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

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Originally Posted by grav
...

This same general formula applies at all r, inside the sphere and outside.
The same error as for orbital speed?
Newton's law of gravitation F = GMm/r^2 that gives that acceleration is for the force outside of a sphere. The equations for force, acceleration and orbital speed are different inside the sphere. A textbook example which Newton worked out (Shell theorem) is that the force on an object anywhere inside a hollow sphere is zero.
ETA: The other part of the shell theorem is that the field from a sphere is the same as the field from a point mass. If we replace a sphere with its equivalent point mass then r can go to zero and the equation is the same as r -> 0.

ETA: Also M(r) should be zero outside of a sphere so what looks like your version of Newton's law of gravitation has no gravity between objects! You need a textbook source for that equation that states what it is for and what M(r) is. My guess is sum of the mass inside a radius.

That is an obvious test case for your calculations. Plug in the valid density of a thin shell, e.g. a delta function at R the radius of the sphere. Birkhoff's theorem (the GR equivalent of the shell theorem) says "a spherically symmetric thin shell, the interior solution must be given by the Minkowski metric; in other words, the gravitational field must vanish inside a spherically symmetric shell". You need to get the same result.
Last edited by Reality Check; 2019-Oct-17 at 08:50 PM.

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Originally Posted by grav
That wiki page for deriving the Schwarzschild solution also uses the same A and B ....
Deriving the Schwarzschild solution article (external solution) uses A(r) and B(r) equations and then gets a valid metric for the outside of a sphere. These A(r) and B(r) equations do not appear out of nowhere as your A and B "variables" seem to. The symmetries of the problem are used to diagonalise the metric which sets most components to zero. Spherical symmetry gives that 2 of remaining four components of the metric only depends on r, thus A(r) and B(r). The A(r) + B(r) metric is used to get the Christoffel symbols in terms of A and B. The EFE are then solved. The Schwarzschild metric is the end result of the derivation.

The above is what is missing from your calculations. You start with stating a A+B metric when you need to to derive that metric as in the A(r) + B(r) metric in the article.

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Originally Posted by grav
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.
Your H is invalid (uses outside the sphere orbital velocity), grav.

The internal conditions of the specific body are an equation of state which is thermodynamics. In the case of the already fully solved since 1939 EFE for the internal metric (Tolman–Oppenheimer–Volkoff equation), the equation state relates pressure and density.

4. Originally Posted by Copernicus
Grav, I don't understand. Why wouldn't have been solved before.
It is because the integrations that are involved in solving for all three variables in the metric are extremely difficult, almost impossible in fact. It seems that no matter what coordinate choice we make, or what our given condition is, we can solve for all the variables minus one. It can get very frustrating to come so close to solving each time and then thwarted on the final variable due to impossible integrations. There are algorithms that can solve for the integrations that come closer and closer to the correct result so they can be solved numerically that way, but I wanted something more direct. Usually what thwarts the integrations is a polynomial in the denominator, so rather than make a coordinate choice right away and cross my fingers I found I needed to work through the equations without making a coordinate choice in such a way that we can just multiply the givens to each of the variables so that they only appear in the numerator, leaving just a single common polynomial in the denominator. Then we make a coordinate choice which eliminates that denominator altogether and everything becomes fully integratable. The only problem now is that the coordinate choices I have found so far simply replace that polynomial with one that is more easily managable, one that will allow us to transfer it to the numerator in the form of a Taylor series, but one that is very simple and converges extremely quickly.So quickly in fact that in general one only needs the first term of the series (which is 1 actually) to numerically solve the metric for bodies like the earth and sun to many digits.

5. Originally Posted by Reality Check
Your H is invalid (uses outside the sphere orbital velocity), grav.
It is perfectly valid. The equation itself applies internally just as well as externally, anywhere that the metric applies which is everywhere in space-time, in the same way that the other equations I posted here for Newton do, but they will have different functions that depend upon the local density and pressure. So since the functions will vary for density and pressure internally from that externally, they would have to be used as different ranges, yes, the internal and external each with their own individual set of functions, but those functions still apply to the same equations however.

