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Need to find the Ricci scalar curvature of this metric

  1. May 27, 2012 #1
    Need to find the Ricci scalar curvature of this metric:

    ds2 = e2a(z)(dx2 + dy2) + dz2 − e2b(z)dt2


    I tried to find the solution, but failed to pass the calculation of Riemann curvature tensor:

    <The Christoffel connection> Here a'(z) denotes the first derivative of a(z) respect to z.
    [itex]\Gamma\stackrel{x}{xz}[/itex]=[itex]\Gamma\stackrel{x}{zx}[/itex]=a'(z)
    [itex]\Gamma\stackrel{y}{yz}[/itex]=[itex]\Gamma\stackrel{y}{zy}[/itex]=a'(z)
    [itex]\Gamma\stackrel{z}{tt}[/itex]=b'(z)e2b(z)
    [itex]\Gamma\stackrel{z}{xx}[/itex]=[itex]\Gamma\stackrel{z}{yy}[/itex]=-a'(z)e2a(z)
    [itex]\Gamma\stackrel{t}{tz}[/itex]=[itex]\Gamma\stackrel{t}{zt}[/itex]=b'(z)
    [itex]\Gamma\stackrel{}{either}[/itex]=0

    <The Riemann curvature tensor>
    [itex]R\stackrel{x}{zxz}[/itex]=[itex]R\stackrel{y}{zyz}[/itex]=-a''(z)-[a'(z)]2
    [itex]R\stackrel{z}{tzt}[/itex]=b''(z)+[b'(z)]2

    I tried to find the Ricci scalar curvature(R) from current result, but it gave a function depend on z. Is there any problem in my calculation?

    Thanks for answering this question~!
     
  2. jcsd
  3. May 27, 2012 #2

    phyzguy

    User Avatar
    Science Advisor

    I didn't check your calculation, but why do you think the Ricci scalar shouldn't depend on z?
     
  4. May 27, 2012 #3

    Dale

    Staff: Mentor

    I got the same for the Christoffel symbols, but I got a lot more non-zero elements for the Riemann curvature tensor.
     
  5. May 27, 2012 #4
    Sorry for that I did not write down the other non-zero terms of Riemann curvature tensor which can be deduced by symmetry and anti-symmetry properties.
    However, I still have a contradiction that
    Rt _ztz-b''(z)-[b'(z)]2
    but
    Rz_tzt=[b''(z)+[b'(z)]2]e2b(z)

    Did you also get the same result?
     
  6. May 28, 2012 #5

    Mentz114

    User Avatar
    Gold Member

    I get

    Rztzt = -[b''(z)+[b'(z)]2]e2b(z)
     
    Last edited: May 28, 2012
  7. May 28, 2012 #6
    Thank you, I will check my result again
     
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