Why isn't an infinite, flat, nonexpanding universe filled with a uniform matter distribution a solution to Einstein's equation?

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
33
down vote

favorite
10












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. )










share|cite|improve this question



















  • 5




    Infinite, flat and non-expanding metric is surely a solution to Einstein's equations: Minkowski metric $eta_mu nu$
    – Avantgarde
    20 hours ago










  • @Avantgarde: Is it still considered Minkowski space when the mass distribution is included?
    – D. Halsey
    20 hours ago






  • 8




    @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.
    – Ben Crowell
    20 hours ago














up vote
33
down vote

favorite
10












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. )










share|cite|improve this question



















  • 5




    Infinite, flat and non-expanding metric is surely a solution to Einstein's equations: Minkowski metric $eta_mu nu$
    – Avantgarde
    20 hours ago










  • @Avantgarde: Is it still considered Minkowski space when the mass distribution is included?
    – D. Halsey
    20 hours ago






  • 8




    @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.
    – Ben Crowell
    20 hours ago












up vote
33
down vote

favorite
10









up vote
33
down vote

favorite
10






10





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. )










share|cite|improve this 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. )







general-relativity cosmology space-expansion cosmological-constant






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 13 mins ago









tparker

21k42112




21k42112










asked 21 hours ago









D. Halsey

532411




532411







  • 5




    Infinite, flat and non-expanding metric is surely a solution to Einstein's equations: Minkowski metric $eta_mu nu$
    – Avantgarde
    20 hours ago










  • @Avantgarde: Is it still considered Minkowski space when the mass distribution is included?
    – D. Halsey
    20 hours ago






  • 8




    @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.
    – Ben Crowell
    20 hours ago












  • 5




    Infinite, flat and non-expanding metric is surely a solution to Einstein's equations: Minkowski metric $eta_mu nu$
    – Avantgarde
    20 hours ago










  • @Avantgarde: Is it still considered Minkowski space when the mass distribution is included?
    – D. Halsey
    20 hours ago






  • 8




    @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.
    – Ben Crowell
    20 hours ago







5




5




Infinite, flat and non-expanding metric is surely a solution to Einstein's equations: Minkowski metric $eta_mu nu$
– Avantgarde
20 hours ago




Infinite, flat and non-expanding metric is surely a solution to Einstein's equations: Minkowski metric $eta_mu nu$
– Avantgarde
20 hours ago












@Avantgarde: Is it still considered Minkowski space when the mass distribution is included?
– D. Halsey
20 hours ago




@Avantgarde: Is it still considered Minkowski space when the mass distribution is included?
– D. Halsey
20 hours ago




8




8




@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.
– Ben Crowell
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.
– Ben Crowell
20 hours ago










5 Answers
5






active

oldest

votes

















up vote
19
down vote













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 $rho = Lambda$.






share|cite|improve this answer


















  • 1




    This doesn't answer the question. The OP specifically mentions the stability issue and says that's not what the question is about.
    – Ben Crowell
    18 hours ago






  • 4




    @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.
    – knzhou
    18 hours ago






  • 4




    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.
    – Ben Crowell
    15 hours ago






  • 4




    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.
    – Empischon
    8 hours ago










  • Comments are not for extended discussion; this conversation has been moved to chat.
    – ACuriousMind♦
    44 mins ago

















up vote
15
down vote













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.






share|cite|improve this answer


















  • 5




    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.
    – Owen
    11 hours ago

















up vote
8
down vote













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
$$
R_munu - frac12Rg_munu + Lambda g_munu = frac8pi Gc^4T_munu
$$






share|cite|improve this answer






















  • I think this is the answer that OP is after!
    – tfb
    4 hours ago










  • 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.
    – ja72
    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?
    – Arthur
    2 hours ago







  • 1




    @ja72: the field equations for GR apply in finite and infinite universes.
    – tfb
    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.
    – ja72
    1 hour ago

















up vote
0
down vote














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.






share|cite|improve this answer








New contributor




Kolaru is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
























    up vote
    0
    down vote













    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.






    share|cite|improve this answer








    New contributor




    elliot svensson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.

















