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Conformal symmetry of FRW spacetime

  1. May 22, 2015 #1
    The standard spatially flat FRW metric in Cartesian co-moving co-ordinates is given by:

    $$ds^2=dt^2-a(t)^2(dx^2+dy^2+dz^2)$$
    As far as I understand it the fact that the metric can be written in a form that is independent of ##x,y,z## implies that the Universe has the physical qualities of being spatially homogeneous and isotropic.

    But by writing the FRW metric in another way one can see that the Universe has another symmetry.

    $$ds^2 = a(t)^2(\frac{dt^2}{a(t)^2}-dx^2-dy^2-dz^2)$$

    If we introduce conformal time ##\tau## defined by:

    $$d\tau = \frac{dt}{a(t)}$$

    we get:

    $$ds^2 = a(\tau)^2(d\tau^2-dx^2-dy^2-dz^2)$$
    This way of writing the metric displays a conformal symmetry: it stretches equally in (conformal) time and space.

    What physical quality does this conformal symmetry imply for the Universe?

    Here are some of my speculations:

    Photons obey Maxwell's source free equations which are conformally invariant.

    Does the conformal symmetry of the FRW metric imply that photons actually propagate with constant energy/frequency?

    Maybe the time we actually measure is conformal time and the redshift that we see is due to our energy scales increasing with the age of the Universe. If expanding photon wavelengths have constant energy then as our atoms have a fixed size their energies will increase with the scale of the Universe.
     
  2. jcsd
  3. May 22, 2015 #2

    Chalnoth

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    Science Advisor

    The time we actually measure is the integral ##\int ds## of the metric over a path where ##dx = dy = dz = 0##. So in these coordinates, ##s = \int a(\tau)d\tau = \int dt##.

    Essentially, if we were to use conformal time to do physics in local systems, we'd end up with a whole bunch of extraneous factors that depend upon ##a(\tau)## that unnecessarily complicate the equations.
     
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