# Tag Info

0

The law of energy conservation certainly also applies here. What is not reflected will be absorbed. R+A=1 has to hold. And according to Kirchoff's law E=A (absorptivity is emissivity) you could write R=1-E. If you have to consider transmission through the glass as well, then R+A+T=1 would hold. You do not specify what type of glass you are talking about, ...

1

Maxwell's equations for electromagnetism are linear so if only photons are involved they cannot interact and exchange momentum. Once we we include interactions with virtual electrons that emerge from the vacuum, however, then photons can interact with these electrons and exchange momentum. This process is called Delbrück scattering. It is hard to see ...

2

I've heard sinusoidal EM waves described as self-sustaining, so that suggests (to me) that their strength is maintained. Is this true? The determining factor for whether a wave dies down is not whether it's sinusoidal; it's whether or not you can treat it as as plane wave rather than a spherical one. If you take the expression you've written down and ...

0

Your derived equation $LC = \mu\epsilon$ in order to hold, $L$ and $C$ characteristic values of the transmission line must be expressed as per-unit length. Link. So the equality is not geometry independent.

1

First note that if an electromagnetic wave's frequency is above the local upper hybrid frequency, $f_{uh}$, it is called a free mode because it stops interacting with the plasma (for the most part, ignoring Faraday rotation). If the electromagnetic wave's frequency is below the local $f_{uh}$, then it will interact with the plasma. If the wave propagates ...

1

One needs to solve the Maxwell equations with the appropriate boundary conditions on the screen and the slits - this is also to say that thinking in terms of plane waves passing through a slit is misleading, since such waves extend to infinity in all directions. Huygens-Fresnel approch is actually an approximate way of solving the Maxwell equations, derived ...

0

I like the simplification of taking 1/2 of circuit out. Another simplification is take out the load replacing with wire. Now suppose we give a range of behavior for unknown C and unknown L and based on this unknown fr=12π√LC Where: fr = resonant frequency (Hz) L = circuit inductance (H) C = circuit capacitance (F) V is unknown but battery so DC This is a ...

0

There is some ambiguity about what one calls black body radiation: although introductory textbook treatments speak about a "black body" that absorbs all the radiation and so on, what is really meant is radiation (quantum photon gas) in thermal equilibrium. The Planck formula then folows immediately for open space, but the solution may be different ...

0

There is not any displacement of current as you have said it here There is only oscillations of free electrons in conductors when an electric field is applied, and its repulsive force by negatively charged terminal where electrons have been accumulated push the immediate electrons in the directly connected conductor, while these free electrons are in their ...

8

Does the plate feel an unbalanced force towards any direction? Yes. The change of EM momentum on the reflective side is twice the change of EM momentum on the absorptive side. So the radiation pressure is twice as large on the reflective side, leading to a net force pointing from the reflective to the absorptive side. Doesn’t the answer to the previous ...

1

$\sigma T^4$ is the amount of energy the surface of a black body emits per square meter per second. The total energy emitted by a spherical black body with radius R per second is therefore $4\pi R^2\sigma T^4$. The energy flux at distance $r$ is thus $\frac{4\pi R^2\sigma T^4}{4\pi r^2} = \frac{R^2}{r^2}\sigma T^4$ so it does depend on r.

0

Regarding 1) and the first part of 2): The area increases with distance. Therefore, the absolute power (dP) which you measure per $cm^2$ radial area decreases with increasing radius.

2

In the morning and evening the sunlight has to travel longer distance through the atmosphere to reach our eyes. By the time it reaches our eyes, the blue light waves, due to it's higher frequency, get's scattered almost completely. Only the red waves, due to its lower frequency and higher wavelength, remains. During the day time, light has to travel ...

1

This phenomenon is most prominent just before sunrise or just after sunset. It happens because as the rocket goes higher and higher, the sun's rays might interfere and cause the exhaust color to change. If the exhaust color was red at the beginning, as it would go higher and higher, it's color might have changed to whitish color because some amount of ...

0

Three different energies are needed to remove the $1^{st}$, $2^{nd}$, and $3^{rd}$ electrons. This means photons with three different wavelengths are needed.

0

You can only neglect resistance of conductors for good conductors with short distances. Power source itself has resistance too and it is not that small depending on source. It is all very very complex. Electrons are particles, electromagnetic waves are waves, however at higher frequency electrons will behave more like a wave and less like particles. ...

