# Radiation dominated universe in Newton's approximation (no )

• Mishra
In summary, the conversation discusses the use of Newton's approximation (without GR) to find Friedman's equation by assuming a homogeneous and isotropic universe and using F=ma and conservation of energy. It is found that the scale factor goes as t^2/3 for a matter dominated universe. The same approach is attempted for a universe dominated by radiation, but it is noted that Newtonian physics does not account for the gravitational effect of radiation. It is suggested that a modification to Newtonian physics is needed to introduce this interaction. However, this modification gives a different result than General Relativity, which accounts for both pressure and energy density in the gravitational force. The conversation ends with a discussion about the factor of 2 difference between the "Newton
Mishra
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 ?

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.

Mishra said:
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 ?
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.

mfb said:
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.
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##.

mfb
Chalnoth said:
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).

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.

PeterDonis said:
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.
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.

Chalnoth said:
You're right that the math is a bit different in the two cases. But the concept is the same.

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.

PeterDonis said:
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.
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.

Chalnoth said:
the photon also has its own impact on the curvature of space-time

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.

Chalnoth said:
by the equivalence principle, its impact on the curvature must be the same as its response to that curvature.

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.)

## What is a radiation dominated universe?

A radiation dominated universe is a theoretical model in cosmology which suggests that in the early stages of the universe, radiation was the dominant form of energy. This means that the energy density of radiation was much higher than that of matter or dark energy.

## How does Newton's approximation play a role in a radiation dominated universe?

In Newton's approximation, the effects of gravity are calculated using his laws of motion and his law of universal gravitation. In a radiation dominated universe, the effects of radiation on the expansion of the universe are also taken into account using Newton's laws. However, this approximation is only applicable in the early stages of the universe and is not accurate for later stages.

## What is the significance of a radiation dominated universe in the study of cosmology?

A radiation dominated universe is important in cosmology because it helps us understand the early stages of the universe and how it evolved over time. It also helps us understand the role of radiation in the expansion of the universe and its effects on the formation of structures such as galaxies and clusters of galaxies.

## How does a radiation dominated universe differ from a matter dominated universe?

In a matter dominated universe, the energy density of matter is higher than that of radiation. This means that matter has a stronger influence on the expansion of the universe compared to radiation. This model is more accurate for later stages of the universe, whereas a radiation dominated universe is more applicable in the early stages.

## What evidence supports the idea of a radiation dominated universe?

One of the main pieces of evidence for a radiation dominated universe is the cosmic microwave background radiation, which is leftover radiation from the early stages of the universe. This radiation is uniform in all directions, which supports the idea of a homogeneous and isotropic universe, as predicted by a radiation dominated model.

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