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What happens to EM waves outside the universe?

  1. May 7, 2013 #1
    If the universe is expanding at less than the speed of light, what happens to EM radiation
    that is emitted from stars at the edge of the universe?
     
  2. jcsd
  3. May 7, 2013 #2
  4. May 8, 2013 #3
    But if the universe were finite, wouldn't it then have an edge?
     
  5. May 8, 2013 #4

    WannabeNewton

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    Not necessarily. The simplest k = +1 geometry, the 3-sphere, is compact ("finite") but has empty manifold boundary ("unbounded"). Such objects are called closed manifolds lending to why you may see the term closed universe.
     
  6. May 8, 2013 #5
    Is there a more in depth and simple explanation? I am very confused
     
  7. May 8, 2013 #6
    several I've already posted the FAQ sub forum and Ned wrights FAQ.

    here is a useful balloon analogy

    http://www.phinds.com/balloonanalogy/

    this site describes loosely a few finite shapes with no edge.

    http://abyss.uoregon.edu/~js/cosmo/lectures/lec15.html

    this on going thread also has tons of descriptions and information on edge and shape

    https://www.physicsforums.com/showthread.php?t=684006

    also keep in mind the shape of what is called the Observable universe is finite and spherical. The observable universe is simply defined as the furthest we can measure. Due to speed of light and expansion.

    http://en.wikipedia.org/wiki/Observable_universe
     
    Last edited: May 8, 2013
  8. May 10, 2013 #7
    Not it doesn't. If the Universe was finite and unbounded it just means that if you travel in one direction for long enough you will eventually end back at your starting point. In other words the Universe "loops" back on itself. Why is this so hard for people to understand?
     
  9. May 10, 2013 #8
    Its hard to grasp at first, but I think I get it. So there is no outside the universe, it instead might be curved in on itself right? Does gravity cause this?
     
  10. May 10, 2013 #9
    Do you have a vested interest Flatland?
     
  11. May 10, 2013 #10
    astrophysics is complicated for the casual reader (like me) because there are usually several different theories illustrating a single concept, example the shape of the local universe or the shape of the entire universe. to make communication easier many of the theories are reduced to initials or a tag name. this complicates thing a lot for the casual reader.

    Keep reading. It took me a week and a half to get thru this article the first time I read it. I had 40 pages of notes that I looked up to get through this one page

    http://en.wikipedia.org/wiki/Shape_of_the_Universe
     
  12. May 10, 2013 #11
    I agree, I can normally read books in a day or so, but just getting through one book on relativity took me a month. Have you read "Relativity Demystified"? its a great book if you havn't read it or don;t know much on relativity
     
  13. May 10, 2013 #12

    phinds

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    I've never seen ANY confusion about the shape of the "local" universe, assuming you mean what is more commonly called the "observable" universe. It is a sphere with a radius based on the speed of light and the age of the universe and it is centered on you. Right now it's about 47 billion light years in radius. There is no confusion or contention about that and I'm not aware of any other theories about it.
     
  14. May 10, 2013 #13
    yes of course. I should have given more thought in selecting my example.

    Edit - maybe my point was to read astrophysics you have to have a vast underpinning of knowledge. how I got here is I was reading quantum physics. read how the quantum vacuum was related to the cosmological constant. Thus came in to search around the forum. Interesting place
     
    Last edited: May 10, 2013
  15. May 11, 2013 #14

    Chronos

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    You are missing the point. And making a point that has no basis in fact. Show the math.
     
  16. May 11, 2013 #15
    A major outstanding problem is that most quantum field theories predict a huge value for the quantum vacuum. A common assumption is that the quantum vacuum is equivalent to the cosmological constant. Although no theory exists that supports this assumption, arguments can be made in its favor.

    Such arguments are usually based on dimensional analysis and effective field theory. If the universe is described by an effective local quantum field theory down to the Planck scale, then we would expect a cosmological constant of the order of M_{\rm pl}^4. As noted above, the measured cosmological constant is smaller than this by a factor of 10−120. This discrepancy has been called "the worst theoretical prediction in the history of physics!".

    Some supersymmetric theories require a cosmological constant that is exactly zero, which further complicates things. This is the cosmological constant problem, the worst problem of fine-tuning in physics: there is no known natural way to derive the tiny cosmological constant used in cosmology from particle physics. Structural Quantum Gravity is an approach of Quantum Gravity that predicts Einsteins field equations with cosmological constant as the classical limit of the action of Structural Quantum Gravity.



    Sorry to step on your thread , yoyopizza
     
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