[Q] Photospheres from a distance

[freeztar]

This is a basic question I guess, but one that never seems to come up.

When I was taking Astronomy in College, I remember stopping my professor in the middle of her discussion on photons transmitted by distant stars to ask her why we see so many distant stars when photons must "spread out" over galactic distances.

 

I only remember her first response, which was to think of a star as a photo-"sphere". That helped my understanding immensely, but it still left me with the same question.

 

Let's imagine a star about 1 billion light years away. As it goes through its process (let's assume a star like the sun for the sake of this thought experiment), it is emitting photons in every direction. Figure 1 is where my initial question comes into play...

 

As the photons linearly travel through space, it seems that at a certain distance, the amount of photons that could be detected on Earth would diminish to the point of obscurity.

 

In the Photonic-sphere model, this is simply not the case (fig 2).

 

If we consider the light travelling from a distant star to have a wave pattern, it seems that the amplitude of the waves would have to increase over time/space. (fig 3)

 

Would this amplitude be enough to "fill the gaps"?

 

I have this bad feeling that I'm missing something stupid here, but I hope not. :doh:

[modest]

I think things do vanish pretty much to the point of obscurity. Brightness follows the inverse square law so that something twice as far as the sun is 1/4 as bright. At 3 times the distance, it is 1/9th as bright. The fractions get pretty large pretty quick which is why something as bright as the entire Andromeda galaxy is barely visible to the naked eye. It’s also why Hubble has to take exposures of the deep field that last months. It’s relatively rare that any photons hit the detector over those distances.

 

Where space is not expanding (or where expansion can be ignored), brightness can be derived using your linear photon diagram like so:

 

A star emits X photons in one second. Those photons spread out in three dimensions. At any given time, those photons will make a sphere where its center is the source that emitted them. Kind of like this, only in three dimensions:

 

 

The number of photons, X, stays the same, but the sphere which they make up gets larger as they radiate outward. The area of the red sphere is [math]4 \pi r^2[/math] where r is the distance any given photon has traveled.

 

So, the flux we measure at one small spot on the red sphere (let’s say over a meter squared and for one second) can be multiplied by the area of the whole red sphere to get the total number of photons. The total number of photons can be called the absolute luminosity. This gives us,

[math]L = F \times (4 \pi r^2)[/math]

Radius, r, is nothing more than the distance to the star, so changing it to d and rearranging,

[math]F = \frac{L}{4 \pi d^2}[/math]

This is the equation that’s used to find distance to astronomical objects if you measure it’s flux (or apparent brightness) and know its absolute brightness, L (or, also called, absolute luminosity).

 

This is depicted really well on wikipedia’s inverse square law page:

 

 

To use this with real numbers... We detect 1370 watts per meter squared from the sun. In other words, the energy of all the photons crossing a meter squared in one second is 1370 joules. If we also knew the sun emits [math]3.86 \times 10^{26}[/math] total watts, then we could find the distance to the sun:

[math]F = \frac{L}{4 \pi d^2}[/math]

rearranging,

[math]d = \sqrt{\frac{L}{4 \pi F}}[/math]

or,

[math]d = \sqrt{\frac{3.86 \times 10^{26}}{4 \times 3.1416 \times 1370}} = 1.49 \times 10^{11} \ \text{meters}[/math]

So, the distance to the sun should be about 149 million kilometers :) The same thing should work for anything that is not so far away that expansion of space is a factor, which is pretty much what a standard candle is.

 

Here's another page that's probably a little better than wikipedia's inverse square law page for an explanation of apparent luminosity:

Luminosity and Distance

Does this sound about right?

 

~modest

[Jay-qu]

Thanks Modest, that sounds good but Im going to take it that little bit further by making some approximations.

 

Our sun emitts [math]3.86 \times 10^{26}[/math]Watts, with a peak wavelength of about 500nm.

 

Each 500nm photon has Energy= h*c/wavelength [math]=4 \times 10^{-19}[/math]

 

So the Sun emits approximately [math]10^{45}[/math] photons per second.

 

Thats a lot of photons! If they all go in equally space random directions from a point, there would be an angle of 360/10^45 [math]=3.6 \times 10^{-43}[/math] degrees between each photon. There would be 10^45/4pi=8*10^43photons per steradian.

