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I have some questions about the work in this paper

  1. May 13, 2015 #1
    I have been using this paper to study the properties of the Godel metric:

    I have some questions that start from Theorem 3.1 and go from there.

    1. On page 5, it says that Gμν = Rμν - ½ gμνR = 8πGρuμuν + gμνΛ (note: Λ is supposed to be the cosmological constant, also uμ is a covariant 4- velocity vector component and the expression ρuμuν equals the stress energy momentum tensor Tμν ) (There must have been a typo in one part of the paper where it has ρuμuν instead of ρuμuν. Please correct me if you think they actually meant to use superscripts the first time for whatever reason instead of subscripts.)

    This basically boils down to writing the Einstein field equations as:
    Rμν - ½ gμνR = 8πGTμν + gμνΛ

    instead of the usual
    Rμν - ½ gμνR + gμνΛ = 8πGTμν

    Is there some kind of special circumstance where you are supposed to put the cosmological constant on the right hand side of the equations and a special circumstance for the left hand side? For example, is there some rule such as: If the cosmological constant is negative then you put it on the right, but if the constant is positive then you put it on the left? If there is not some kind of rule like that, then why and how did the writer of the paper know to put it on the right instead of the left?

    2. The paper says that Tμν= ρuμuν . Now, according to my calculations, if you do some algebraic manipulation of this version of the EFEs:

    Rμν - ½ gμνR = 8πGTμν + gμνΛ

    then you can derive the stress energy momentum tensor as follows:

    T00 = 1/(8πG)
    T02 and T20 = ex1/(8πG)
    T22 = e2x1/(8πG)
    All other elements are 0.

    Note: This uses the c= 1 convention, Λ = -1/(2a2) , ρ = 1/(8πGa2)

    Now if you set ρuμuν equal to Tμν then you should be able to derive the covariant 4-velocity vector components:

    u0 = a
    u2 = aex1
    The other two components are 0.

    Here is an example of how I calculated these components:
    T00= ρu0u0 =
    1/(8πGa2) * u0 * u0 = 1/(8πG)

    For this equation to hold true, then u0 * u0 must equal a2, so then u0 = a

    That is how I calculated my covariant 4-velocity vector. Now I successfully derived the covariant 4-velocity vector. However, the conclusion of this section of the paper said that the particles in the Godel space time actually have velocities that correspond to the contravariant version of this 4 velocity vector:

    uμ = <1/a , 0 ,0 ,0>

    Now it is a simple step to simply raise an index for a vector, so I understand how this was derived.

    However, why did the velocity vector get changed from the covariant to the contravariant version in the first place? In other words, why do the particles in a Godel space-time have velocity vector uμ instead of uμ? After all, it is uμ (and not uμ) that plays a part in calculating Tμν. Would it be wrong to say that particles in a Godel space-time travel with velocity vector uμ?

    3. Finally, what exactly is a? How do I determine its value?
  2. jcsd
  3. May 13, 2015 #2


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    Gold Member

    I stopped reading at the point I've cut your post.

    The subscripts and superscripts are space-time indexes, also known as tensor indexes. They are legal in the upper ( contravariant ) position and the lower (covariant) position.

    I fear you do not know this which means the calculations could not make sense to you.

    Have a look at this http://en.wikipedia.org/wiki/Einstein_notation
    [/PLAIN] [Broken][/PLAIN] [Broken]
    I apologise if I've misunderstood the situation.[/PLAIN] [Broken][/PLAIN] [Broken]
    Last edited by a moderator: May 7, 2017
  4. May 13, 2015 #3


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    Staff: Mentor

    No. It can be put on either side, whichever is more convenient. The physics is the same either way.

    The covariant version is not a vector; it's a covector. A covector is a linear mapping of vectors to numbers. The metric provides a one-to-one correspondence between vectors and covectors, so they are often used interchangeably in the math, whichever is more convenient; but they are physically distinct things. So the quantity ##u_{\mu}## that appears in the stress-energy tensor is not the velocity vector; it's the covector that corresponds to the velocity vector (using the metric to lower the index).

    See here:


    This article uses ##1 / 2 \omega^2## in place of ##a##, but that's an easy substitution to make.
  5. May 14, 2015 #4
    You have misunderstood. I know about tensor notation and contravariance and covariance. I was saying that those 4-velocity vectors should have been covariant. I said correct me if have reason why it was correct to make it contravariant.
  6. May 14, 2015 #5


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    Staff: Mentor

    Covariant "vector" quantities (i.e., quantities with one lower index) are not vectors; they're covectors. See post #3.
  7. May 15, 2015 #6
    Even if it can be put on either side, there is one problem:

    According to my calculations, if I put the gμνΛ term on the left side of the Einstein field equations, then I get the following Tμν:

    T11 = 1/(8πG)
    T22= e2x1/ (16πG)
    T33 = 1/(8πG)

    Every other element is 0 , This uses the c = 1 convention

    The above stress energy momentum tensor is totally different from the one I derived in the OP, and the co-vector that corresponds to the 4-velocity that I derived for this particular stress energy momentum tensor is also totally different from the one in the OP. Naturally, the 4-velocity vector itself (the contravariant one) is also different from the one in the OP.

    Therefore, it does seem to make a difference which side you put the cosmological constant on.

    What do you think about this? How would I know which side is more convenient or which to choose?
  8. May 15, 2015 #7


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    Staff: Mentor

    In that case, you've done something wrong. ##T_{\mu \nu}## must be the same, because it's defined the same; it's still ##\rho u_{\mu} u_{\nu}##. Moving the ##\Lambda## term from one side of the equation to the other doesn't change the fluid's 4-velocity; it can't. Shuffling terms around in the equation doesn't change any physics.
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