# Tag Info

10

It's a rather late answer, but the paper Visible optical beats at the hertz level has just appeared on the Arxiv and this describes exactly the phenomenon you ask about. This image from the paper shows the experimental setup: The light frequency is modified using acousto-optic modulators (labelled AOM in the diagram), and to make it look pretty a lens is ...

0

I too am confused on lensing. When light is emitted onto an object outside of space, normally it is absorbed like on earth. I'm guessing since objects creates a curvature in space is what affects it? So lensing really only works outside of a planetary object? Am I just reading too much into it?

1

I agree with @DavidHammen here, the 'high' and 'low' most likely refer to intensity. A microwave oven works by causing water molecules to oscillate (i.e., shake back-and-forth), which results in the generation of energy (i.e., heat) due to intermolecular "friction." Every molecule in the universe has a resonant frequency. Think of "rocking" on a swing ...

1

It is believed that high-level microwave radiation is harmful to human ... It is not just believed but well known that extremely intense microwave radiation will cook people. We use microwaves to cook meat, after all, and a good portion of our bodies is in the form of meat. Lesser intensities can cause survivable burns, even lesser intensities might ...

1

An important factor in determining how much energy electromagnetic radiation carries is its intensity, which is just the power per area. In fact, it can be found in the wikipedia article that intensity is sometimes taken to be synonymous with 'level' (even though it is not really correct), so surely this is the right variable to look at.

0

Electricity can be static, like the energy that can make your hair stand on end. Magnetism can also be static, as it is in a refrigerator magnet. A changing magnetic field will induce a changing electric field and vice-versa—the two are linked. These changing fields form electromagnetic waves. Electromagnetic waves are formed by the vibrations of electric ...

1

On this diagram, why is the atomspheric opacity shaped as it is? Different parts of the atmosphere are responsible for the shape of that curve. Electromagnetic radiation impinging on some object can be reflected by the object, absorbed by the object, or transmitted through the object. An ideal black body absorbs all incoming radiation, regardless of ...

2

Observable simply means that with the right detector, we could see it. As you have realised, opacity refers to what percentage of the radiation is blocked. The thing is, we can 'see' radio waves - just not with our eyes. Our eyes are only sensitive to a very small range (on the diagram it's the rainbow). If you have something that can detect radio waves, ...

1

You can think of light as the carrier of the electromagnetic interaction. The particles interact with light, not directly with each other. It is an experimental fact that light does not interact with itself. Note that this is not the case with quantum chromodynamics (the theory of nuclear matter). This theory is built along the same lines as quantum ...

1

Unfortunately, in quantum mechanics "ordinary" reasoning does not get you anywhere. The photon, like any other particle, is neither a particle nor a wave; it is an entity that we can only describe mathematically. It's only when we observe it that it shows up as either particle or wave. Or senses, and hence our logic, evolved to make sense of the real ...

2

What constituent of internal energy does an electron excitation represent? You can think of electrons as just like planets orbiting the sun and get the correct answer to this question. An electron in a higher energy level has less kinetic energy, but more potential energy as it is (generally) farther from the nucleus. The net result is more energy. ...

2

The change in electronic excitation represents both a potential and a kinetic energy term in classical physics, but there is no simple correspondence to classical physics terms, when you are looking at quantum systems. All we really care about is the total energy difference between electronic states. Those energy differences correspond to the energies at ...

1

I am going to address only your second point: It seems that the maximum temperature an object could have is when: B) When the speed of the electrons nears the speed of light. In fact this does not impose any limit to temperature. When you add more and more heat to a body, its atoms and molecules move faster. At this stage, the electrons are not ...

7

The entire premise of this question is false, neither do electrons orbit atoms with a well-defined speed, nor does this, in any way, correspond to the temperature, since that is a property of systems in thermal equilibrium, not of single particles. Also, whether the Planck length signifies really a shortest achievable wavelength is...debatable.

0

When deriving Planck's law, the setting is that photons at all frequencies are in a thermal equilibrium. This is not a small assumption, although it is more or less valid for various macroscopic objects. For such a thermal equilibrium to be possible, there has to be a black body, an object that can absorb and emit all frequencies. Ideally, a black body ...

2

Photons have spin 1, hence they can be treated inside the cavity as a Bose gas. Their distribution obeys the Bose-Einstein statistics: $$\bar{n}_r = \frac {1}{e^{\beta\epsilon_r}-1}$$ where $\bar{n}_r$is the mean occupation number of the energy level $r$, $\epsilon_r$ is the energy of the energy level $r$, and $\beta = \frac{1}{k_bT}$, $k_b$ being Boltzmann ...

2

In the case of frequency modulation, the information ("data") is contained in slight modulations of the carrier frequency over time. A "1" could mean that the frequency gets a bit higher and a "0" could mean it gets a bit lower. The modulation in frequency must be tiny enough as to not overlap with the next carrier frequency. And as long as the next higher ...

2

Normally the "data" that is modulated onto the radio frequency of the carrier has a lower frequency than the carrier (this is not an absolute requirement, but few systems exceed it). In case of cell phones, the carrier wave is around 800-1900MHz, while the data rate is on the order of a few Mbit/s, i.e. two orders of magnitude slower than the carrier ...

1

We cannot multiply light by mere reflections, because the very definition of "reflection" means that the same light comes out. We can however multiply light by letting it pass through special materials which we "pumped" into a certain state, that's called Laser. And yes, a Laser design includes a sequence of reflections, but it is not the series of ...

1

Electromagnetic waves travelling in chiral media will have some parallel component of their electric and magnetic fields, although they are usually not exactly parallel. The quantity $\mathbf{E}\cdot\mathbf{B}$ is proportional to the helicity of light, which basically tells you how close the beam is to having left or right-handed circular polarisation. Tang ...

0

I've already provided two possibilities in the comment above, and now I'm expanding my answer because I realize that this is achievable even in free space (though by a mathematical trick instead of physical reasons): Imagine two identical planewaves except that they have opposite wave vector and polarization, then the E and B fields are parallel where the ...

0

The process of light propagation is described by the Maxwell equations. $$\nabla\cdot{\bf D} = \rho$$ $$\nabla\cdot{\bf B} = 0$$ $$\nabla\times{\bf E} = - {{\partial{\bf B}}\over{\partial t}}$$ $$\nabla\times{\bf H} = {\bf J} + {{\partial{\bf D}}\over{\partial t}}$$ These equations say (in simple terms) that: change in the electric field is causing ...

1

This is primarily a biological question. We (humans in particular, mammals specifically) can't see radio waves because our bodies do not have the sensors to detect them. We can detect light in the visible spectrum because the rods and cones of our retina that constitute our light sensors absorb photons in that spectrum, ultimately activating neurons in our ...

1

The statement "group velocity represents energy or information transmission" is not entirely accurate. And that is a big can of worms, in fact. But let us start from the beginning. What is the dispersion relation? Suppose that you have some quantity $u$ that depends on a coordinate $x$ and on time $t$. Dispersion relation $\omega(k)$ is basically a ...

1

If we have a wave of a well-defined direction and frequency, the dependence of the field $F$ (something that is waving) on position and time is $$F = F_0 \cos (\omega t - k_x x - k_y y - k_z z )$$ Adult physicists would use complex exponentials instead of the cosine but I decided to remove this potentially difficult piece of maths. The argument of the ...

2

Vacuum magnetic birefringence basically involves the same loop diagram as light-light elastic scattering except that two of the four photons come from a magnet. Detecting this effect is the aim of the PVLAS experiment in Ferrara, Italy. See arXiv:1406.6518 and references within. The experiment is running at the moment but the sensitivity is not good enough ...

2

Your math does check out: \begin{align} r&=vt \\ &=0.05\cdot2.9979\times10^{10}\frac{cm}s\cdot4\cdot604800\,s\\ &=3.63\times10^{15}\,cm\\ &=36.3\times10^9\,km\\ &=0.012\,pc \end{align} When a supernova explodes, it enters the free expansion phase, it's position is linear in time ($r=vt$, as used above). It stays in this phase for a few ...

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No. There is no void left by the lack of an aether. The very notion of aether should serve as a warning as to how catastrophically analogical reasoning can fail. "Water waves are in water, sound waves are in air, therefore there must be something in which light propagates." This is flawed logic, and decades of physics were arguably hindered by adhering to ...

0

Just another perspective: Since the sphere is non-ergodic, your observation depends on your and the source locations inside the sphere. For ergodic shapes (ellipsoid, etc), you will see an evenly lit world.

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As most people know, "let there be light" is a famous biblical quote, from Genesis. Now, on to the teacher's shirt. Those equations on his back are Maxwell's equations. "Let there be light" is a joke, because Maxwell's equations describe electromagnetic fields, and light is a form of electromagnetic radiation, so the equations can be used to describe ...

1

I don't know specifically what you're looking for, but I can give you the basic idea of the relation between the cross section and the differential cross section. Generally, the cross section $\sigma$ is defined as the integral of the differential cross section $\frac{d\sigma}{d\Omega}$ over the entire solid angle. Here, $d\Omega$ is the spherical surface ...

0

Yes, all radio waves have inthe nearfield the same sequence of the E and the B field. It's the right hand grip rule (conventional direction of current) because there is no principal difference to a straight wire.

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I think the key to the paradox is that you can't ignore the reflection coefficent. Lets say you have a lump of coal inside a shphere of polished steel. Yes, the coal emits much more heat than the steel; but that doesn't mean there is a net transfer of heat from the coal to the steel. Because the steel reflects back the excess heat. There is a fixed ...

2

Your picture is ok, so long as you realise that this is a snapshot in time. If you looked "from above" the electric field of the waves (by your definition of the polarisation direction) would be pointing towards or away from you - so this becomes rather difficult to draw in the same way. It might be better to think of the E-field as a field of arrows, ...

3

For the benefit of brevity I'll give a straight answer: the second Weyl scalar is not related to electromagnetic radiation in any way. As per my previous comment, a simple proof is the Schwarzschild (or Kerr) spacetime. It describes a stationary black hole, without electromagnetic (or even gravitational) radiation, yet there is a basis in which all Weyl ...

5

Take a look at the conventional form of Maxwell equations. They tell us that Gauss's law actually applies every time. However, to get the field $\vec{E}$ from the charge distribution by the usual methods, we also need to know that $$\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t} = 0$$ Because otherwise the field could not be generated by the ...

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