- #1

- 218

- 4

1. He started with an ansatz: ds

^{2}= -B(r)c

^{2}dt

^{2}+A(r)dr

^{2}+ r

^{2}(dθ

^{2}+ sin

^{2}(θ)d∅

^{2}) (I used the - + + + signature here). He also set this solution from the start to be a vacuum solution so that T

_{mv}= 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 R

_{μν}= 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:

ds

^{2}= -(1- 2a/r)c

^{2}dt

^{2}+dr

^{2}/(1-2a/r) + r

^{2}(dθ

^{2}+ sin

^{2}(θ)d∅

^{2}) where a = GM/c

^{2}

This is how Schwarzschild derived the exterior schwarzchild metric. Basically, this whole process stemmed from starting with an assumption that T

_{mv}= 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 ds

^{2}= -c

^{2}dt

^{2}+ dl

^{2}+ (b

^{2}+ l

^{2})(dθ

^{2}+ sin

^{2}(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:

ds

^{2}= -c

^{2}dt

^{2}+ dl

^{2}+ A(l)(dθ

^{2}+ sin

^{2}(θ)d∅

^{2})

(I chose A(l) as the unknown function because b

^{2}+ l

^{2}is a function of l where b is just a constant)

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

Γ

^{1}

_{22}= -A' /2 (Note that A' is the derivative of A(l) with respect to l)

Γ

^{1}

_{33}= -A'sin

^{2}(θ) /2

Γ

^{2}

_{12}& Γ

^{2}

_{21}= A' /2A

Γ

^{3}

_{13}& Γ

^{3}

_{31}= A' /2A

Γ

^{2}

_{33}= -sin(θ)cos(θ)

Γ

^{3}

_{23}& Γ

^{3}

_{32}= cot(θ)

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

R

_{11}= (-4AA'' + 2(A' )

^{2})/ 4A

^{2}

5. Now, since the inverse metric tensor element g

^{11}= 1 and R

_{11}was the only non-zero Ricci tensor element, then the curvature scalar R = the same as R

_{11}.

6. Now just like the exterior Schwarzschild metric, I assumed a stress energy momentum tensor of T

_{mv}= 0. Therefore, if you take the equation:

R

_{mv}- (1/2)g

_{mv}R = (8πG/c

^{4})T

_{00}

and multiply both sides of the equation by the inverse metric tensor g

^{mv}, then you end up with:

R= 0 (since summing g

^{mv}with g

_{mv}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})/ 4A

^{2}= 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 vacuum solution nevertheless).

ds

^{2}= -c

^{2}dt

^{2}+ dl

^{2}+ K(dθ

^{2}+ sin

^{2}(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:

Γ

^{2}

_{33}= -sin(θ)cos(θ)

Γ

^{3}

_{23}& Γ

^{3}

_{32}= 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:

R

_{22}= 1

R

_{33}= - sin

^{2}(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:

G

_{22}= 1

G

_{33}= - sin

^{2}(θ)

Multiplying this Einstein tensor by c

^{4}/ 8πG in order to derive a stress energy momentum tensor does not yield a T

_{mv}= 0. Instead:

T

_{22}= c

^{4}/ 8πG

T

_{33}= - c

^{4}sin

^{2}(θ)/ 8πG

This goes against my earlier setting of a vacuum 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 T

_{mv}= 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?