1. Oct 12, 2007

### BosonJaw

Hello guys!

Why must an observer concentrate on one particular volume of space for a finite amount of time, rather than receiving the information/light instantaneously from it? E.G The Hubble UDF image? Why was the light not immediately received similar to when I look in my binoculars at the moon?

Is this fundamentally due to the C+ universe expansion velocity? Can someone please explain to me the principals of the observation of light over extremely long distances?

How much radiant energy from the sun is lost on its journey to earth? What mechanism is responsible? When sunlight on the way to earth interacts with interstellar matter what type of energy transfer is this? Assuming a radiant/thermal energy transfer (which I have no idea if this occurs)How can light be of infinite range, given conservational laws?

2. Oct 12, 2007

### Integral

Staff Emeritus
There is nothing mysterious about this. The amount of light we recieve from a distant galaxy is less then that we recieve from the moon by many orders of magnitude. The long exposure times reflect this greatly reduced amount of light energy we recieve.

3. Oct 12, 2007

### BosonJaw

So what process/mechanism causes light intensity to diminish over distance?

In space why isn't the total amount of light radiated from an object conserved regardless of the distance traveled?

I never said there was anything mysterious about it, I asked a question.

Last edited: Oct 12, 2007
4. Oct 12, 2007

### Staff: Mentor

No, as light travels away from an object, it spreads out because it is radiated in all directions. Consider a point source of light inside a baseball-sized sphere. It has a certain surface area to shine its light on. Now consider the same point source of light inside a basketball-sized object. It has a much larger surface area on which it spreads out the same amount of light, resulting in a lower intensity.

The relationship is an inverse square of the distance.

Last edited: Oct 12, 2007
5. Oct 12, 2007

### BosonJaw

So the loss of light intensity over distance/time is a result of divergence only? I thought that radiative energy could be transfered to other forms as well?

6. Oct 12, 2007

### Staff: Mentor

I'm not sure what you mean. Depending on where you look, dust gets in the way, but other than that, a photon will just keep going forever.

7. Oct 13, 2007

### BosonJaw

Ok, what happens if this photon encounters matter, in which it interacts?

8. Oct 13, 2007

### Integral

Staff Emeritus
When we look at the spectrum of distant stars dark bands are observed. These dark bands are called adsorption bands and are due the adsorption of some light photons by inter stellar gases. The adsorption bands are fingerprints of the different molecules in space. This is the method we have to identify inter stellar material.

9. Oct 13, 2007

### NerfMonkey

I'm sorry if this is an obvious question, but what exactly does that mean? At first I assumed you meant that you would square the distance from the light source and then flip that number (2 miles*2 miles=4 miles, inverse=1/4), but then when you're 1 mile from the source you would get 1 as the final answer, and that can't be right. Also using 0 as the distance you would get 0 light.

I also considered that you meant inverse square to mean ^-2 instead of ^2, but then using 1 for the distance again you get 1, and using 0 you get 0.

So what do you mean?

EDIT: I did some Googling on the subject and found the inverse square law. It makes sense I suppose, but one thing still confuses me; the final answer is the amount of light that still reaches you compared to if you were right at the light source, so why do you get an answer of 0 when you're 0 distance from the light? That leads me to believe that the final answer is the decrease in light intensity, but then when you're 2 miles from the light source you'd get 3/4 of the original light, at 3 miles you'd get 8/9, at 4 miles you'd get 15/16. That's completely nonsensical, so can you tell me what I'm doing wrong here?

Last edited: Oct 13, 2007
10. Oct 13, 2007

### BosonJaw

I dont think you can use the numbers 1 and 0 arbitrarily like that, can you?

1 mile could also mean 5,280 ft or x yards or x inches etc. 0 in terms of distance from a source is mathematically undefined. 0 would be the source.....Right?

11. Oct 13, 2007

### Integral

Staff Emeritus

He means that the intensity of the light drops off as $\frac 1 {r^2}$. That is only the dependence on distance there are other factors to arrive at the exact expression. Are you familiar with the expression for the surface area of a sphere? Look it up if not.

consider that at some point in time a fixed number of photons is generated by a source. The intensity is essentially the number of photons per unit area. All of the photons generated travel at the same speed so after a bit of time has passed they have all traveled the same distance. Can you visualize the sphere of photons moving out from the source, ever expanding? The number of photons does not change but the surface area always increases. At any time the intensity of light at any point on the surface of the photon sphere is like :

$$\frac N {4 \pi r^2}$$

Where N is the Initial number of photons and r is the distance from the source. Once again this is not the complete expression, to really get a meaningful number for the intensity you would need to know something about the energy of each photon (which is not necessarily constant)

12. Oct 13, 2007

### NerfMonkey

I was just using miles as pretty much a random unit as I figured most everyone would be familiar with it. I was thinking purely in mathematical terms, so it really doesn't matter what unit, if any, you use, does it?

I suppose you're right. If you sat down inside a light bulb that was radiating light only in an outward direction you wouldn't get any light inside, so I guess the answer of 0 makes perfect sense. I was thinking about the light bulb as a 3D object, not just a 2D object with only surface area, so that was why it didn't make sense to me to have 0 light at 0 distance.

But still, is the final answer the decrease in light intensity, or the total intensity? The fraction gets smaller as you increase your distance from the source, so it would make no sense for it to be the decrease in intensity. So it must be the total intensity of the light at that distance, right?

EDIT: Integral's post seems to be in agreement with my train of logic. I think I've got it now. Thanks.

Last edited: Oct 13, 2007
13. Oct 13, 2007

### BosonJaw

Hmmm, according to Mr. Integral it is the decrease in concentration of photons per volume of space. Ok, so are you saying the Hubble had to stare at one point for extended time to allow for the diminished quantity of incoming photons in the UDF for example?

Also, If I may refer back to one of my original questions that I still feel a bit unsure of. Is the overall energy content of EM radiation conserved over distance? Or does matter interaction convert some of the radiative energy to heat? The reason I ask is because you say light travels for infinite distances.

For instance, Maybe you could explain the dynamics of a laser pointer being projected into the sky, where does the light from this device terminate? I realize the divergence causes it to become invisible to the naked eye at some point, but lets say hypothetically the beam has a divergence rating of 0, would this beam be infinitely visible?

14. Oct 13, 2007

### dilletante

The UDF photo seen here had an exposure time of about 1 million seconds:

http://hubblesite.org/newscenter/archive/releases/2004/07/

15. Oct 14, 2007

### BosonJaw

Wow!

So the Hubble stared at this image for over 1 week continuously to get this image? At that magnification (what was that magnification anyway?) wouldn't the slightest movement compromise the image? What was seen as this exposure was taken? Did the image become progressively brighter? Or was the image actually formed during the exposure time? (please excuse my ignorance)

16. Oct 14, 2007

### dilletante

From the NASA Hubble site:
You can explore their site and get answers to most of your questions:

http://hubble.nasa.gov/index.php [Broken]

Last edited by a moderator: May 3, 2017
17. Oct 14, 2007

### BosonJaw

Thank you!!!

18. Oct 14, 2007

### Staff: Mentor

The equation refers to a change in distance. So just saying "one mile" hasn't told you anything about a change in intensity. Moving out from one mile to two is what gives you 1/4. If you moved from 1/2 mile to 2, the intensity would have dropped by 1/16.
There is no term "total intensity". Intensity is the number of photons per unit area. It is not conserved (it is what varies by the inverse square law). What is conserved is the total energy. Intensity times area would give you energy.

Last edited: Oct 14, 2007
19. Oct 14, 2007

### Staff: Mentor

EM radiation is radiated heat energy. It can be converted to heat energy in an object (kinetic) if it is absorbed by an object, but it does not dissipate in any other way besides the inverse square law - the total energy is conserved.
A laser is unidirectional, not omnidirectional like a light bulb. Theoretically, the light from a perfectly coherent laser would travel forever at the same intensity as it started. If it makes a 1" diameter spot on your wall, it would make a 1" diameter spot on the moon.

20. Oct 14, 2007

### Staff: Mentor

The provided link has much of that information. It isn't quite as simple as taking one exposure since the telescope orbits the earth and the earth gets in the way. A week is the total exposure. And a ccd chip works similar to film - you don't see anything while it is being exposed, you develop it (read it) when the exposure is finished. So once an orbit it would take an exposure an hour or so long and at the end of the hour, the astronomers would see the result. And they could combine exposures after a few hours or days and watch the image build.

The link doesn't say what the magnification is, but magnification doesn't mean anything without a reference. I have a poster of the UDF on my wall and if you stand across the room from it it is a different magnification than if you walk up to it. The link talks about the field of view - it is a slice of the sky about 1/10th the width of the full moon. So depending on how big you would display it, it would probably be 10-100x magnification. Pictures of planets you might see in a magazine (or on my website!) are up to about 500x magnification.

The high precision of the tracking shouldn't be too surprising - in space, it doesn't actually have to move like a telescope on earth does. Basically, it just has to use its reaction control system (flywheels) to keep it stable. And flywheels are great for making tiny adjustments to keep it pointed. Tracking for an earth-based telescope (like mine!) is a very big problem.

Last edited: Oct 14, 2007