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Starting w/ an ansatz vs. Starting with a complete solution

  1. Dec 16, 2014 #1
    Recently, I have been studying the methodology used by Schwarzchild to derive the exterior Schwarzchild solution to the Einstein field equations. Now, here was the process:

    1. He started with an ansatz: ds2= -B(r)c2dt2 +A(r)dr2 + r2(dθ2 + sin2(θ)d∅2) (I used the - + + + signature here). He also set this solution from the start to be a vaccum solution so that Tmv = 0.

    2. He then used the metric tensor and inverse metric tensor that this metric would make to derive the Christoffel symbols.

    3. He then derived the Ricci tensor (with the unknown functions still in place).

    4. Due to some algebraic manipulation of the Einstein field equations, he was able to get it so that the equations reduced to Rμν= 0.

    5. Because the Ricci tensor equals 0, this creates a set of homogenous ordinary differential equations where you solve for the two unknown functions.

    6. Solving for said functions yields: A(r) = 1/(1- 2a/r) and B(r)= 1- 2a/r where a is an integration constant.

    7. Finally, he added some physical significance to the integration constant a in order to make the solution reduce to Newton's laws under ordinary circumstances. This yielded:

    ds2= -(1- 2a/r)c2dt2 +dr2/(1-2a/r) + r2(dθ2 + sin2(θ)d∅2) where a = GM/c2

    This is how Schwarzchild derived the exterior schwarzchild metric. Basically, this whole process stemmed from starting with an assumption that Tmv = 0 and an assumption that there would be those two unknown functions A(r) and B(r) in the metric.

    Now, having analyzed this methodology, I tried to use the same method with the Morris/Thorne wormhole metric and see if I would get the proper result. Here is how things went:

    1. You should first know that the wormhole metric is ds2 = -c2dt2 + dl2 + (b2 + l2)(dθ2 + sin2(θ)d∅2)

    2. Since I was reverse engineering this metric and trying to re-derive it starting with its original ansatz, I predicted the ansatz to be as follows:

    ds2 = -c2dt2 + dl2 + A(l)(dθ2 + sin2(θ)d∅2)

    (I chose A(l) as the unknown function because b2 + l2 is a function of l where b is just a constant)

    3. The next thing I did was derive the Christoffel symbols:

    Γ122 = -A' /2 (Note that A' is the derivative of A(l) with respect to l)
    Γ133 = -A'sin2(θ) /2
    Γ212 & Γ221 = A' /2A
    Γ313 & Γ331 = A' /2A
    Γ233= -sin(θ)cos(θ)
    Γ323 & Γ332 = cot(θ)

    4. After this, I derived the Ricci tensor (fortunately, there was only one element)

    R11 = (-4AA'' + 2(A' )2)/ 4A2

    5. Now, since the inverse metric tensor element g11 = 1 and R11 was the only non-zero Ricci tensor element, then the curvature scalar R = the same as R11.

    6. Now just like the exterior Schwarzchild metric, I assumed a stress energy momentum tensor of Tmv = 0. Therefore, if you take the equation:

    Rmv - (1/2)gmvR = (8πG/c4)T00

    and multiply both sides of the equation by the inverse metric tensor gmv, then you end up with:

    R= 0 (since summing gmv with gmv equals 4 and the whole left side of the equation turns into R - 2R which = - R. This yields - R = 0 which is just simply R = 0)

    7. Now that it is established that R = 0, I have a simple ordinary differential equation to solve:

    (-4AA'' + 2(A' )2)/ 4A2 = 0

    8. Upon solving for the function A(l), I derived a constant (K) as the solution to the differential equation. (This works as long as K does not equal 0)

    This leaves my final solution to the Einstein field equations as: (I know that this is not the Morris-Thorne wormhole solution, but it should be some random vaccum solution nevertheless).

    ds2 = -c2dt2 + dl2 + K(dθ2 + sin2(θ)d∅2)

    Now here is where my problem and question comes in:

    I plugged this newly derived metric into the various general relativistic tensors and tensor-like objects in order to see if the stress energy momentum tensor I derived was 0. Here were my Christoffel symbols:

    Γ233= -sin(θ)cos(θ)
    Γ323 & Γ332 = cot(θ)
    Every other Christoffel symbol was 0. Remember that the unknown function did turn out to be just a constant after all.

    Here was my Ricci tensor:

    R22 = 1
    R33 = - sin2(θ)
    Every other element was 0.

    My curvature scalar R = 0

    When plugging this stuff into the Einstein field equations to get an Einstein tensor, you get:

    G22 = 1
    G33 = - sin2(θ)

    Multiplying this Einstein tensor by c4/ 8πG in order to derive a stress energy momentum tensor does not yield a Tmv = 0. Instead:

    T22 = c4/ 8πG
    T33 = - c4sin2(θ)/ 8πG

    This goes against my earlier setting of a vaccum solution. This is what led me to note that starting with an ansatz and a pre-determined stress energy momentum tensor leads to a space-time interval solution that will lead to a different stress energy momentum tensor (from what you pre-determined) if you plug that solution into the general relativistic tensors.

    Is it supposed to be that way, or did I just do something wrong? Perhaps I missed a solution to that ordinary differential equation, but a constant solution does work to solve the equation and I could not find any other means to derive another solution. When I plugged my solution into the relativistic tensors, was I supposed to get a stress energy momentum tensor of Tmv= 0, or would it be expected for me to get a non-zero tensor as I did?

    If I am supposed to get a different tensor from my pre-determined stress energy momentum tensor, then how can one really call the space time interval a solution to the Einstein field equations when plugging in the solution yields something different than what is expected?
  2. jcsd
  3. Dec 16, 2014 #2


    Staff: Mentor

    This is where you went wrong. You can't just assume this; or rather, you can, but you have to allow for the possibility that the assumption is not valid. Schwarzschild just happened to be lucky: the metric ansatz he wrote down turned out to actually be consistent with the vacuum assumption. Your ansatz is not (i.e., there is no vacuum solution that has a metric line element of the form you wrote down), and that's why you got the results you got.

    You are aware, btw, that the Morris-Thorne wormhole solution is not a vacuum solution, right? I believe we verified in a thread a while back that its Einstein tensor does not vanish.

    This is not how you compute R. You compute R by contracting the Ricci tensor with the (inverse) metric tensor, i.e., ##R = g^{mv} R_{mv}## . The equation you wrote (which, btw, should have a Tmv on the RHS, not T00) only serves as a check to see what kind of stress-energy tensor is consistent with the Ricci tensor and Ricci scalar that you've already computed. If there's no way to make the RHS zero (which there isn't for the wormhole metric you wrote down), that means the metric you wrote down, with the Ricci tensor you computed, is not consistent with a vacuum stress-energy tensor.
    Last edited: Dec 16, 2014
  4. Dec 16, 2014 #3
    So you are basically saying that the validity of an ansatz is luck based?

    By the way: I know that you contract the inverse metric with the Ricci tensor to get R. When I got the R = 0 thing, I was just saying how if you take:

    gmv(Rmv - (1/2)gmvR) then you will get R - 2R.

    The T00 was simply a mistake. I meant Tmv.
  5. Dec 16, 2014 #4


    User Avatar
    Science Advisor
    Gold Member

    I would put it differently. An ansatz needs certain minimal features to be a valid GR metric: it must be symmetric, and its value at every point must be related to the Minkowski metric by some linear transform, all with the same signature. I don't think much more is needed. Where luck (or skill) comes in is whether it is consistent with additional assumptions you want to make. A trivial example is if you take a constant metric as your ansatz and seek to find how to make R > 0. There is nothing 'wrong' with the ansatz, or the assumption, only requiring these both at the same time.
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