Logic of GR as mathematically derived

[grav-universe]

Here's a rather simple one I have found so far.

z^2 = 1 - 2 m (L_t / r)

So if we make a coordinate choice for z or L_t, we can easily find the other. Then we can find L with one of the other relationships found before, such as

L = m (L_t / r)^2 / (dz / dr)

I still need the logical foundation for it, though.

 

[grav-universe]

Here's something interesting. We can re-arrange that last one to get

(L_t / r) = (1 - z^2) / (2 m)

and combine it with

L = m (L_t / r)^2 / (dz / dr)

to get

L = (1 - z^2)^2 / (4 m (dz / dr))

which gives us the relationship between L and z, but if we then combine that with

a' = - c^2 (dz / dr) L / z

from the OP, we find

a' = - c^2 (1 - z^2)^2 / (4 m z)

which gives us a direct relationship between the local acceleration and time dilation, independent of r or any other factors other than the mass of the gravitating body M, since m = G M / c^2. So if one only knows the numerical time dilation at some shell (regardless of the value of r for that shell) and the mass of the gravitating body, they can find the local acceleration at that shell by applying this formula which is coordinate independent. Of course that also works vice versely, one can find the time dilation according to whatever the local acceleration is, rather than work with gravitational potential for a specific coordinate dependent r or whatever, but finding the solution for z in terms of a' in that manner appears to be long and messy.

 

[PeterDonis]

So if one only knows the local time dilation and the mass of the gravitating body M, they can find the local acceleration, independent of r.

This doesn't look right. I haven't worked through the details of your derivation, but just looking at the standard formula for the proper acceleration of a static observer, I don't see how knowing the time dilation z and the mass M, but not r, let's you find the acceleration:

a = \frac{M}{r^2 \sqrt{1 - 2M / r}} = \frac{M}{r^2 z}

So if I know z and M, I still can't find a without knowing r as well.

 

[grav-universe]

This doesn't look right. I haven't worked through the details of your derivation, but just looking at the standard formula for the proper acceleration of a static observer, I don't see how knowing the time dilation z and the mass M, but not r, let's you find the acceleration:

a = \frac{M}{r^2 \sqrt{1 - 2M / r}} = \frac{M}{r^2 z}

So if I know z and M, I still can't find a without knowing r as well.

Right. With this relationship you couldn't, simply because it includes r, but work through any specific numerical examples you like for the relationship in the last post with SC, GUC, EIC, or any other of the same form and it should work out (I haven't yet but I'm confident because I can see how it relates and that it is invariant). The local acceleration appears to be directly tied to the time dilation only (and M).

(By the way, just a quick mention that that would actually need to be a' = - G M L_t^2 / (r^2 z) to be coordinate independent, but of course in SC, L_t = 1 so amounts to the same thing for that coordinate system only.)

 

[PeterDonis]

Right. With this relationship you couldn't, simply because it includes r, but work through any specific numerical examples you like for the relationship in the last post with SC, GUC, EIC, or any other of the same form and it should work out (I haven't yet but I'm confident because I can see how it relates and that it is invariant). The local acceleration appears to be directly tied to the time dilation only (and M).

This doesn't make sense to me. The formula I posted shows that if you only know z and M, not r, you can't know a. In other words, given a fixed z and M, there are multiple possible values of a. So I don't see how it's possible for a to only be tied to z and M.

 

[PeterDonis]

This doesn't make sense to me. The formula I posted shows that if you only know z and M, not r, you can't know a. In other words, given a fixed z and M, there are multiple possible values of a. So I don't see how it's possible for a to only be tied to z and M.

Never mind; I see now how to rewrite the formula. We take the formula for z:

z = \sqrt{1 - \frac{2M}{r}}

and rearrange to:

r = \frac{2M}{1 - z^2}

and then substitute into the formula for a to obtain:

a = \frac{M}{r^2 z} = \frac{M \left( 1 - z^2 \right)^2}{4 M^2 z} = \frac{\left( 1 - z^2 \right)^2}{4 M z}

which is what you wrote a few posts ago.

 

[grav-universe]

Well sure, you could do it like that too I suppose, but only if you like doing things the easy way.

 

[grav-universe]

Ah, okay. For the past week I have been trying to incorporate assumption C for the conservation of angular momentum m_p v'_t r', where m_p here is the mass of the particle, which is considered invariant so divided out for convenience, leaving the constant of motion P = v'_t r', which is measured the same locally at every shell depending upon how a particle is originally set in motion. Okay, so by reverse engineering the solution to the SC metric in another thread, I had gained P = v'_t r' = v'_t r / sqrt(1 - 2 m / r), which I took to mean r' = r / L, using the inferred radius by extending a local ruler at r all the way along r radially that is contracted radially by a factor of L = sqrt(1 - 2 m / r). One problem I have had with that, though, upon seeing that the quantities z, L / dr, and L_t / r are invariant for a particular shell (or coordinate independent), is that P is not invariant as it is currently expressed, although it should be.

