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Question

In Newtonian gravity, an infinite volume filled with a uniform distribution of mass would be in perfect equilibrium. At every point, the gravitational forces contributed by masses in one direction would be exactly counterbalanced by those in the opposite direction.

But when Einstein tried to apply General Relativity to possible cosmologies, he found it necessary to include the cosmological constant in order to get a static universe.

In qualitative terms, it seems to me that the gravitational stresses that the masses would impose on the spacetime should all cancel out, and likewise, that the resulting flat spacetime should have no effect on the motion of the masses.

However, the math of the situation is beyond my current skills, so I'm asking how it produces the nonequilibrium condition?

(I realize that such an equilibrium solution might not be stable, and that there are many other very good reasons to believe in an expanding universe, so I'm not trying to promote any alternative theories. I'm just curious about this particular point. )

Infinite, flat and non-expanding metric is surely a solution to Einstein's equations: Minkowski metric $eta_mu nu$ — 20 hours ago
@Avantgarde: Is it still considered Minkowski space when the mass distribution is included? — 20 hours ago
@Avantgarde: The OP is asking about a cosmology with a uniform mass distribution. Minkowski space isn't a solution to the Einstein field equations when the stress-energy tensor is nonzero. — 20 hours ago

score 19 · Answer 1 · 2018-09-23 22:28:28Z

This is a rather subtle question, which confused even Newton. It is very tempting to think that a Newtonian universe with perfectly uniform mass density is stable because the gravitational force cancels everywhere by symmetry. This is wrong.

Here's an analogous question: suppose a function $f$ obeys
$$f''(x) = 1$$
and we want to solve for $f(x)$. Since every point on the real line is the same as every other point, we might think that by symmetry,
$$f(x) = textconstant.$$
But this is completely wrong, because the second derivative of a constant is zero. And stepping back, the whole question doesn't make any sense, because there isn't enough information. To solve a general differential equation, you need boundary conditions.

For concreteness, let's take $f'(infty) = -f'(-infty)$. This by itself is enough to specify
$$f(x) = fracx^22 + textconstant.$$
But now the translational symmetry has been broken: not every point is equivalent anymore, because we have a minimum at $x = 0$. This is inevitable. You can't solve the differential equation without boundary conditions, and any choice of boundary conditions breaks the symmetry.

Similarly in Newton's infinite universe we have
$$nabla^2 phi = rho$$
where $rho$ is the mass density and $phi$ is the gravitational potential. Without boundary conditions, the subsequent evolution is not defined. With any set of boundary conditions, you will have a special point towards which everything collapses. So the answer to your question is that both the Newtonian and relativistic universes immediately start to collapse; the symmetry argument does not work in either one, so there is nothing strange to explain.

We often assume the gravitational potential goes to zero at infinity (in Newtonian gravity) or that the metric is asymptotically flat (in relativity). But this boundary condition doesn't work when the mass distribution extends to infinity as well, which leads to the pitfall here. The same point holds in electrostatics. In the end, both the Newtonian and relativistic universes can be made stable with a cosmological constant. In the Newtonian case, this is simply the trivial statement that $nabla^2 phi = rho - Lambda$ has constant solutions for $phi$ when $rho = Lambda$.

This doesn't answer the question. The OP specifically mentions the stability issue and says that's not what the question is about. — 18 hours ago
@BenCrowell That's because the answer is exactly the same for both the Newtonian and GR cases: the symmetry argument that all the gravitational pulls "cancel out" doesn't work. — 18 hours ago
No, the logic is completely different for newtonian gravity than for GR. Your answer doesn't have any GR in it, and cannot be adapted to work as an argument in GR. — 15 hours ago
This is a 'frame challenge' answer, pointing out to a flaw in the reasoning in the original question. The frame challenge answers are perfectly ok on stack exchange. — 8 hours ago
Comments are not for extended discussion; this conversation has been moved to chat. — 44 mins ago

score 15 · Answer 2 · 2018-09-23 20:48:50Z

Nice question!

