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Calculating Area and Volume of a sphere through line element

  1. Jun 25, 2015 #1
    1. The problem statement, all variables and given/known data
    Flat space-time in polar coordinate is considered. The line element is
    ds2= -dt2+dr2+r2(dθ2+sin2θdΦ2)

    The actual answers are given below, but I cant come up to them. Need urgent help.

    2. Relevant equations
    dA = √g11g22 dx1 dx2
    dV = √g11g22g33 dx1 dx2dx3

    3. The attempt at a solution
    g1= 1
    g2= r2
    g3= r2sin2θ

    ⇒dA = √1.r2 dr.dθ
    dA= r dr dθ
    (ACTUAL ANSWER= dA = r2 sinθ dθ dΦ)

    Cant calculate Volume
    (ACTUAL ANSWER= dV= r2sinθ dθ dΦ dr)
     
  2. jcsd
  3. Jun 25, 2015 #2

    RUber

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    The area is surface area, not area of the projection. That is why the answer is in ##d\theta d\phi##.
    What variables do your g1, g2, g3 correspond to? It seems like g1 is the dr term, g2 is the d\theta term, and g3 is the d\phi term.
    Why can't you calculate the volume?
     
  4. Jun 25, 2015 #3
    Why would the answer be dΘdΦ instead of drdΦ???
    G1,G2,G3 are basis four vector, then how could they be dr,dθ and dΦ???
    Explanation needed.
     
  5. Jun 26, 2015 #4

    RUber

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    Perhaps I am unfamiliar with your application. However, these look like spherical coordinates to me, and the surface area does not require a change in r.
    g1 is 1, which corresponds to the square of the factor used for a change in r dr. g2 is r^2, which corresponds to the square of the factor used for a change in theta d theta.
    g3 is r^2 sin^2 theta which corresponds to the square of the factor used for a change in phi d phi.
    So, from what you have shown, it seems clear that you should be using sqrt(g2 g3) d theta d phi to calculate dA.
     
  6. Jun 26, 2015 #5

    vela

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    What is the problem statement you were given, word for word? You seem to be omitting important details.
     
  7. Jun 27, 2015 #6
    According to the question, I had to calculate the area and volume of line element described above.
     
  8. Jun 29, 2015 #7
    May I have certain more specification into the mistake I am doing...:bow:
     
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