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Horizon problem - why do we need inflation?

  1. Dec 13, 2009 #1
    Why do we need an inflation to solve the horizon problem?
    http://en.wikipedia.org/wiki/Horizon_problem

    If O(t) is omnuim ((c) Frederik) - wavefunction of the whole Universe then initial conditions like

    O(t=0) = const

    solve the problem?
     
  2. jcsd
  3. Dec 13, 2009 #2

    BillSaltLake

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    I had wondered about this myself. GR by itself doesn't predict any particular spectrum for how energy was distributed in early times. How then can someone claim that the observed smoothness of the CMB is "too smooth" ? It doesn't seem to contradict GR.

    However, if QM is added, small deviations could occur around the Planck time or earlier that would be now be of comparable length scale to a large fraction of the Universe and larger. These deviations would have been greatly amplified by gravity and might be manifested now as a much greater variation in mass density than is observed, although I haven't actually tried to calculate that hypothetical density spectrum.
     
  4. Dec 13, 2009 #3

    Haelfix

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    The horizon problem simply states that regions in space that are seperated by vast distances and moving away from each other rapidly, ought not to share similar properties. They are not in causal contact with each other at any time in the universe's history(and we know how much time they had to communicate, given roughly by the age of the universe and we know the speed of light), so its damn peculiar that their observed temperatures are so similar.

    So for a long time scientists thought that ultimately we would see some amount of departures from homogeneity at very large scales, but the observations kept piling in, and they realized that the tension with theory was growing rapidly out of control.
     
  5. Dec 13, 2009 #4

    BillSaltLake

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    The idea that remote regions of space "ought not to share similar properties" is vague. How similar should the properties of two separated regions be? Without a model that predicts a specific spatial energy density spectrum, we have nothing to compare the observed (smooth) spectrum against, and so the "smooth" observation wouldn't contradict anything.

    In other words, the opinion that it's "too uniform" is just a value judgment until the uniformity is compared against a model that has a specific prediction.
     
  6. Dec 13, 2009 #5

    Nabeshin

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    It seems you are almost arguing that the uniformity in the CMB is not that surprising.

    Considering two causally disconnected regions of spacetime, would you expect them to have the same temperature to one part in a hundred thousand? I do not think one needs a precise model of what one expects these things to look like to say "in general, no." After reaching that conclusion you can say one of two things:
    1) Well that's fortuitous! I'll call it a day.
    2) Since that's extremely unlikely, what made it that way?

    Inflation is obviously the theory of someone in camp #2.
     
  7. Dec 13, 2009 #6

    BillSaltLake

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    Suppose I told you that a certain ball bearing factory was producing steel bearings whose density varied from piece to piece with a variance of 1/100,000. Without any guideline or standard to consult, would you call that "high" or "low" variance? 1/100,000 sounds like a small number, but for all we know this may be an out-of-spec batch, and 1/1,000,000 or less would be normal.

    GR by itself (without QM or inflation) doesn't give us any guideline as to what variance to expect. Someone who examines GR alone might conclude that there is not necessarily any conflict with GR theory vs. the observed uniformity.

    However, I think that adding QM to GR makes it clear that there is a conflict, which inflation (I presume) solves.

    I haven't examined another possibility, however: that gaussian statistical fluctuations in early photon numbers (inverse square root of N) might have caused large-scale fluctuations which could have been amplified by expansion. In this case, GR by itself might predict too much variation, and inflation would be needed.
     
    Last edited: Dec 13, 2009
  8. Dec 13, 2009 #7

    Nabeshin

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    I think your example is an incorrect representation of the situation. I think the appropriate thing to say would be this:
    I find 10,000 random ball bearings from all over the place. I find that their densities are all the same to one part in one hundred thousand. Would I consider this odd? I think anyone would say "Certainly! These ball bearings must have been produced in the same factory/by the same company/from the same mould/etc" But would have to admit some correlation between them. No such correlation exists in the standard framework of GR, which is why we appeal to inflation to solve the problem.
     
  9. Dec 13, 2009 #8

    BillSaltLake

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    Agreed. 1/100,000 is a small number. A question should be raised here-- what order of magnitude of fractional variation in temperature (to give a specific example, let's say over about a 10 degree angle) would you expect to see with GR only if there were no inflation, at the last scattering surface (CMB)?
     
  10. Dec 13, 2009 #9

    Nabeshin

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    Right. That's the logical next question to ask, and I can't give a calculation or reference to one, but I think the point is: significantly larger than 10^-5. My point is that I don't believe we need an exact prediction from GR to say that the homogeneity is more than could be expected.

    I feel like we should be able to do a back-of-the-envelope calculation as to what kind of distribution could be expected, but I don't know of one off the top of my head and am not strong enough in this type of thing to produce one.

    Edit: Spent half an hour looking for something to no avail.
     
  11. Dec 13, 2009 #10
    If initial conditions were the same everywhere (the simplest form of initial conditions, otherwise extra information is encoded in these initial conditions) and laws of physics are the same, then there is absolutely no surprise that from the same initial conditions spacetime+matter everywhere we get identical states later

    For me it is much more difficult to explain that there are still some variations.
     
