# Producing a metric with only a single coordinate function

1. Dec 4, 2015

### grav-universe

Given the metric

$$c^2 d\tau^2 = c^2 B(r) dt^2 - A(r) dr^2 - C(r) r^2 d\phi^2$$

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

$$(-4 r A^2 B C) R_{00} = 2 r A B C B" - r A C B'^2 - r B C A' B' + 4 A B C B' + 2 r A B B' C'$$

$$(4 r A B^2 C^2) R_{11} = 2 r A B C^2 B" - r A C^2 B'^2 - r B C^2 A' B" + 8 A B^2 C C' + 4 r A B^2 C C" - 4 B^2 C^2 A' - 2 r B^2 C A' C' - 2 r A B^2 C'^2$$

$$(-4 A^2 B) R_{22} = - r^2 A B' C' - 2 r^2 A B C" - 8 r A B C' + r^2 B A' C' + 4 A^2 B + 2 r B C A' - 2 r A C B' - 4 A B C$$

$$R_{33} = R_{22} (sin \phi)^2$$

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

$$\frac {2 B"} {B'} - \frac {B'} {B} - \frac {A'} {A} + \frac {4} {r} + \frac {2 C'} {C} = 0$$

and finding the indefinite integrals for this gives

$$log(B'^2) - log(B) - log(A) + log(r^4) + log(C^2) = k$$

$$log(\frac {B'^2 r^4 C^2} {A B}) = k$$

$$\frac {B'^2 r^4 C^2} {A B} = e^k = k_2$$

where k and k_2 are constants. In this thread, Pervect works out the local proper acceleration of a static observer as

$$a = \frac {c^2 B'} {2 B \sqrt{A}}$$

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

$$a = \frac {c^2 B'} {2 B \sqrt{A}} = \frac {c^2 \sqrt{k_2}} {2 \sqrt{B} r^2 C}$$

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

$$\frac{2 r C^2 B"} {B'} - \frac {r C^2 B'} {B} - \frac {r C^2 A'} {A} + \frac {8 B C C'} {B'} + \frac {4 r B C C"} {B'} - \frac{4 B C^2 A'} {A B'} - \frac {2 r B C A' C'} {A B'} - \frac{2 r B C'^2} {B'} = 0$$

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

$$c^2 d\tau^2 = c^2 B(r) dt^2 - A(r) dr^2 - D(r) d\phi^2$$

The derivative and double derivative for C are now

$$C = D / r^2$$

$$C' = (r D' - 2 D) / r^3$$

$$C" = (r^2 D" - 4 r D' + 6 D) / r^4$$

and plugging those into the terms for the r_11 component we just gained, it reduces to
simply

$$\frac {2 B" D^2} {B'} - \frac {B' D^2} {B} - \frac {A' D^2} {A} + \frac {4 B D D"} {B'} - \frac {2 B A' D D'} {A B'} - \frac {2 B D'^2} {B'} = 0$$

and rearranging, we get

$$D^2 (\frac {A'} {A} + \frac {B'} {B} - \frac {2 B"} {B'}) = \frac {4 B D D"} {B'} - \frac {2 B D'^2} {B'} - \frac {2 B A' D D'} {A B'}$$

Earlier we gained from the R_00 component, the relation

$$\frac {A'} {A} + \frac{B'} {B} - \frac {2 B"} {B'} = \frac {4} {r} + \frac {2 C'} {C}$$

which now works out in terms of D to simply

$$\frac {A'} {A} + \frac{B'} {B} - \frac {2 B"} {B'} = \frac {2 D'} {D}$$

So plugging that into what we gained from the R_11 component now gives

$$2 D' D = \frac {4 B D D"} {B'} - \frac {2 B D'^2} {B'} - \frac {2 B A' D D'} {A B'}$$

and multiplying these terms with B' / (2 B D D') gives

$$\frac {A'} {A} + \frac {B'} {B} = \frac {2 D"} {D'} - \frac {D'} {D}$$

Finding the indefinite integrals in the same way as we did before for R_00, we get

