So am I right in thinking that the birth of a star, as witnessed from nearby, would not be an instantaneous ‘burst’ into life (even though perhaps this is what happens at the core, as fusion starts), but that it would be a gradual ‘warming up’?
I wish I were capable of such a calculation, and would appreciate any attempt by the student to produce it, although more to the point of my question would be the calculation of how many photons would be generated to cover a spherically concentric field with a several light years radius of densely packed arpetures (since there exists the potential for the photons to be seen from any place around the star), and then that number extrapolated to account for at least roughly the number of stars that can be (potentially) seen.
Well, if you had a good neutrino detector you could see the star burst into life as far as its neutrino emissions are concerned since they are unimpeded (mostly) in their journey from the core.
For a new star one light year away:
t[sub]0[/sub] - fusion begins
t[sub]1yr[/sub] - neutinos arrive at Earth
t[sub]~100,000yr[/sub] - first photons arrive at Earth
And I suspect that you’re right that it would take some time for the star to reach thermal equilibrium and achieve a constant brightness. I have no idea how long that would take.
Actually, this isn’t hard. Think of it this way: the star is emitting photons in all directions, so they expand in a sphere. The surface area of that sphere is 4 x pi x r^2, where r is the distance to that star. Calculate the surface area in square centimeters for a star at some distance. Now divide that by the size of your pupil in square cm. That number is the ratio of the total photons emitted versus what your eye sees. Multiply it by 20 and that’s your emitted flux.
This is actually a simplification. That’s really the number of photons the star emits that your eye detects. Also, it neglects atmospheric absorption, which is actually kinda important, since some IR and UV don’t make it down through the air. For the Sun, you see very roughly 1/2 or so the number of emitted photons that hit the top of the Earth’s atmosphere in an area the size of your pupil.
The stellar nursery shown in this (slow loading warning!) photograph may illustrate the problem in your question. It isn’t just the surface of the star that this new light has to reach, it also has to penetrate a huge cloud of gas, and dust that surrounds the entire region where stars are born. By the time light pressure, and stellar wind have cleared away enough of the cloud, the stars themselves are long stabilized, in most cases, and shining brightly.
Tris
“The road to truth is long, and lined the entire way with annoying bastards.” ~ Alexander Jablokov ~
Yeah, I know it’s not hard. I’m just too lazy to do it myself. (And anyway, I always seem to misplace an exponent or three when I try to do this sort of calculation.) 
And I notice that you didn’t work it out either 
So, next question:
What is the distance of the most distant stars we can detect with the naked eye?
I think that is S Doradus. But I don’t know, off hand what the distance is.
Tris
Ok, 169,000 light years.
Magnitude 8.6 to 11.7 Is that visible to the naked eye? I think so.
Man you have to type fast to be first on this board.
Tris
Yes, it’s S Doradus in the Large Magellanic Cloud, 169,000 light years away. However, it is possible that S Doradus is not a single star.
The most distant object visible to the naked eye is the Andromeda Galaxy, at over 2 million light years away.
You said it.
I think it’s about 6 or 7.
Usually a magnitude of about 6 or maybe 6.5 is considered to be at the limit of visibility to the human eye.
Okay, so this source says that Andromeda is 2.9 million light years away. That would make the surface area about 91,608,841,778,678.37 lightyears. This source says that there are 9.4637x10^12 kilometers in a light year and 100,000 centimeters in a kilometer. So, that makes the surface area about 86,695,859,594,087,849,016,900,000,000,000 centimeters. Divided by the diameter of a pupil (given by Bad Astronomer as .5 centimeters), that gives 173,391,719,188,175,698,033,800,000,000,000 centimeters. Times 20, that’s 3,467,834,383,763,513,960,676,000,000,000,000 photons coming from Andromeda galaxy that comprise the surface of the sphere. Is that right?