The internal conditions of the specific body are an equation of state which is thermodynamics. In the case of the already fully solved since 1939 EFE for the internal metric (Tolman–Oppenheimer–Volkoff equation), the equation state relates pressure and density.
That is not a solution for the variables in the metric. They have made the coordinate choice D = r^2 and replaces A with another unknown variable m(r) that would have to be solved for or given. Then they solve for b (where b = B' / B) in the second EFE for p, and set it equal to the isotropic equation that relates p and rho and the derivative of p since that contains a single b itself. I have tried that many times myself since it is always very tempting to eliminate the single b in both equations by solving for b and setting them equal. The problem is that if we are eliminating b to gain that equation, we cannot then solve for b (or its integral B) which is the only unknown left in the metric, so they are making no attempt to actually solve the metric itself, only showing the equation they came up with using the EFE's.

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Hi Grav, What are the benefits to knowing the interior metric?

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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. 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
grav, in your equation what is r?

8. Originally Posted by Copernicus
Hi Grav, What are the benefits to knowing the interior metric?
Well to be frank, I'm not sure specifically, but I bet there are lots of uses. I mostly just enjoy working through the math trying to find a way to solve for something that isn't previously known

grav, in your equation what is r?
That is the radius r where the unknown variables will be found for a spherical shell at that radius. For the internal metric, r lies somewhere between r=0 at the center of the body and r = R where R is the surface of the body.

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Lets say that you were modeling the internal metric that was not static, in fact moving at relativistic velocity, would the mass distribution just be modeled my the lorentz mass equivalent. What I mean is could it still be modeled as static if the lorentz mass was used at each point.
Last edited by Copernicus; 2019-Oct-20 at 03:03 PM.

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Originally Posted by grav
It is because the integrations that are involved in solving for all three variables in the metric are extremely difficult, almost impossible in fact...
That isn't correct, grav. As you have been told several times, the "now fully solved" part of the is wrong because the EFE's have been fully solved since 1939 by the Tolman–Oppenheimer–Volkoff equation. The main issue in a solution for the EFE is simply that there are 10 differential equation to solve. That needs conditions to be imposed so that the independence of the equations are removed. For a static spherical solution for the inside of a body, there are not enough conditions to solve the EFE. What is left over is an unsolved differential equation. But we can solve that differential equation by adding an equation of state relating the pressure and density inside the sphere. That is the solution that has been around for 80 years!

It is mathematically impossible to solve a set of differential equations just by expanding 1 as a polynomial. If there are unknown functions, e.g. an A(r) or B(r), you just get polynomial involving those unknown functions.

It is textbook mathematics that not every function can be expanded as a polynomial. Thus even if the EFE are expanded as polynomials, this is not a general solution.

Your H is invalid (uses outside the sphere orbital velocity).

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The orbital velocity outside of a sphere is not the same as inside

Originally Posted by grav
It is perfectly valid. The equation itself applies internally just as well as externally....
Once more: 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).
This is textbook physics that has been known for over 300 years for Newtonian gravitation. GR reduces to Newtonian gravitation for low gravity and so gives the same result in that limit.

What makes this even less valid is that you are looking for a non-rotating solution. There cannot be any "orbits" inside the sphere! The Sun without its various rotations has plasma going in all directions. The Earth without its rotation has magna moving outward and inward and liquid rock rotating. There are bodies that are just rock or ices, e.g. small moons, asteroids, probably Ceres.

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Originally Posted by grav
That is not a solution for the variables in the metric. ....
When supplemented with an equation of state, F(ρ, P) = 0, which relates density to pressure, the Tolman–Oppenheimer–Volkoff equation completely determines the structure of a spherically symmetric body of isotropic material in equilibrium.
It is a solution for the functions in the EFE. That solution is a metric. To solve the EFE, we take the geometry of the situation and use that to give the general form of the metric. There will be unknown functions, e.g. A(r). Mathematics (zero out non-diagonal elements of the proposed metric, calculate the Christoffel symbols, etc.) and physics (the equation of state ) are applied to derive these unknown functions. All of the functions in that equation are known. Thus it is a solution for all of the "variables in the metric". I gave you an example of how solving the EFE gives a metric. Deriving the Schwarzschild solution (external solution).

What the Tolman–Oppenheimer–Volkoff equation shows is that it is impossible to solve the EFE for this case using only mathematics. If we do valid mathematics as Tolman, et. all did then there remains an unknown function v(r).