So to make the distance between photons comparable ~1mm (of the order of your pupil size) you would have to be [math]2.7 \times 10^{39}[/math] metres away. Which is around 10^23 parsecs. Much larger than the observable universe.

[belovelife]

Doesn't this directly relate to the fact that any direction we look, we see galaxies at that distance.

the distance being equal to the observable universe. In this case we are at the center. Which this

would mean that the observable universe from our perspective is limited because of

the disance luminosity formula given in post #1

[modest]

Doesn't this directly relate to the fact that any direction we look, we see galaxies at that distance.

the distance being equal to the observable universe. In this case we are at the center. Which this

would mean that the observable universe from our perspective is limited because of

the disance luminosity formula given in post #1

 

No. The distance is limited because photons have only had a limited amount of time to reach us because the universe is only so old. It is not a limitation of luminosity and is not related to the question. See,

Cosmological horizon - Wikipedia, the free encyclopedia

and

Observable universe - Wikipedia, the free encyclopedia

~modest

[freeztar]

Perfect answers, thanks! :)

 

That is a lot of photons! :)

[modest]

That is a lot of photons! :)

 

I wish we were getting a few more of 'em in Kansas this week :)

 

~modest

[modest]

So the Sun emits approximately [math]10^{45}[/math] photons per second.

 

Thats a lot of photons! If they all go in equally space random directions from a point, there would be an angle of 360/10^45 [math]=3.6 \times 10^{-43}[/math] degrees between each photon.

 

So to make the distance between photons comparable ~1mm (of the order of your pupil size) you would have to be [math]2.7 \times 10^{39}[/math] metres away. Which is around 10^23 parsecs. Much larger than the observable universe.

 

I have a quick followup question.

 

Are you sure you don't have the photons falling off linearly rather than with the square of the distance? I get about 288 parsces.

 

1 photon per mm^2 is 1E6 per m^2

 

[math]d=\sqrt{\frac{\text{total photons}}{4 \ \pi \ \text{Flux}}} = \sqrt{\frac{10^{45}}{4 \ \pi \ 10^6}} = 8.92 \times 10^{18} \ \text{meters} = 288 \ \text{parsecs}[/math]

[freeztar]

I have a quick followup question.

 

Are you sure you don't have the photons falling off linearly rather than with the square of the distance? I get about 288 parsces.

 

1 photon per mm^2 is 1E6 per m^2

 

[math]d=\sqrt{\frac{\text{total photons}}{4 \ \pi \ \text{Flux}}} = \sqrt{\frac{10^{45}}{4 \ \pi \ 10^6}} = 8.92 \times 10^{18} \ \text{meters} = 288 \ \text{parsecs}[/math]

 

But 288 parsecs is less than 1000 ly. That can't be correct, can it? :confused:

[modest]

But 288 parsecs is less than 1000 ly. That can't be correct, can it? :confused:

 

I might be too tired for math, but I believe it's correct. To check the #'s... If each millimeter squared had one photon per second then each meter squared would have a million photons per second.

 

The surface area of a sphere with a radius of [math]8.9206 \times 10^{18}[/math] meters is,

[math]S = 4 \pi r^2 = 4 \times 3.14159 \times (8.9206 \times 10^{18})^2 = 10^{39} \ m^2[/math]

Multiply that by the number of photons in each meter squared and we should get the total number of photons the sun makes in a second,

[math]\mbox{Photons} = 10^{39} \times 10^6 = 10^{45}[/math]

So, yeah, that should be right. If you were standing 943 lightyears away from the sun, and space were not expanding between the two of ya (edit, it wouldn't be at that distance :doh:), and your pupil was one mm^2, then you should catch about a photon a second.

 

~modest

 

EDIT:

A fully dilated pupil is more like 15mm^2, so there would be closer to 66,666 photons in a meter squared so that one would hit the pupil every second. This makes the distance to the star [math]r = \sqrt{10^{45}/4/3.14/66,666} = 3.45 \times 10^{19} \ m[/math] or... 3,651 lightyears so that we get 1 photon per pupil per second.

