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54

Most electromagnetic radiation is of very high frequency - the magnetic field changes many times per second. This means that the compass just doesn't have time to "follow" the magnetic field changes. The only thing that does affect a compass is a DC magnetic field - usually this is a large piece of iron etc. that gets magnetized (e.g. by the earth's ...


26

The frequency is one very good argument (and I guess the most important factor) but it might be worth also looking at the magnitudes of the fields. The earth's magnetic field has a strength of roughly $31µT$. The intensity of the sunlight hitting the earth is about $1300W/m^2$. Since the intensity is related to the electric field $E$ of an electromagnetic ...


20

It is a standard exercise in quantum electrodynamics to find the angular dependence of the differential cross section. Which more or less means how probable it is for the photons to scatter at a certain angle, given the energy of the incident particles. So assuming the spins of the electron-positron pair is averaged, and that you don't care about the photon ...


19

Basically the same reason as what Floris said, but this also has another important aspect: Visible light has a far too small wavelength to affect a compass. Not only does the field oscillate too quickly around an average of zero – even at any single “snapshot” in time of the electromagnetic wave, there would nowhere be a large region where ...


11

Absolutely. If Compton scattering occurred in first order in $e$, the only contributing diagram would be the obvious one. Say we're in a frame with the electron initially at rest and an incoming photon in the $z$ direction. Then the electron 4-momentum is $$p^\mu_{\text{in}} = (m,0,0,0)$$ while the photon 4-momentum is $$k^\mu_{\text{in}} = ...


8

It's tempting to think of the light as a little ball (the photon), and since little balls have a definite position the little ball has to be in a superposition of a state where it goes through one slit and a state where it goes through the other. However this is not a good description of what actually happens. The light is not a photon, and it's not a wave ...


6

Update: According to this paper, "On the Interpretation of the Redshift in a Static Gravitational Field", the answer I give below is a common but misleading interpretation. The classical phenomenon of the redshift of light in a static gravitational potential, usually called the gravitational redshift, is described in the literature essentially in ...


5

If we assume you are a sphere in space, at the same distance from the sun as Earth, then we can calculate the heat absorbed - and we can calculate how hot you need to be so heat in = heat out (assuming uniform surface temperature, and radiative heat transfer only). For this, we need the Stefan-Boltzmann expression for total emission at a given temperature: ...


5

The E=mc^2 formula only applies to an object at rest, and light is never at rest. You want to use the more general formula: $E^2={m_0}^2c^4+p^2c^2$ Then you can set the mass to zero. $E=pc$ What this says is that light has momentum, which is related to its energy.


5

The claim that during the experiment they let the detector on but did not stored the data so it showed waves, only when they stored the data it showed as particle. is inaccurate and inconsistent with quantum mechanics. In a double-slit experiment, any device that can even in principle provide which-way information will destroy the interference ...


4

you see the beam of light because it lightens the molecules in the air. The photons you see have been diffused by the molecules of air. In the vacuum photons won't change of direction (they go at the speed of light) and you wouldn't see the beam passing in front of you.


4

Have a look at my answer to Slit screen and wave-particle duality because this covers a lot of topics relevant to your question. You're correct that if we imagine the photon as a little ball then if the arms of the interferometer are different lengths the two "halves" of the little ball cannot arrive at the detector at the same time. But this is not how the ...


3

You are correct in one thing: if an atom in an isotropic medium spontaneously emits a photon, it can do so in any direction at all, and the overall emission will be evenly spread over the unit sphere. However, lasers work using stimulated emission, which is slightly different: if an atom is excited, you can induce it to emit its energy by shining an initial ...


3

The longitudinal mode decouples from all physical processes as a consequence of gauge invariance, which in turn forces the Ward identity $$ k^\mu \mathcal{M}_\mu = 0$$ where the S-matrix element decomposition $\mathcal{M}^\mu$ is obtained from the polarization vector $\epsilon^\mu(k)$ by $\mathcal{M} = \epsilon^\mu(k) \mathcal{M}_\mu$. This decoupling (and, ...


3

Hurl it several yards away Rather than a simple yes/no, let's figure out some numbers - as usual I work with "round" numbers. Mass of golf ball = 50 gram "several yards" = 5 meters Projectile launched at 45° with initial velocity $v$ travels for a time $$t = 2\frac{v}{\sqrt{2}g}$$ during which time it will travel a distance $$x = v_x t = ...