Here's a possible statement of the logic in the newtonian case. (1) In newtonian mechanics, we assume that inertial reference frames exist (this is one popular modern way of restating Newton's first law), we assume that such frames are global, and we assume that we can always find such a frame by observing a test particle that is not acted on by any force. (2) In the newtonian homogeneous cosmology, we could assume that the force on a chosen test particle P can be found by some limiting process, and that the result is unique. (This is basically a bogus assumption, but I don't think that ends up being the issue here.) (3) Given that the result is unique, it must be zero by symmetry. (4) By assumptions 1 and 2, P defines an inertial frame, and by assumption 1, that frame can be extended to cover the entire universe. Therefore all other particles in the universe must have zero acceleration relative to P.

In general relativity, assumption 1 fails. Test particles P and Q can both be inertial (i.e., no nongravitational forces act on them), but it can be false that they are not accelerated relative to one another. For example, we can make an FRW cosmology in which, at some initial time, $dota=0$, but then it will have $ddotane0$ (in order to satisfy the Einstein field equations for a uniform dust). (In this situation, the Einstein field equations can be reduced to the Friedmann equations, one of which is $ddota/a=-(4pi/3)rho$.)

This shows that the newtonian argument (or at least one version of it) fails. It does not prove that there is no other semi-newtonian plausibility argument that explains why an initially static universe collapses. However, I'm not sure what criteria we would be able to agree on as to what constitutes an acceptable semi-newtonian plausibility argument. Some people have developed these semi-newtonian descriptions of cosmology at great length, but to me they appear to lack any logical foundations that would allow one to tell a correct argument from an incorrect one.

I may be missing something, but I feel like this is answering a slightly different question than the OP is asking. This answer seems to be directed to the uniqueness of the solution, whereas I interpret OP's question to be about its existence. — 11 hours ago

score 8 · Answer 3 · 2018-09-24 13:48:56Z

The equation governing the curvature of spacetime in general relativity is
$$
R_munu - frac12Rg_munu = frac8pi Gc^4T_munu
$$
Or, really, it's $16$ equations in one: $mu$ and $nu$ can both take values from $0$ to $3$, representing the four components of any given coordinate system, and for each such choice you get a new equation (with the small caveat that it is symmetric in $mu$ and $nu$, so really, it's only $10$ distinct equations).

The symbols $R_munu$, $R$, $g_munu$ and $T_munu$ are so-called tensors, which for now you can think of as real-valued functions on spacetime. (The actual values you get will be dependent on your choice of coordinate system, but if you fix one coordinate system for the region you're interested in they do become just functions. $R$ is a single function while the remaining three, again, are collections of $16$ functions: one for each $mu, nu$ pair.)

The left-hand side represents the curvature of spacetime. In flat spacetime we have $R_munu = 0$ for any $mu, nu$, and we also get $R = 0$, so the left-hand side will be $0$.

The right-hand side is one big constant multiplied by $T_munu$, which represents the energy at each point in space. In a "sensible" coordinate system (where the $0$-component represents time, and the three remaining coordinates represent space), $T_00$ will represent the energy density (including mass density; the other components of $T_munu$ represent things like pressure and momentum density). If there is uniform non-zero mass everywhere, then $T_00$ will be non-zero. That means that $R_00 -frac12Rg_00$ will also be non-zero, which means that we do not have flat spacetime as either $R_00$ or $R$ must be non-zero.

In order to regain flat spacetime in this case, and allow both $R_munu$ and $R$ to be zero, it is necessary to add a third term to the left-hand side: the cosmological constant $Lambda$, giving us
$$
R_munu - frac12Rg_munu + Lambda g_munu = frac8pi Gc^4T_munu
$$

Maybe this equation assumes a finite universe or some sort of boundary where all mass-energy $T_mu nu$ is contained. But with infinite universe the above quation would not apply. — 2 hours ago
@ja72 I thought the Einstein equation (like most differential equations) had just local terms: the curvature of space at any point (which is just a bunch of derivatives) equals some combination of the energy density, pressure and momentum density at that point (which is just a bunch of local measurements). Which part of the equation would care about the global structure of the universe? — 2 hours ago
@ja72: the field equations for GR apply in finite and infinite universes. — 1 hour ago
That was my point. historically this equation was developed before the Hubble parameter $H$ and the Friedmann equations which do require a "radius" and a finite universe. Maybe the equation above isn't valid for a non-finite universe. — 1 hour ago

Kolaru 1092 · Answer 4 · 2018-09-24 02:51:07Z

In qualitative terms, it seems to me that the gravitational stresses that the masses would impose on the spacetime should all cancel out, and likewise, that the resulting flat spacetime should have no effect on the motion of the masses.

