# Thread: EFE's now fully solved for the internal metric

1. ## EFE's now fully solved for the internal metric

I started working on this four years ago and have had this solution worked out for a couple of years now and am surprised I have allowed that much time to pass. I took some time to delve into it further to see what extra insights might be had and to attempt to reduce it further or to find a way to integrate it more directly as well as experimenting with different coordinate choices to see where they might lead but I always ended back at the original solution with the single best coordinate choice to be made that allows us to ingrate and solve for all of the variables. As such, the original solution is still the simplest I have managed to come up with so I have decided to present it finally as is. I have taken a couple of days from work to do this and there is a lot to cover and it may be somewhat hurried so please bear with me. I will attempt to present it as concisely and clearly as I can.

Given the metric

we will find solutions for the variables A, B, and D for the internal metric of a static symmetrical body. These solutions will work for all nonrelativistic bodies. That is, all bodies other than black holes, so bodies that do not contain an event horizon, whereas the surface of the body lies above where the event horizon would otherwise be, normal stars and planets for instance. These equations could hypothetically be applied to a black hole as well, since the solutions are taken directly from EFE's, but they are found with the idea that an observer can remain static at some r within a body and make measurements, all the way to the center, which is not feasible within an event horizon. In this thread, peterdonis graciously gave me EFE's for the energy density and radial and tangent pressures, respectively, reworked in post #13 as

I will simplify this further using

to gain

The only assumption we shall make is that the radial and tangent pressures are isotropic so that p = s. Setting the last two equations equal, they can then be rewritten as

We shall now make our coordinate choice. We want to choose one that places the center of our coordinate chart at the center of the body where r = 0 in both cases and is as close to the Newtonian solution as possible and such that b and d can be integrated easily to determine B and D. I find that

works quite well, where k is a constant. In fact it is the only coordinate choice i have found so far that will allow us to integrate properly. There are a couple of other slight variations I have found that work also but they tend to be much more complicated. It appears that we are very limited on our coodinate choice in that case although I am still experimenting with other possibilities. So now we have

We shall now solve for the variables with what we have gained so far.

which gives us

\int
Last edited by grav; 2019-Oct-11 at 07:46 PM.

2. We now have all of the variables solved in terms of the single remaining unknown variable H. Okay, so here's the thing. I originally spent months trying to come up with another assumption besides isotropic pressures to fill in the blank for the final variable before I realized it wasn't possible. If it were, then the same solution would exist for all of the variables within all bodies. That would mean that all bodies are essentially the same, only varying primarily in size. However, while the external metric depends upon only the total mass/energy of a body, the internal metric would vary depending upon the internal mass/energy distribution of the body. So what we require then is not an assumption at all, but a given internal condition of the particular body we are solving for.

This is true also in Newtonian physics. Ideally for Newtonian physics we would want to know the function for the internal mass distribution as our given condition and the total mass M to know where the limit lies, or the surface radius R if the mass distribution is in terms of r. From this we can determine the acceleration of gravity and orbital speed at any radius r, for example. However, if we don't know the mass distribution we could just as easily start with another condition such as the function for the acceleration at any r or the orbital speed to find the other two. This is true also with GR. So H will be our initial given condition which varies from body to body and whose function will depend upon the makeup of the particular body being considered. So far only one solution to the internal metric has been found with the given condition being that of a body with constant energy density. Here we will attempt to find for all other possible internal conditions in terms of H and the total mass of the body M, or E / c^2 in terms of the total mass/energy.

In this thread by pervect, we find that

where vo is the locally measured orbital speed, although found there in terms of the variables f and h. This is taken directly from the metric and is true internally as well as externally. From this we can determine that

So the physical value for H is such that if we were to dig a ring cavity around the circumference at some radius r within the body and let an object freely orbit through the ring, v_o would be the locally measured orbital speed that would allow the object to do so. If we did this for all r, we could find the function for H in terms of r for that particular body. This function will be the condition we are given about the body.