P.S. On Massive Neutron Cores by J. R. Oppenheimer and G. M. Volkoff Phys. Rev. 55, 374 – Published 15 February 1939. PDF.
It has been suggested that, when the pressure within stellar matter becomes high enough, a new phase consisting of neutrons will be formed. In this paper we study the gravitational equilibrium of masses of neutrons, using the equation of state for a cold Fermi gas, and general relativity.
Equation 1 is the known result for the most general static line element (metric) with spherical symmetry.
Equation 2 is a constraint on the energy momentum tensor (no traverse stresses and no mass motion).
Equation 3 is the resulting 3 EFE differential equations. These have an unknown function that is determined by the equation of state.

You have an unknown function. Expanding that unknown function as a polynomial gives an infinite number of unknown coefficients.
Last edited by Reality Check; 2019-Oct-20 at 08:44 PM.

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Originally Posted by Reality Check
Once more: 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).
This is textbook physics that has been known for over 300 years for Newtonian gravitation. GR reduces to Newtonian gravitation for low gravity and so gives the same result in that limit.

What makes this even less valid is that you are looking for a non-rotating solution. There cannot be any "orbits" inside the sphere! The Sun without its various rotations has plasma going in all directions. The Earth without its rotation has magna moving outward and inward and liquid rock rotating. There are bodies that are just rock or ices, e.g. small moons, asteroids, probably Ceres.
I see that the orbital speed would be different because the mass at distance greater than r would also affect the orbit. I don't see how the magma or plasma have an effect, because grav is just talking about static bodies.

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Originally Posted by Copernicus
I see that the orbital speed would be different because the mass at distance greater than r would also affect the orbit. I don't see how the magma or plasma have an effect, because grav is just talking about static bodies.
That is actually my point- a static (e.g. non-rotating) body has no orbits inside it so there are 2 errors
1. Using an orbital speed for outside of a sphere inside the sphere when they are different.
2. Using an orbital speed at all.

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Originally Posted by Reality Check
That is actually my point- a static (e.g. non-rotating) body has no orbits inside it so there are 2 errors
1. Using an orbital speed for outside of a sphere inside the sphere when they are different.
2. Using an orbital speed at all.
I was agreeing with your orbit position.

16. Originally Posted by Copernicus
Lets say that you were modeling the internal metric that was not static, in fact moving at relativistic velocity, would the mass distribution just be modeled my the lorentz mass equivalent. What I mean is could it still be modeled as static if the lorentz mass was used at each point.
I haven't tried to model a rotating body yet. I suppose that would be the next step. Something along the lines of the Kerr metric, which i haven't studied. Baby steps

17. [QUOTE=Reality Check;2495472]That isn't correct, grav. As you have been told several times, the "now fully solved" part of the is wrong because the EFE's have been fully solved since 1939 by the Tolman–Oppenheimer–Volkoff equation. The main issue in a solution for the EFE is simply that there are 10 differential equation to solve. That needs conditions to be imposed so that the independence of the equations are removed. For a static spherical solution for the inside of a body, there are not enough conditions to solve the EFE. What is left over is an unsolved differential equation. But we can solve that differential equation by adding an equation of state relating the pressure and density inside the sphere. That is the solution that has been around for 80 years![quote]That is an equation derived directly from the EFE's. It is not a solution to the metric nor is it an attempt at a solution.

It is mathematically impossible to solve a set of differential equations just by expanding 1 as a polynomial. If there are unknown functions, e.g. an A(r) or B(r), you just get polynomial involving those unknown functions.

It is textbook mathematics that not every function can be expanded as a polynomial. Thus even if the EFE are expanded as polynomials, this is not a general solution.
One unknown is found using a coordinate choice, ideally D = r^2. Another can be the given internal function. So we just have one final unknown to actually solve for using the EFE's, which is the challenge. It seems it should be a simple task, but it is not.

Internal to the body one just uses the orbital speed that applies internally.

18. Originally Posted by Reality Check
Once more: 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).
This is textbook physics that has been known for over 300 years for Newtonian gravitation. GR reduces to Newtonian gravitation for low gravity and so gives the same result in that limit.
Whatever value v_o has, whether it falls with r inside or sqrt(1/r) outside, that the value that is used in H = 1 / (1 + c^2 / v_o^2) in either case.