I think there is a subtle misunderstanding here.

In general relativity, a universe uniformly filled with mass (or really with energy), will see all the forces acting on a massive object cancel, as in the Newtonian case. This is due the the symmetries of an uniform universe.

However, this does not mean that the universe is flat.

In Newtonian mechanics, forces cancelling each other implies a motion following a straight line. Then intuitively it seems like an object could definitely not follow a straight line in a curved space time. But here is the catch: in the context of general relativity the concept of straight line does not make much sense.

To understand why, we must ask ourselves what is a straight line. In Newtonian mechanics it is easy: it is the path followed by an inertial object i.e. an object on which we can define a reference frame such that this object in this frame seems to be static and affected by no forces.

The principle of equivalence of general relativity, however, tells us that such frame can in fact be defined for any free falling object (object only affected by gravitational forces). The trajectories of such objects are called geodesics and that is the best we can do to extend the newtonian concept of straight line. In other words, in general relativity "straightness" is an effect of reference frame.

Then, not very suprisingly, in a universe were we consider only gravitation, all objects follow geodesics and thus are moving following "generalized straight lines".

What all of that indicate is that considerations on objects motion do not give us (to my knowledge) any informations about the structure of space and time. Unfortunately this answer does not explain the necessity of an expanding universe, just point out the Newtonian argument is limited. The reason is that as far as I know there is no simple intuitive explanation on the reason behind the expansion of the universe.

elliot svensson 1011 · Answer 5 · 2018-09-24 15:08:05Z

With appreciation for contributor knzhou above and his excellent textbook-quality answer, I would like to add a few shouts from the gallery:

When knzhou says that the evolution of the universe with no boundary conditions is not defined, you must not misunderestimate the implication: that is math boiling over, not "sure, why not?" math.

In functions, when the value goes to infinity we don't try to graph the point--- it's a place where the math ain't working. But with gravity fields, we don't have the luxury of showing tidy values at all places that aren't blowing up to infinity because that's actually happening at all locations.

The static equation for an infinite constant density field is like Sammy in Sideways Stories from Wayside School: When Mrs. Jewls finally gets the last smelly raincoat off of him, it turns out that he's actually a dead rat.

score 19 · Answer 6 · 2018-09-23 22:28:28Z

This is a rather subtle question, which confused even Newton. It is very tempting to think that a Newtonian universe with perfectly uniform mass density is stable because the gravitational force cancels everywhere by symmetry. This is wrong.

Here's an analogous question: suppose a function $f$ obeys
$$f''(x) = 1$$
and we want to solve for $f(x)$. Since every point on the real line is the same as every other point, we might think that by symmetry,
$$f(x) = textconstant.$$
But this is completely wrong, because the second derivative of a constant is zero. And stepping back, the whole question doesn't make any sense, because there isn't enough information. To solve a general differential equation, you need boundary conditions.

For concreteness, let's take $f'(infty) = -f'(-infty)$. This by itself is enough to specify
$$f(x) = fracx^22 + textconstant.$$
But now the translational symmetry has been broken: not every point is equivalent anymore, because we have a minimum at $x = 0$. This is inevitable. You can't solve the differential equation without boundary conditions, and any choice of boundary conditions breaks the symmetry.

Similarly in Newton's infinite universe we have
$$nabla^2 phi = rho$$
where $rho$ is the mass density and $phi$ is the gravitational potential. Without boundary conditions, the subsequent evolution is not defined. With any set of boundary conditions, you will have a special point towards which everything collapses. So the answer to your question is that both the Newtonian and relativistic universes immediately start to collapse; the symmetry argument does not work in either one, so there is nothing strange to explain.

