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I Does the Planck length expand in a FRW universe?

  1. Jun 5, 2017 #1
    By using particle physics natural units with ##\hbar=c=1## so that Planck's length ##l_P=(8\pi G)^{1/2}## we can express Einstein's field equations as
    $$G_{\mu\nu}=l_P^2\ T_{\mu\nu},$$
    where ##G_{\mu\nu}## has dimension ##[\hbox{proper length}]^{-2}##, ##l_P## has dimension ##[\hbox{proper length}]##, ##T_{\mu\nu}## has dimension ##[\hbox{proper length}]^{-4}##.

    In cosmology we assume the expanding FRW metric. If we assume flat space for simplicity and cartesian coordinates then we have the following line element
    $$ds^2=-dt^2+a^2(t)\left(dx^2+dy^2+dz^2\right).$$
    Therefore an interval of proper length in the x-direction for example is given by
    $$ds=a(t)dx$$
    If ##l_P## is a proper length then should it expand with the scale factor ##a(t)## or should it remain constant?

    In order for it to remain constant then its corresponding comoving interval ##dx \sim 1/a## which seems unnatural to me.

    Therefore I think that as ##l_P## is a proper length it should expand with the scale factor ##a(t)##.

    Does this make sense?
     
  2. jcsd
  3. Jun 5, 2017 #2

    PeterDonis

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    Staff: Mentor

    ##l_P## isn't a proper length. It's a physical constant that happens to have units of proper length.

    The proper length ##ds = a(t) dx## is the proper distance between two events happening at the same time ##t## and separated by a spacelike coordinate interval ##dx##. If we hold ##dx## constant (i.e., we have two comoving objects), this proper distance increases with ##t##, since ##a(t)## does. So the ratio of ##ds## to ##l_P##, i.e., the number of Planck lengths between the two comoving objects, will increase.
     
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