What makes this even less valid is that you are looking for a non-rotating solution. There cannot be any "orbits" inside the sphere! The Sun without its various rotations has plasma going in all directions. The Earth without its rotation has magna moving outward and inward and liquid rock rotating. There are bodies that are just rock or ices, e.g. small moons, asteroids, probably Ceres.
Right, the body itself is not rotating, but we could hypothetically dig a ring cavity around the circumference of the body at r and let an object orbit there. The speed that the object would have at r is v_o that will be used in H.

19. Originally Posted by Reality Check

It is a solution for the functions in the EFE. That solution is a metric. To solve the EFE, we take the geometry of the situation and use that to give the general form of the metric. There will be unknown functions, e.g. A(r). Mathematics (zero out non-diagonal elements of the proposed metric, calculate the Christoffel symbols, etc.) and physics (the equation of state ) are applied to derive these unknown functions. All of the functions in that equation are known. Thus it is a solution for all of the "variables in the metric". I gave you an example of how solving the EFE gives a metric. Deriving the Schwarzschild solution (external solution).

What the Tolman–Oppenheimer–Volkoff equation shows is that it is impossible to solve the EFE for this case using only mathematics. If we do valid mathematics as Tolman, et. all did then there remains an unknown function v(r).

P.S. On Massive Neutron Cores by J. R. Oppenheimer and G. M. Volkoff Phys. Rev. 55, 374 – Published 15 February 1939. PDF.

Equation 1 is the known result for the most general static line element (metric) with spherical symmetry.
Equation 2 is a constraint on the energy momentum tensor (no traverse stresses and no mass motion).
Equation 3 is the resulting 3 EFE differential equations. These have an unknown function that is determined by the equation of state.

You have an unknown function. Expanding that unknown function as a polynomial gives an infinite number of unknown coefficients.
The EFE's are the equations that they have rearranged and combined to form another equation that is useful. The EFE's must be used to solve for the unknowns in the metric. They have made a coordinate choice for one, D = r^2, same as for the external Schwarzschild metric. They also have A = 1 / (1 - 2 m(r) / r) where m(r) could be the given internal function to solve for A. They have not nor have they attempted to solve for final unknown B in the metric, however, where they use B = e^v, which also gives b = v'.
Last edited by grav; 2019-Oct-21 at 10:21 PM.

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Originally Posted by grav
One unknown is found using a coordinate choice, ....
That is obviously wrong, grav. If we have an unknown function A(r) then selecting a different coordinate system still has an unknown function A(z) where z is r transformed into that coordinate system. To determine A(r) the EFE have to be actually solved (emphasized so that you take note of this basic point about solving differential equations).

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Originally Posted by grav
Whatever value v_o has...
Nothing to do with the textbook physics I cited: 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).
"v_o" outside of a sphere is not "v_o" inside a sphere.

A non-rotating sphere has no orbits inside it and no orbital speeds of anything inside it. If we imagining digging a circular tunnel in it then an object in that tunnel either cannot move (the not rotating sphere is rigid) or can move at any speed (a non-rigid not rotating sphere).

ETA: This is your second post where you explicitly use an equation derived only for outside of a sphere in GR.
Originally Posted by grav
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.
That is taken directly from the external metric and is only true externally.

The orbital velocity measured by a co-located static observer is:
c⎷h(df/dr)/f(dh/dr)

Last edited by Reality Check; 2019-Oct-22 at 12:23 AM.

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Originally Posted by grav
The EFE's are the equations that they have rearranged and combined to form another equation that is useful....
That is correct, grav. The Tolman–Oppenheimer–Volkoff equation is where the "EFE's now were fully solved for the internal metric back in 1939"! Applying symmetry and constraints is how the EFE are solved as in the example I gave you of Deriving the Schwarzschild solution (external solution) and as in the derivation of this equation for the "internal metric" solution.
Originally Posted by Reality Check
...
P.S. On Massive Neutron Cores by J. R. Oppenheimer and G. M. Volkoff Phys. Rev. 55, 374 – Published 15 February 1939. PDF.

Equation 1 is the known result for the most general static line element (metric) with spherical symmetry.
Equation 2 is a constraint on the energy momentum tensor (no traverse stresses and no mass motion).
Equation 3 is the resulting 3 EFE differential equations. These have an unknown function that is determined by the equation of state.
You have an unknown function. Expanding that unknown function as a polynomial gives an infinite number of unknown coefficients.
Changing the coordinate system gives an unknown function in that coordinate system. This is totally obvious. Say we have an unknown function in Cartesian coordinates, A(x, y, z). Transform to any other coordinate system, e.g. spherical coordinates. It becomes an unknown function in those coordinates, e.g. A(r, polar angle, azimuthal angle).

