# Tag Info

## Hot answers tagged luminosity

26

The ball is probably glowing because it has strontium aluminate in, which produces light by phosphoresence. It's a characteristic of phosphorescence that the light emission is quite long lived. This happens because when you shine light onto a phosphor the light promotes it into an excited state that subsequently decays by interactions with the solid lattice ...

11

The jargon for what you are looking for is "standard candles": things whose luminosities we can determine without knowing their distance. They are of particular interest to astronomers because they can be used to measure distances. There are many such objects, but all of them should be treated with some caution. In no case is our knowledge of the luminosity ...

10

Moonlight is not almost as bright as mid-dawn. Moonlight is really fairly low illumination, and human eyes don't detect color well in low illumination. Moonlight is reflected sunlight. If there's enough of it, it produces color mostly the same way sunlight does. This page, found through a Google search, shows a spectrum of moonlight. All the colors are ...

9

The bigger dip comes when the cooler star passes in front of the hotter object. The reason the dip is larger in this case is the amount of light given off from the area of the hotter star which is covered by the cooler star is much larger than the amount of light given off by the same area on the cooler star. Thus when the cool star passes in front of the ...

8

The biggest thing about this supernova is how CLOSE it is. A mere 21 million light years away (as opposed to being a billion light years away). The folks at John Hopkins think that studying a ype Ia supernova is valuable for several reasons. SNe Ia are also very bright compared to other standard candles, which means they can be seen at high redshifts ...

8

I believe you should have googled first: google hits Especially the second link very clearly explains the main reason: The primary reason why the color red is used for danger signals is that red light is scattered the least by air molecules. The effect of scattering is inversely related to the fourth power of the wavelength of a color. Therefore blue ...

7

If you think about it logically, it should be easy to visualize. In fact, the brighter star does not have to be larger necessarily. It could very well be smaller- perhaps the larger star is a red giant, while the smaller star is a blue main sequence, which has higher luminosity. In any case, the middle point of the M occurs when the star with a lower ...

6

Yes, It's true... We know that our eyes have three types of cone cells - S (short), M (medium) & L (large). The naming is done in order to differentiate the cells from "which cell absorbs which color". S to Blue, M to Green and L to red. The peak wavelength of L is 564 nm, yellowish-green. The peak of M is 534 nm, bluish-green. The peak of S is 420 nm, ...

5

What will be the proportion of this correct number hitting a 1 meter solar panel will depend on the atmosphere and there is a geographic dependence: clouds (albedo is on average 30% but higher the more northern the country), amount of daylight available, inclination of sun and locality ... . In Greece the maximum in summer at noon is about 800 out of ...

5

A Type Ia supernova is different from many others because of the extremely close similarity of the circumstances leading up to it. First you begin with a white-dwarf star which is composed of the Carbon and Oxygen remnant core of a former main sequence star under extreme pressures and held up by electron degeneracy pressure. A white-dwarf which happens to be ...

4

My hypothesis is that the use of red for danger signals is because we instinctively recognize red things as potentially dangerous. Aposemitism (Wikipedia) is the use of bright colors by prey to signal the presence of secondary defense mechanisms, such as toxicity. Aposematism only functions as a defense mechanism because predators, such as humans, ...

4

That's exactly right (in the ideal case). Of course, this energy is spread out all over the EM spectrum (with some blocked by the atmosphere), and photovoltaic cells can't make much use of photons that are too low-energy (they don't excite electrons into the conduction band), and they can't fully utilize too-high-energy photons either. There are other ...

3

The idea is just to make use of the relationship between luminosity (the amount of energy emitted per second from the star - in other words, the power) and flux (the amount of power hitting the surface). Because, the flux can be modeled as a large number of (imaginary) spherical energetic wavefronts emerging from the star in 3-D space as a function of time. ...

2

Dear Michael, with the energies and luminosities you described (which are not necessarily those that may be quickly achieved by either machine, and the LHC is still not there), the SSC would be a more potent machine to discover all the things you mentioned, especially the physics associated with the very heavy scales - a few TeV (which is surely extra ...