We can see that the value for H at any r in a nonrelativistic body is extremely small. Even if the body were very close to relativistic at some r, whereby vo becomes very close to c, the largest H becomes is 1/2. But in most cases this value is very small indeed. To see just how small, let's take another equation from pervect's thread,

where acc is the locally measured proper acceleration of a static observer at some radius r, alos taken directly from the metric so is true in all cases, internally as well as externally. Since for a nonrelativistic body, A ≈ 1, we have

At the surface of the earth, for example, acc is about 10 m/sec^2, giving b ≈ 2 * 10^(-16). At the surface of the sun, it is about 5.5 * 10^(-14). Since D ≈ r^2 for a nonrelativistic body also, whereby d ≈ 2 / r, with the radius of the earth being 6.4 * 10^6 m, we would gain a value for H of H ≈ 6.4 * 10^(-10) for the earth and H ≈ 2 * 10^(-5) for the sun at their surfaces. These values will become important again later. The values may rise some internally to the body if most of the mass/energy is concentrated near the center but otherwise will eventually begin to fall until the proper static acceleration at the center becomes zero since there is no direction there to accelerate, whereby the value of b becomes zero at the center also and therefore so does H.

Now we have one further task before we can begin to solve for each of the variables. We must determine the values of each of the constants k, kB, and kD. We can do this by finding their values at the limit at the surface R of the body. It is important to note that so far these same solutions work equally well for the external metric as for the internal metric. We have assumed only isotropic pressure and made a coordinate choice, both of which can be made the same for the internal and external metric alike, so we have therefore so far solved for the entire metric in general. The only difference between them is that the function for H will be different internally than externally, whereby the functions and values for A, B, and D will be different. This may be true for ranges of r within the body as well where different materials that make up the body produce different functions for the mass/energy distribution and therefore different functions of H. This cannot be helped as it is determined by the specific makeup of the body.

Nonetheless, at the places where different ranges meet, the upper limit of the lower range and the lower limit of the upper range, and at the limit between the internal and external metric at the surface R, the actual values of A, B, and D must be identical. We can see this by taking the values for A, B, and D at R + dr, an infinitesimal distance above the surface in the vacuum. Then as we pass slightly below the surface to R - dr (one might visualize digging a small hole there so that there is no obstruction), the values of A, B, and D must remain relatively the same, even though the functions for the internal and external metric are different, so that these values are identical at R. The reason is that the values of A, B, and D will vary with the mass/energy distribution of the body, but an infinitesimal distance below R, that mass/energy distribution will have varied only infinitesimally as well, passing only through an infinitesimally thin spherical shell of the mass/energy of the body, so that the values of the variables will change only infinitesimally, allowing for a smooth transition between the external and internal metric as well as different ranges within the body.

Okay, so from the first post of the first thread that was linked to here, we can find general relationships between the variables for the external vacuum metric such as

where m = G M / c^2 and M is the total mass of the body, or m = G E / c^4 using the total energy. Using these relations along with our coordinate choice, we can gain

and

and setting these two equal and solving for k r^2, we find

so since r = R at the surface and using the value for H at the surface once we know the function for H in terms of r, we get k = - HR^2 / R^2. If we don't know the value for R but rather only the total mass/energy of the body, we can instead use k r^2 = - HR (r / R)^2 = - H^2 x^2, where x is now a fraction of r to R with a value of zero to one and we can substitute this into our equations in place of k r^2 if we wish and find H in terms of x for the internal metric. Moving on, from what we gained for B earlier, we can now find

The integral here is still in terms of H because we would first have to plug in the function for H in terms of r in general, then integrate, and finally we can plug in r = R or x = 1 to the result of the integration. Last we have

Now for the integrations themselves. Unfortunately the equations

and

do not integrate directly which forces us to use a Taylor series for -[log(1 - 2 H^2 - k r^2)]', which becomes

At first I wondered if using a converging series meant that the equations would not be considered exact solutions. But I realize now that they still are. Many functions such as log, sin, cos, etc. are really just condensed notation for an infinite series such as this. It is only when we are trying to find the actual numbers for the values of A, B, and D at some r that the series must be used and it converges extremely rapidly. We can find the numerical solutions to each of the variables to as many digits as we need by running through additional terms. Even equations that contain numbers such as √2 and π can only be brought out to a given number of digits practically because these are irrational numbers which contain an infinite number of digits, but that does not mean that the equations themselves that contain these are not exact solutions.

