# B Deriving GR without EFE's. Could it be this simple?

1. Nov 30, 2015

### grav-universe

For three years now i have been attempting to find some fundamental properties of gravity from which the three relativistic variables of GR, time dilation and the radial and tangent length contractions, could be determined without incorporating the EFE's. There is no rule or reason that i know of that says the EFE's absolutely must be used in order to determine GR, so I set out to find another, perhaps simpler method. I think I may have finally found it. So simple in the end it appears it was right under my nose the whole time.

$$c^2 dτ^2 = z^2 c^2 dt^2 - dr^2 / L^2 - dθ^2 r^2 / L_t^2$$

where z is the time dilation, L is the radial length contraction, and L_t is the length contraction in the tangent direction. One fundamental relationship

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

was determined in this thread. m here is just G M / c^2, where M is the central mass/energy and G is the gravitational constant. The derivatives of this equation give us the R_00 component of the Ricci tensor, but this is not enough to solve for all three variables. We can make a coordinate choice for one of them and find the resulting relation between the other two for that particular coordinate system, but we will need another equation to solve for each of the last two individually, and here it is at last I believe.

In this thread, pervect provides an equation for the velocity of a circular orbit as measured by a local observer, regardless of the coordinate system used, as

$$v_t^2 / c^2 = h f' / ( f h')$$

I have some reservations about how this was derived but I have seen similar derivations so we will accept it for the purposes of this thread. In terms of the variables in the metric used here, that works out to

$$v_t^2 / c^2 = z' r / (z (1 - L_t' r / L_t))$$

Okay, so for an object freefalling from rest at infinity, as demonstrated in the first thread, we have

$$z / sqrt(1 - v^2 / c^2) = K$$

where K is a constant of motion. Freefalling from rest with v=0 where z = 1 at r = infinity in the low Newtonian limit, this gives K = 1 for all r when the object passes. In the low Newtonian limit, we would have an escape velocity of v^2 = 2 m c^2 / r at r = infinity where all relativistic factors work toward unity, but we don't know how those factors figure in for finite r so we will combine this with an unknown function g(r) to cover the relativistic effects, giving

$$v^2 = 2 m g c^2 / r$$

as measured by a local observer. It can be demonstrated that regardless of whatever the relativistic equation for v^2 turns out to actually be, there will be some function for g that compensates. Okay, so with K = 1 for the constant of motion for escape velocity, combining the last two equations gives

$$z = sqrt(1 - 2 m g / r)$$

Now let's try something with that. We are allowed to make one coordinate choice, so we will say g = L_t^n, where n is numerical and we can choose what we want n to be. Solving for r, then, we get

$$r = [2 m (L_t / r)^n / (1 - z^2)]^{1 / (1 - n)}$$

where L_t / r and z so therefore 1 - z^2 are invariants and 2 m is constant. This means that regardless of the coordinate system used, these values always remain the same for a particular spherical shell, so if we know what these values are for that shell, we can then apply them to this equation and our chosen value for n will tell us at what r that shell is placed within our coordinate system. We can choose any value for n that we wish for our coordinate choice except one, n = 1. The reason is that it would cancel r from the left side of the equation and leave just

$$r^0 = 1 = 2 m (L_t / r) / (1 - z^2)$$

The resulting equation is composed entirely of invariants in this case so it is either true or it is not, independent of the coordinte system applied, and we cannot say yet which is the case. We can however, come as extremely close to n = 1 as we wish for our coordinate choice, so let's see where that leads us.

Let's start with pervect's equation for circular orbit. From z = sqrt(1 - 2 m g / r), we gain for its derivative,

$$z' = m (g - g' r) / (r^2 z)$$

giving

$$v_t^2 / c^2 = m (g - g' r) / (z^2 r (1 - L_t' r / L_t))$$

Dividing that by the escape velocity we have

$$v_t^2 / v^2 = (1 - g' r / g) / (2 z^2 (1 - L_t' r / L_t))$$

With g = L_t^n, whereas its derivative is g' = n L_t^(n - 1) L_t', we gain

$$2 z^2 v_t^2 / v^2 = (1 - n L_t' r / L_t) / (1 - L_t' r / L_t)$$

$$2 z^2 v_t^2 / v^2 = 1 + (1 - n) (L_t' r / L_t) / (1 - L_t' r / L_t)$$

$$2 z^2 v_t^2 / v^2 = 1 + (1 - n) / (L_t / (L_t' r) - 1)$$

Now we know that in the low Newtonian limit with large r, L_t approaches closer and closer to unity, so as this occurs the difference d L_t per dr, or L_t' becomes smaller and smaller, making the denominator L_t / (L_t' r) - 1 very large with greater r. This may be seen more clearly perhaps when L_t is expressed in polynomial form in terms of m and r. This can be done with any of the unknown variables, providing all of the possible terms involved. So we have