I realize now that with the tangent motion of a particle, the local shell is not using some locally inferred radius r', but rather C' / (2 pi), where C' is the locally measured circumference of the shell at r. C' / (2 pi) ≠ r / L here of course, so with the distant observer inferring a circumference of the shell of C = 2 pi r, the local static observer with a tangent contraction of rulers of L_t, will physically measure C' = (2 pi r) / L_t by placing infinitesimal rulers end to end around the circumference, which is invariant. So to conserve the locally measured angular momentum of a particle, we would have

p'_angular = p'_t r' = [m_p v'_t / sqrt(1 - (v'/c)^2)] (C' / 2 pi) = constant

where from assumption B we gained sqrt(1 - (v'/c)^2) = z / K, so

= m_p v'_t K ((2 pi r / L_t) / 2 pi) / z

= m_p v'_t K (r / L_t) / z

and upon dividing out the invariant m_p and constant of motion K, we gain another constant of motion

P = v'_t r / (L_t z)

which still works out to P = v'_t r / sqrt(1 - 2 m / r) as found by reverse engineering SC, but is now invariant. This is the new corrected equation for assumption C. I will have to go back to a couple of other threads where I got this far and then got stuck at this point also.

 

[grav-universe]

Let me ask this. This Wiki link has the derivation for the Schwarzschild solution. For the 3 lines contained in the link under "Using the field equations to find A(r) and B(r)" and shown below, what is each saying physically?

\rm{4 \dot{A} B^2 - 2 r \ddot{B} AB + r \dot{A} \dot{B}B + r \dot{B} ^2 A=0}

\rm{r \dot{A}B + 2 A^2 B - 2AB - r \dot{B} A=0}

\rm{- 2 r \ddot{B} AB + r \dot{A} \dot{B}B + r \dot{B} ^2 A - 4\dot{B} AB=0}

Well, since L_t / r is an invariant, and those equations should represent invariants, I tried simply replacing r with r / L_t in the equations. They all work out to zero with Schwarzschild, as they should anyway since L_t is just 1, but they also all work out to zero with GUC, where L_t is non-unity. GUC is the same as SC but all of the shells are moved uniformly closer to the center while keeping the same coordinate distance between the shells and cutting out the volume inside the event horizon, so it has the same value for L at any r as SC, so the same z and L for any shell, and so the same A and B, only represented with the conversion for r1 instead of r. Apparently then, L_t / r is the correction factor for A and B when changing coordinates systems of the form r1 = r - n m from SC, simply sliding the shells uniformly inward or outward, where n has any numerical value. They didn't work out for EIC for some reason though, not sure why. I need those same equations but including C for tangent length.

 

[grav-universe]

Ahah! Yay. :) I can now see where one of the solutions to the Ricci tensors comes from. From the relationship we found before

m L_t^2 = (dz / dr) L r^2

which for convenience I will write dz / dr as just z' with second derivatives double primed.

We can re-arrange to gain

L_t^4 = z'^2 L^2 r^4 / m^2

The variables in the tensors are

A = 1 / L^2

B = - z^2

and B' = d(-z^2) = - 2 z z'

z' = - B' / (2 z)

z'^2 = B'^2 / (4 z^2) = - B'^2 / (4 B)

so we can rewrite the relationship once more to

L_t^4 = - B'^2 r^4 / (4 m^2 A B)

Finding the derivative for that using Wolfram, we get

d(L_t^4) = d[- B'(r)^2 r^4 / (4 m^2 A(r) B(r))] = r^3 B' (r B A' B' + A(r B'^2 - 2 B (r B" + 2 B')) / (4 m^2 A^2 B^2)

Now, if as a coordinate choice, we make L_t = 1, then its derivative is zero. So now we have

(0) (4 m^2 A^2 B^2) / (r^3 B') = 0 = r B A' B' + A(r B'^2 - 2 B (r B" + 2 B'))

r B A' B' + r A B'^2 - 2 r A B B" - 4 A B B' = 0

This is the same as the third Ricci tensor solution in the last post. :) Now I just need something similar to find the other two.

 

[grav-universe]

In terms of solving for C(r), for cases where C(r) is non-unity, with the metric of the form

ds^2 = A(r) dr^2 + B(r) c^2 dt^2 + C(r) dO^2 r^2 {as Wiki has it but with c^2 drawn out of B}

we would have

C(r) = 1 / L_t^2, giving

d[L_t^4] = d[1 / C^2] = - 2 C' / C^3

and

d[- B'^2 r^4 / (4 m^2 A B)] = r^3 B' (r B A' B' + A(r B'^2 - 2 B (r B" + 2 B'))) / (4 m^2 A^2 B^2) from before, so

- 2 C' / C^3 = r^3 B' (r B A' B' + A(r B'^2 - 2 B (r B" + 2 B'))) / (4 m^2 A^2 B^2)

- 8 m^2 A^2 B^2 C' / (r^3 C^3 B') = r B A' B' + A(r B'^2 - 2 B (r B" + 2 B'))

 

[grav-universe]

Actually, here's a simpler one involving C. From the last post, we had

1 / C^2 = -B'^2 r^4 / (4 m^2 A B)

1 = -B'^2 C^2 r^4 / (4 m^2 A B)

d(1) = 0 = d[-B'^2 C^2 r^4 / (4 m^2 A B)]

= r^3 C B'(r B C A' B' + A(r C B'^2 - 2 B(r C B" + r B' C' + 2 C B'))) / (4 m^2 A^2 B^2)

r B C A' B' + r A C B'^2 - 2 r A B C B" - 2 r A B B' C' - 4 A B C B' = 0