      Your Answer




      StackExchange.ifUsing("editor", function ()
      return StackExchange.using("mathjaxEditing", function ()
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      );
      );
      , "mathjax-editing");

      StackExchange.ready(function()
      var channelOptions =
      tags: "".split(" "),
      id: "151"
      ;
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function()
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled)
      StackExchange.using("snippets", function()
      createEditor();
      );

      else
      createEditor();

      );

      function createEditor()
      StackExchange.prepareEditor(
      heartbeatType: 'answer',
      convertImagesToLinks: false,
      noModals: false,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: null,
      bindNavPrevention: true,
      postfix: "",
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      );



      );













       

      draft saved


      draft discarded


















      StackExchange.ready(
      function ()
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f430419%2fwhy-isnt-an-infinite-flat-nonexpanding-universe-filled-with-a-uniform-matter%23new-answer', 'question_page');

      );

      Post as a guest






























      5 Answers
      5






      active

      oldest

      votes








      5 Answers
      5






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      19
      down vote













      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 $rho = Lambda$.






      share|cite|improve this answer


















      • 1




        This doesn't answer the question. The OP specifically mentions the stability issue and says that's not what the question is about.
        – Ben Crowell
        18 hours ago






      • 4




        @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.
        – knzhou
        18 hours ago






      • 4




        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.
        – Ben Crowell
        15 hours ago






      • 4




        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.
        – Empischon
        8 hours ago










      • Comments are not for extended discussion; this conversation has been moved to chat.
        – ACuriousMind♦
        44 mins ago














      up vote
      19
      down vote













      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 $rho = Lambda$.






      share|cite|improve this answer


















      • 1




        This doesn't answer the question. The OP specifically mentions the stability issue and says that's not what the question is about.
        – Ben Crowell
        18 hours ago






      • 4




        @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.
        – knzhou
        18 hours ago






      • 4




        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.
        – Ben Crowell
        15 hours ago






      • 4




        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.
        – Empischon
        8 hours ago










      • Comments are not for extended discussion; this conversation has been moved to chat.
        – ACuriousMind♦
        44 mins ago












      up vote
      19
      down vote










      up vote
      19
      down vote









      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 $rho = Lambda$.






      share|cite|improve this answer














      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 $rho = Lambda$.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited 17 hours ago

























      answered 19 hours ago









      knzhou

      35.1k899172




      35.1k899172







      • 1




        This doesn't answer the question. The OP specifically mentions the stability issue and says that's not what the question is about.
        – Ben Crowell
        18 hours ago






      • 4




        @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.
        – knzhou
        18 hours ago






      • 4




        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.
        – Ben Crowell
        15 hours ago






      • 4




        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.
        – Empischon
        8 hours ago










      • Comments are not for extended discussion; this conversation has been moved to chat.
        – ACuriousMind♦
        44 mins ago












      • 1




        This doesn't answer the question. The OP specifically mentions the stability issue and says that's not what the question is about.
        – Ben Crowell
        18 hours ago






      • 4




        @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.
        – knzhou
        18 hours ago






      • 4




        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.
        – Ben Crowell
        15 hours ago






      • 4




        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.
        – Empischon
        8 hours ago










      • Comments are not for extended discussion; this conversation has been moved to chat.
        – ACuriousMind♦
        44 mins ago







      1




      1




      This doesn't answer the question. The OP specifically mentions the stability issue and says that's not what the question is about.
      – Ben Crowell
      18 hours ago




      This doesn't answer the question. The OP specifically mentions the stability issue and says that's not what the question is about.
      – Ben Crowell
      18 hours ago




      4




      4




      @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.
      – knzhou
      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.
      – knzhou
      18 hours ago




      4




      4




      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.
      – Ben Crowell
      15 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.
      – Ben Crowell
      15 hours ago




      4




      4




      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.
      – Empischon
      8 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.
      – Empischon
      8 hours ago












      Comments are not for extended discussion; this conversation has been moved to chat.
      – ACuriousMind♦
      44 mins ago




      Comments are not for extended discussion; this conversation has been moved to chat.
      – ACuriousMind♦
      44 mins ago










      up vote
      15
      down vote













      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.






      share|cite|improve this answer


















      • 5




        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.
        – Owen
        11 hours ago














      up vote
      15
      down vote













      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.






      share|cite|improve this answer


















      • 5




        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.
        – Owen
        11 hours ago












      up vote
      15
      down vote










      up vote
      15
      down vote









      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.






      share|cite|improve this answer














      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.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited 19 hours ago

























      answered 19 hours ago









      Ben Crowell

      44.7k3146272




      44.7k3146272







      • 5




        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.
        – Owen
        11 hours ago












      • 5




        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.
        – Owen
        11 hours ago







      5




      5




      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.
      – Owen
      11 hours ago




