# Tag Info

1

My teacher told me that speed of light is constant in the universe. Your teacher is mistaken I'm afraid. The speed of light varies with gravitational potential. You can see Einstein talking about this in 1920 in the Einstein digital papers: Also see Irwin Shapiro saying the same in his Shapiro Delay paper dating from 1961: There's also ...

1

The equations that describe how light, and all other electromagnetic radiation, works have a couple of constants which mean that the speed of light is constant. The equations don't depend on where you are, what temperature it is, which way you are looking etc so we assume they apply everywhere. It is entirely possible (although experimentally unlikely) that ...

0

You are essentially observing an interference effect. This is equivalent to what you would see in a single slit experiment. The fringes are caused as the light rays that passes between your fingers are travelling slightly different distance to reach you eye. This causes their phases to be somewhat different. As a result the light rays partially cancel, ...

1

speed is $$\frac{\text{distance}}{\text{time interval}}$$ but at the event horizon of a black hole, time interval becomes $0$. Imagine a flashlight flashes periodically 1 flash/s (in flashlight's reference frame). As flashlight getting close to the event horizon, someone far away from the event horizon will see flashlight flashing $0.1$ flashes/s, ...

1

As I know, Field Theory, that to what appeals the topic creator cannot explain the very powerful gravitation fields . So trying to understand what happens with a photon there are inside the Black Hole in meaning of Field Theory, or Special Relativity, isn't a good idea. The Nature has no the alone space , and the alone time , you can abstractly image ...

1

So, it's probably more boring than you're thinking. What has happened is: within a certain speck of substance, if you fire two photons in, they come out together. Now, light is known to behave differently in substances, with the clearest example being refraction: light "slows down" in certain substances like water and glass, behaving a little bit as if it ...

2

For better clarity, let's define the following: Axial direction = the direction the person & light beam are drawn into the BH. Radial direction = the direction perpendicular to the axial direction. If we, looking in the same direction as the person & light are being drawn into the BH, watch the light beam as it is drawn into the BH, we will see the ...

4

Now there is a light ray moving outward at the speed of light. I'm afraid that isn't the case; within the event horizon of a Schwarzschild black hole, the radial coordinate is timelike and so, moving 'outward' toward the horizon is as impossible as moving 'backward' in time. This plain to see in the Kruskal–Szekeres coordinates: Image credit See ...

2

I think a possible analogy would be to imagine that the singularity is a waterfall. By emitting light, you are trying to send a signal upstream using a tame fish. Outside the event horizon the fish is able to make headway against the current. But the river flows so fast within the event horizon as it approaches the waterfall, that your fish ends up going ...

-10

How does light behave within a black hole's event horizon? It doesn't behave at all. If the event horizon of a black hole is the distance from the center from within which light cannot escape, imagine a person with a flashlight falls into the black hole. I've explored this with a variety of relativists, and posed this question. The answer comes ...

1

Be careful with the terminology. The emission of light depends from the temperature of the emitting body. The bandwidth goes from infrared to red and blue to ultraviolet than higher the temperature. Red and blue shift means, that the well known spectrum from elements like hydrogen is shifted to the blue or red side. And yes, this happens due to the doppler ...

-2

Yes, the ratio is for example $4/5$ and they are asking for minimum distance $n_1=4k$; $n_2=5k$ (for some $k$), the distance will be minimum when $n$ is minimum, so $k=1$ and $n_1=4$ and $n_2=5$.

1

The star emits red light because it has a surface temperature somewhere around 3,500 Kelvin. Our sun has a temperature of about 6,ooo Kelvin so it emits yellow/white light. Like any black body radiation, the temperature determines the frequency of the emitted photons. As for the gravity potion of your question, I may be wrong but I think you talking about ...

1

Can I see the edges of the visible spectrum at home? The edges of the visible spectrum do not exist. They are like the edges of the audible spectrum which you can try to find using this Online Tone Generator. Can you identify with precision the lowest and the highest frequency you are able to hear?

1

Take a light-source spreading over the visible spectrum (e.g. light bulb, sun light) and decompose this light with a prism. You can as well wait for the next rainbow or make it yourself on a sunny day.

-2

You have a misconception, photons are massless, but there is a momentum term that is usually left out of the energy equation. Since a photon has no mass, so it can travel at the speed of light with no problems. You're thinking of e = mc^2, and since a photon has non zero energy, obviously it must have mass? No! That equation is the simple version for the ...

1

This depends on the reflectivity of the objects and the size in the sky. The visual albedo of the moon when near (but not at) full, is about $0.12$. Earth's is around $0.39$. Being in low-earth orbit, the actual amount will vary based on the terrain and atmosphere. If it's overcast below, the value could be much higher. You can assume that a given area ...

