Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Why is the density infinite at the singularity?

  1. Jul 21, 2016 #1
    I always has the impression that the density of the universe is infinite at the singularity because its just amount of stuff divided by volume and if the distance between stuff is 0 then the volume is 0. So divide by zero and you get infinity. But I have been told by others that dividing by zero is not infinity but undefined. So why not say the density of the universe at the singularity is undefined rather than infinite?
    Im not assuming the singularity is real just want to understand why this description of the singularity is given.
  2. jcsd
  3. Jul 21, 2016 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member
    2017 Award

    You are implicitly assuming the singularity to be real by asking for its density.

    The Big Bang singularity is not part of the space-time and it is likely that our known physics break down before you get there. Asking for its physical properties, based on our physical models, is therefore not scientifically sound and you could make up any answer you want - so I am going to say "blue".
  4. Jul 21, 2016 #3
    As I understood, the reason ( or at least one of the reasons) one assumes the singularity is not real is because numbers like density and other features come out as infinity. if that were not the case perhaps you wouldn't say it wasn't real? So my question is not , is it real? Which is the one you seem to be answering. Thats just inst my question . But how is the infinite value arrived at (that perhaps leads you to the conclusion that it isn't real)?
  5. Jul 21, 2016 #4


    User Avatar
    Science Advisor

    Whether a quantity comes out "infinite" or "undefined" is not particularly relevant. Our equations that model space-time are differential equations. They only work at all over regions where the quantities involved are well defined. They are only predictive over regions where the quantities involved are suitably continuous and differentiable.

    We apply our models on "manifolds". A manifold is, roughly speaking, a collection of (pieces of) cartesian coordinate systems that cover a region of interest. Each of these coordinate patches covers an "open" region. Open in this case means that every point within a patch is surrounded by other points in the patch. If a particular patch has a boundary, that boundary will not fall within that patch (it may fall within a neighboring patch). We pick patches so that within each patch all of the physical quantities are well defined, continuous and suitably differentiable.

    Some relevant physical quantities diverge as one approaches the singularity. One cannot preserve continuity if the singularity is included in the patch. So it is not included. It amounts to a boundary that is not part of any coordinate patch.
  6. Jul 21, 2016 #5
    Dear windy miller,

    Sometimes it seems to me that almost all Physics students really are Mathematics lovers. They forget that Mathematics deals rather well with infinities, but in Physics they generally signify that a theory is incomplete, has reached the limit of validity. So, singularities, as in general all absolutes, can't have any physical significance, they are just mind teasers.
  7. Jul 21, 2016 #6

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    This is standard shorthand for "in a limiting process, become unboundedly large", which has a precise mathematical definition.

    Let ##a \left( t \right)## be the scale factor of the universe, with ##t## cosmological time. Define ##V \left( a \left( t \right) \right)## by
    $$V \left( a \left( t \right) \right) = a \left( t \right)^3.$$
    It is true that if ##a \left( t \right) =0##, then ##1/V \left( a \left( t \right) \right)## is undefined, but what is actually meant by the above is
    $$ \lim_{t \rightarrow 0} \frac{1}{V \left( a \left( t \right) \right)} = \lim_{t \rightarrow 0} \frac{1}{a \left( t \right)^3} = \infty.$$
    This means that given positive any number ##L## (think very large number), there exists a positive number ##\epsilon## (think very small number) such that ##1/V \left( a \left( t \right) \right)## is larger than ##L## whenever ##t## is less than ##\epsilon## (but still positive).

    This is an example where the precise mathematical definition of "limit", which many students hate, has easily visualizable physical content.

    $$\lim_{t \rightarrow 0} \rho \left( t \right) = \infty,$$
    i.e., standard cosmological models predict that for small enough ##t##, density ##\rho## becomes larger than standard physics can handle. Either we need to consider new comsological models which fit all known observations, and for which density does not become too large for standard physics to handle, or we need new physics (and new cosmolgical models). Most physicists think the latter.

    We just don't know. This is the nature of scientific research; we try to push the boundary between the known and the unknown.
    Last edited: Jul 21, 2016
  8. Jul 21, 2016 #7


    User Avatar
    Science Advisor

    Sort of. Inflation, which is the current most popular family of pre-big bang cosmological models, is intended to do the former (it doesn't quite work, but that's a different story). These models do, however, also require new physics.
  9. Jul 28, 2016 #8
    To give a bit of a different perspective that is presented here, what you're looking at in a Friedmann model is a cosmology with some matter in it. For homogeneous, isotropic matter, we assume that matter is some form of a perfect fluid, so the density is a function of the scale factor, usually a power law

    \rho = 1/a^(1+w)

    Where the pressure and density are related by P=w \rho. For 'dust' (regular matter) w is zero, and we see that as a becomes zero, \rho asymptotes to infinity. There are different values of w for different types of matter - radiation has w=1/3 for example, as the energy density of a photon is given by E=h/lambda (wavelength) and getting compressed therefore increases energy density of each photon, as well as the crunching effect of putting more photons in the same volume.

    Now you question actually contains an error: There are plenty of singularities which do not have infinite energy density. A simple example is the Kasner solution to GR - this is a vacuum solution (no matter/energy density at all) and yet there is a singularity at some point in the evolution. You can think of this as being a cube which is expanding along one axis and contracting on the other two. As the two axes collapse to zero length, the Ricci scalar goes to infinity and we have a singularity (a strong one, in the sense of Tipler and Krolak).

    There are several other examples of finite density singularities - I think John Barrow has a review somewhere. So the real answer to your question is "It doesn't necessarily have to, but in when it does so it happens as a limiting process, and thus the infinity that you arrive at it the limit of some process like lim epsilon-> 0+ 1/epsilon."
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted