New answers tagged

1

Aside from simply looking at the Lorentz transformation, seeing a divergence and concluding "meh, it doesn't work", another way to gain insight into the divergence is through the statement: no finite sequence of finite boosts will get you to a speed $c$ relative to your beginning inertial frame. Imagine yourself in a spaceship with orientation controls ...


3

To see what's going on, it's enough to do this in two dimensions, with the Lorentz form $\pmatrix{-1&0\cr 0&1\cr}$. (I've set $c=1$.) The Lorentz group is the group that preserves this form. A typical element is $$\pmatrix{\pm\sec\theta&\tan\theta\cr \tan\theta&\pm\sec\theta\cr}$$ where $\theta$ runs through the open interval from $-\pi/...


-1

Lorentz transformations apply to objects with nonzero mass. For an object with mass, it would require an infinite amount of energy to reach light speed.


0

The color matching functions (in the first chart) tell you in absolute quantities how much light from each primary is needed to match a reference light. The scalar quantities are known as tristimulus values. The chromaticity coordinates are the color matching functions normalized so that the sum of each value adds up to 1. This gives you a better way to ...


0

When we talk about a (non-spinning) black hole we normally mean the spacetime geometry called the Schwarzschild metric. This is one of the simpler solutions to the Einstein equations, but it's important to understand that the solution is based on a number of assumptions and as a result it is only an approximation to a real black hole. The assumption that is ...


-3

If you mix pure red spectrum light with pure blue spectrum light, you don't get purple light. You get green light. Purple light does not exist, except as a construct of your perceptions when red and blue cones are both activated in your eye, but green cones are not. In other words, if all cones except green ones are triggered in your eyes, you perceive ...


0

As has been commented by fractalspawn this is almost certainly a photochromic effect (https://en.wikipedia.org/wiki/Photochromism). In this process a molecule in form A can absorb a visible or UV photon and be isomerized into another form B. This will have a different absorption spectrum than the starting material and if this absorbs in the visible part of ...


2

Consider the light that originates at a period (full stop, '.') on a page. If the face plate is right up against the page, the light from that period will enter only a few fibers, those that are in contact with the period. If you lift the plate from the page, the light from the period spreads out before it hits the plate. If you move the plate far enough (...


1

The colours are formed by thin film interference, much like a soap bubble or an oil slick. Light is partially reflected off of the front surface of the coating and also off of the back surface. The two reflected light waves experience a difference in path length of twice the thickness of the film. This leads to light interference which can be constructive,...


3

Based on my own anecdotal evidence, it doesn't. Several years ago there was a partial solar eclipse in my area. I don't remember precisely how much of the sun's disk was covered - it wasn't much, surely nowhere near 90% - but I do remember getting out of the house in the morning, thinking "hmm, it's quite dark today", then having the eerie realization that ...


0

This actually contains two questions. 1) The sun seems brighter (more dazzling) if there is more scattering in the atmosphere. The sun would actually look very small to us in the sky if there were no atmosphere (it's the same angular size as the moon) and most of the brightness seen in the direction of the sun is from small deflection rayleigh scattering. ...


79

Human perception is generally logarithmic. For example, the perceived loudness of a sound is measured using decibels, where an decrease of $10 \text{ dB}$ divides the sound intensity by $10$. So if the eclipse were heard instead of seen, "90% coverage" might mean reducing the intensity from $120 \text{ dB}$ to $110 \text{ dB}$, a small change. Perceived ...


42

The graph looks exponential because the vertical axis is logarithmic! If you were to re-plot it as linear lumens per square meter, it would be much more v-like, or even u-like. It so happens that a logarithmic plot matches our subjective perception of light intensity better than a linear one would. That's a result of our eyes having evolved to work well in ...


0

From the wiki article on the sun's radiation: Solar irradiance spectrum above atmosphere and at surface. Extreme UV and X-rays are produced (at left of wavelength range shown) but comprise very small amounts of the Sun's total output power. You ask: My Question is, why exactly does the intensity vanish below 240 nm ? If i look at the plank's law, ...


3

It is an optical illusion that uses stroboscopic light. It is the same principle that allows you to see still pictures into a movie. For instance, when you see the drops floating still, the drop is not the same, the light frequency is synchronized with that of the the falling drops, so that every time the light is on it shows you the image of a different ...


3

If an object is black, it should absorb all visible frequencies This is too simplistic. Objects appear black if they absorb "most" of the light that hits it. We can't make materials that absorb 100%. What happens to that small fraction that is not absorbed is critical to how you perceive it. Most surfaces we encounter scatter reflected light well. ...


3

The Doppler effect proves that light can behave like a wave under some circumstances, but it doesn't prove that light is a wave. Likewise the photoelectric effect proves that light can behave like a particle under some circumstances, but doesn't prove that light is a particle. Light is described by quantum field theory, specifically quantum electrodynamics,...


0

In a polaroid type of polarising material the molecules are aligned in the same direction throughout the sheet when the material is manufactured. Their electric dipoles are therefore also aligned and thus absorb photons whose electric field is parallel with the electric dipoles (to an extent that depends on cos(theta)^2 where theta is the angle between ...


1

It is incomprehensible to me that the most up-voted answers, and the ones posted by the most seeminglly knowledgeable people, all attempt to treat this question in terms of individual photons striking individual electrons. In fact, the phenomenon of transparency is all but incomprehensible in terms of such quasi-QM explanations. The natural, sensible ...


1

This answer is a little circular and like Burley's. Transparent materials have uniform electromagnetic coupling between its molecules. Think of glass as a uniform array of tiny capacitors.


1

One can rationally ask what one could possibly see from 27 light-years away. Assuming that spacetime is neither grainy nor foggy (something that it very possibly is), it's mostly a matter of the size of the telescope and how well the scene is lit. One square meter of ground on Earth (roughly the area of a human sized "pixel", if we are generous and account ...


4

The simple answer is no, or, more precisely, the arguments that you cite give no weight to the idea that past states "travel" through the Universe. Let's look at the 13.5 light year away mirror. Yes, in theory you could see your own birth through it, if you could overcome all the optical resolution problems associated with such an undertaking. But this is ...


0

Body with higher temperature will produce the EM waves of higher energy than the EM waves produced by the body at lower temperature. All the things in the universe emits EM waves all the time. Since the body at higher temperature emits EM waves of higher energy so when they will fall on a body of lower temperature they will increase the average Kinetic ...


20

There isn't a simple answer to that. Colour arises when the light absorption or emission of a system is dependent on the wavelength. For example chlorophyll (i.e. plants) is green because it absorbs red and blue light so only the green light is reflected and reaches our eyes. So the question would be how does the light absorption and emission of trillions ...


10

The 'color' would be an ultra-bright burst of gamma rays as the trillions of electrons rush apart, frying both you and your eyes to a crisp. More seriously, if you confined the cloud of electrons, it wouldn't emit any particular color on its own -- for instance, there could be no optical transitions since there are no nuclei. If you shined light on it, it ...


1

This depends upon how you want light to solid. If you want light to be solid in the way the "Star Wars" movies have light sabers, I would say no. There are however materials that trap photons so they have zero velocity. Photons are in a sense trapped, and these are sometimes called artificial black holes. The energy of the photons then contribute a tiny ...


6

No because solid is a state of matter. Light cannot be considered matter since it is made up of particles which have no mass and I'm pretty sure occupy no space (i.e. photons have no volume). Edit: Since photons are at the quantum level, we can't actually fathom what it would mean for them to occupy space. But on this thread someone pointed out that there ...


5

The translucent sheets and optical fibers are being lit from below. Light enters one end with a relative small angle incidence. Once inside the sheet/fiber the light experiences total internal reflections multiple times because it hits the sides of the sheet/fiber with a large angle incidence. It's not until the light gets to the far end of the sheet/...


0

When you see a rainbow is when the incoming white light comes from a very small bright source, like when the sky is very clear in one direction and the sun shines in from that direction to the rain under the clouds on the opposite side from where you are looking. If the light source is diffuse, then you get lots of rainbows all jumbled together into uniform ...


1

Light is multiply scattered in clouds, i.e. it is refracted by one water droplet or ice crystal, then refracted again by the next, then again by the next and so on. By the time the light reaches you eyes it will have been scattered many times. This means all the light hitting the cloud gets thoroughly mixed up and all the rainbows are jumbled up together and ...


19

I'll add a theoretical limit to the actual record put forward by John Rennie. To image an object as more than a featureless "point source", it must be resolved by the telescope. The angular resolution $\theta$ of a telescope is: $$\theta\sim1.22\frac{\lambda}{D_{\rm aperture}}$$ $\lambda$ is the wavelength of light, $D_{\rm aperture}$ is the diameter of ...


22

To address your last point, there are several stars of which we have been able to resolve images i.e. see the star as more than just a featureless point. There is a list of these stars on Wikipedia (I love that they put the Sun at the top of the list - true but pedantic :-). The farthest away of the stars in the list is Epsilon Aurigae at about 2000 light ...


2

Not surprisingly, physicists have looked for variations in the speed of light as a function of frequency in vacuum. The state of the art in 1972 can be found in Z. Bay and J. A. White, 'Frequency Dependence of the Speed of Light in Space', Phys. Rev. D 5(4) 796-799 (1972). Using data from pulsar emissions (radio, visible, x-ray) and other sources (see paper),...


0

As John stated, the relation is not constant. Where does the need to "show that $\frac{i}{r}$ is constant come from? You know that if $i=0\rightarrow r=0$. You know that $$n_i\sin{i}=n_r\sin{r}$$ From this you should be able to see clearly that there's no linear relation except when $i$ approaches $0$, or $n_r$ approaches $n_i$. You should never go into ...


1

A simple answer is that light has energy, whereas heat is a form of energy. It is similar to the difference between a billiard ball and the kinetic energy of the billiard ball Electromagnetic radiation classically is described by fields, but in the underlying quantum mechanical framework it is emergent from a confluence of an enormous number of photons. ...


3

In a vacuum all frequencies and amplitudes of light travel at the same speed of c = 299 792 458 m/s. Frequency is equivalent to colour. Amplitude relates to intensity. When light travels in material mediums (air, water, glass, etc) it travels at a slower speed v < c which depends on frequency. The ratio of c/v is what we measure as the refractive ...


7

In physics, heat is energy that spontaneously passes between a system and its surroundings in some way other than through work or the transfer of matter. When a suitable physical pathway exists, heat flows spontaneously from a hotter to a colder body.The transfer can be by contact between the source and the destination body, as in conduction; or by radiation ...


0

Heat and light are different but they are both forms of energy. Heat is a form of kinetic energy contained in the random motion of the particles of a material. Light is a form of electromagnetic energy. As with other forms of energy, heat energy can be transformed into light energy and vice versa. The particles of a material vibrate or oscillate because ...


2

Also they are closely related in empirical life, they are different and the concepts come, I would claim, from different theories. Light is electromagnetic radiation in the visible spectrum which is produced by moving charges according to Maxwell's equations. Heat is a thermodynamical concept and usually describes energy transfer that is not work (or ...


1

There are devices called Monochromators, which can produce narrow band illumination with pretty good accuracy, so if implemented correctly, one may measure the visual range of a person, I would believe.


0

What comes to mind is separating white light (e.g. from the sun) through a prism, and using the colors in the spectrum with corresponding wavelength to determine what range you can see.


1

Light from the focus, when reflected to a parabolic mirror, will all be reflected in parallel rays.


1

What is the difference between "observed" and "true" angles of rotation? What I think you mean is that the "true" angle is "specific" to the compound. It is defined like absorption coefficient $\beta$ as a constant of proportionality : $absorption \space A = coefficient \space \beta \times concentration \space C \times path \space length \space L$ $...


1

For your first question you want to find a way to express the observered angle in polarization in a form that is independent of the concentration of the molecule and pathlength of the light. In other words you are looking for units of angle per concentraion pathlength and you need to divide by the concentration and pathlength. Of course when you want to ...


0

The best intuitive explanation I can come up with is that the Fresnel equations, which perfectly describe the reflexion of light from a plane, optical quality interface, are an expression of the requirement that the field vectors should be continuous across the interface. This requirement is itself an expression of the absence of either static or flowing ...


0

This figure helps to get an intuition of how light is made up of photons Left and right handed circular polarization, and their associate angular momenta. The purple sticks in the middle are the photons which build up the macroscopic light. The individual photons, as elementary particles are point like, and are characterized by their spin , +/-1, and ...


3

There are two effects at work here to form the Color Afterimage. The blue light stimulates the S-type cone cells most, and they simply "tire out". Local supplies within the cell of ATP become run down, and the cell cannot therefore signal as often or as effectively. When you looked away into a more "balanced" light, white light that would normally fire all ...


1

Unpolarized light is more properly called light with random polarization. That makes it more clear what it means: the polarization state (circular, linear, elliptic) varies randomly over space, wavelength, and time. Consider the scenario below, where a diffuse light source is converted to a collimated beam with a narrow range of wavelengths $\lambda\pm\...


1

I tried to confirm the suggestion of @Peter Shor in the comments by creating a simple program that generates vectors by randomizing the angle $\phi$ from $0$ to $2\pi$ of a vector $\vec A_i = A\hat r_i$ that lies on a x-y plane. All the vectors have the same amplitude $A$. The program adds (by vector addition: summing all x-components and y-components) n = ...


0

You cannot immobilize light, but you can absorb it, convert it to heat. A first step in measuring light intensity (watts/square meter) is to make a black body to absorb the light. Then, by conservation of energy, the heating of that black body tells you how much light is being absorbed, in watts. The illuminated area of the black body is the "square ...



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