To see how fast this series converges, consider the values of H we came up with at the surface of the earth and sun earlier. Since H ~ 6.4 * 10^(-10) for the earth, then since the integral contains a series for H^2, and k r^2 is also on the order of H^2, we should gain almost 19 digits of accuracy for each term we include when finding the actual numbers. For the sun we would gain almost 10 digits of accuracy for each term included. A computer could run through hundreds of terms within seconds and gain thousands of digits of accuracy if desired. Most calculations that we would actually use only require a few digits of accuracy, however, in which case a single term is enough. The value of H^2 falls at the rate of (r/R)^4 in general as we shall see shortly, so integrates even faster in that case at smaller r, although depending upon the precise function for H. If most of the mass/energy is concentrated toward the center for example, it might continue to rise some before falling to zero at the center of the body as stated earlier.
Last edited by grav; 2019-Oct-11 at 07:49 PM.

3. Now let's find the actual functions we shall use. As far as I know, any function can be stated in terms of r as

where the C's are coefficient constants to each term of r. As for the function of b, we know that the proper acceleration falls to zero at the center of the body so b does as well, as also stated earlier. Therefore its function cannot contain any negative exponents of r and C_0 is zero as well, whereby b is at least r1 to first term so that b=0 at r=0. For D, we know that D = r2 to first term in the Newtonian approximation which we are trying to come close to for nonrelativistic bodies, so d is r-1 to first term. Since H = b / (b + d), that means H is r2 to first term. So let's say we have a body with a function for H that happens to contain three terms.

Since H is unitless, here I have included terms of the constant R also to each of the coefficients so that the units work out correctly so that the C's become unitless numerical coefficients. We can rewrite this to

where x = r/R and has a value between zero and one as discussed earlier. First of all, we can now find HR easily with this when considering that x = r/R = 1 at the limit of R, so that HR simply becomes HR = C2 + C3 + C4, merely the sum of the numerical coefficients. Also, since k R^2 = -HR^2 as determined earlier, then k r^2 at some other radius r becomes k r^2 = - HR^2 (r/R)^2 = - HR^2 x^2 as determined earlier in the thread. Applying these to our integrations, then, for the integration involved in B, for example, we have

[LaTeX ERROR: Image too big 1001x31, max 650x600]

[LaTeX ERROR: Image too big 1333x19, max 650x600]

I will stop there as the calculations are already beginning to become clustered, and the actual integration would become much more so, but the intent was to show that the integration is possible and one can tell by examining what we have gained so far that that is indeed the case. A computer can handle these calculations with ease and I would highly recommend it. We have now obtained the equations for A, B, and D for the internal metric and demonstrated how they can be integrated to gain the numerical values for each. In conclusion, we will now move on to one additional extra bonus, that of obtaining the function for the internal mass/energy distribution itself. I would rather have started with that as our given condition and found the values for the variables using it, but as of yet I have not found a way to do so, so we will have to find it after the fact, but at least it can be gained. One finds that the EFE for the energy density found at the beginning of this thread can be rearranged to

I find it is much easier to demonstrate this by working it backward to obtain the original form of the EFE than the reverse. So we have

and finding the derivatives gives us

then multiplying times (8 A) / (d D3/2) gives

which matches the original EFE. Now let's consider what a distant observer would infer for the volume from r=0 to r. The distant observer says that the volume is simply Vd = (4 π / 3) r^3. We will apply the Schwarzschild coordinate system for this where D = r^2, so that we now have Vd = (4 π / 3) D3/2. Since D = C r^2 as shown in the other thread, where C is the contraction of surface area in the tangent direction, then since C = 1 in this case, both the distant observer and a local observer at r will agree upon the area of a spherical shell at r. However, √A is the contraction in the radial direction at r, so an infinitesimal distance dr as inferred by a distant observer will be locally measured as √A dr by the local observer. Let's consider what the distant observer will infer for the volume of an infinitesimally thin shell at r. That becomes Vd' = dVd / dr = (2 π) d D3/2. Now let's look at what the local observer measures. The tangent dimensions for the volume are the same, so the volume only differs along the radial dimension which is measured √A times greater, so the local observer measures the volume of the infinitesimally thin spherical shell to be √A times larger. However, this is measured over a radial distance that is √A larger as well, so we have V' = (dV √A) / (√A dr) = dV / dr = Vd', so this value is the same for both the distant observer and the local observer in the Schwarzschild coordinate system. Multiplying this by the energy density, we get

So this becomes the amount of energy of the body that lies below r when integrated from r=0 to r. However, the value for the energy density in terms of rho that is used in the metric is actually multiplied by the constant -G / c^4 so that its units become those on the order of 1 / r^2, so we must do the same to E here to make it proportional, giving

