# Tag Info

69

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.

24

It would be possible to see the progress of photons through space if the light pulse were exceedingly intense, and if the dust cloud from which they reflect were positioned and shaped to reflect the light toward us. Rather than shooting a beam from Point A to Point B, it would be better if the light source were between us and the dust cloud, as light ...

12

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), ...

10

In principle, yes. You can reverse any decay process and the corresponding synthesis will be valid - in this case, since $H_0\to\gamma\gamma$ happens, then $\gamma\gamma\to H_0$ will also happen, assuming the kinematics work out. However, the corresponding probability is very small. Out of all the possible things that could happen when two photons cross ...

9

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) Of course a camera that literally takes one trillion full frames per second is totally impossible with today's ...

8

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 ...

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 ...

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

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 ...

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

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 ...

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

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

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 ...

1

Some background. You want to detect the image of an object. First, either 1) you illuminate it with some light source [a lamp, the sun] or 2) the object itself radiates light out [a star, a fluorescent sample]. Imagine to divide the object in many small parts (voxels): to reconstruct the image of the object, you need to detect independently the light coming ...

1

The speed of light is constant in a vacuum. However, it can change direction in the presence of gravity so in a sense it does accelerate.

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