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This may seem silly, but how do they know that the effects we are seeing from dark energy, (which we only assume must be there due to observed acceleration), are not from gravity pulling us out from the other side?

Say there's all of this fast moving energy like light, and it extends further out than slower moving galaxies in the sphere of the entire universe.

Would the gravity of all those particles be enough to produce the observations that we assume are due to a pushing effect?

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  • $\begingroup$ you are making the mistake of visualizing the expansion like a sphere in three dimensions expanding. Even though the gravity we measure creates trajectories and orbits in our three dimensions, dark energy is a phenomenon in the four dimensional space time. "all those particles" are all around us symmetrically in three dimensions. see hyperphysics.phy-astr.gsu.edu/hbase/mechanics/earthole.html as an example of how only the mass internally affects the acceleration, not the external one to the trajectory. $\endgroup$
    – anna v
    Commented Apr 3, 2014 at 6:41
  • $\begingroup$ Possible duplicate: physics.stackexchange.com/q/15546/2451 $\endgroup$
    – Qmechanic
    Commented Apr 3, 2014 at 6:41
  • $\begingroup$ See the question Qmechanic has linked. If our bit of the universe is below the average density this could account for the acceleration. However this requires us to be at the centre of the underdense region, which seems a bit of a coincidence. Also there is other evidence for dark energy. $\endgroup$ Commented Apr 3, 2014 at 7:07
  • $\begingroup$ See for example Void or Dark Energy? and Local Void vs Dark Energy: Confrontation with WMAP and Type Ia Supernovae $\endgroup$ Commented Apr 3, 2014 at 7:28
  • $\begingroup$ It would depend on how much gravity all that light and other light speed moving particles has, and where it is now. Did it continue in a straight line, or curve around in an orbit of the gravity of the universe? By the way, I didn't like the answers to that other question, gravity extends beyond the visible horizon. $\endgroup$
    – rowanman28
    Commented Apr 3, 2014 at 7:49

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