# Radiation pressure on a Dyson sphere

To find the outward pressure from the sun's light on an enveloping spherical shell (Dyson sphere), one can simply divide the insolation by $c^2$. Using the entire system, we can specify the power of the sun in watts, $W$, leading to a pressure of $P=W/(4 \pi R^2 c^2)$.

The problem is that a Dyson sphere structure emits its own thermal radiation. So then one would think that the pressure from the thermal radiation on the outside surface of the Dyson sphere could just cancel with the inside radiation pressure from the sun since the power rates are the same.

There's one more problem, the outer surface's thermal radiation has a different directionality. There would be a net pressure pushing outward since the sun's light is preferentially directed outward. What would that be?

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Great question. Are you considering a vanishingly thin sphere, where light is absorbed and the re-emitted isotropically, with possible paths back into the sphere? Or are you imagining that the sphere has finite, non-negligible thickness, with different temperatures on the inner and outer surface? – kleingordon Jan 24 at 6:55
@kleingordon Ah ha, you ask of the addition outward radiation pressure due to the exchange of photons from points on the inside the sphere to other points on the inside of the sphere.  That should be included to be most realistic, because you can't make the inner surface cooler except with active cooling.  I would approximate the inner and outer surfaces to be the same temperature myself.  However, I think the central challenge of the problem will be solved with or without that detail. – AlanSE Jan 24 at 12:57

## 2 Answers

The basic assumption here is isotropic emission on the surface of the sphere, which is exactly $W$, the power production of the star. I thought that would be a shockingly simple conversion, and I was right. Using spherical angles of $\theta$ for azimuthal and $\phi$ for polar angle, the fraction the photon momentum is reduced by due to the angle factor is $\cos{(\phi)}$. Let it be known here that I'm using the Wolfram Mathworld spherical convention. As per that reference, a unit of solid angle will be $dA = \sin{(\phi)} d\phi d\theta$.

Now, I care nothing about the total number of photons emitted as I'm calculating the correction factor for angle. So for this factor, I divide the outward momentum by the integral of $1$ over the solid angular interval. This solid angle interval is a $2 \pi$ half circle pointing outward on the surface of the Dyson sphere, and $\phi=0..\pi/2$, $\theta=0..2 \pi$.

$$\frac{\text{momentum}}{\text{number}} = \frac{ \int_{2\pi} \cos{(\phi)} dA }{\int_{2\pi} dA }= \frac{\int_0^{2\pi} \int_0^{\pi/2} \cos{(\phi)} \sin{(\phi)} d\phi d\theta }{\int_0^{2\pi} \int_0^{\pi/2} \sin{(\phi)} d\phi d\theta} = \frac{1}{2}$$

What is this arguing? It is arguing that the inward pressure from the emission of $W$ on the outside surface of the Dyson sphere is $1/2$ the outward pressure from the absorption of sunlight on the inside surface.

Let's assume that the inner surface is the same temperature as the outer surface. In that case, both sides of the thin Dyson Sphere will exert a pressure of $W/(2 c^2 A)$ from thermal emission. These would cancel each other out, but the inside surface has re-absorbs 100% of what it emits (as per comments). So here are the additions and subtractions to outward (or lifting) pressure.

• outside surface radiation: $-W/(2 c^2 A)$
• inside surface radiation: $W/(2 c^2 A)$
• inside surface re-absorption:$W/(2 c^2 A)$
• sun's radiation absorption: $W/(c^2 A)$

Now, for the total, we have a correction factor of $-1/2+1/2+1/2+1=1.5$. This is the factor to correct the $P$ value I gave in the question in order to get the correct radiation lifting pressure on a Dyson Sphere.

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Well-reasonsed. By the way, I think both here and in the question itself, you should be dividing by $c$ instead of $c^2$ (check the units). I'm currently trying to come up with an answer that works at the photon-by-photon level to see if I get the same answer. If I don't, I want to try to figure out what the error in my reasoning is. More to come soon – kleingordon Jan 28 at 0:02
@kleingordon That's right. I should have been thinking $E=pc$ whereas I was stupidly thinking some form of E=mc2 that didn't apply. – AlanSE Jan 28 at 3:02

I suspect there's a fallacy lurking somewhere in the following argument, so I want people to point it out for me in the comments. I post this purely for self-pedagogical purposes.

Let's think about this in a photon-by-photon manner. We assume the temperature of the sphere remains constant in time. The only heating and cooling mechanisms are radiative. Then, for a given interval in time, the sphere must emit exactly as many photons as it absorbs. I therefore assume that the time between an absorption event and the corresponding emission event to be effectively zero.

Consider a photon emitted by the sun. It will deposit its full momentum $p$ on the sphere over a time interval $R/c$, resulting in a force of $pR/c$.

The absorption of this photon will lead immediately to a re-emission, as discussed above. With a probability of $1/2$, the emitted photon will be sent out of the sphere, carrying average outward momentum $p/2$ as discussed in Alan's answer. The net momentum transferred to the sphere is then $p(1 - 1/2)$ = $p/2$, over a time interval of $R/c$, so the force is $pc/(2R)$. If we let $n$ denote the number of re-emissions into the sphere as the result of the absorption of an initial stellar photon, then in this case $n=0$, the force is $pc/(2R)$, and this outcome has probability $Pr_0 = 1/2$.

Now consider the $n=1$ case: the absorption of a photon from the sun results in a re-emission into the sphere, this re-emitted photon is then absorbed back by the sphere, and that the subsequent emission is out of the sphere. I'm treating the chance that the photon runs into the star at the center to be negligible. Note that $Pr_1 = 1/4$ because there's a 1/2 chance that $n=0$, then a $1/2 * 1/2$ chance that $n>1$. To find the momentum per time deposited in the sphere when the inward-directed photon is emitted and re-absorbed, one must integrate over all cords of the sphere emanating from the emission point, taking into account the angle of the photon with respect to the normal direction of the sphere. I can go into more detail for this step if requested, but the result is an average momentum/time (force) of $2 p c /R$. Then, the total force exerted by the photon for the $n =1$ event is $pc/(2R) + 2 p c /R$.

We can now sum up the forces over all $n$, weighted by $Pr_n$. I claim that $Pr_n = (1/2)^{n+1}$ ; you can think of it as the probability of flipping a coin $n+1$ times and getting tails the first $n$ times, then heads on the last flip. The formula is \begin{equation} {\rm Force \:per \:photon} = \frac{pc}{R}\sum_0^\infty(1/2)^{n+1}(1/2 + 2n) = \frac{5}{2}\frac{pc}{R} \end{equation}.

Let $N_\ast$ denote the number of photons emitted by the star per time. We can then relate $N$ to the luminosity of the star, $W$, by $N_\ast = W/(p c)$. The rate of increase of force on the sphere with time is then $N_\ast \frac{5}{2} \frac{pc}{R} = \frac{5}{2} \frac{W}{R}$. If we imagine instantaneously constructing the sphere at time $t = 0$, or similarly "turning on" the star at $t=0$, then the total outward force on the sphere would be \begin{equation} F_{\rm total} = \frac{5}{2} \frac{W}{R} t \end{equation}

In summary, this argument seems to suggest that the force on the sphere grows in an unbounded manner over time. The temperature of the sphere remains constant, but the temperature of the radiation field inside the sphere continues to grow with time, eventually cooking everything inside (unless they figure out a way to store/use the radiative energy quickly enough). What went wrong in this analysis?

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For one to give the force per photon, they must answer "at what time", and because of this I think one would do better to quote the impulse per photon.  The infinite series is a neat way to look at it, but this non-steady-state force summing may result in a huge math problem.  The time of first wall hit is deterministic (R/c), but subsequent hits are distributed at times 0 to 2R/c.  This results in infinite sums of piecewise functions familiar to particle transport.  Even if you can find a way around that, force per photon should be a decreasing function of time and not a constant. – AlanSE Jan 28 at 3:24
Every time I use the word force, I'm talking about an average momentum deposition over a time interval. I can see how this might be a problem considering that for some of these averages, the time interval must extend to an arbitrarily large value of time (for photons that remain trapped in the sphere). This might indeed pose a problem for my analysis. However, what if one were to focus on energy for a moment, and not momentum? What is wrong with the photon-by-photon argument that the radiation energy density inside the sphere grows without bound over time? – kleingordon Jan 28 at 3:35
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