$$\frac {A B D} {D'^2} = k_3$$

The relation we gained earlier for R_00 was

$$A B = \frac {B'^2 C^2 r^4} {4 m^2} = \frac {B'^2 D^2} {4 m^2}$$

so plugging that into what we have now gives

$$\frac {B'^2 D^3} {4 m^2 D'^2} = k_3$$

$$B' = \frac {2 m \sqrt{k_3} D'} {D^{(3/2)}}$$

and the indefinite integral of that is

$$B = - \frac{4 m \sqrt{k_3}} {\sqrt{D}} + k_4$$

and switching back to C for a moment,

$$B = - \frac{4 m \sqrt{k_3}} {\sqrt{C} r} + k_4$$

The constant of motion for energy is

$$E = \frac {\sqrt{1 - v^2 / c^2}} {\sqrt{B}}$$

where E = 1 for an object falling from rest at infinity, so we have

$$B = 1 - v^2 / c^2 = - \frac{4 m \sqrt{k_3}} {\sqrt{C} r} + k_4$$

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

$$B = 1 - \frac{2 m} {\sqrt{C} r} = 1 - \frac{2 m} {\sqrt{D}}$$

as our general relation for B. The derivative of B is

$$B' = \frac{m D'} {D^{(3/2)}}$$

and plugging these into the relation we gained from R_00 earlier, we now get

$$B'^2 = \frac{4 m^2 A B} {D^2} = \frac{m^2 D'^2} {D^3}$$

$$4 m^2 A (1 - \frac{2 m} {\sqrt{D}}) = \frac {m^2 D'^2} {D}$$

$$A = \frac {D'^2} {4 D (1 - \frac {2 m} {\sqrt{D}})}$$

And plugging those back into the metric finally gives us

$$c^2 d\tau^2 = c^2 (1 - \frac{2 m} {\sqrt{D(r)}}) dt^2 - \frac {D'(r)^2} {4 D(r) (1 - \frac {2 m} {\sqrt{D(r)}})} dr^2 - D(r) d\phi^2$$
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

$$c^2 d\tau^2 = c^2 B(r) dt^2 - \frac {4 m^2 B'(r)^2 dr^2} {(1 - B(r))^4 B(r)} - \frac {4 m^2 d\phi^2 } {(1 - B(r))^2}$$

If we wish to find a solution for an isotropic metric, where A = C or A = D / r^2, then we would have

$$A = \frac {D'^2} {4 D (1 - \frac {2 m} {\sqrt{D}})} = D / r^2$$

$$r D' = 2 D \sqrt{1 - 2 m / \sqrt{D}}$$

with a solution for D according to Wolfram of

$$D = \frac {(k_5 r + m)^4} {4 k_5^2 r^2}$$

$$C = \frac{D} {r^2} = \frac {(k_5 r + m)^4} {4 k_5^2 r^4}$$

$$C = \frac{(k_5 + m / r)^4} {4 k_5^2}$$

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

$$C = \frac{(2 + m / r)^4} {16}$$

$$C = (1 + m / (2 r))^4$$

2. Dec 4, 2015

### Staff: Mentor

Strictly speaking, the EFE's use the Einstein tensor; but in the vacuum case the Einstein tensor being zero is equivalent to the Ricci tensor being zero, so it amounts to the same thing. The Einstein tensor components don't look any neater, except that $G_{22}$ and $G_{33}$ are exactly equal (the factor of $\sin^2 \theta$ difference goes away).

Just to sanity check, let's evaluate how this works out for the Schwarzschild case, where $D = r^2$ (you have already partially evaluated the isotropic case, where $D = r^2 \left( 1 + m / 2r \right)^4$--but it would be nice to confirm that your formula for the metric gives the right isotropic metric when the formula for $D$ is plugged in). The metric becomes

$$c^2 d\tau^2 = c^2 (1 - \frac{2 m} {\sqrt{r^2}}) dt^2 - \frac {(2r)^2} {4 r^2 (1 - \frac {2 m} {\sqrt{r^2}})} dr^2 - r^2 d\Omega^2$$

On evaluating the square roots and canceling, this gives the correct Schwarzschild vacuum metric. (Note, btw, that I used the more usual notation of $d\Omega^2$ to represent the angular part of the metric.)