Okay, I’m not sure exactly where your error lies, if you have one, but I keep getting that the surface area is 1.057×10[sup]14[/sup] lyr[sup]2[/sup], as opposed to your 0.916×10[sup]14[/sup]. It’s not a huge difference, though. But the thing to note is the unit is lyr[sup]2[/sup], not lyr. So, even though there are 9.4637×10[sup]17[/sup] cm in a lyr, you have to multiply by this number squared. That makes the surface area 9.4652×10[sup]49[/sup] cm[sup]2[/sup]. You then have to divide by the area of a pupil. If the diameter is 0.5 cm, the area is about 0.2 cm[sup]2[/sup]. Thus the ratio is about 4.7×10[sup]50[/sup]. Times 20 photons/sec is 9.47×10[sup]51[/sup] photons/sec.
Now, if, just for fun, we take the solar luminosity to be 4×10[sup]26[/sup] W, and assume every radiated photon is 500 nm (E = 4×10[sup]-19[/sup] J), then the Sun emits around 10[sup]45[/sup] photons/sec. Thus Andromeda would require 10 Million Suns to produce that kind of output.
Of course, Andromeda has a lot more than that. Two major factors get in the way. First, not all of the light emitted by a star is visible. I think the average is around 0.1%. So, we’d have to multiply the number by 1000, to get 10 Billion. Also, Andromeda is a lot brighter than the minimum threshold. I think it’s around 3 mags, or 16 times brighter. We’d have to multiply again, this time by 16, to get 160 Billion Suns. That sounds right to me, but my numbers in this paragraph may well be way off.
Well, Lib, it’s time to introduce the dreaded concept of probability waves. If you are constructing a mental picture of an expanding sphere of tiny bb’s rushing off into space, there is an inherent flaw in your picture. What there is, sort of, is for each photon there is an expanding spherical wave of probable locations, which, when an individual probable location coincides with the probable location of an electron in that chemical on your retina, is collapsed into a photon, to interact with the also collapsed electron.
Damn that Heisenburg, anyway.
Tris
“It should be possible to explain the laws of physics to a barmaid.” ~ Albert Einstein ~
“You should see the place where Einstein used to drink!” ~ Triskadecamus ~
I’m glad you raised the BBs image, Tris, because that is pretty much what I was thinking, and wondering how they could possibly not disperse more than it seems they would. But you say that the waves “expand”. Could you explain that a bit? Does that mean that they expand in size like pond ripples, or that they increase in amplitude, or what exactly?
From this, and from our discussion in Great Debates, it sounds like the universe might best be described as a probability field.
No, amplitude, if anything, decreases with more range. That is because the amplitude (squared) is the probability. When Tris says they expand, Tris means that the incidental nature of the wave is propagating outward through time. This is the same as the ripples on the pond effect, and is also incidentally where the bb-analogy holds. If you think of every packet as a bb, you can think of these packets travelling in some direction through time determined in part by their momentum (which is also a probability wave). All of this assumes some very basic and not necessarily true things about the nature of these waves, as well as solving the Time-Dependent Schrondinger Equation, but for now just think of expanding in terms of the ripples on a pond effect.
I’ve got to quit reading my popularized physics books at the bar.
Isn’t there also a Time-Independent Schrondinger equation? How would that affect our conception of the expanding photon sphere?
How does configuration space affect the picture?
Yaeah.
The probability wave (which is the locus of points where the probability is greatest that the photon exists) is a spherical surface, which expands at C, until such time that the particular probability collapses into an actual photon. It has the same energy and frequency (not considering Doppler effects, and a few other esoteric things) as it did when it was emitted during some other interaction between particles which had their probabilities collapsed by interacting, out there in the Great Nebula of Andromeda, which is not a point source of photons, and may well include, among the forty or so photons needed for you to see it in any given sixteenth of a second, photons emitted thousands of years apart, and from similarly divergent locations in the rather large assembly of stars, gas clouds, dust, and other cosmic debris we call an object.
We are treading on the crumbling edges of my understanding of such matters. It is entirely possible that some damned super physics hot shot has invalidated every single thing I have said in this thread, while I was typing it. Cosmology is not for the faint of heart.
Tris
If I was in a bar, explaining physics, I could drink a beer now.