2

I think the best solution is to use stellar evolutionary tracks such as those provided by the Geneva group: http://webast.ast.obs-mip.fr/equipe/stellar/. In particular, this (enormous) package of stellar evolutionary tracks: http://webast.ast.obs-mip.fr/equipe/stellar/database/all_tracksiso.UBVRIJHKLM.tar.gz has colors and magnitudes in many bands for each ...

2

The area of a circle is calculated using its radius instead of its diameter: $$A = \pi r^2 = \pi \left(\frac{d}{2}\right)^2 = \pi \frac{d^2}{4}$$ which is where your missing factor of 4 reappears.

2

I was just about to post this, but John Rennie beat me to it, but I'd like to add that the model that he.and I use does appear to explain your observation that the equalization time $\tau$ increases as the voltage is decreased. The problem can be crudely modeled as $$T'(t)=-\alpha T(t)^4+\alpha T_0^4+\gamma \frac{V^2}{R(T(t))}$$ where $\alpha$ gives the ...

2

In the graph given in your question, the middle of the 'M' dip represent the eclipse of the brighter star against the dimmer one. I see an animation from Wikipedia at http://en.wikipedia.org/wiki/Binary_star#Eclipsing_binaries that shows a small and large stare eclipsing each other. When the small one is in front you get big dip in magnitude and when the ...

2

It's a really complicated relationship that depends on the metallicity of the star. There is a paper that does show this though: See New grids of stellar models from 0.8 to 120 solar masses at Z = 0.020 and Z = 0.001 Here are the Geneva Grids: http://obswww.unige.ch/~mowlavi/evol/stev_database.html An extensive and homogenous database of stellar ...

2

It's been a while for me but the standard candles include Cepheid variable stars whose periods tell us their luminosity Type 1A supernovae which have a highly uniform peak luminosity

2

There's no way to calculate such a thing from first principles: the luminosity of a galaxy depends on how efficiently it managed to form stars out of the matter available to it. Star formation is far too complicated to admit of a first-principles calculation. But there are empirical relationships between mass and luminosity, which have been used and refined ...

2

Stepping down in luminosity, go from the outdoors in sunlight, move indoors to a room with the light on, then step outside into you front yard with a full moon. In each of these cases, even though eventually the environment seems quite bright, it still takes time for your eyes to adapt to the lower light levels. I have seen moonlit nights here (Australia) ...

2

This question confuses two frameworks, the classical thermodynamic one with the quantum mechanical one. A perfectly homogenous radiating body belongs to classical thermodynamics. Photons belong to quantum mechanics, as well as the uncertainty principle. Lets look at the problem classically. Electromagnetic radiation is continuous from a classical body ...

1

There are two things. More photons means a brighter beam. power (Energy/sec) is proportional to the number of photons/sec. Photons with shorter wavelengths and higher frequencies have more energy. That is a bluer beam has more power. So $P = nh\nu$ where n is the number of photons/sec.

1

The apparent brightness of an incandescent bulb is a very strong function of the temperature of the filament, because it behaves approximately like a black body. Thus, much of the power emitted will be in the IR. The black body spectrum for different temperatures can be found, for example at wikipedia: Note this is a visual representation of Planck's law. ...

1

I think there is at least one mistake in what you've written above. The radiative diffusion equation is $$\frac{d(aT^4)}{dr} = -\frac{3 \kappa \rho}{c} F_{rad}$$ $$\frac{dT}{dr} = -\frac{3 \kappa \rho}{4acT^3} F_{rad}$$ (I think you have mixed up $F$ and $L$ in your second equation, and the answer may be as simple as that). Radiation pressure is P_{rad} ...

1

This relation is quite important, non trivial, and mathematical, and was proved by Etherington along with the other closely related theorem in this paper I. M. H. Etherington (Philosophical Magazine ser. 7, vol. 15, 761 (1933)) This theorem only depends on photon conservation and the fact that photons only travel in null geodesics in Reimannian ...

1

See D. Hogg's Distance measures in cosmology, 2000 http://arxiv.org/abs/astro-ph/9905116 Section 7, Luminosity Distance, p. 6 $D_L=(1+z)^2 D_A$ follows because the surface brightness of a receding object is reduced by a factor $(1+z)^{−4}$, and the angular area goes down as $D^{-2}_A$.

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