- #1
grav-universe
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- 1
Given the metric
[tex]c^2 d\tau^2 = c^2 B(r) dt^2 - A(r) dr^2 - C(r) r^2 d\phi^2[/tex]
and solving only for a static, spherically symmetric vacuum spacetime, I want to reduce the number of coordinate functions A, B, and C from three to only one using the EFE's. We can then make a coordinate choice for that one function in terms of r and automatically produce the entire metric from our choice for that single function. The four components of the Ricci tensor for a vacuum spacetime are all zero and these are
[tex](-4 r A^2 B C) R_{00} = 2 r A B C B" - r A C B'^2 - r B C A' B' + 4 A B C B' + 2 r A B B' C'[/tex]
[tex](4 r A B^2 C^2) R_{11} = 2 r A B C^2 B" - r A C^2 B'^2 - r B C^2 A' B" + 8 A B^2 C C' + 4 r A B^2 C C" - 4 B^2 C^2 A' - 2 r B^2 C A' C' - 2 r A B^2 C'^2[/tex]
[tex](-4 A^2 B) R_{22} = - r^2 A B' C' - 2 r^2 A B C" - 8 r A B C' + r^2 B A' C' + 4 A^2 B + 2 r B C A' - 2 r A C B' - 4 A B C[/tex]
[tex]R_{33} = R_{22} (sin \phi)^2[/tex]
as graciously worked out for me by Markus Hanke in this thread. Okay, so we will start with the R_00 component. Dividing all of its terms by r A B C B" gives simply
[tex]\frac {2 B"} {B'} - \frac {B'} {B} - \frac {A'} {A} + \frac {4} {r} + \frac {2 C'} {C} = 0[/tex]
and finding the indefinite integrals for this gives
[tex]log(B'^2) - log(B) - log(A) + log(r^4) + log(C^2) = k[/tex]
[tex]log(\frac {B'^2 r^4 C^2} {A B}) = k[/tex]
[tex] \frac {B'^2 r^4 C^2} {A B} = e^k = k_2[/tex]
where k and k_2 are constants. In this thread, Pervect works out the local proper acceleration of a static observer as
[tex]a = \frac {c^2 B'} {2 B \sqrt{A}}[/tex]
where B = f and A = g for the metric he uses (with an opposite sign convention for the metric itself). With the equation we just gained then, that gives
[tex]a = \frac {c^2 B'} {2 B \sqrt{A}} = \frac {c^2 \sqrt{k_2}} {2 \sqrt{B} r^2 C}[/tex]
Since we know that in the low Newtonian limit, B and C work toward unity and the proper acceleration works toward a = G M / r^2 = m c^2 / r^2 and k_2 is a constant for all r, then k_2 = 4 m^2 as we can see in the last part of the equation.
Okay, so let's explore the R_11 component. If we divide all of the terms on the right side by A B B', we gain
[tex]\frac{2 r C^2 B"} {B'} - \frac {r C^2 B'} {B} - \frac {r C^2 A'} {A} + \frac {8 B C C'} {B'} + \frac {4 r B C C"} {B'} - \frac{4 B C^2 A'} {A B'} - \frac {2 r B C A' C'} {A B'} - \frac{2 r B C'^2} {B'} = 0[/tex]
Now we are going to change the metric slightly to make the mathematics easier. We will say D = C r^2, so that the metric now becomes
[tex]c^2 d\tau^2 = c^2 B(r) dt^2 - A(r) dr^2 - D(r) d\phi^2[/tex]
The derivative and double derivative for C are now
[tex]C = D / r^2[/tex]
[tex]C' = (r D' - 2 D) / r^3[/tex]
[tex]C" = (r^2 D" - 4 r D' + 6 D) / r^4[/tex]
and plugging those into the terms for the r_11 component we just gained, it reduces to
simply
[tex]\frac {2 B" D^2} {B'} - \frac {B' D^2} {B} - \frac {A' D^2} {A} + \frac {4 B D D"} {B'} - \frac {2 B A' D D'} {A B'} - \frac {2 B D'^2} {B'} = 0[/tex]
and rearranging, we get
[tex]D^2 (\frac {A'} {A} + \frac {B'} {B} - \frac {2 B"} {B'}) = \frac {4 B D D"} {B'} - \frac {2 B D'^2} {B'} - \frac {2 B A' D D'} {A B'}[/tex]
Earlier we gained from the R_00 component, the relation
[tex]\frac {A'} {A} + \frac{B'} {B} - \frac {2 B"} {B'} = \frac {4} {r} + \frac {2 C'} {C}[/tex]
which now works out in terms of D to simply
[tex]\frac {A'} {A} + \frac{B'} {B} - \frac {2 B"} {B'} = \frac {2 D'} {D}[/tex]
So plugging that into what we gained from the R_11 component now gives
[tex]2 D' D = \frac {4 B D D"} {B'} - \frac {2 B D'^2} {B'} - \frac {2 B A' D D'} {A B'}[/tex]
and multiplying these terms with B' / (2 B D D') gives
[tex]\frac {A'} {A} + \frac {B'} {B} = \frac {2 D"} {D'} - \frac {D'} {D}[/tex]
Finding the indefinite integrals in the same way as we did before for R_00, we get
[tex]\frac {A B D} {D'^2} = k_3[/tex]
The relation we gained earlier for R_00 was
[tex]A B = \frac {B'^2 C^2 r^4} {4 m^2} = \frac {B'^2 D^2} {4 m^2}[/tex]
so plugging that into what we have now gives
[tex]\frac {B'^2 D^3} {4 m^2 D'^2} = k_3[/tex]
[tex]B' = \frac {2 m \sqrt{k_3} D'} {D^{(3/2)}}[/tex]
and the indefinite integral of that is
[tex]B = - \frac{4 m \sqrt{k_3}} {\sqrt{D}} + k_4[/tex]
and switching back to C for a moment,
[tex]B = - \frac{4 m \sqrt{k_3}} {\sqrt{C} r} + k_4[/tex]
The constant of motion for energy is
[tex]E = \frac {\sqrt{1 - v^2 / c^2}} {\sqrt{B}} [/tex]
where E = 1 for an object falling from rest at infinity, so we have
[tex]B = 1 - v^2 / c^2 = - \frac{4 m \sqrt{k_3}} {\sqrt{C} r} + k_4[/tex]
where v is now the escape velocity, which in the low Newtonian limit is just v^2 = 2 m c^2 / r and C works toward unity, so k_4 = 1 and k_3 = 1/4. We now have
[tex]B = 1 - \frac{2 m} {\sqrt{C} r} = 1 - \frac{2 m} {\sqrt{D}}[/tex]
as our general relation for B. The derivative of B is
[tex]B' = \frac{m D'} {D^{(3/2)}}[/tex]
and plugging these into the relation we gained from R_00 earlier, we now get
[tex]B'^2 = \frac{4 m^2 A B} {D^2} = \frac{m^2 D'^2} {D^3}[/tex]
[tex]4 m^2 A (1 - \frac{2 m} {\sqrt{D}}) = \frac {m^2 D'^2} {D}[/tex]
[tex]A = \frac {D'^2} {4 D (1 - \frac {2 m} {\sqrt{D}})}[/tex]
And plugging those back into the metric finally gives us
[tex]c^2 d\tau^2 = c^2 (1 - \frac{2 m} {\sqrt{D(r)}}) dt^2 - \frac {D'(r)^2} {4 D(r) (1 - \frac {2 m} {\sqrt{D(r)}})} dr^2 - D(r) d\phi^2[/tex]
The metric is now described solely in terms of the single function D. Once we make a coordinate choice for D in terms of r, we will automatically gain the entire metric. It is interesting that we only needed the R_00 and R_11 components of the Ricci tensor to solve for the metric. We could also rearrange using the couple of relations we found to describe a metric in terms of only the time dilation, for example, with
[tex]c^2 d\tau^2 = c^2 B(r) dt^2 - \frac {4 m^2 B'(r)^2 dr^2} {(1 - B(r))^4 B(r)} - \frac {4 m^2 d\phi^2 } {(1 - B(r))^2}[/tex]
If we wish to find a solution for an isotropic metric, where A = C or A = D / r^2, then we would have
[tex]A = \frac {D'^2} {4 D (1 - \frac {2 m} {\sqrt{D}})} = D / r^2[/tex]
[tex]r D' = 2 D \sqrt{1 - 2 m / \sqrt{D}}[/tex]
with a solution for D according to Wolfram of
[tex]D = \frac {(k_5 r + m)^4} {4 k_5^2 r^2}[/tex]
[tex]C = \frac{D} {r^2} = \frac {(k_5 r + m)^4} {4 k_5^2 r^4}[/tex]
[tex]C = \frac{(k_5 + m / r)^4} {4 k_5^2}[/tex]
where C works toward unity in the low Newtonian limit with large r and this relation would reduce to 1 = k_5^4 / (4 k_5^2), so our constant k_5 = 2, giving
[tex]C = \frac{(2 + m / r)^4} {16}[/tex]
[tex]C = (1 + m / (2 r))^4[/tex]
[tex]c^2 d\tau^2 = c^2 B(r) dt^2 - A(r) dr^2 - C(r) r^2 d\phi^2[/tex]
and solving only for a static, spherically symmetric vacuum spacetime, I want to reduce the number of coordinate functions A, B, and C from three to only one using the EFE's. We can then make a coordinate choice for that one function in terms of r and automatically produce the entire metric from our choice for that single function. The four components of the Ricci tensor for a vacuum spacetime are all zero and these are
[tex](-4 r A^2 B C) R_{00} = 2 r A B C B" - r A C B'^2 - r B C A' B' + 4 A B C B' + 2 r A B B' C'[/tex]
[tex](4 r A B^2 C^2) R_{11} = 2 r A B C^2 B" - r A C^2 B'^2 - r B C^2 A' B" + 8 A B^2 C C' + 4 r A B^2 C C" - 4 B^2 C^2 A' - 2 r B^2 C A' C' - 2 r A B^2 C'^2[/tex]
[tex](-4 A^2 B) R_{22} = - r^2 A B' C' - 2 r^2 A B C" - 8 r A B C' + r^2 B A' C' + 4 A^2 B + 2 r B C A' - 2 r A C B' - 4 A B C[/tex]
[tex]R_{33} = R_{22} (sin \phi)^2[/tex]
as graciously worked out for me by Markus Hanke in this thread. Okay, so we will start with the R_00 component. Dividing all of its terms by r A B C B" gives simply
[tex]\frac {2 B"} {B'} - \frac {B'} {B} - \frac {A'} {A} + \frac {4} {r} + \frac {2 C'} {C} = 0[/tex]
and finding the indefinite integrals for this gives
[tex]log(B'^2) - log(B) - log(A) + log(r^4) + log(C^2) = k[/tex]
[tex]log(\frac {B'^2 r^4 C^2} {A B}) = k[/tex]
[tex] \frac {B'^2 r^4 C^2} {A B} = e^k = k_2[/tex]
where k and k_2 are constants. In this thread, Pervect works out the local proper acceleration of a static observer as
[tex]a = \frac {c^2 B'} {2 B \sqrt{A}}[/tex]
where B = f and A = g for the metric he uses (with an opposite sign convention for the metric itself). With the equation we just gained then, that gives
[tex]a = \frac {c^2 B'} {2 B \sqrt{A}} = \frac {c^2 \sqrt{k_2}} {2 \sqrt{B} r^2 C}[/tex]
Since we know that in the low Newtonian limit, B and C work toward unity and the proper acceleration works toward a = G M / r^2 = m c^2 / r^2 and k_2 is a constant for all r, then k_2 = 4 m^2 as we can see in the last part of the equation.
Okay, so let's explore the R_11 component. If we divide all of the terms on the right side by A B B', we gain
[tex]\frac{2 r C^2 B"} {B'} - \frac {r C^2 B'} {B} - \frac {r C^2 A'} {A} + \frac {8 B C C'} {B'} + \frac {4 r B C C"} {B'} - \frac{4 B C^2 A'} {A B'} - \frac {2 r B C A' C'} {A B'} - \frac{2 r B C'^2} {B'} = 0[/tex]
Now we are going to change the metric slightly to make the mathematics easier. We will say D = C r^2, so that the metric now becomes
[tex]c^2 d\tau^2 = c^2 B(r) dt^2 - A(r) dr^2 - D(r) d\phi^2[/tex]
The derivative and double derivative for C are now
[tex]C = D / r^2[/tex]
[tex]C' = (r D' - 2 D) / r^3[/tex]
[tex]C" = (r^2 D" - 4 r D' + 6 D) / r^4[/tex]
and plugging those into the terms for the r_11 component we just gained, it reduces to
simply
[tex]\frac {2 B" D^2} {B'} - \frac {B' D^2} {B} - \frac {A' D^2} {A} + \frac {4 B D D"} {B'} - \frac {2 B A' D D'} {A B'} - \frac {2 B D'^2} {B'} = 0[/tex]
and rearranging, we get
[tex]D^2 (\frac {A'} {A} + \frac {B'} {B} - \frac {2 B"} {B'}) = \frac {4 B D D"} {B'} - \frac {2 B D'^2} {B'} - \frac {2 B A' D D'} {A B'}[/tex]
Earlier we gained from the R_00 component, the relation