When placed back into the equation, we now have

which matches what we find with the external Schwarzschild metric using the total mass/energy m, which would be all of the mass/energy that lies below a radius r that is external to the body. However, this is a general equation, simply a rearrangement of the first EFE, so it applies to the metric internally as well as externally. The only difference is that while externally we would use the constant m for the total mass/energy, internally we would use m(r), which is only that amount of mass/energy that lies below r internal to the body. In this equation, D and 1 and the entire right side are invariants, making m(r) invariant as well. So as an invariant, although we used the Schwarzschild metric to find it, if it is invariant in one coordinate system, then it is invariant in all coordinate systems, making this equation true for all coordinate systems in general. So we shall now apply it to the coordinate system we have chosen to find the function for m(r).

Last edited by grav; 2019-Oct-11 at 07:53 PM.

4. What are EFE's? All I get from a Google search are the names of a Spanish news agency and and NFL football player. If it is an acronym for something in physics, a statement of the form "E(whatever) F(whatever) E(whatever), or EFE for short" at the beginning of your presentation would have been helpful.

5. Originally Posted by Hornblower
What are EFE's? All I get from a Google search are the names of a Spanish news agency and and NFL football player. If it is an acronym for something in physics, a statement of the form "E(whatever) F(whatever) E(whatever), or EFE for short" at the beginning of your presentation would have been helpful.
Oh geesh, I didn't think of that. EFE's stands for Einstein's field equations which apply to general relativity and gravity.

6. Didn't Karl Schwarzschild do this in 1919?

7. Originally Posted by Strange
Didn't Karl Schwarzschild do this in 1919?
That was a solution for the external vacuum metric where the given condition there is zero energy density and zero pressures. So far only one solution exists for the internal metric of a body, that of a constant energy density within the body, which is highly unlikely to occur in nature. We require a given condition such as this that describes the internal mass/energy distribution of the body, but the possibilities are endless, and I wanted to be able to solve for any possible internal mass/energy distribution. With a given function that describes the interior of a particular body in some way, a coordinate choice, and an assumption such as isotropic pressure, we could hypothetically solve for all of the unknown variables in the metric completely and gain numerical values for any properties we wish to know at some radius r within the body. The only thing stopping us is the math. It seems no matter what we do, when solving for the variables using the EFE's, we are always left with one final variable to solve for that won't integrate in order to gain a solution. Usually this is simply because we are left with a polynomial in the denominator that won't cooperate in the integrating process, which can become very frustrating when solving for all variables minus one and knowing the full solution with the final unknown variable is right in front of us if we could just get it to integrate properly.

However, I have figured out a trick to get the math to work. Usually our instinct might be to make a coordinate choice first that looks simple and then attempt to work through the math using that coordinate choice and cross our fingers in the meantime that it works out all of the way through the calculations, but it seems we always run into something non-integratable toward the end. If we hold off on making a coordinate choice, however, and work through the math in a certain way, then when we find solutions toward the end that must be integrated but contain a polynomial in the denominator, we can set that polynomial as our coordinate choice, something simple like k r^2, where k is a constant and r^2 is on the same order as the denominator. Then if everything in the numerator is in terms of r, we can now easily integrate and find full solutions to each of the variables. This in a nutshell is what I have done.

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The math is admittedly over my head, but this does sound interesting and I do love that unlike most ATM threads, in this one you have plenty of math!

Can you give us an example of something that was not solvable before, but you can now solve using your new technique? This may help me follow along a bit better with what you are talking about here.

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Originally Posted by grav
I started working on this four years ago and have had this solution worked out for a couple of years now and am surprised I have allowed that much time to pass....
Some posts with lots of math but there seems to be no ATM idea, grav
The mainstream has been solving the Einstein Field Equations since they were written. The solution for the outside of a static symmetrical body is the Schwarzschild metric found in 1915 and the interior Schwarzschild metric was found a year later.

What I read is math that seems to result in the wrong answer which suggests a mistake in your mainstream calculation. On the other hand, if you are using different conditions for the interior of a body in the EFE then you are doing mainstream physics. It may be right. It may be wrong. But it seems not ATM.
Last edited by Reality Check; 2019-Oct-14 at 11:42 PM.

