# Tag Info

30

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 ...

24

To be a little pedantic, nobody has yet done precision spectroscopy of antihydrogen, though the recent success in trapping it at CERN (all over the news this week, paper here) is an early step toward that. It's possible that there are small differences in the spectrum of antihydrogen and hydrogen, though these differences can't be all that large, or they ...

20

Suppose you are analysing the weights of people in the UK to see what the distribution of weights looks like. Suppose also you can measure the weight to arbitrary precision, so that no two people's weights will be exactly the same. When you're finished you plot your data on a histogram, but the trouble is that because everyone has a different weight you get ...

18

The full quantum analysis of the hydrogen atom is a quantum two body problem, however, one of those bodies is extremely massive compared to the other, so that this problem, as a first approximation, is analysed by solving either the first quantised (i.e. for one quantum particle in a classical environment) Schrödinger or Dirac equations for inverse square ...

16

When astronomers started to get spectra of stars and began classifying them, the initial classification was based on the strength of the Balmer absorption lines in the spectra. The Balmer lines are created by electons in hydrogen atoms that are currently in the second energy level (N=2) absorbing energy and jumping up to higher levels. The stars with the ...

16

There's a few ways the temperature can be measured remotely. The easiest way is to measure the amount, and for bonus, spectrum, of the radiated heat. All objects greater than Absolute Zero radiate a certain amount of energy. The wavelength spectrum can be determined by Planck's Law, and the amount of energy by the Stefan-Boltzmann law, both of which are ...

16

The resolving power of a prism is given by the formula $$\frac{\lambda}{\Delta \lambda} = b\ \frac{dn}{d\lambda},$$ where $b$ is the base length of the prism, $\lambda$ is the wavelength and $n(\lambda)$ is the refractive index. You don't say, but let's assume you are using a crown glass prism. According to this useful document, crown class has $dn/d\... 14 Given that most green pointers are frequency doubled from a 281.8 THz infrared laser ($c$/1064 nm), it's possible that you have a two frequencies$f_1$and$f_2$in the original infrared laser (i.e., it is multimode). After passing through the "frequency doubling" nonlinear crystal you see three frequencies:$2 f_1$,$2f_2$, and$f_1 + f_2$. It looks like ... 13 Almost all exoplanets observed are near F, G, and K stars. In part, this is because astronomers are looking for earth-like planets, so they look at stars similar to our Sun, but there are also some physical reasons. Sahu et al (2006) have provided some evidence that red dwarfs (class M) are more likely to have planets than other spectral types, though it is ... 12 Good quantum number are associated with operators that commute with the Hamiltonian. They correspond to conseved quantities. Overall angular momentum is conserved, but the portions of it due to orbital motion and due to spins are not themselves conserved. n is 'bad' in that there's no conserved physical quantity related to radial motion. A decently ... 12 What you're describing does happen to some extent. It's called "Doppler broadening": absorption and emission lines in hot materials are wider (in wavelength) than absorption and emission lines in cool materials, because atoms in the hot material occupy a broader range of velocities. An atom at rest can't absorb any old photon and convert the extra energy ... 11 I think that it is important to recognize the practical difference between Raman scattering and fluorescence. If the energy of the photon is resonant with some molecular transition (meaning that it is equal to the energy difference between ground energy state and one of the excited states of the molecule), that the molecule can absorb this photon undergoing ... 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. 11 The broadening of emission lines is not due to something that is happening to each individual star, but rather something that affects the whole population. As stars in a galaxy get older, their orbits change relative to the orbits of other stars of the same age. Most relevant for the discussion here, the "velocity dispersion" of a stellar population ... 10 Just compare the resolution of the two: Prism depending on n, there is no good material n>1.7 (besides diamond) depending on base length if you use a equilateral triangle have to use more than one to overcome this prism absorb light, you have got scattering (stray light) too Now a grating: optimize it for your wavelength choose lines per milimeter ... 10 Given what we know about planetary formation (Link 1, Link 2, Link 3 and Link 4), and the theories around it, it would probably be a safe bet to say that ALL stars end up having some left over material that might become planets. I think the bigger question is how many of those planetary orbits stay stable enough throughout the life of the star? All these ... 9 In theory, perhaps. It is possible, using multilayer dielectric coatings, to produce a surface which is reflective in very narrow bands (in this case, the Sun's dark lines)and transmissive (or absorptive) elsewhere. In practice, the spectral "blurring" caused by atmospheric transmission/absorption/re-emission effects would make this effect pretty much ... 9 The Aethrioscope (see Wiki page with this name) was invented in 1818 by Sir John Leslie and the basic idea for a pyrometer (see Wiki pahe with this name) was conceived in the late 1700s by Josiah Wedgewood. These were calibrated by comparing observed colour with that of hot metals / clays (as appropriate) of known temperature. The idea was to heat a small ... 9 Lasers by definition only emit a single wavelength of light. You use one if you want that wavelength or if you want your photons to be in phase. You don't care about the photon phases, and you want to sample all wavelengths, so a laser is very much the wrong tool. If you just want collimation of the light, mirrors, lenses, or even just well-separated ... 8 Blackbody radiation for a white hot object emits the spectrum from infrared to ultraviolet. See: http://en.wikipedia.org/wiki/Black_body Graphite is a decent approximation of a blackbody radiator. So graphite heated to white hot will emit the full spectrum of visible light. Note however, that the spectrum will not be flat. There will be more energy on ... 8 You've asked a lot of questions there, and I'll try to answer them one by one. First, though, I want to ask what post you're reading about metallicity in the core vs. out here in the 'burbs because I don't think it is correct. Obviously, for example, we exist and we're ~26,000 light-years (half-way out) from the galactic center and we have a fair amount of ... 8 The major tool in investigating the composition of an astronomical body is spectroscopy. This technique makes use of the fact that a body composing certain elements/compounds will shine more brightly(or less) at particular wavelengths. The pattern of 'lines' taken by a spectrograph can then be used to infer the original composition. This technique is used ... 8 A galaxy spectrum is a quite complex and complicated topic, and many entire careers are fully devoted to understanding them, so this can only be a simplified answer. It is still quite lengthy, though, so if you're impatient, I've summarised it at the bottom. A blend of starlight of different spectral types makes up the continuum. The light is emitted by ... 7 the physical reason why this is happening is that absorption of a medium is frequency dependent. Mathematical description The most prominent example of a natural description might be the Lambert-Beer law that states that the change of a quantity$q$,$dq/dx$is related to its value at$x$multiplied by a scalar factor$\lambda\$, which might depend on some ...

7

While most measurements in astronomy are better in space, precision spectroscopy can actually do quite well on the ground. One of the best spectrographs (some would say the best) is HARPS, the High-Accuracy Radial Velocity Planetary Searcher used for finding extrasolar planets. As described in its instrument paper (pdf; note that the sole purpose of this ...

7

The atmosphere obscures data in three main ways; it absorbs light, it emits light in the infrared, and finally it diffracts light leading to distorted images. Observers have ways to deal with all three things, but I'll focus on the first two since they are more directly related to your question: 1) Atmospheric absorption. This plot gives a rough idea of ...

7

According to Bohr model, the absorption and emission lines should be infinitely narrow, because there is only one discrete value for the energy. There are few mechanism on broadening the line width - natural line width, Lorentz pressure broadening, Doppler broadening, Stark and Zeeman broadening etc. Only the first one isn't described in Bohr theory - it's ...

6

This question was asked a couple of years ago and things have changed since then. We now know that small planets are found around stars across a broad range of metallicities and that it is only the existence of giant planets that are affected by low metallicity. Nature article here. It was previously thought that small planets were more common around small ...

Only top voted, non community-wiki answers of a minimum length are eligible