2

I'm not sure about the “pick it up with a laser” part, but let's simplify by assuming that we can hit the golf ball from below the ground with photons. Let's further assume that the mass of the ball is $m_{ball} = 46\,\mathrm g$ and we want to hurl it five yards away ($\approx 4.6\,\mathrm m$). If the ground is level and we neglect air drag and wind, the ...


2

A photon that strikes the metal either has enough energy to release a valence electron from its bond with the atoms in the metal, from a certain frequency and up, or its energy is just not enough to eject the electron from the metal. The surplus of energy over the energy needed to eject the electron is used as kinetic energy of the electron and determines ...


2

In the two-dimensional rubber sheet visualization, it is wrong to think that things fall towards the massive object because they are "rolling down the hill" of the curved spacetime. There is no perpendicular gravity pulling things down into the well. What happens is that you are moving along your world line at a constant velocity, "into the future at one ...


2

The maxwellian wave is an emergent phenomenon from a great multitude of photons with the frequency of the maxwellian wave. This is explained in this blog entry by Lubos Motl. I will give you my experimentalist's interpretation of this: A photon as a quantum mechanical entity has a wavefunction. This wavefunction is a solution of a form of Maxwell's ...


2

When you have only one electron then $\Delta S=0$ makes intuitive sense: you can change the angular momentum $l$ of the atom by changing it's internal structure (by pushing the electron in "another orbit" if you will), while you certainly can't change the internal structure of the electron to change $s$. Would it be possible to change $s$ then you could ...


2

Your argument that the energy should radiate away would be true if your inductor were a good antenna, in which case it would be a bad inductor! The problem is an impedance mismatch: The inductor produces a magnetic field (which stores the energy you inquire about), but little electric field. That is the wrong ratio, or impedance, to couple to the vacuum ...


2

There's an old argument by Newton and Wigner, that the photon as a massless particle can't have a position operator and therefore no position space wave function. The paper you're thinking of is T. Newton and E. Wigner, "Localized States for Elementary Systems," Rev. Mod. Phys. 21, 400–406 (1949) doi:10.1103/RevModPhys.21.400. Photons are ...


2

The energy of a photon is given by the equation E = hf where h is Planck's constant and f is frequency. The energy would decrease, making the frequency decrease (since h is constant). So, if the photon was blue light, then it would get redder and redder as time when on. There is a point, however, when your system eventually stops working. This is because the ...


2

In light propagation, oscillation does not mean any movement in space. It is the value of the electromagnetic field, at one given point in space, that oscillates. The picture that you quote does not represent the movement in space, but the electromagnetic field value as a function of time. Compare to waves in water: if you put a little boat on the water, ...


2

A laser is just a thin slice from the spectrum of light. Is it more efficient compared to the visible spectrum of light? It depends on the frequency of the laser and how efficiently the solar panel can turn light of that frequency into electrical energy. If a solar panel would operate better/best with light of a certain frequency, using a laser with that ...


2

It would certainly require a material that allows electron release from energies lower than those of the visible spectrum. The energy of a wave is given by E=hf where h is the planck constant (6.63 x 10^-34) and f is the frequency. The wavelengths of IR light range from 0.001 m to 750 x 10^-9 m. (Hyperphysics.com, infrared) Using this knowledge you can get ...


2

When one says "photon" one is in the quantum mechanical frame. Quantum mechanics does not follow the rules of classical mechanics if one tries to consider the photon one classical entity, like a bag of energy flowing. The photon is a point like elementary particle in the standard model, it has no extent and when it hits the detector it registers at a ...


2

We know that solar cells generate electricity by utilizing the energy of the photon, This is an every day language, electricity. It means things electrical in general every day language. but how does it generate electricity forever? What is generated when the photons hit any material, is heat, and the sun's energy is at maximum 1300Watts per ...


2

Q1: For photons of energies much less gamma rays, the quantum mechanical photon-photon interaction is negligible. This is consistent with the classical electrodynamic description where the principle of superposition holds (electromagnetic waves pass through each other unchanged, as well as through electric/magnetic fields). Q2: in reality, charge is defined ...


2

Theoretically yes, the laser principle does not consume any material. There is a light source that excites the electrons in the material to higher levels, they deexcite to some intermediate one, here the avalanche of photons appears producing the laser light and leaving the electrons in the ground state. And you can repeat the process without a loss.



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