmetric, I believe, with a strong amplification of the emission into the forward cones, which should also become narrower). I think you are mistaking your ... 1 The wavelength does of course remain the same. Think about two circles (representing wavefronts) evolving over time. Their respective radii increase at the same rate, such that the distance between them (the wavelength) always stays the same. Now think about drawing a box of fixed size and placing it over the wavefronts. This represents your detector (or ... 1 The prism doesn't reflect light, it refracts the light. Different wavelengths will be refracted (bent) by different amounts. In the visible spectrum, this will produce the familiar rainbow pattern we're all familiar with. If you direct the refracted light at a linear detector, each pixel of the detector will measure the optical power at a different ... 1 This solution: \varphi(x,t)=(a_1\cosh(\lambda t)+a_2\sinh(\lambda t))(a_3\cosh(c \lambda x)+a_4\sinh(c \lambda x)) is about as useful as this solution: \varphi(x,t)=(a_1+a_2 t)(a_3+a_4 x). They're both valid, they're just not useful in most physical problems. (Except in the exponentially decreasing case (a_1\cosh(\lambda t)+a_2\sinh(\lambda t))=c_5 ... 1 I am just trying to answer your question. I understand that you expect an answer of "yes" or "no". I am not sure if I'll succeed. I start from the formula in my comment. The variance for the initial electron energy is<(E - \ E_0)^2> = \Sigma_E P_{initial}(E) \ E^2 - [\Sigma_E P_{initial}(E) \ E]^2 .  The variance for the final electron energy ...

1

I guess the question is whether the LED will also act as a photodiode, i.e. whether incident light will excite electrons and therefore generate a current. If so, then reflect the diodes light back onto it will indeed reduce the current flowing in the diode by generating an opposing EMF. This turns out to be surprisingly hard to Google, but Wikipedia ...

1

John Rennie's Answer, that complex numbers simplify calculations with sinusoidally varying quantities by letting you do linear operations with complex exponentials and then reverting back to sinusoids at the end of your calculation, is altogether correct and a summary of what is called the phasor method for dealing with any quantity that varies sinusoidally ...

1

The speed of an electromagnetic wave is indeed independent of the speed of both the source and the receiver. However, this does not mean that the relative motion between the source and the receiver has no effect on the wave's properties. The effect that is being used is called the Doppler effect, and it is the fact that the received frequency of a wave will ...

1

Space weather events, such as CMEs, can cause rapid variations (seconds to 10s of minutes) in the Earth's geomagnetic field which can induce an electric field on the surface of the Earth. This electric field induces electrical currents in the power grid and other (grounded) conductors. A consequence of this could be transformer burn out or other network ...

1

This is a kind of optical amplifier -- see 1, 2. Probably the main effect of the molecules on the beam will be scattering / refraction, just like if you put anything into of a beam of light. But a secondary effect, especially if the beam is much much larger than the clump of molecules, is that the profile of the beam will change a bit. The part of the beam ...

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