  12. Dec 14, 2009 #11

    Chalnoth

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    The problem is this: parts of the universe separated by distance scales that, in standard big bang theory, have never been in contact, are highly-correlated. Postulating initial conditions for which this is so doesn't solve the problem, it merely ignores it.
     
  13. Dec 14, 2009 #12
    Why? Initial conditions are also the part of the laws of physics and they have the same status.

    Do you find surprising that Fine structure constant is 1/137.035999679... everywhere?
    Did different parts of the Universe need to be 'in casual contact' in order to 'synchronize' the value?
     
  14. Dec 14, 2009 #13

    Chalnoth

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    In what sense do they have the same status? For every physical theory I am aware of, initial conditions are entirely separate from the theory, and dependent upon which system you're studying.

    That's not the same thing as initial conditions, unless you want to posit a theory wherein alpha is allowed to vary.

    Initial conditions are a particular configuration of the system in question, a configuration that, in general, won't be the same afterwards.

    By contrast, alpha is a parameter in the theory for Quantum Electro-Dynamics (as well as a few others). It is just a number, a number which the theory does not predict but which must be measured experimentally.

    With that little bit of semantics out of the way, yes, I strongly suspect that alpha is actually a parameter which can vary from place to place, that there is a set of physical laws more fundamental than QED of which we are not yet aware that allows this parameter to take various values, and ensures that it remains relatively constant at low energies. Therefore I strongly suspect that parts of the universe that are not in causal contact with one another take different values for the fine structure constant.

    Finally, let me just point out that even if you don't like my semantics here, the point remains that positing initial conditions doesn't solve the horizon problem: in order to explain something, by Occam's Razor, your explanation must have fewer parameters than that which it explains. But simply stating what we think needs explaining is actually just the initial conditions doesn't do this, so it isn't an explanation, and we are still left with the problem.
     
  15. Dec 14, 2009 #14

    Say, equations of toe are:

    TOE(Omnium(t)) = 0
    (1)

    Now we want to define initial conditions at t=0:

    Omnium(t=0) = 0
    (2)

    So we just have 2 equations defining the behavior of Omnium. We can call (1) 'physical laws' and (2) 'initial conditions' but in fact, there is no difference.
     
  16. Dec 14, 2009 #15

    Chalnoth

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    I don't think you answered my post at all.
     
  17. Dec 14, 2009 #16
    I had answered why they have the same status.

    When we talk about some subsystem of the Universe, there is a difference between 'laws' and 'boundary conditions'. At least the difference is obvious.

    When we talk about the WHOLE UNIVERSE there are no boundaries, so 'boundary' or 'initial' conditions are just additional limitations. So if TOE equations do not give the unique solutions for the Omnuim, then we add additional conditions, in a form of additional equations.

    So we have a longer list of equations - nothing more.

    Regarding your point about the Ocamms razor. It is on MY side. Compare 2 cases:
    a) void initial conditions Omnium(x,y,z,0)=0
    b) void except 73 pre-existing particles forming 4 triangles. Why 73? Not 74? Why triangles? isnt that obvious that only a) is the simples case and is resistent to Ocamms razor?
     
  18. Dec 14, 2009 #17
    Additional note.

    'Initial' conditions should not necessarily define the state of the Universe at t=0.
    For example, because time might be not well defined at the Big Bang.
    So again, they are just additional equations.
     
  19. Dec 14, 2009 #18

    Chalnoth

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    Except initial conditions are just a particular type of boundary condition. So you're contradicting yourself.

    There's a fundamental difference between an equation that lists a relationship between components of the universe at any place or time, and one that specifically lays out a particular configuration of the universe at one time (or place, for that matter).

    First you'd have to show that those 73 pre-existing particles extend necessarily from setting that particular initial condition, instead of some other. As near as I can tell, you gain nothing at all in the theory from setting that initial condition.
     
  20. Dec 14, 2009 #19
    No. Take newtonian Universe.

    Fg = G * M1*M2 / r^2

    This is a physical law. Now we can examine how 2 bodies will behaive:

    M1=1, M2=0.1
    at t=0: X1=.. Y1=.. Z1=.. Vx1=.. Vy1=.. Vz1=...
    X2=.. Y2=.. Z2=.. Vx2=.. Vy2=.. Vz2=...

    These are initial conditions.

    The boundary (t=0) is artificial of course. Boundary conditions can define the state of the system at t=0, or any other property.

    For example, "and at least 2 masses of N will ever collide". This condition is also a boundary one, even it does not define the state of the system at some t, it just claims that system reaches some property at some t.

    Now whats about QM where we can not in principle know the complete state of the system? Or when no 'global' physical t can be defined (like in cosmology).

    Lets say that in TOE the very notion of TIME is emergent. What are you going to do if you dont have t=0? In TOE very likely you have neither time nor spacial boundary. How, in principle, you can define 'boundary' conditions?

    The very words 'boundary' or 'initial' (=boundary at earlier time) in general do not make any sense if we think about the whole universe. They are 'boundary' or 'initial' only in mathematical sense, defining some additional restrictions of the solution.
     
  21. Dec 14, 2009 #20

    George Jones

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