[tex]\frac {A'} {A} + \frac{B'} {B} - \frac {2 B"} {B'} = \frac {4} {r} + \frac {2 C'} {C}[/tex]
which now works out in terms of D to simply
[tex]\frac {A'} {A} + \frac{B'} {B} - \frac {2 B"} {B'} = \frac {2 D'} {D}[/tex]
So plugging that into what we gained from the R_11 component now gives
[tex]2 D' D = \frac {4 B D D"} {B'} - \frac {2 B D'^2} {B'} - \frac {2 B A' D D'} {A B'}[/tex]
and multiplying these terms with B' / (2 B D D') gives
[tex]\frac {A'} {A} + \frac {B'} {B} = \frac {2 D"} {D'} - \frac {D'} {D}[/tex]
Finding the indefinite integrals in the same way as we did before for R_00, we get
[tex]\frac {A B D} {D'^2} = k_3[/tex]
The relation we gained earlier for R_00 was
[tex]A B = \frac {B'^2 C^2 r^4} {4 m^2} = \frac {B'^2 D^2} {4 m^2}[/tex]
so plugging that into what we have now gives
[tex]\frac {B'^2 D^3} {4 m^2 D'^2} = k_3[/tex]
[tex]B' = \frac {2 m \sqrt{k_3} D'} {D^{(3/2)}}[/tex]
and the indefinite integral of that is
[tex]B = - \frac{4 m \sqrt{k_3}} {\sqrt{D}} + k_4[/tex]
and switching back to C for a moment,
[tex]B = - \frac{4 m \sqrt{k_3}} {\sqrt{C} r} + k_4[/tex]
The constant of motion for energy is
[tex]E = \frac {\sqrt{1 - v^2 / c^2}} {\sqrt{B}} [/tex]
where E = 1 for an object falling from rest at infinity, so we have
[tex]B = 1 - v^2 / c^2 = - \frac{4 m \sqrt{k_3}} {\sqrt{C} r} + k_4[/tex]
where v is now the escape velocity, which in the low Newtonian limit is just v^2 = 2 m c^2 / r and C works toward unity, so k_4 = 1 and k_3 = 1/4. We now have
[tex]B = 1 - \frac{2 m} {\sqrt{C} r} = 1 - \frac{2 m} {\sqrt{D}}[/tex]
as our general relation for B. The derivative of B is
[tex]B' = \frac{m D'} {D^{(3/2)}}[/tex]
and plugging these into the relation we gained from R_00 earlier, we now get
[tex]B'^2 = \frac{4 m^2 A B} {D^2} = \frac{m^2 D'^2} {D^3}[/tex]
[tex]4 m^2 A (1 - \frac{2 m} {\sqrt{D}}) = \frac {m^2 D'^2} {D}[/tex]
[tex]A = \frac {D'^2} {4 D (1 - \frac {2 m} {\sqrt{D}})}[/tex]
And plugging those back into the metric finally gives us
[tex]c^2 d\tau^2 = c^2 (1 - \frac{2 m} {\sqrt{D(r)}}) dt^2 - \frac {D'(r)^2} {4 D(r) (1 - \frac {2 m} {\sqrt{D(r)}})} dr^2 - D(r) d\phi^2[/tex]
The metric is now described solely in terms of the single function D. Once we make a coordinate choice for D in terms of r, we will automatically gain the entire metric. It is interesting that we only needed the R_00 and R_11 components of the Ricci tensor to solve for the metric. We could also rearrange using the couple of relations we found to describe a metric in terms of only the time dilation, for example, with
[tex]c^2 d\tau^2 = c^2 B(r) dt^2 - \frac {4 m^2 B'(r)^2 dr^2} {(1 - B(r))^4 B(r)} - \frac {4 m^2 d\phi^2 } {(1 - B(r))^2}[/tex]
If we wish to find a solution for an isotropic metric, where A = C or A = D / r^2, then we would have
[tex]A = \frac {D'^2} {4 D (1 - \frac {2 m} {\sqrt{D}})} = D / r^2[/tex]
[tex]r D' = 2 D \sqrt{1 - 2 m / \sqrt{D}}[/tex]
with a solution for D according to Wolfram of
[tex]D = \frac {(k_5 r + m)^4} {4 k_5^2 r^2}[/tex]
[tex]C = \frac{D} {r^2} = \frac {(k_5 r + m)^4} {4 k_5^2 r^4}[/tex]
[tex]C = \frac{(k_5 + m / r)^4} {4 k_5^2}[/tex]
where C works toward unity in the low Newtonian limit with large r and this relation would reduce to 1 = k_5^4 / (4 k_5^2), so our constant k_5 = 2, giving
[tex]C = \frac{(2 + m / r)^4} {16}[/tex]
[tex]C = (1 + m / (2 r))^4[/tex]