      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.
      – Owen
      11 hours ago










      up vote
      8
      down vote













      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
      $$
      R_munu - frac12Rg_munu + Lambda g_munu = frac8pi Gc^4T_munu
      $$






      share|cite|improve this answer






















      • I think this is the answer that OP is after!
        – tfb
        4 hours ago










      • 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.
        – ja72
        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?
        – Arthur
        2 hours ago







      • 1




        @ja72: the field equations for GR apply in finite and infinite universes.
        – tfb
        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.
        – ja72
        1 hour ago














      up vote
      8
      down vote













      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
      $$
      R_munu - frac12Rg_munu + Lambda g_munu = frac8pi Gc^4T_munu
      $$






      share|cite|improve this answer






















      • I think this is the answer that OP is after!
        – tfb
        4 hours ago










      • 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.
        – ja72
        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?
        – Arthur
        2 hours ago







      • 1




        @ja72: the field equations for GR apply in finite and infinite universes.
        – tfb
        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.
        – ja72
        1 hour ago












      up vote
      8
      down vote










      up vote
      8
      down vote









      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
      $$
      R_munu - frac12Rg_munu + Lambda g_munu = frac8pi Gc^4T_munu
      $$






      share|cite|improve this answer














      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
      $$
      R_munu - frac12Rg_munu + Lambda g_munu = frac8pi Gc^4T_munu
      $$







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited 2 hours ago

























      answered 9 hours ago









      Arthur

      1,1081813




      1,1081813











      • I think this is the answer that OP is after!
        – tfb
        4 hours ago










      • 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.
        – ja72
        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?
        – Arthur
        2 hours ago







      • 1




        @ja72: the field equations for GR apply in finite and infinite universes.
        – tfb
        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.
        – ja72
        1 hour ago
















      • I think this is the answer that OP is after!
        – tfb
        4 hours ago










      • 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.
        – ja72
        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?
        – Arthur
        2 hours ago







      • 1




        @ja72: the field equations for GR apply in finite and infinite universes.
        – tfb
        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.
        – ja72
        1 hour ago















      I think this is the answer that OP is after!
      – tfb
      4 hours ago




      I think this is the answer that OP is after!
      – tfb
      4 hours ago












      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.
      – ja72
      2 hours ago




      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.
      – ja72
      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?
      – Arthur
      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?
      – Arthur
      2 hours ago





      1




      1




      @ja72: the field equations for GR apply in finite and infinite universes.
      – tfb
      1 hour ago




      @ja72: the field equations for GR apply in finite and infinite universes.
      – tfb
      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.
      – ja72
      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.
      – ja72
      1 hour ago










      up vote
      0
      down vote














      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.






      share|cite|improve this answer








      New contributor




      Kolaru is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





















        up vote
        0
        down vote














        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.






        share|cite|improve this answer








        New contributor




        Kolaru is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.



















          up vote
          0
          down vote










          up vote
          0
          down vote










          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.






          share|cite|improve this answer








          New contributor




          Kolaru is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.










          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.







          share|cite|improve this answer








          New contributor




          Kolaru is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.









          share|cite|improve this answer



          share|cite|improve this answer






          New contributor




          Kolaru is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.









          answered 13 hours ago









          Kolaru

          1092




          1092




          New contributor




          Kolaru is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.





          New contributor





          Kolaru is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          Kolaru is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.




















              up vote
              0
              down vote













              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.






              share|cite|improve this answer








              New contributor




              elliot svensson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.





















                up vote
                0
                down vote













                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.






                share|cite|improve this answer








                New contributor




                elliot svensson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.



















                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  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.






                  share|cite|improve this answer








                  New contributor




                  elliot svensson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.









                  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.







                  share|cite|improve this answer








                  New contributor




                  elliot svensson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.









                  share|cite|improve this answer



                  share|cite|improve this answer






                  New contributor




                  elliot svensson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.









                  answered 1 hour ago









                  elliot svensson

                  1011




                  1011




                  New contributor




                  elliot svensson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.





                  New contributor





                  elliot svensson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.






                  elliot svensson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.



























                       

                      draft saved


                      draft discarded















































                       


                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function ()
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f430419%2fwhy-isnt-an-infinite-flat-nonexpanding-universe-filled-with-a-uniform-matter%23new-answer', 'question_page');

                      );

                      Post as a guest













































































                      Comments

                      Popular posts from this blog

                      What does second last employer means? [closed]

                      Installing NextGIS Connect into QGIS 3?

                      One-line joke