Solving Math Problem: Get Help from Our Dudes - Kevin

  • Thread starter homology
  • Start date
In summary, if one were to look at an infinite universe with an average radius and average density of stars, they would be able to "see" a star approximately every two points in space.
  • #1
homology
306
1
Okay, let's say that stars have an average radius and an average density in an infinite universe (infinitely old as well). How far on average could you look before "seeing" a star in your line of sight?

Its a homework problem, all I want is a little nudge. I've blanked out and can't even begin to figure out how to work it (I know... I'm ashamed of myself). So if one of your dudes could get me started I'd be pretty stoked.

Thanks,

Kevin
 
Astronomy news on Phys.org
  • #2
Look up Olber's paradox. It's a good starting point. Newton also suggested the universe could not be both static and infinitely old because it would collapse under its own gravitation.
 
  • #3
Hmm, I should have mentioned that we did study Olber's paradox. Figured out that something was wrong with the cosmology of Newton's time. These exercises follow that section. I'm positive I've done something like this before but for the life of me I can't remember.
 
  • #4
Very interesting problem. I will propose some steps for a simplified solution in two dimensions.

Assuming euclidean space, the angular size of the stars as a function of the distance D and their radius R is [tex]\theta = \frac{2R}{D}[/tex].

Then, knowing the star density (which gives an average number of stars in a “volume” [tex]N_V = \rho \pi D^2[/tex]), one can calculate the average number of stars on a “surface” [tex]N_S = \rho 2 \pi D dD[/tex].

The angular size covered by stars on this “surface” will be [tex]\phi_s = \frac{(2R) (\rho 2 \pi D dD)}{D} [/tex]. The total covered angular size throughout the whole “volume” will be [tex]\phi_T = \int \phi_s = \rho 4 \pi R D[/tex].

To cover the whole sky, the condition [tex]\phi_T = 2 \pi[/tex] holds and therefore [tex]D = \frac{1}{2 R \rho}[/tex]

If you assume that the number of stars N in a given volume follows a Poisson distribution, there will be a probability to see farther away. Note that none of both assumptions are acceptable: the distribution is not a Poisson distribution (correlation functions are used) and the space is not euclidean due to expansion (which has an influence on the angular size, as discussed here: https://www.physicsforums.com/showthread.php?t=59595).
 
Last edited:
  • #5
:cry: this solution is wrong, sorry. It would be right if one could simply add the [tex]\inline \phi_s[/tex] from different "surfaces" in the "volume" to impose the condition [tex]\inline \phi_T = 2 \pi[/tex], but one cannot, because the area covered by stars in one surface may overlap de area from the previous or from others. At the moment I have no better idea how to proceed.
 

FAQ: Solving Math Problem: Get Help from Our Dudes - Kevin

1. How can Kevin help me solve my math problem?

Kevin has a strong background in mathematics and is able to provide step-by-step guidance and explanations for solving various types of math problems. He can also offer tips and strategies to help improve your problem-solving skills.

2. Is there a limit to the types of math problems Kevin can help with?

No, Kevin is well-versed in a variety of math subjects, including algebra, geometry, calculus, and more. He is equipped to assist with any type of math problem you may have.

3. How does Kevin communicate with me while helping with my math problem?

Kevin can communicate with you via email, messaging, or video chat, depending on your preference. He will work with you to find the most effective way to help you understand and solve your math problem.

4. Will Kevin solve the math problem for me or just guide me through it?

Kevin's goal is to guide and assist you in solving the math problem on your own. He will provide explanations and tips, but ultimately encourage you to work through the problem yourself to fully understand the solution.

5. How long does it typically take for Kevin to help me solve a math problem?

The time it takes for Kevin to help you with a math problem may vary depending on the complexity of the problem and your understanding of the subject. However, he will work with you at your own pace and ensure that you have a clear understanding of the solution before moving on.

Back
Top