Another note: the physical meaning of the function $D$ isn't immediately obvious. It would be interesting to try expressing everything in terms of $B$, since that has an obvious physical meaning--it's the "time dilation factor" for a static observer, or, in more coordinate-free language, it's the invariant squared norm of the timelike Killing vector field.

Now for the obvious next question:

How many independent functions do you need to describe a static, spherically symmetric spacetime that is not vacuum everywhere (i.e., the stress-energy tensor components are nonzero in at least some region)? Note that for this case, you will need the Einstein tensor components, since those are what are equal to the SET components (with a factor of $8 \pi$).

The non-vacuum case is also of interest because we know from Birkhoff's Theorem that there is only one spacetime geometry that is spherically symmetric and vacuum, namely, the Schwarzschild geometry. (Note that you don't have to assume staticity to prove the theorem; the fact that the spacetime is static can actually be derived from the assumptions of spherical symmetry and vacuum. That was the original point of Birkhoff's Theorem.) So the vacuum case you've analyzed is of limited range, since it only works for one particular solution to the Einstein Field Equation. The case of a static, spherically symmetric spacetime that is not vacuum is more general, since there are multiple possible solutions meeting those conditions.

Last edited: Dec 4, 2015
3. Dec 5, 2015

### grav-universe

Thanks PeterDonis. The metric in terms of the time dilation is also given. It is

$$c^2 d\tau^2 = c^2 B(r) dt^2 - \frac {4 m^2 B'(r)^2} {(1 - B(r))^4 B(r)} dr^2 - \frac{4 m^2} {(1 - B(r))^2} d\Omega^2$$

I would like to move forward to a more general solution, but I'm not sure what that would entail. I would have to see specifically how all of the functions in a metric are laid out like they were done for me with the components of the Ricci tensor.

4. Dec 5, 2015

### Staff: Mentor

You equate the components of the Einstein tensor with the components of the stress-energy tensor. The components of the Einstein tensor are expressions in A, B, C and their derivatives, just like the Ricci tensor components (but slightly different expressions), and the components of the stress-energy tensor are the energy density and the pressures in the radial and tangential directions. For the case you are looking at, the Einstein tensor (like the Ricci tensor) is diagonal, so you have three equations:

$$G^t{}_t = - \frac{4 r^2 A C C'' - r^2 A (C')^2 - 2 r^2 A' C C' + 12 r A C C' - 4 r A' C^2 + 4 A C^2 - 4 A^2 C}{4 r^2 A^2 C^2} = - 8 \pi \rho$$
$$G^r{}_r = - \frac{r^2 B (C')^2 + 2 r^2 B' C C' + 4 r B C C' + 4 r B' C^2 + 4 B C^2 - 4 A B C}{4 r^2 A B C^2} = 8 \pi p$$
$$G^\theta{}_\theta = G^\phi{}_\phi = - \frac{2 r A B^2 C C'' - r A B^2 (C')^2 + r A B B' C C' - r A' B^2 C C' + 4 A B^2 C C' + 2 r A B B'' C^2 - r A (B')^2 C^2 - r A' B B' C^2 + 2 A B B' C^2 - 2 A' B^2 C^2}{4 r A^2 B^2 C^2} = 8 \pi s$$

where $\rho$ is the energy density (the minus sign is there because we are using the mixed Einstein tensor, with one upper and one lower index; if both indexes were lower, there would be extra factors on the RHS of these equations that would make them messier), $p$ is the radial pressure, and $s$ is the tangential pressure (note that the pressure must be the same in both tangential directions, by spherical symmetry, and the last two diagonal components of the Einstein tensor are identical for the same reason). I used Maxima to compute the specific expressions for the Einstein tensor components that are given above. Note that $\rho$, $p$, and $s$ are all functions of $r$.