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Originally Posted by grav
Now let's find the actual functions we shall use. As far as I know, any function can be stated in terms of r as...
That "any" seems wrong, grav.
For example there are valid mathematical functions such as the Dirac delta function and the Heaviside step function. We could have a body with constant density (step function) or even a point mass (delta function). Analytic functions are less common than non-analytic functions. Also see Can any continuous function be represented as an infinite polynomial? (answer = No).

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Originally Posted by grav
That was a solution for the external vacuum metric where the given condition there is zero energy density and zero pressures. So far only one solution exists for the internal metric of a body, that of a constant energy density within the body, which is highly unlikely to occur in nature.
This looks like the interior Schwarzschild metric which "is an exact solution for the gravitational field in the interior of a non-rotating spherical body which consists of an incompressible fluid (implying that density is constant throughout the body) and has zero pressure at the surface.". It is density that is constant, not energy density. This almost happens in nature, e.g. it is approximately the Earth (rotates relatively slowly and density does not vary greatly). The approximation will be worse for a star. This is the first of several static spherically symmetric perfect fluid (the Wikipedia article needs some attention though) solutions, some of which are simple models for stars such as white dwarfs and neutron stars.
Last edited by Reality Check; 2019-Oct-15 at 12:15 AM.

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Originally Posted by grav
We now have all of the variables solved in terms of the single remaining unknown variable H. ...
Let us assume that you have a correct "H". The issue with this post is that you are deriving the metric internal to a body. So introducing a "locally measured orbital speed" into H automatically invalidates the derivation.
The thread you link to is Some general formulae for circular orbits in symmetric spacetimes. The thread is about the Schwarzschild metric which is valid outside of a body.

13. Originally Posted by Dave241
The math is admittedly over my head, but this does sound interesting and I do love that unlike most ATM threads, in this one you have plenty of math!

Can you give us an example of something that was not solvable before, but you can now solve using your new technique? This may help me follow along a bit better with what you are talking about here.
The entire interior metric was not previously solvable except in the special case of constant interior mass/energy density. With a variable form for H, we can now solve for any form of H in terms of r.

14. Originally Posted by grav
The entire interior metric was not previously solvable except in the special case of constant interior mass/energy density. With a variable form for H, we can now solve for any form of H in terms of r.
So why are you posting this here, rather than submitting it to a journal for review and possible publication?

It is math, so there would seem to be a pretty clear yes/no decision that could be made.

15. Originally Posted by Reality Check
Some posts with lots of math but there seems to be no ATM idea, grav
The mainstream has been solving the Einstein Field Equations since they were written. The solution for the outside of a static symmetrical body is the Schwarzschild metric found in 1915 and the interior Schwarzschild metric was found a year later.

What I read is math that seems to result in the wrong answer which suggests a mistake in your mainstream calculation. On the other hand, if you are using different conditions for the interior of a body in the EFE then you are doing mainstream physics. It may be right. It may be wrong. But it seems not ATM.
Right, it is not really ATM, but I wasn't sure where else to place it. It is a new solution that must be demonstrated so I figured the proving grounds.

There exists only one solution to the interior metric so far, that for a constant interior mass/energy density, while the forms that H can take with this new solution are endless.

You are right about the math. The math itself works fine but the coordinate choice I made using k r^2 had an unexpected problem when I wrote a program for it. I should have done that before posting this thread. I am now using one of my slight variations for the coordinate choice that works perfectly. I am currently working on finding forms of H that provide closed form solutions to the variables. So far I have only found a very limited form to which I am attempting to apply more freedom. I will make a very brief post showing the new coordinate system, the closed form solutions, and demonstrate how any forms of H in terms of r can be found in general within the next few days.

16. Originally Posted by Reality Check
Let us assume that you have a correct "H". The issue with this post is that you are deriving the metric internal to a body. So introducing a "locally measured orbital speed" into H automatically invalidates the derivation.
The thread you link to is Some general formulae for circular orbits in symmetric spacetimes. The thread is about the Schwarzschild metric which is valid outside of a body.
The locally measured orbital speed is the physical interpretation of H, as if a ring cavity were dug around the circumference at some radius r within the body and an object were allowed to orbit freely through the ring at that internal radius.

The solutions in that thread are just general formulas taken directly from the metric without assuming anything about pressures or densities or whether the metric is internal or external.

