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Radiation dominated universe in Newton's approximation (no )

  1. Jun 7, 2015 #1
    Hello,

    I just recently found out that one could find the Friedman's equation in Newton's approximation (without GR) by assuming that the universe in homogeneous and isotropic simply by using F=ma and the conservation of energy.
    On can then find that the scale factor goes as t^2/3, as expected for a matter dominated universe.

    I am trying to do the same thing for a universe dominated by radiation.
    The energy inside a box filed with radiation would be ##E=h/\lambda\sim h/R## where ##R## is the scale factor. The energy density would be ##\rho=E/V\sim E/R^3 \sim h/R^4##. One can see the extra dilution factor 1/R when dealing with radiation. If this expression for ##\rho \sim 1/R^4## is put in the Friedman's equation we find ##R\sim t^{1/2}##, as expected for a radiation dominated universe.

    What I would like to do is to find the Frieman's equations using Newton and this idea (##E_{rad}\sim 1/R^4##). I But I cannot workout the conservation of the energy nor the Newton's equation F=ma.

    Would anyone have a lead ?
     
  2. jcsd
  3. Jun 7, 2015 #2

    mfb

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    In Newtonian physics, radiation does not have mass. It can fall down, but it does not attract other stuff. I guess you have to modify Newtonian physics to introduce such an interaction.
     
  4. Jun 7, 2015 #3

    Chalnoth

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    For normal matter, the first Friedmann equation does not change if derived from Newtonian gravity. The result is exact in either case. Here is one derivation I found, for example:
    http://www.astronomy.ohio-state.edu/~dhw/A5682/notes4.pdf

    This shouldn't be too terribly surprising, as General Relativity reduces to Newtonian gravity in the weak field regime, and the uniform-density fluid in a homogeneous, isotropic universe has no strong gravitational fields.

    But, as mfb mentioned, Newtonian gravity has no notion of the gravitational effect or response of radiation.
     
  5. Jun 7, 2015 #4

    Chalnoth

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    There is a way to do this, but it gives the wrong answer.

    What you can do is take a gravitational potential well, and find the path of an object passing through that potential well in the limit as its mass approaches zero and its speed approaches c. It turns out that the result does converge to a specific value, but that value is different from the General Relativity value by a factor of two.

    The reason here is that in General Relativity, both pressure and energy density contribute to the force. In Newtonian gravity, if you have a gas with some energy density ##\rho## and pressure ##p##, it's going to have a gravitational force proportional to ##\rho## alone (note: energy density and mass density can be treated identically here, as ##E = mc^2##: it's just a unit conversion factor that differs). In General Relativity, the force is proportional to ##\rho + 3p## (the three comes in because of the three dimensions of space, and the pressure pushes out in each of those three directions). With photons, which have ##p = \rho/3##, the force is proportional to ##2\rho##.
     
  6. Jun 7, 2015 #5

    PeterDonis

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    This is how you derive the stress-energy tensor due to radiation, which is what's relevant for the Friedmann equation, but it's not how you derive the bending of light by a massive object like the Sun.

    The factor of 2 difference between the "Newtonian" value for light bending and the GR value is not because of the light's pressure; it's because in GR, the "acceleration due to gravity" in the weak field limit has a velocity-dependent factor. This factor is 1 in the limit of zero velocity, but it is 2 in the limit of velocity going to the speed of light. This factor cannot depend on the stress-energy tensor of the test object (the light beam or anything else); if it did, it would violate the equivalence principle.
     
  7. Jun 8, 2015 #6

    Chalnoth

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    You're right that the math is a bit different in the two cases. But the concept is the same. In the case of the single photon, you have the momentum which gravitates. In the case of a gas of photons, the momentums of the individual photons cancel out (assuming you're at rest with respect to the gas), but contribute to the pressure of the gas instead.
     
  8. Jun 8, 2015 #7

    PeterDonis

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    No, it isn't. In the case of the single photon having its path bent by a mass like the Sun, the fact the photon's momentum gravitates (i.e., contributes to its stress-energy tensor) is irrelevant. The equation of motion for the photon does not include the photon's stress-energy tensor, or its "mass", or anything like that. It can't, by the equivalence principle.
     
  9. Jun 8, 2015 #8

    Chalnoth

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    Sure, but the photon also has its own impact on the curvature of space-time, and by the equivalence principle, its impact on the curvature must be the same as its response to that curvature.
     
  10. Jun 8, 2015 #9

    PeterDonis

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    Which is negligible for light bending; that's why light bending is not a good analogy for understanding the dynamics of a radiation dominated universe.

    No, that's not what the EP says. It's true that "inertial and gravitational mass must be equal" is a fairly common misstatement of the EP, but it's still a misstatement, based on leftover Newtonian intuitions that are not valid in GR. In GR, there is no such thing as "gravitational mass"; there is no single property of an object that determines the spacetime curvature it produces. The stress-energy tensor has ten components. Also, gravity is not a force in GR, so there is no property of an object that determines its "response to the gravitational force". Once again, look at the equation of motion for a freely falling object in Schwarzschild spacetime; you will not see the object's "mass" anywhere. And, as I said, it's that last statement that really embodies the EP--or at least the part of it that talks about free-fall motion. (There's also a part that talks about accelerated motion, but that's proper acceleration and doesn't come into play in what we're currently discussing.)
     
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