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Volume of a spacetime metric

  1. Sep 17, 2015 #1
    I want to calculate two things (This is not a homework question so I am posting here or actually I dont have homework like this)

    First question is finding universe volume using spacetime metric approach.The second thing is find a smallest volume of a spacetime metric (related to plank distances I guess ).

    I looked wikipedia and there said the volume of a matric equation is ##vol_g=√(detg)dx^0∧dx^1∧dx^2∧dx^3## (In local coordinates ##x^μ## of a manifold, the volume can be written like this)

    ##g=diag(-c^2,1,1,1)## the metric will be ##ds^2=-c^2dt^2+dx^2+dy^2+dz^2##.

    The determinant of the flat spacetime metric is , ##detg=c^2## and then ##√(detg)=c##

    The volume will be ##vol_g=cdx^0∧dx^1∧dx^2∧dx^3##

    Lets start with my first question.How can we calculate universe volume using spacetime metric ?
    ##dx^0∧dx^1∧dx^2∧dx^3## this part is complicated for me.


    I want to make a assumption.In my first question (Universe volume)
    ##dx^0=(13.9,0,0,0)##
    ##dx^1=(0,46,0,0)##
    ##dx^2=(0,0,46,0)##
    ##dx^3=(0,0,0,46)##
    Now the exterior product is complicated.If my assumption is true then How I can calculate the other exterior product ? If its not true what are the true numbers. Here the units are light year for distance and year for time.

    And Also I want to make a assumption about my second question(The smallest volume of spacetime metric)
    Here again ##vol_g=cdx^0∧dx^1∧dx^2∧dx^3##

    ##dx^0=(5.39 10E(-44),0,0,0)##
    ##dx^1=(0,1.61 10E(-35),0,0)##
    ##dx^2=(0,0,1.61 10E(-35),0)##
    ##dx^3=(0,0,0,1.61 10E(-35))##
    here 5.39 10E(-44)=planck time (second) and 1.61 10E(-35)=planck lenght (meter)
     
  2. jcsd
  3. Sep 17, 2015 #2

    vanhees71

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    I don't know what you mean by "volume of the universe". Does it perhaps refer to the Robertson-Walker spacetime and it's asked about the three-dimensional spatial volume in the sense of the standard foliation induced by the choice of standard comoving (Gaussian) coordinates?
     
  4. Sep 17, 2015 #3

    martinbn

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    Looking at these numbers, I am guessing that you are taking the radius of the observable universe 46 and the age 13.9 and you want to find the volume of that portion of spacetime.
     
  5. Sep 17, 2015 #4
    I didnt choose Robertson-Walker spacetime metric cause I want to calculate "flat spacetime".Without any mass on spacetime.Or any other object or energy (DM,BM,Dark energy)

    "Volume of the universe"=Volume of Observable universe.

    We know observable universe volume but we are using 3d not 4d.

    Yes, but I dont know 46 or 13.9 are right numbers to answer my question
     
  6. Sep 17, 2015 #5
    The main idea must be "Calculating observable unverse radius in 4d" and I thought I should use the flat spacetime metric.And also I want to calculate the smallest volume in 4d.So agai I used flat spacetime metric
     
  7. Sep 17, 2015 #6

    Nugatory

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    I'm not sure what you're trying to do here. If you're planning to use the flat (Minkowski) metric, then ##dx\,dy\,dz\,dt## is a perfectly satisfactory volume element for integration purposes (measure time in years and distance in light-years and you don't have to mess with an annoying factor of ##c##). However, the universe isn't flat so that's not the right metric to use if you expect that the calculation will tell yoiu anything about the universe; and even if you use a more appropriate metric, the four-dimensional "volume" that results of the integration can't be interpreted as the volume of anything because the geometry still isn't Euclidean.

    You can integrate ##dx\,dy\,dz\,dt## across those bounds if you want, but you have the same problems as above, and the added annoyance that there is no particular reason to believe that the Planck time and length are actually going to give you any sort of minimum possible volume. You may want to give this Insights articke a try: https://www.physicsforums.com/insights/hand-wavy-discussion-planck-length/
     
  8. Sep 18, 2015 #7
    Lets suppose I use FRW metric.4d is not an Euclidean but how its affects to find a volume ?
     
  9. Sep 18, 2015 #8

    Nugatory

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    I'm sorry, I shouldn't have said "not Euclidean" because of course it is possible to find meaningful surface areas and volumes in a non-Euclidean geometry.
     
  10. Sep 18, 2015 #9
    I want to find the universe volume without any gravitational affect.So I should use Flat Spacetime Metric or FRW metric.I thought to use Flat spacetime metric cause FRW metric depends mass and cosmological constants.Flat spacetime depends nothing.And our Universe is flat.

    I think there is no reason to think volumes which are less then planck volume.The thing is in QG these sclae are important.Thats a thing whiich I want to thing about it.The planck scales are also important in information theory.If we use planck length and also time what should be the 4d volume
     
    Last edited: Sep 18, 2015
  11. Sep 18, 2015 #10

    PeterDonis

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    There's no such thing. There is gravity in the universe.

    It's spatially flat. It's not a flat spacetime. You need to use the FRW metric.
     
  12. Sep 18, 2015 #11
    If I use FRW metric the idea will be true isnt it ?
     
  13. Sep 18, 2015 #12

    PeterDonis

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    What idea?
     
  14. Sep 18, 2015 #13
    The numbers and my question.I tried to answer my question using Flat spacetime metric.And I learned that , thats wrong so FRW metric should give the right answer about the volune of universe ?
     
  15. Sep 18, 2015 #14

    PeterDonis

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    Yes, provided you do the calculation correctly and you know what portion of the "universe" you want the volume of. From your OP, it looks like you want the 4-volume of the currently observable portion of the universe, from the Big Bang to "now", i.e., to the spacelike surface of constant commoving time corresponding to our present moment on Earth. Your OP doesn't appear to describe a correct way of doing that. You can't just plug in values like ##13.9##, ##46##, etc., for ##dx^0##, ##dx^1##, etc, into an integral; those values aren't coordinate differentials, they're limits of integration. (In flat spacetime, you can get away with confusing the two, but not in curved spacetime.)

    Also, the spatial radius of the observable universe wasn't always 46 billion light-years; it is a function of time, so what you really need to do is find a way to express the spatial 3-volume of a spacelike slice of constant coordinate time, as a function of time, and then integrate that function over the age of the universe (from time zero to time 13.9 billion years).
     
  16. Sep 19, 2015 #15
    Lots of work to do.But unfortunately I cant do this myself.Again thanks for help
     
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