2

What you are confusing here is speed and velocity. Light speed is constant, but the velocity, which takes into account the direction as well as the speed is not. As an example of how something can accelerate without changing speed, consider the case of circular motion, where the acceleration of an object moving at a speed $v$ in a circle of radius $r$ is ...

0

The disconnect is between the first and second clauses of your first sentence: Light speed is constant, therefore experiences no acceleration Yes, the speed of light is a constant, but it experiences no acceleration in its direction of travel. Light definitely accelerates laterally when gravity pulls on it, which is why it curves when passing near ...

1

This Q boils down to How bright is Earthshine? According to the usual source Oceans reflect the least amount of light, roughly 10%. Land reflects anywhere from 10–25% of the Sun's light, and clouds reflect around 50%. So the amount of sunlight reflected (i.e. albedo) depends on what part of the earth is facing your observer and how cloudy it is. ...

0

I would say light sometimes ACTS like a wave but always acts like a particle. I love this subject and like many others I too find duality to be very unintuitive. Its not that I can't imagine something being two things at once its just that I wonder why we need the wave theory at all. I know that's blasphemous and I'm truely not trying to overtake this ...

0

It is always both - but one or other character may be less obvious in a given situation, as is nicely shown. In the answer to the question Which side of wave-particle duality to choose on a given situation" My point is - a photon is never purely one or the other - but physics is all about "good approximations". But just like relativistic effects can be ...

2

If you actually look at (atmospherically) scattered light, you will see that even at ground-level, there is a distinct blue tint. How do you see that, you ask? Any old shadow will do - find some building that throws a nice big shadow, make a photo with a decent camera, and analyze it thoroughly - you'll find that there's indeed a blue tint, as expected. Why ...

0

Here's my addition. Many of the answers above use the erroneous argument that frequency is the determining quantity, on the basis that the same object viewed in different media appears to be the same colour. This is meaningless, since the light has to travel through the vitreous humor (with refractive index 1.33) immediately prior to reaching the retina. ...

11

It's because you're not looking far enough. From personal experience, it takes at least 10 km of atmosphere to build up a really obvious blue (see, for example, this picture), and if you're not in hilly country, the horizon is only 5km away. In contrast, most of the sky has distances to space on the order of hundreds of kilometers.

3

Two reasons: The scattering separates red/orange and blue in different directions. At sunset you'll see the red parts that are missing from the blue skies by day. This isn't noticeable for objects close by, because those objects surround you. The blue from some objects mixes with the red from others. Secondly, there is a lot of air between you and the ...

2

Building on prior answers, the facts are: Color is determined by the energy of the EM Wave that reaches your eyeball. Energy is defined as $E = hf$, where $h$ is Planck's constant and $f$ is the light's frequency. Thus, the color of an EM Wave is defined by its frequency. In other words, measuring the frequency of an EM Wave is sufficient to identifying the ...

28

If I understand you right, you're referring to the phenomenon seen in this picture (from the first Google hit), that near the horison the color of the sky is more light-blue (not exactly white): Rayleigh scattering The scattering in the atmosphere is for a large part Rayleigh scattering off of nitrogen and oxygen molecules, which are much smaller than ...

0

I assume that in the quadrupole transition there is a 2 photon interaction, since photons only have spin 1. Hence why it is a less probable transition.

15

You say: For a non-transparent object like a brick, when the light is absorbed by an electron it will eventually be re-emitted. but this isn't true. In a solid the excited state can decay by transferring energy to lattice vibrations instead of emitting a photon. This means the energy of the incident photon is converted to heat and the photon is lost ...

41

For an object to be transparent, the light must be emitted in the same direction with the same wavelength as initially. When light strikes a brick, some is reflected in other directions, and the rest is re-emitted in longer, non-visible wavelengths. That is why a brick is opaque to visible light. Some materials we consider transparent, like glass, are ...

0

Time appears to slow down as light passes a strong object of gravity. Time is not really a physical property, time is simply a unit of measure invented by earthlings. Time appears to change because gravity bends light and it actually travels further to get to the observer so it takes longer to get there. Time is actually distance travelled. The light that ...

0

Seems like they've found a way to do it! http://phys.org/news/2015-11-device-theoretically-bit-infinite-amount.html The problem with my initial idea about using an ordinary container is described succinctly in the above article: When light is put inside a cavity, it basically interacts with the matter that surrounds it (e.g., glass or metal), and this ...