[Jay-qu]

I am ashamed of my back of the envelope calculation. Modest you are correct. The mistake I made was too consider all of the photons leaving in a single plane :doh:

 

I did it a way similar to you not long after:

 

[math]

d=\sqrt{\frac{3 \times 10^{26}}{4 \ \pi \times{\text{photon energy} \times 10^6}}}[/math]

 

which is more like ~10^19m as you calculated. Very interesting.

 

Clearly this is why you dont see distant galaxies with your eyes, since our eyes are very poor integrators. While if you leave the hubble staring for 48 hours at a patch of space, you can get a great image.

[freeztar]

It's late and I'm about to retire, but a quick question.

 

If a star is 10,000ly away from us, how spread apart are the photons? It seems like there would reach a point where the distance would be several cm which leads to the seemingly ludicrous idea that one could be looking at a star, move their head a foot to the left and not see it anymore. I feel like I'm still missing something. :D

[Jay-qu]

firstly your question depends on the stars luminosity - there are many stars that are brighter than our own (and plenty less luminous too..).

 

You have to remember that these photons are travelling bloody fast, so when you say you dont see it, you mean you havent received enough photons to form an image. So moving your head to the right will not help at all, because on average you will still get the same amount of photons per second.

[freeztar]

firstly your question depends on the stars luminosity - there are many stars that are brighter than our own (and plenty less luminous too..).

 

Indeed. For the sake of example, let's assume it is exactly like the Sun.

You have to remember that these photons are travelling bloody fast, so when you say you dont see it, you mean you havent received enough photons to form an image.

 

Yes, except I'm stumbling on the fact that in my example, you wouldn't be receiving any.

 

So moving your head to the right will not help at all, because on average you will still get the same amount of photons per second.

 

Take a look at my first pathetic drawing in the first post (or better yet, Modest's depiction). If the photons are travelling linearly and they do indeed "fan out" the further away an object is, it seems that at a certain distance your eyes could be in one position and then moving slightly left or right would put you in a position where your eyes are not receiving any photons. Hopefully that makes sense. :D

[Jay-qu]

What you are saying makes sense, but you are thinking about the photons as a stream on EM radiation - like a beam. But the photons are discrete in both length and width. So at some sufficient distance, you may not receive a photon for a second, but there is a good (statistical) chance you will get one the next second. Does that make sense?

[modest]

The mistake I made was too consider all of the photons leaving in a single plane

 

Yeah, I think you were just keepin' us on our toes :)

 

Take a look at my first pathetic drawing in the first post (or better yet, Modest's depiction). If the photons are travelling linearly and they do indeed "fan out" the further away an object is, it seems that at a certain distance your eyes could be in one position and then moving slightly left or right would put you in a position where your eyes are not receiving any photons. Hopefully that makes sense. :shrug:

 

It's like standing in the rain. Moving around doesn't really avoid the rain, it just means you'll catch a different drop than you would have had you not moved.

 

To illustrate how dim things get at large distances. If you are 1 billion lightyears from a star like our sun (ignoring expansion) and you're looking at it with a 1 meter telescope...

[math]F = \frac{L}{4 \pi r^2} = \frac{3.86 \times 10^{26} W}{(4) (3.14) (9.46 \times 10^{24} m)^2} = 3.43 \times 10^{-25} \mbox{Watts} = 2.14 \times 10^{-6} eV/s[/math]

A 500 nm photon is 2.48 eV. So, from 1 billion lightyears away, and with a 1 meter telescope, you would have to look at the sun for an average of (2.48/2.14E-6) just over a million seconds or 13.5 days to catch just one photon!

 

It would clearly be very difficult to resolve something as dim as a single star over that great a distance.

 

~modest

[freeztar]

What you are saying makes sense, but you are thinking about the photons as a stream on EM radiation - like a beam. But the photons are discrete in both length and width. So at some sufficient distance, you may not receive a photon for a second, but there is a good (statistical) chance you will get one the next second. Does that make sense?

 

Yes, and I think that was part of what was tripping me up. Hence my third diagram.

It's now pretty easy to see how this works out if we consider a detector plate. I think the final piece of the puzzle, for me anyway, is to figure out how the photons are interpreted by the brain if the photon detection plates are our eyes. :naughty: