# How would radiation pressure inside of the Sun affect an indestructible person?

I went down something of a rabbit hole regarding whether an indestructible person trying to walk on the Sun would fall in, and sink. I want to know how radiation pressure would affect that person.

As I understand it, a person would have to sink approximately 0.45 the solar radius in order to experience a density approximately equal to the density of the ground on Earth.

But how would radiation pressure effect that person? I am vaguely forming the understanding that the effects of radiation pressure vary based on surface area, but I am unsure of how to perform the necessary calculations. People are saying that a person would sink in, so I am guessing the surface area would be too great for radiation pressure to have an effect, but I do not understand why that is true.

Additionally, is there any depth into the Sun at which point a person would stop sinking due to gravity (without rising) purely as a result of radiation pressure instead of density?

• Just FYI: to find how far someone would sink you need to find the depth at which they are neutrally buoyant, rather than the point where the density is "approximately equal to the density of the ground on Earth", unless you're saying that your "indestructible person" has a density that's similar to some specific type of ground on Earth, rather than the average density of a normal human. Feb 4 at 18:21
• more about sinking: At what depth below the Sun's surface does the density reach that of water? and answers & links therein
– uhoh
Feb 5 at 6:16

Radiation force is totally negligible at the surface of the Sun - about one hundred thousandth the force due to gravity.

The radiation force does grow as you move into the solar interior, but only as the inverse square of the radius considered, but will not increase much inside the radius at which the solar luminosity begins to be produced (the core). As a result, the radiation force will still be about a ten thousandth of gravity or less inside the Sun.

In order to increase the influence of radiation you need to increase the surface area exposed to that radiation whilst keeping the mass (and hence gravitational force) of the object constant.

(Note added: I took you question to be literally about radiation pressure due to light. If you replace "radiation pressure" with just "pressure", then indeed the point where you would "float" is where the density is about that of the human body, halfway into the solar interior. This is because at that point the pressure gradient balances the gravitational force for material of that density. However, the gas pressure at this depth is about a billion atmospheres, so you would be crushed.)

Details:

The flux of radiation (power per unit area) from the surface of a blackbody is $$\sigma T^4$$, where $$T$$ is the temperature and $$\sigma$$ is the Stefan-Boltzmann constant.

The force due to radiation is the radiation flux integrated over the surface area $$A$$ that an object presents to the radiation, and then divided by the speed of light, $$c$$.

The force due to gravity is just $$mg$$, where $$m$$ is the mass of the body and $$g$$ is the local gravity.

The ratio of the two is given algebraically by $$\frac{F_r}{F_g} = \frac{ A \sigma T^4}{mg c}$$ and for a spherical object of mass $$M$$ inside radius $$R$$, then the surface gravity is $$g= GM/R^2$$, so $$\frac{F_r}{F_g} = \frac{ A \sigma T^4 R^2}{GMm c}\ .$$

At the surface of the Sun $$T=5800$$ K, $$\sigma = 5.7\times 10^{-8}$$ in SI units and we could assume that a person presents an area $$A \simeq 1$$ m$$^2$$ and has a mass of 80 kg. The mass of the Sun $$M=2.0\times 10^{30}$$ kg and the radius $$R = 7.0\times 10^{8}$$ m. Putting these numbers in, we find $$\frac{F_r}{F_g} \simeq 10^{-5}$$ i.e. That the radiation force is a very small fraction of the gravitational force and could probably be ignored for most purposes.

You might then have thought that because the temperature in the interior of the Sun is much hotter (approaching 15 million K at the centre), that the radiation force/radiation pressure would become much more important compared to gravity. That is not the case. The radiation flux coming from the centre is just the luminosity of the Sun divided by the surface area of a sphere at the solar interior radius considered (note that this is true when the energy is carried solely by radiation and takes full account of the fact that radiation is hitting you from all sides). That is because the solar luminosity is all generated by fusion reactions in the core. In fact if you get very close to the centre then only a fraction of the luminosity should be considered becase some of the energy generation would be "above" you. The solar gravity would also change because you need only consider the mass interior to your radius (which goes down) and because the radius you are at is getting smaller. The ratio of radiation force to gravity becomes $$\frac{F_r}{F_g} = \frac{A L R^2}{4\pi GM( where $$M( is now the mass inside radius $$R$$. You can see that the $$R^2$$ terms cancel and so we see that this ratio will grow as you move into the Sun, since $$M( would get smaller, at least until you get to the core at about $$R= 0.2R_\odot$$, when $$L$$ would start to decrease too.

Now, you should be able to convince yourself that if you use $$R=7\times 10^8$$ m, $$L= 3.8\times 10^{26}$$ W and $$M( kg, appropriate for the surface, then you get the same answer as before. The mass of the Sun is however quite centrally concentrated so even if you move close to the solar core with $$R \simeq 10^7$$ m, you are only going to increase this fraction by a factor ofa few and thus the force due to radiation is still a small fraction of that due to gravity. An exact answer would need a detailed model of the run of $$M( with $$R$$ and of $$L( with $$R$$.

A further complication is that in the outer 20% of the Sun, the energy is mostly carried by convection rather than radiation (see https://astronomy.stackexchange.com/a/43196/2531). In this convective zone, the radiative forces will be even lower than estimated above by factors of a few.

• There might be a typo in the equation above i.e. That the radiation force is a very small fraction of the gravitational force and could probably be ignored for most purposes., since it seems to show that the gravitational force is very much smaller than radiation force. Feb 5 at 3:49
• @Allure correct, thanks. Fixed. Feb 5 at 6:53
• Question. The flux $\sigma T^4$ is the absolute flux emenating from a blackbody, right? The resulting radiation pressure inside the Sun though depends on the "net flux", the delta between the "out-flux" and the "in-flux" from the direction of the surface. That delta should be very small inside the sun where it's almost isotropic radiation from all sides; 10 inches behind me, the blackbody has almost (but, of course, not quite) the same temperature as 10 inches in front of me, when I look towards the center. Feb 5 at 15:12
• What I mean is that a body hovering somewhere above the Sun's surface (with little radiation coming from "above") feels the full impact of the photosphere's radiation. A body embedded in the photosphere though would experience almost no net force from radiation pressure. -- A little voice mumbling "preservation" in my ear says that the net radiation emanating must have made its way through the underlying layers, that is, the "net flux" must be the same integrated over the spherical surface at that depth at all depths. If we neglect convection transport, that is, which is probably a stretch. Feb 5 at 15:26
• @Peter-ReinstateMonica ignoring convection for a moment, the entire luminosity of the star must equal the net outward radiative flux multiplied by the surface area, unless you are inside the energy generation zone. I have not used the $\sigma T^4$ prescription inside the star, but you can see that $L/4\pi R^2 = \sigma T^4$ at the surface. Your point about convection is fair enough. The radiative flux is lower in a convection zone, so makes outward radiation force even less important. Feb 5 at 16:18