0

I've heard a tragic misconception that blackbody radiation is somehow related to the apparent color of an object. The uninformed reader might read these answers and get the idea that if you paint an object black, no infrared light will come off of it. The key here is that you may reduce the amount reflected - the object would be very cold indeed if it ...

0

It's simply because water is much flatter and smoother than most surfaces. You see reflections in water but not, say, sand, for the same reason you see your reflection in a polished piece of steel but not a rough-sanded piece of steel. All materials reflect light to some extent, but a rough surface scatters the reflected rays in all directions, so ...

2

The answer to your question can be found in the description of the engineering of the monument You can read the whole story at that link; I will just quote the most pertinent statement: Using the statistical mean of the 100-year data, the altitude and azimuth angles for the structure were adjusted to provide time/error fluctuation of plus or minus 12 ...

1

In optics, you rather speak of intensity of light which is the energy per time and per surface. The brighter the light the more intensity it has. The energy transported by a light beam per minute is proportional to the squared amplitude of a wave or the number of photons times their energy. $E\propto |E_0|^2$ for waves and $E=n \cdot E_{photon}=n \cdot h ... 1 For practical purposes we can say that laser light only moves in one direction and that its intensity at any point that is in line with the laser light is independent of distance. That is not the case, depending on what exactly you mean by "for practical purposes". If you are using a laser beam to illuminate a target that is larger than the divergence ... 0 There are two different answers which might be "reasonable" to this question. Taking your question very literally, laser light, by definition, is light amplified by stimulated emission from a pumped gain medium with population inversion. In practice, one of the easiest ways of exploiting amplification by spontaneous emission is through multiple passes of ... 2 (1) it's the energy in the sense that the photon oscillates at a certain frequency. (2) i'm not sure you can physically explain a light wave. More light is just more photons, more energy is photons with higher frequencies. (3) when it comes to the particle nature of light the photon has a frequency. It frequently oscillates through positive and negative ... 0 There are plenty of algorithmic sky models. see for instance paper (and previous work section) https://hal.inria.fr/inria-00288758 But this is not related at all to "color temperature", i.e. Plank's law (or black body). (well, there is the Sun's one at the begining, but after it's about selective absorption, Rayleigh and Mie (multiple) scattering, etc). -3 I guess wearing dark coloured clothes just makes us feel warmer. They actually have nothing to do with the body temperature of a person. They retain our temperature but do not increase it, do they? So "dark clothes for winter" might not be relevant after all. Often by wearing light clothes and by doing heavy exercise, one can increase one's body temperature ... 2 So yes, if you compute the Poynting vector (energy flux density),$\vec E \times \vec H$, for an exponentially decaying evanescent wave, you indeed find zero time-averaged energy flux perpendicular to the reflecting plane. Ask you say, this leads to a conundrum --- how do evanescent waves transfer energy across barriers? For sure, we know they can transfer ... 1 I did a quick calculation of the single-slit diffraction patterns for three different slit widths of 100, 60, and 40 microns. Wavelength was that of green light (about 5000 Angstroms). Intensity plot is shown below for a progression of decreasing slit widths from 100 microns width (blue curve), to 60 microns width (green curve), and then to 40 microns width ... 1 Short answer: On axis or central lobe field increases with the area of the slit (and thus the intensity$I\propto area^2$). Long answer: If you have a monochromatic plane wave traveling along the$z$axis incident on a diffraction slit of width$w$(as shown in the image below), then the initial field is$E(r,t) = E_0 e^{ik_0 z- \omega t}\$. Now ...

0

The only reason I could imagine a frequency shift is fluorescence (ok, this is cheating :-) ) ultra slight Doppler effect due to the thermal motion of scatterer/reflector atoms. (at macroscopic scale for many photons and atoms, it's more like an ultra slight frequency blur).

0

Now generally, it is true that light incident upon an object is slightly shifted in frequency, but when we come to point of your question there are other factors weighing in. In these cases it is better to look at the phenomena in Maxwell's wave model of light. When light falls upon a body, it energizes the lattice to vibrate (this can be thought of as ...

14

To augment Rennie's answer with a graphical representation let me post this diagram: Imagine, if you will, not a single beam of light but a series of wavefronts. When part of the wavefront slows due to a different density, the wavelength also compresses, thus introducing the characteristic bend.

0

A hot object will glow like a light bulb, even in the absence of an electric current. As such, I think that the glowing is a "heat effect".

2

The Earth does look blue, even continents, from satellite. (Still, in normal direction the amount of atmosphere is not that large). Almost all the satellite photographs that you see are retreated to improve the contrast and restore the white balance. (and many other issues such as band-variations of illumination, hot spot, etc).

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