Note also that it will probably be simpler, rather than trying to solve the last of the three equations above, to use the fact that the covariant divergence of the stress-energy tensor is zero. The covariant divergence is a set of four equations $\nabla_a T^a{}_b = 0$; the $r$ component of this equation reads

$$\nabla_a T^a{}_r = \partial_a T^a{}_r + \Gamma^a{}_{ad} T^d{}_r - \Gamma^e{}_{ar} T^a{}_e = \partial_r T^r{}_r + \Gamma^t{}_{tr} \left( T^r{}_r - T^t{}_t \right) + \Gamma^\theta{}_{\theta r} \left( T^r{}_r - T^\theta{}_\theta \right) + \Gamma^\phi{}_{\phi r} \left( T^r{}_r - T^\phi{}_\phi \right) = 0$$

The $\Gamma$s are the Christoffel symbols, which can also be easily computed with Maxima, and filling in the results along with the stress-energy tensor components gives

$$\frac{dp}{dr} + \left( \rho + p \right) \frac{B'}{2B} + \frac{2}{r} \left( p - s \right) = 0$$

This, when it is rearranged to put everything but $dp / dr$ on the RHS, is a well-known equation called the Tolman-Oppenheimer-Volkoff equation (in a slightly more general form--the original TOV equation assumed isotropic pressure, so the $p - s$ term was zero). Combined with the first two of the EFE equations above, it gives a complete solution for the general static, spherically symmetric case.

Last edited: Dec 5, 2015
5. Dec 5, 2015

### grav-universe

That is so cool. Thank you very much for laying all of that out for me. Looks like I have a new project to begin working on. Awesome

6. Dec 8, 2015

### grav-universe

Well, from the first three sets you give, I have so far managed to come up with the relation

$$p' + (\rho + p) \frac{B'} {2 B} + (p - s) \frac{D'} {D} = 0$$

or

$$\frac{dp} {dr} + (\rho + p) \frac{B'} {2 B} + (\frac{2} {r} + \frac{C'} {C}) (p - s) = 0$$

so it looks like the coordinate choice C = 1 has already been made for what you have at the end of your post. I also found the relation

$$A = \frac{[(4 s + \rho - 3 p) (D' / D) - 4 p'] (D' / D)} {4 (\rho + p) (1 / (8 \pi D) - p)}$$

so if we make a coordinate choice for D and we know the pressures, we could potentially find A. The problem here is that for a vacuum solution, the pressures all becomes zero and divide by each other so we would end up with A = 0 / 0 which could be anything, and I want to directly see the result for a vacuum, so I'll keep working on it.

For an isotropic pressure where p = s, we would have

$$A = \frac{(D' / D)^2 / 4 - p' (D' / D)/ (\rho + p)} {1 / (8 \pi D) - p}$$

and since all three pressures are zero for a vacuum solution, that could be considered isotropic, which perhaps brings us a little closer.

Last edited: Dec 8, 2015
7. Dec 8, 2015

### Staff: Mentor

No; that equation does not assume any coordinate choice other than the general form of the metric we have been using; i.e., it does not assume any specific form for A, B, C, or D.

8. Dec 8, 2015

### grav-universe

I just double checked by placing each of the values you give for $\rho$, p, s, and finding for p' into the last relation in the post directly and I still get

$$p' + (\rho + p) \frac{B'} {2 B} + \frac{2} {r} (p - s) = -\frac{C'} {C} (p - s)$$

so if that last relation really does equal zero without a coordinate choice, then the pressure must always be isotropic, which would be an interesting discovery.

9. Dec 8, 2015

### Staff: Mentor

I don't understand; where is the $C' / C$ term coming from on the RHS? The LHS is just the last equation I wrote down in post #4, with nothing substituted at all; the RHS should be zero.

10. Dec 8, 2015

### Staff: Mentor

Perhaps I should amplify somewhat on what is necessary for a solution, given the equations I posted in post #4. These are three equations, so if we consider the functions $\rho$, $p$, and $s$ to be known functions of $r$, we can use those three equations to solve for $A$, $B$, and $C$ (or $D$). But that's not really what we want, because in general (and in fact in the usual case), we do not know $\rho$, $p$, or $s$ in advance; they also have to be solved for. So we need more equations, and we may also need more assumptions/choices to narrow things down.

One obvious equation we can add is an equation of state for the matter: this is an equation relating $p$ and $\rho$. (In principle, it could be two equations, relating both $p$ and $s$ to $\rho$, but in every treatment I've seen, the pressure is assumed to be isotropic, which amounts to eliminating one variable, $s$, by assumption, and adding one equation to the set of equations to be solved.) It is less obvious, but can be shown (IIRC), that we can also always eliminate $A$ or $C$ by a proper choice of coordinates--in other words, these two functions are not actually independent. (In the vacuum case, this is obvious because the $G^t{}_t$ equation only contains $A$ and $C$; it doesn't contain $B$. But, IIRC, it turns out to be possible even in the general static, spherically symmetric case.)

