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39

Sometimes we do, and the phenomenon is called a light echo. What you're looking at there is NOT moving gas. It's an "echo" exactly as you describe. The problem is that you need a pulse of light. If you have a constant stream of light, the "light echos" will be exactly like what you see in fog on earth.


19

The link which NeuroFuzzy provided appears to be time lapse photography of a pulse of light from V838 Monocerotis, the most spectacular light echo in the history of astronomy, according to the European Space Agency. It is much brighter than even a supernova, but it is not exactly an explosion. The light from the initial pulse reached the earth in 2002. ...


6

It's a good question, and one that puzzled me for a while as well. However the answer is very simple. For a massive particle like an electron the total energy is given by: $$ E^2 = p^2c^2 + m^2 c^4 $$ where $p$ is the momentum and $m$ is the rest mass of the electron. Electrons can obviously have any momentum you want, so the total energy can be any value ...


6

If some of the light is reflected off the dust at such an angle that it is diverted to reach the observer, the observer will see that light. However, those specific photons reaching the observer will not reach B (unless they are reflected there by the observer). Similarly, unless the observer is at point B (which is not the case in the question as asked), ...


5

We don't really have a good perspective on what a photon "feels" or, indeed, anything about what its universe would look like. We're massive objects; even the idea of "we must travel at the speed of light because we're massless" makes little sense to us. But we can talk, if you like, about what the world looks like as you travel faster and faster: it's just ...


4

No, it is sufficient for the photon energy to exceed the band gap. Any excess energy is transformed into kinetic energy for the electron in the new band. You get exactly the same effect when ionizing an atom - the excess energy simply powers the electron into a faster continuum state. You should also take into account that photon energies are never exactly ...


3

First, webcams do use CCD or CMOS sensors, usually whichever chip is cheapest at the time. You can catch photons, but not reliably. In other words, for every photon that you catch, you will miss several. There will also be a noise signal, typically equivalent to many photons, Consider a CCD sensor. When a photon arrives, it may successfully excite an ...


3

You haven't said precisely what instrument you are using, but what you are seeing is normal for grating monochromators. Suppose we have light of one and only wavelength $\lambda$ incident normally on a grating. Scattered light will be seen at the following angles, $\theta$: $$ d \sin \theta = m \lambda \qquad \text{for } m=0, \pm 1, \pm 2, \pm 3, \ldots$$ ...


3

Actually, you have it backwards. The magnetic field isn't made of photons. Photons are made of magnetic (rather, electromagnetic) fields. To be specific, photons are ripples in the electromagnetic field. So, a magnet is surrounded by a magnetic field. If the magnet is not moving, then the field is stationary, and there are no photons. Wiggle the magnet, and ...


2

When a wave travels through a rope, the rope goes up and down, the position of all the 'rope-particles' changes, they oscillate and this makes up the wave. With light, it is indeed the electromagnetic field oscillating, but you shouldn't think of the arrows that represent that field in your first picture of light as 'extending into the rest of the space'. ...


2

Let's say you build a ping pong ball counter. It increments the count every time a ping pong ball hits the sensor. You throw a ball, and it hits the sensor: Detected! You throw a ball across the sensor from left to right... no detection, because you didn't hit the sensor. Your eyeball is a light sensor, which creates pictures from the light that hits ...


2

Using a camera that can capture "Motion at a Trillion Frames Per Second", this can be done at the laboratory scale. The technique used has been called femto-photography. (Image credit to Ramesh Raskar, Associate Professor, MIT Media Lab) More info from MIT here.


2

My masters project was on something like this (though with hydrogen alpha emission lines for gas clouds between galaxies rather than dust between stars) and the answer in that case (and almost certainly in this one too) is that you can't see it because it's just too dim, though using hundreds of telescope hours can get you close (maybe). Again, this is not ...


1

this is a description of an interaction between the electron and photons, which would collapse the wavefunction (right?). No this isn't right. As long as the system stays isolated, the interaction simply means that there are cross terms in the relevant Hamiltonian and that you have a two-particle quantum system, whose state space is the tensor product ...


1

Let's look at where the electromagnetic interaction comes from in hydrogen. At first quantization you have a multiparticle system so the wavefunction is defined as $\psi=\psi(x_1,y_1,z_1,x_2,y_2,z_2,t)$ and the point is to write the Hamiltonian. And the Hamiltonian comes from the Lagrangian. For a single particle of charge $q$ in an external ...


1

The key point here is the notion of distinguishability. If the two paths are made distinguishable, by associating different polarization to each path, then there is no way for the two amplitudes to interfere when they meet again because the "which-path" information is present. It doesn't matter whether or not the observer decide to actually look at the ...


1

In Minkowski spacetime, the spacetime interval of lightlike movements is zero. That means, from the (hypothetical) point of view of a massless particle such as a photon, it does not even exist one Planck time. At a proper time zero, any wavelength becomes meaningless, even if the physical process is the same that we observe. For the answer you have to take ...


1

I'd like to continue from Jims Bond's already comprehensive answer. Suppose that we had conclusive proof of our observation of a $V>c$ and that we've ruled out alternative (5) of Jim's answer. I agree with Jim that the next most likely of Jim's alternatives is: 2) We would be forced to conclude that $c$ is not the limiting speed of information ...


1

The possible room for the nasty alternative scenario mentioned in WetSavannaAnimal's answer has been worked out in detail in this article. When the OPERA experiment due to a flawed cable connection appeared to have seen faster than light neutrinos, it was possible to calculate that this was impossible, because the room there exists for Lorentz invariance ...


1

You misunderstand. All objects have some escape velocity, which is the velocity needed for anything (photons or matter) to escape from that object's gravitational field. And that's not the velocity it needs to maintain under some sort of constant thrust, but the initial velocity it needs to, shall we say, coast away from the object. For a black hole that ...


1

The equation $E = m^2c^2 + p^2 c^2$ is restricted to Special Relativity. However, in classical physics we have $$ \vec{F} = m \vec{a}, $$ and $$\vec{F} = m \vec{g},$$ whence $$ m \vec{a} = m \vec{g}. $$ This can be written as $$ m \big( \vec{a} - \vec{g}\big) = \vec{0}. $$ From a mathematical point of view we have $$ \big( m = 0 \big) \vee \big( \vec{a} - ...


1

However it did pass within Δx of the electron. The Δx is not the difference in space with the electron, as the electron is bound to a nucleus with a potential simulated by "an infinite potential well" . The Δx is related to the whole system, from the center of its mass as a possible location to start with. So the problem is : "photon + atom" as a ...


1

I tried answering this by going to the XCOM database where you can get a calculation of the stopping power of elements and compounds. First - pick a few likely candidates. I found a table of elements with density which is a good place to start. The highest density elements are also among the highest Z ones: proton density ...



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