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Originally Posted by grav
There exists only one solution to the interior metric so far, that for a constant interior mass/energy density, while the forms that H can take with this new solution are endless.
That is a "only one solution" misconception that I addressed in a later post, grav. This is the first of several static spherically symmetric perfect fluid (the Wikipedia article needs some attention though) solutions.

18. Originally Posted by Strange
So why are you posting this here, rather than submitting it to a journal for review and possible publication?

It is math, so there would seem to be a pretty clear yes/no decision that could be made.
Yes, I suppose that would be the next step. I submitted it here for peer review first however I have already obtained some good suggestions like finding forms of H that provide closed form solutions.

19. Actually I will go ahead and make a very brief post now showing the new coordinate system. Later I will post again with the closed form solutions.

20. Originally Posted by Reality Check
Those are for a perfect fluid solution only, which means a solution for constant mass/energy density. That single solution may take many forms depending upon the coordinate choice, but it is still the same solution. My solutions apply to all possible mass/energy density distributions within the body.

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Originally Posted by grav
The locally measured orbital speed is the physical interpretation of H, ....
Your source says that is wrong. You cannot use the orbit speed for an object outside of a sphere inside a sphere: Some general formulae for circular orbits in symmetric spacetimes.

A formal question to illustrate my point. Consider Newtonian gravitation for a spherical mass of constant density. Is the orbital speed for an object inside a sphere the same as outside the sphere the same, grav? The answer is obviously no (Orbit Speed Inside and Outside a Mass Distribution).
IF01: Why should the GR orbital speed law for an object inside a sphere be the same as outside the sphere, grav?
The answer should be that it would only be the case for a very specific density distribution. But you are trying to get a solution for any density distribution. Thus plugging in an outside the body orbital sped into a general inside the body inside H cannot be right.
Last edited by Reality Check; 2019-Oct-16 at 08:44 PM.

22. Originally Posted by grav
Originally Posted by Reality Check
Some posts with lots of math but there seems to be no ATM idea, grav [...]
Right, it is not really ATM, but I wasn't sure where else to place it.
Closed pending moderator discussion.

ETA:

Reopened and moved to S&T. If any ATM ideas do come up, please don't share them here.

Please remember, the ATM forum is not intended as a developmental environment...nor for ATM Q&A...nor for checking one's math.
Last edited by PetersCreek; 2019-Oct-16 at 04:49 PM.

23. Here is a very brief post that summarizes the topic using the new coordinate system.

Given the metric

we will find solutions for the variables A, B, and D (where D = C r^2) for the internal metric of a static spherically symmetrical body. The EFE's are

I will simplify these further using

to gain

Assuming isotropic pressures, p = s, so setting them equal gives

We shall now make our coordinate choice.

whereby solving for F gives

We shall now solve for the variables.

which gives us

Each of the unknowns in the metric now have fully integratable solutions and we can find the numerical values for each at some radius r, given the function H in terms of r that is specific to the body we are considering.

24. Originally Posted by Reality Check
Your source says that is wrong. You cannot use the orbit speed for an object outside of a sphere inside a sphere: Some general formulae for circular orbits in symmetric spacetimes.

A formal question to illustrate my point. Consider Newtonian gravitation for a spherical mass of constant density. Is the orbital speed for an object inside a sphere the same as outside the sphere the same, grav? The answer is obviously no (Orbit Speed Inside and Outside a Mass Distribution).
IF01: Why should the GR orbital speed law for an object inside a sphere be the same as outside the sphere, grav?
The answer should be that it would only be the case for a very specific density distribution. But you are trying to get a solution for any density distribution. Thus plugging in an outside the body orbital sped into a general inside the body inside H cannot be right.
Yes, that is correct. The same formula applies internally and well as externally. The same is true for Newtonian gravity. So for a constant density, we would have

The acceleration acc becomes greater external to the body with smaller r and v_o larger as well. Below the surface, however, it becomes smaller proportionally to r. So at r = R / 10, say, the acceleration is 1/10 as great as at the surface and r is 1/10 as great also, so v_0 is 1/10 as great than at the surface. The same is true with GR. b is approximately 2 acc / c^2, and d is approximately 2 / r, so

The precise equation being

This is a relation taken directly from the metric without any reference to the EFE's. The EFE's are not necessary to find it. So that relation is true regardless of whether the energy density and pressures are zero as with the external vacuum solution or if those quantities are nonzero as with the internal metric, since both use the same form of the metric.