We often assume the gravitational potential goes to zero at infinity (in Newtonian gravity) or that the metric is asymptotically flat (in relativity). But this boundary condition doesn't work when the mass distribution extends to infinity as well, which leads to the pitfall here. The same point holds in electrostatics. In the end, both the Newtonian and relativistic universes can be made stable with a cosmological constant. In the Newtonian case, this is simply the trivial statement that $nabla^2 phi = rho - Lambda$ has constant solutions for $phi$ when $rho = Lambda$.

This doesn't answer the question. The OP specifically mentions the stability issue and says that's not what the question is about. — 18 hours ago
@BenCrowell That's because the answer is exactly the same for both the Newtonian and GR cases: the symmetry argument that all the gravitational pulls "cancel out" doesn't work. — 18 hours ago
No, the logic is completely different for newtonian gravity than for GR. Your answer doesn't have any GR in it, and cannot be adapted to work as an argument in GR. — 15 hours ago
This is a 'frame challenge' answer, pointing out to a flaw in the reasoning in the original question. The frame challenge answers are perfectly ok on stack exchange. — 8 hours ago
Comments are not for extended discussion; this conversation has been moved to chat. — 44 mins ago

score 15 · Answer 7 · 2018-09-23 20:48:50Z

Nice question!

Here's a possible statement of the logic in the newtonian case. (1) In newtonian mechanics, we assume that inertial reference frames exist (this is one popular modern way of restating Newton's first law), we assume that such frames are global, and we assume that we can always find such a frame by observing a test particle that is not acted on by any force. (2) In the newtonian homogeneous cosmology, we could assume that the force on a chosen test particle P can be found by some limiting process, and that the result is unique. (This is basically a bogus assumption, but I don't think that ends up being the issue here.) (3) Given that the result is unique, it must be zero by symmetry. (4) By assumptions 1 and 2, P defines an inertial frame, and by assumption 1, that frame can be extended to cover the entire universe. Therefore all other particles in the universe must have zero acceleration relative to P.

In general relativity, assumption 1 fails. Test particles P and Q can both be inertial (i.e., no nongravitational forces act on them), but it can be false that they are not accelerated relative to one another. For example, we can make an FRW cosmology in which, at some initial time, $dota=0$, but then it will have $ddotane0$ (in order to satisfy the Einstein field equations for a uniform dust). (In this situation, the Einstein field equations can be reduced to the Friedmann equations, one of which is $ddota/a=-(4pi/3)rho$.)

This shows that the newtonian argument (or at least one version of it) fails. It does not prove that there is no other semi-newtonian plausibility argument that explains why an initially static universe collapses. However, I'm not sure what criteria we would be able to agree on as to what constitutes an acceptable semi-newtonian plausibility argument. Some people have developed these semi-newtonian descriptions of cosmology at great length, but to me they appear to lack any logical foundations that would allow one to tell a correct argument from an incorrect one.

I may be missing something, but I feel like this is answering a slightly different question than the OP is asking. This answer seems to be directed to the uniqueness of the solution, whereas I interpret OP's question to be about its existence. — 11 hours ago

score 8 · Answer 8 · 2018-09-24 13:48:56Z

The equation governing the curvature of spacetime in general relativity is
$$
R_munu - frac12Rg_munu = frac8pi Gc^4T_munu
$$
Or, really, it's $16$ equations in one: $mu$ and $nu$ can both take values from $0$ to $3$, representing the four components of any given coordinate system, and for each such choice you get a new equation (with the small caveat that it is symmetric in $mu$ and $nu$, so really, it's only $10$ distinct equations).

The symbols $R_munu$, $R$, $g_munu$ and $T_munu$ are so-called tensors, which for now you can think of as real-valued functions on spacetime. (The actual values you get will be dependent on your choice of coordinate system, but if you fix one coordinate system for the region you're interested in they do become just functions. $R$ is a single function while the remaining three, again, are collections of $16$ functions: one for each $mu, nu$ pair.)

The left-hand side represents the curvature of spacetime. In flat spacetime we have $R_munu = 0$ for any $mu, nu$, and we also get $R = 0$, so the left-hand side will be $0$.