The above leaves four equations (three EFE, or two EFE and the covariant divergence equation, plus the equation of state) and four unknowns ($A$ or $C$, $B$, $\rho$, and $p$), so it should in principle be solvable. But, as I think I said before, in the general case no closed-form solution is known; the equations have to be integrated numerically. The only case for which I know there is a closed-form solution is the (highly unrealistic) case of constant density (and isotropic pressure).

11. Dec 13, 2015

### grav-universe

Well, here's something interesting so far. For the non-vacuum solution, we could use still use the same form of the functions in the RHS of the components of the Ricci tensor and just consider that the LHS of those components is now non-zero, while still denoting them $R_{00}$, $R_{11}$, and $R_{22}$. We find that the relations to the pressures in the non-vacuum solution then become simply

$$2 R_{00} = (8 \pi) (\rho + p + 2 s) B \text{ giving } B = \frac {2 R_{00}} {(8 \pi) (\rho + p + 2 s)}$$

$$2 R_{11} = (8 \pi) (\rho + p - 2 s) A \text { giving } A = \frac {2 R_{11}} {(8 \pi) (\rho + p - 2 s)}$$

$$2 R_{22} = (8 \pi) (\rho - p) D \text { giving } D = \frac {2 R_{22}} {(8 \pi) (\rho - p)}$$

Placing these values for the functions into the metric now gives us

$$c^2 d\tau^2 = c^2 \frac {2 R_{00}} {(8 \pi) (\rho + p + 2 s)} dt^2 - \frac {2 R_{11}} {(8 \pi) (\rho + p - 2 s)} dr^2 - \frac {2 R_{22}} {(8 \pi) (\rho - p)} d\Omega^2$$

Last edited: Dec 13, 2015
12. Dec 13, 2015

### Staff: Mentor

Nonzero, but not the same as the RHS of the Einstein Field Equation components--i.e., not the simple expressions $8 \pi \rho$, $8 \pi p$, and $8 \pi s$. That's why I wrote down the Einstein tensor components for the non-vacuum case. So I don't see how you can arrive at the equations you wrote here.

13. Dec 13, 2015

### grav-universe

Perhaps it would help for visualization purposes if I lay these out similar to the way I have been using them. If we convert all of the C's to D's, it reduces the number of terms considerably as many of them cancel and the r's fall out completely. So we would have

$$C = \frac{D}{r^2}$$

$$C' = \frac{D'}{r^2} - \frac{2 D}{r^3}$$

$$C" = \frac{D"}{r^2} - \frac{4 D'}{r^3} + \frac{6 D}{r^4}$$

Plugging those directly into each of the components of the Ricci tensor and EFE's and cancelling out terms, we gain

$$\frac{4 A}{B} R_{00} = \frac{B'^2}{B^2} + \frac{A' B'}{AB} - \frac{2 B"}{B} - \frac{2 B' D'}{B D}$$

$$4 R_{11} = \frac{2 B"}{B} - \frac{B'^2}{B^2} - \frac{A' B'}{A B} + \frac{4 D"}{D} - \frac{2 A' D'}{A D} - \frac{2 D'^2}{D^2}$$

$$\frac{4 A}{D} R_{22} = \frac{B' D'}{B D} + \frac{2 D"}{D} - \frac{A' D'}{A D} - \frac{4 A}{D}$$

$$4 A (8 \pi \rho) = \frac{4 D"}{D} - \frac{D'^2}{D^2} - \frac{2 A' D'}{A D} - \frac{4 A}{D}$$

$$4 A (8 \pi p) = - \frac{D'^2}{D^2} - \frac{2 B' D'}{B D} + \frac{4 A}{D}$$

$$4 A (8 \pi s) = - \frac{2 D"}{D} + \frac{D'^2}{D^2} - \frac{B' D'}{B D} + \frac{A' D'}{A D} - \frac{2 B"}{B} + \frac{B'^2}{B^2} + \frac{A' B'}{A B}$$

Now we can see the relationships between them much more easily. For instance, if we subtract the terms on the RHS of the $p$ row from the terms on the RHS of the $\rho$ row, we gain twice the same values on the RHS of the $R_{22}$ row, and so forth.