2

For "normal" materials that you find in everyday life and labs, the refractive index $n$ is positive. These materials will always behave as in your Case 1. However, it is possible to produce materials which have a nontrivial structure at scales shorter than the wavelength, in which case the light only sees an "effective" blurred-out ...

0

Radiation pressure: (source) $$P_{\text {incident }}=\frac{\langle S\rangle}{c}=\frac{I_{f}}{c}$$ Where where ${\displaystyle P}$ is pressure (usually in Pascals), ${\displaystyle I_{f}}$ is the incident irradiance (usually in W/${m^2}$) and ${\displaystyle c}$ is the speed of light in vacuum. Irradiance of perpendicular to the sail beam depends on intensity ...

1

Questions: Is radiation pressure wavelength dependent does a blue photon move the solar sail more then a red photon? For 1, one has to go through the derivations of the Poynting vector and the energy density and momentum density of the classical electromagnetic field. To answer it I would go through the number of photons in the classical beam. I would say ...

3

They are both right. On a "per photon" basis, a 100% reflected photon (with $k = 2\pi/\lambda$) imparts a momentum change of: $$\Delta p = -(p_f - p_i) = -(\hbar(-k) - \hbar k)=2\hbar k$$ which clearly depends on wavelength. Meanwhile, given an energy ($E$) per area per time, the momentum transfer is: $$\Delta p = 2E/c$$ which doesn't depend on ...

0

Other answers are already excellent, but another way to think about this: what do we mean by size? In classical mechanics, one way to think about it is the size determines when a collision or interaction takes place. For example, if I'm standing 10 feet from the center of a road and 5 foot wide truck goes down the road, there is no collision, but if the ...

0

The word "photon" is used for the elementary point particle that is the gauge boson of the electromagnetic force, one of the three forces in the standard model of particle physics. point means that as far as our present theory underlying all physics it has no dimensions, it has spin 1, mass zero, and an energy in the energy-momentum four vector ...

2

rearranging we get $L = \lambda$ This is somewhat correct (though note that as pointed out in the comments, your way of arriving here is incorrect), the "length" of a single photon energy quantum is indeed just the wavelength. However, it is more sensible to just talk about the wavelength as this is a far more general concept that remains useful ...

2

The Veratasium video concentrates on the battery and bulb. Those are the only two objects that exchange energy. It assumes there is no loss in the wires. The Science Asylum video assumes the wires have resistance and there are losses in them. Science Asylum shows that the electric field on the wire is not radial to the wire (as it seems is the case in ...

3

You are correct. The video is only partly right. It is true that EM energy flows along the Poynting vector. This means that energy in a circuit is not transported by the electrons but rather by the fields outside the wire. But despite that, it is not correct that the bulb will light after 1 m/c, it will take the the full second. There are a couple of ...

-1

What Veritasium's video is describing is electromagnetic induction and EM power and signal wireless transmission. But the circuit he used in his thought experiment is wrong. Unless the voltage source is a 50-60Hz AC source and in the range of MVolts, a filament lamp or LED at 1m away from the voltage source will never light up! This path of electromagnetic ...

15

This is a transmission line problem. Transmission-line problems can be counterintuitive for people who are used to thinking about the moving charges in circuit but are used to ignoring the fields. (One example here). Let’s imagine that the long cable to nowhere is not pair of parallel wires, but instead that the "outgoing" and "return" ...

9

Let me try to answer this without referencing the actual details of electromagnetism. When you have to worry about the finite speed of information propagation in the system (in this case, the finite speed of light), you always measure time delays due to a perturbation from the point of perturbation. Here, the perturbation to the system (circuit) is the ...

5

General considerations I think it will help to carefully write out all the expressions. The PDE should be read component-wise $$\partial_{tt}\psi_j(x,t)-\nabla^2\psi_j(x,t)=F_j(x,t)$$ Where $x\in\mathbb{R}^3$. I've replaced $J$ with a generic source $\mathbf{F}$, $\mathbf{E}$ with a generic field $\boldsymbol{\psi}$, and set all constants to unity. The ...

0

First of all, there are several equations determining the problem. Important to note that for electric and vector potential you have: $$\Box{\vec{A}=-\mu_0 \vec{j}}$$ The latter is true when assuming a Lorenz gauge, namely: $$div \vec{A}+\frac {1} {c^2}\partial_t \phi = 0$$ The solution of this equation ...