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Originally Posted by grav
Yes, that is correct. The same formula applies internally and well as externally. The same is true for Newtonian gravity.
Read my source where the formula is different inside and outside of the mass distribution: Orbit Speed Inside and Outside a Mass Distribution, grav.
Inside the sphere of uniform density: orbital speed v goes as r. Outside the sphere of uniform density: orbital speed v goes as the square root of 1/r. The graph make this explicit - v rises in a line from r = 0 to the surface of the sphere and decreases outside of the sphere.

The formula being different for inside and outside of the sphere in Newtonian gravity is an argument for the formula being different for inside and outside of the sphere in GR.

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Originally Posted by grav
Given the metric

Can you give a source for this metric, grav?

It is usual to get a metric from the Einstein field equations by applying conditions to solve the 10 independent equations so it is slightly confusing to leap straight to a metric with no derivation. Wikipedia have a Deriving the Schwarzschild solution article (external solution). All of the conditions are needed to solve the EFE. For example throw away that it is a vacuum solution and the EFE cannot be solved. The Tolman–Oppenheimer–Volkoff equation for a body with any density profile solves the issue of not having enough conditions by applying an equation of state for the object in question (white dwarf or neutron star).

27. Originally Posted by Reality Check
Read my source where the formula is different inside and outside of the mass distribution: Orbit Speed Inside and Outside a Mass Distribution, grav.
Inside the sphere of uniform density: orbital speed v goes as r. Outside the sphere of uniform density: orbital speed v goes as the square root of 1/r. The graph make this explicit - v rises in a line from r = 0 to the surface of the sphere and decreases outside of the sphere.

The formula being different for inside and outside of the sphere in Newtonian gravity is an argument for the formula being different for inside and outside of the sphere in GR.
The formulas are the same but the results of the formulas are different for certain quantities as r varies. Varying results for inside and outside of the body do not invalidate the formulas themselves. In the Newtonian version for constant density, both the acceleration and orbital speed inside the body start at zero at the center and rise as a straight line to reach a maximum at the surface, then begin to fall off again, right. The same is approximately true of GR, since it reduces to Newtonian at first order. For example, in the Newtonian version, the formula for the acceleration is

This same general formula applies at all r, inside the sphere and outside. But as it is dependent upon M(r), the mass that lies below r, since M(r) is constant outside the sphere, just M, then G M is constant and the acceleration drops with 1 / r^2. Inside the sphere, however, we would have M(r) / r^3 = constant density, that means that M(r) is variable internally in proportion to r^3, so the acceleration rises in a straight line in proportion to (r^3) / r^2 = r. For vo^2 = acc r, another general Newtonian formula that applies inside and outside of the body, that gives vo in proportion to sqrt((1/r^2) r) = sqrt(1/r) outside the body and sqrt((r) r) = r as a straight line inside the body, as you said.

28. Originally Posted by Reality Check
Can you give a source for this metric, grav?

It is usual to get a metric from the Einstein field equations by applying conditions to solve the 10 independent equations so it is slightly confusing to leap straight to a metric with no derivation. Wikipedia have a Deriving the Schwarzschild solution article (external solution). All of the conditions are needed to solve the EFE. For example throw away that it is a vacuum solution and the EFE cannot be solved.
That wiki page for deriving the Schwarzschild solution also uses the same A and B in the metric. Those are the unknowns they are solving for as I am here, but they have already made the coordinate choice D = r^2, so that part is already in the metric.

The Tolman–Oppenheimer–Volkoff equation for a body with any density profile solves the issue of not having enough conditions by applying an equation of state for the object in question (white dwarf or neutron star).
Right, we need another condition that describes the internal conditions of the specific body we are considering in some way. I do that here with the function H.

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Grav, I don't understand. Why wouldn't have been solved before.
Last edited by Copernicus; 2019-Oct-17 at 03:58 PM.

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Originally Posted by grav
The formulas are the same ....
You are repeating your error, grav. The equation for orbital speed inside a uniform density sphere is different from the equation for orbital speed outside a uniform density sphere (inside = r, outside = square root of 1/r). You even state that later in this post!

That can be written as a single equation with 2 domains (r <= R and r > R where R is the radius of the sphere). But that is not what you do. You take the equation for outside where there is vacuum and apply it inside where there is mass.
Last edited by Reality Check; 2019-Oct-17 at 07:58 PM.

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