14. Dec 14, 2015

### Staff: Mentor

None of these look right, because, once again, you are using the Ricci tensor components instead of the Einstein tensor components. For the non-vacuum case, you can't use the Ricci tensor components, because they are not equal to $8 \pi \rho$, $8 \pi p$, or $8 \pi s$; the Einstein tensor components are equal to those quantities. You can only get away with using the Ricci tensor components in the vacuum case.

15. Dec 14, 2015

### grav-universe

I thought you might say something like that. I'm just saying that it's interesting that the way the functions are laid out in the RHS of the components of the Ricci tensor, although the results on the LHS would now be non-zero for the non-vacuum case, they work out with those simple relations I gave as compared to the way the same functions are laid out in the RHS of the EFE's. I'm curious, though, as to how the components of the Ricci tensor would be laid out for the non-vacuum case. Is it still applicable at all?

Are you saying here that there would be more factors than what you have given for those equations, or a bunch of "partial" factors that when put together work out to exactly the same thing as what you gave? If there are extra factors than what you already gave in the equations, then those extra factors must sum to zero, and that would tell us a lot, so would be very significant, or even with completely different factors in the equations that amount to the same thing, just as long as they don't reduce to precisely the same factors that you already gave.

16. Dec 14, 2015

### Staff: Mentor

The components of the Ricci tensor are the same; the problem is that you don't know what to set them equal to. In the vacuum case, you can set them equal to zero because the Ricci tensor must be zero if the Einstein tensor is zero (which it is in vacuum by the EFE). But if the Einstein tensor is not zero, there is no general formula that tells you what the Ricci tensor components are equal to; the only formula we have is the EFE, which is an equation in the Einstein tensor components, not the Ricci tensor components.

Yes.

17. Dec 14, 2015

### Staff: Mentor

No, they don't, because those relations have $\rho$, $p$, and $s$ in them, and you don't have any equations relating the Ricci tensor components to $\rho$, $p$, and $s$. You only have equations (the EFE components) relating the Einstein tensor components to $\rho$, $p$, and $s$.

18. Dec 14, 2015

### grav-universe

Oh, okay good. Then the equations I gave still apply, but they don't tell us anything new, right, just a curiosity.

Oh wow. Then those are a must have. Those extra factors would be of utmost significance.

They do once we relate them to the same functions involved in each of the components of the Ricci tensor using the EFE's. For instance, we can see from the EFE's that

$$2A(8 \pi) (\rho - p) = \frac{B' D'}{B D} + \frac{2 D"}{D} - \frac{A' D'}{A D} - \frac{4 A}{D}$$

and that is identical to how the functions are laid out for $(4A / D) R_{22}$, so that is how the pressures are related for the $R_{22}$ component of the Ricci tensor (along with the left over function (4A / D) tagged to it), but I agree that doesn't tell us anything new if that is what you are saying since we had to use the pressures according to the EFE's to find it in the first place, so we might as well just use the EFE's altogether. The Ricci tensor is no longer required as it doesn't tell us anything at all about the pressures as you said when using it alone, nor does it add anything to what can already be gained from the EFE's alone.

Last edited: Dec 14, 2015
19. Dec 14, 2015

### Staff: Mentor

The equations describing the components of the Ricci tensor are still the same, yes. But anything you derived by setting those components equal to zero no longer applies if the spacetime is not vacuum. As far as I can tell, that's everything you've done with the Ricci tensor components except write them down.

No, they're not. They don't change the physics at all; they just clutter up the equations to no purpose. The EFE component equations I gave you are the simplest ones possible for the metric we are discussing.

20. Dec 14, 2015

### grav-universe

They wouldn't change the physics, right, which means that those extra factors must sum up to zero. It would give me a relation between the functions, then, extra equations between the functions themselves that equal zero, which I could then use to solve much more easily for the metric in terms of the pressures involved and may even give some direct relation between the pressures themselves. In any case it would add more information. Could you please work those out for me?

Last edited: Dec 14, 2015