0

Hint: 1) these relations are deduced with a delay time: https://en.wikipedia.org/wiki/Li%C3%A9nard%E2%80%93Wiechert_potential with : $\vec{B}=\vec{\nabla} \times \vec{A} = \vec{\nabla} (t-|\vec{r} -\vec{r}_{s}|/c)\times \vec{\dot{A}}=-\frac{\vec{r}}{r}\times\vec{\dot{A}}$ (.) derivation with respect to $\; t_{r}$ For a plane wave, we have $\vec{E}=\vec{B}... 1 Microwave ovens use a magnetron that produces the microwaves, that inside the oven create standing waves. Because of this, there exists points of constructive and destructive interference. So this means there will be hot regions and cooler regions inside the oven. What the turntable does then is allow whatever is being cooked to rotate so that the whole ... 0 The derivation is valid but it misses one very important aspect. The electric field at the interacting molecule/system is different from the macroscopic field in the isotropic medium characterized by$n$. A way to model this is the so called cavity factor. This factor establishes a proportionality between the macroscopic field and the local field responsible ... 1 A scalar would be invariant under rotations. A non-zero vector is not, regardless of how many non-zero components it might have in any particular basis. 0 First, some experimental facts from observation. If you look thru a small hole into a cavity at temperature T, you will see the continuous Planck spectrum as given by his formula. 1) You will not see any variations due to modes of the cavity. 2) The emissivity of the cavity walls also makes no difference. The material on the walls (soot. silver, copper, ... 1 To a first approximation the palm shows a blackbody radiation spectrum of a 37C surface, which would be a smooth continuous function. It peaks around 10 micrometers. However, there are deviations from this curve due to different emissivity at different wavelengths. If the emissivity is low (it is more reflective and releases less radiation on its own) the ... 0 Although dipoles are very popular in EM theory and are well understood by antenna engineers, they are just one of an infinite set of radiating structures. any structure that contains accelerating coulomb charges will radiate. Where and when the conditions that define plane waves occurs is complex and requires a solid knowledge of modern EM theory derived ... 3 The sky is blue because of a phenomena called Rayleigh scattering. Rayleigh scattering occurs whenever radiation of any kind encounters particles (in this case air molecules such as$\text{N}_2$,$\text{O}_2$, or$\text{CO}_2$) that are much smaller than the wavelength of the radiation. In addition, the shorter the wavelenght, the more light gets scattered ... 1 Yeah, the imaginary part (loss) is supposed to be an odd function. For negative frequency, you have gain. Try plotting it! 0 The mean free path of a CMB Wave function (assuming Copenhagen interpretation) would be no more than a few metres on average over the 13.5 billion years of travel, if we assumed an ion or dust mote every metre or so. The wave function would collapse and a new slightly lower energy photon would be created because the ion or dust would gain some tiny amount ... 1 The radiating photons which are emitted from the surface of a hot body have a continuous distribution of different energies (which form the blackbody spectrum) because the electrons which belong to the atoms whose vibrations cause the emission experience a distribution of accelerations. Then that distributions gets shaped smooth by the continuous exchange of ... 4 Microwave radiation spans frequencies from 300 MHz to 300 GHz. Visible light, on the other hand, spans frequencies roughly from 430 THz to 740 THz. As you can see, microwave is too far away from the visible range (several orders of magnitude) to be visible by the human eye. 2 The potential of a neutral but polarizable atom in an electric field is given by $$U=-\frac{\alpha}{2}E^2,$$ where$\alpha$is the polarizability, and$E\$ the electric field. From this, we can see that atoms influenced by such a potential are drawn towards higher magnitude electric fields where the potential minimum exists. And as there are no electric field ...

0

There are generally 2 ways to generate THz radiation: RF and laser. I'm not a electronic engineer but I'd imagine RF oscillator in terahertz range is beyond our currently technology. In you motherboard there are DIMM slots for RAM sticks, which runs at about 2GHz. DIMM slots are through-hole component, which has little solder leads on the back side of the ...

25

A laser carries linear momentum, so when a laser is fired into a black hole, this linear momentum is transferred to the black hole, causing it to accelerate. Of course, for a realistic astrophysical black hole you'd need an unrealistically powerful laser for this to make a significant impact on the motion of the black hole.

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