The right-hand side is one big constant multiplied by $T_munu$, which represents the energy at each point in space. In a "sensible" coordinate system (where the $0$-component represents time, and the three remaining coordinates represent space), $T_00$ will represent the energy density (including mass density; the other components of $T_munu$ represent things like pressure and momentum density). If there is uniform non-zero mass everywhere, then $T_00$ will be non-zero. That means that $R_00 -frac12Rg_00$ will also be non-zero, which means that we do not have flat spacetime as either $R_00$ or $R$ must be non-zero.

In order to regain flat spacetime in this case, and allow both $R_munu$ and $R$ to be zero, it is necessary to add a third term to the left-hand side: the cosmological constant $Lambda$, giving us
$$
R_munu - frac12Rg_munu + Lambda g_munu = frac8pi Gc^4T_munu
$$

Maybe this equation assumes a finite universe or some sort of boundary where all mass-energy $T_mu nu$ is contained. But with infinite universe the above quation would not apply. — 2 hours ago
@ja72 I thought the Einstein equation (like most differential equations) had just local terms: the curvature of space at any point (which is just a bunch of derivatives) equals some combination of the energy density, pressure and momentum density at that point (which is just a bunch of local measurements). Which part of the equation would care about the global structure of the universe? — 2 hours ago
@ja72: the field equations for GR apply in finite and infinite universes. — 1 hour ago
That was my point. historically this equation was developed before the Hubble parameter $H$ and the Friedmann equations which do require a "radius" and a finite universe. Maybe the equation above isn't valid for a non-finite universe. — 1 hour ago

Kolaru 1092 · Answer 9 · 2018-09-24 02:51:07Z

In qualitative terms, it seems to me that the gravitational stresses that the masses would impose on the spacetime should all cancel out, and likewise, that the resulting flat spacetime should have no effect on the motion of the masses.

I think there is a subtle misunderstanding here.

In general relativity, a universe uniformly filled with mass (or really with energy), will see all the forces acting on a massive object cancel, as in the Newtonian case. This is due the the symmetries of an uniform universe.

However, this does not mean that the universe is flat.

In Newtonian mechanics, forces cancelling each other implies a motion following a straight line. Then intuitively it seems like an object could definitely not follow a straight line in a curved space time. But here is the catch: in the context of general relativity the concept of straight line does not make much sense.

To understand why, we must ask ourselves what is a straight line. In Newtonian mechanics it is easy: it is the path followed by an inertial object i.e. an object on which we can define a reference frame such that this object in this frame seems to be static and affected by no forces.

The principle of equivalence of general relativity, however, tells us that such frame can in fact be defined for any free falling object (object only affected by gravitational forces). The trajectories of such objects are called geodesics and that is the best we can do to extend the newtonian concept of straight line. In other words, in general relativity "straightness" is an effect of reference frame.

Then, not very suprisingly, in a universe were we consider only gravitation, all objects follow geodesics and thus are moving following "generalized straight lines".

What all of that indicate is that considerations on objects motion do not give us (to my knowledge) any informations about the structure of space and time. Unfortunately this answer does not explain the necessity of an expanding universe, just point out the Newtonian argument is limited. The reason is that as far as I know there is no simple intuitive explanation on the reason behind the expansion of the universe.

elliot svensson 1011 · Answer 10 · 2018-09-24 15:08:05Z

With appreciation for contributor knzhou above and his excellent textbook-quality answer, I would like to add a few shouts from the gallery:

When knzhou says that the evolution of the universe with no boundary conditions is not defined, you must not misunderestimate the implication: that is math boiling over, not "sure, why not?" math.

In functions, when the value goes to infinity we don't try to graph the point--- it's a place where the math ain't working. But with gravity fields, we don't have the luxury of showing tidy values at all places that aren't blowing up to infinity because that's actually happening at all locations.

The static equation for an infinite constant density field is like Sammy in Sideways Stories from Wayside School: When Mrs. Jewls finally gets the last smelly raincoat off of him, it turns out that he's actually a dead rat.

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