"The EFE's must be used to solve for the unknowns in the metric" makes your solution invalid! You have a metric that appears out of thin air. You manipulate the unknown functions to be related to an unknown function H. The EFE are not used to determine H. Thus your requirement makes your solution wrong.

However there is no such requirement in physics. The Einstein field equations are a set of 10 independent partial differential equations. Any constraints on the system being modeled can be applied. Sometimes just symmetry and boundary constraints solve the EFE (external Schwarzschild solution). Other times other constraints are needed. An obvious constraint for the metric inside a body is the equation of state for that body.
Last edited by Reality Check; 2019-Oct-22 at 12:27 AM.

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How many situations are there exact solutions to the Einstein Field Equations for the internal metric.

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Originally Posted by Copernicus
How many situations are there exact solutions to the Einstein Field Equations for the internal metric.
Static spherically symmetric perfect fluid
1916: Schwarzschild fluid solution,
1939: The relativistic equation of hydrostatic equilibrium, the Oppenheimer-Volkov equation, is introduced,
...
2005: BVW algorithm, apparently the simplest variant now known
The Schwarzschild fluid solution is for an incompressible fluid, the others for a compressible fluid such as the Fermi gas of a neutron star.

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Originally Posted by Reality Check
Static spherically symmetric perfect fluid

The Schwarzschild fluid solution is for an incompressible fluid, the others for a compressible fluid such as the Fermi gas of a neutron star.
How would the isotropic pressure work. Would the sphere have the same pressure at every r in the BVW solution.

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Originally Posted by Copernicus
How would the isotropic pressure work. Would the sphere have the same pressure at every r in the BVW solution.
Isotropic means that there is no direction preference. Isotropic pressure would be that a body does not have a pressure gradient. However the static spherically symmetric perfect fluid solutions are for an isotropic material. I think that allows pressure gradients in the sphere.

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Originally Posted by Reality Check
Isotropic means that there is no direction preference. Isotropic pressure would be that a body does not have a pressure gradient. However the static spherically symmetric perfect fluid solutions are for an isotropic material. I think that allows pressure gradients in the sphere.
This is what wikipedia says, but I don't know what they mean by isotropic pressure. " In metric theories of gravitation, particularly general relativity, a static spherically symmetric perfect fluid solution (a term which is often abbreviated as ssspf) is a spacetime equipped with suitable tensor fields which models a static round ball of a fluid with isotropic pressure."

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Originally Posted by Copernicus
This is what wikipedia says, but I don't know what they mean by isotropic pressure ...
Wikipedia is fairly clear. Static spherically symmetric perfect fluid
In metric theories of gravitation, particularly general relativity, a static spherically symmetric perfect fluid solution (a term which is often abbreviated as ssspf) is a spacetime equipped with suitable tensor fields which models a static round ball of a fluid with isotropic pressure.
isotropic
Isotropy is uniformity in all orientations; it is derived from the Greek isos (ἴσος, "equal") and tropos (τρόπος, "way"). Precise definitions depend on the subject area. Exceptions, or inequalities, are frequently indicated by the prefix an, hence anisotropy. Anisotropy is also used to describe situations where properties vary systematically, dependent on direction. Isotropic radiation has the same intensity regardless of the direction of measurement, and an isotropic field exerts the same action regardless of how the test particle is oriented.
...
Fluid dynamics
Fluid flow is isotropic if there is no directional preference (e.g. in fully developed 3D turbulence). An example of anisotropy is in flows with a background density as gravity works in only one direction. The apparent surface separating two differing isotropic fluids would be referred to as an isotrope.
Similarly for pressure: "Fluid pressure is isotropic if there is no directional preference. An example of anisotropy is in pressure with a background density as gravity works in only one direction.". So my thought that this allows a pressure gradient is probably wrong.

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Originally Posted by Reality Check
Wikipedia is fairly clear. Static spherically symmetric perfect fluid

isotropic

Similarly for pressure: "Fluid pressure is isotropic if there is no directional preference. An example of anisotropy is in pressure with a background density as gravity works in only one direction.". So my thought that this allows a pressure gradient is probably wrong.
Thanks!

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