$$L_t = 1 - a m / r - b m^2 / r^2 - c m^3 / r^3 - ...$$

giving for its derivative,

$$L_t' = a m / r^2 + 2 b m^2 / r^3 + 3 c m^3 / r^4 + ...$$

where a, b, c, etc., are unknown coefficients that are constant for all r and depend upon the coordinate system that has been chosen. We now gain

$$2 z^2 v_t^2 / v^2 = 1 + (1 - n) / ((r / m)(1 - 2 a m / r - 3 b m^2 / r^2 - 4 c m^3 / r^4 - ...) / (a + 2 b m / r + 3 c m^2 / r^2 - ...))$$

One can see here that the denominator will become larger with greater r. Since m and all the coefficients a, b, c, etc. remain constant for all r, then with very large r this works toward

$$2 z^2 v_t^2 / v^2 = 1 + (1 - n) m a / r$$

So regardless of the value of our coefficients, even if there existed an r for which it is small, since m and 'a' remain constant for all r, we can always find some greater r where the denominator r / (m a) becomes large, and larger for all r thereafter. We can also make n anything we want as our coordinate choice excepting unity, so if we make it arbitrarily close to unity, infinitesimally close even, then with an extremely small numerator and a large denominator, the far right side of our equation becomes arbitrarily close to zero. We can make it so infinitesimally small in fact that it really won't even figure into the equation at all anymore and we gain simply

$$2 z^2 v_t^2 / v^2 = 1$$

as a general relation. Now here's the kicker. The left side of our equation is composed entirely of invariants. z is invariant, v_t is the velocity for a circular orbit as measured locally, and v is the escape velocity as measured locally. This means that if these invariants can be made equivalent to unity for one coordinate system, then they must be equivalent to unity in all coordinate systems, so

$$2 z^2 v_t^2 / v^2 = (1 - g' r / g) / (1 - L_t' r / L_t) = 1$$

for all coordinate systems. This occurs with n = 1 and g = L_t, so this relationship is precisely true for at least all large r, and if it is true for all large r then there is no reason to presume that it suddenly becomes untrue at some lesser r so we may conclude with reasonable certainty that it is precisely true for all r that lie within the external scope of our metric. So now looking back at our time dilation, we now have

$$z = sqrt(1 - 2 m L_t / r)$$

as a fundamental relationship for all coordinate systems. Together with our earlier relationship

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

we can now start over with a different coordinate system and solve for all three variables. If for instance we now make L_t = 1 for our coordinate choice, then we gain

$$z = sqrt(1 - 2 m (1) / r)$$

with its derivative

$$z' = m / (r^2 sqrt(1 - 2 m / r))$$

and plugging that and L_t = 1 into our original relationship, we get

$$m (1)^2 / r^2 = (m / (r^2 sqrt(1 - 2 m / r)) L$$

$$L = sqrt(1 - 2 m / r)$$

and so gain the Schwarzschild coordinates as one possible example. Similarly we may gain solutions for any external coordinate system assuming static spacetime and incorporating spherical shells. So how does this look? Does it appear feasible to everybody?

2. Nov 30, 2015

### Staff: Mentor

We've discussed this before. You are not "deriving GR". You are only "deriving" a different way of looking at a particular kind of spacetime--basically a static, spherically symmetric spacetime. GR has a much wider range of application, which your "derivation" does not cover at all.

Also, you seem to be under the impression that what you call the "three relativistic variables of GR" are all that needs to be "derived" in order to "derive GR" without using the EFE. That is not correct. The EFE, even for the restricted case of a static, spherically symmetric spacetime, tells us more than just the spacetime geometry (which is what you are basically groping towards with your "three relativistic variables"): it tells us how that spacetime geometry is related to the matter and energy present. Your "derivation" does not cover that at all. So even in the restricted case you are working with, you are not "deriving" everything that GR and the EFE contain.

Finally, two of your "three relativistic variables of GR", what you call "radial and tangent length contraction", are not even invariants; they are coordinate-dependent. Only what you call "time dilation" has an invariant meaning (and only in a restricted set of spacetimes--a somewhat wider set than the ones you consider, since it includes any stationary spacetime, but still restricted compared to all the cases that GR covers). Since all of the physics in any spacetime is contained in invariants, two of your three variables do not have a direct physical meaning. So they're not good things to focus on anyway.

(And, in any case, all of this is bordering on personal theory, which is off limits on PF.)

In summary, this is not a good thread topic, so this thread is closed.