The Wikipedia article on Radiation Pressure gives an equation for the pressure on a perfect reflector produced by a beam of photons with an energy flux $E_f$ (in units of power/area) as


where $\theta$ is the angle of incidence between the incoming photon beam and the surface normal of the reflector, and $c$ is the speed of light. Maximum pressure is at normal incidence, and it drops to zero at 90° when the incidence is perpendicular to the surface normal.

The $cos^2(\theta)$ behavior can be understood as follows; the component of the photon's momentum perpendicular to the surface scales as $cos(\theta)$, and the projected cross-sectional area of a flat surface exposed to the incoming light also varies as $cos(\theta)$. Together they result in the $cos^2(\theta)$ behavior.

However, Wikipedia's article Solar Sail mentions both the simple $cos^2(\theta)$ shape and also a more complicated, "realistic" shape with the form

$$0.349 + 0.662cos(2\theta) - 0.11cos(4\theta)$$

and states that realistically the force drops to zero at around 60° rather than at 90°.

An actual square sail can be modeled as: F = F0 (0.349 + 0.662 cos 2θ − 0.011 cos 4θ) / R2 Note that the force and acceleration approach zero generally around θ = 60° rather than 90° as one might expect with an ideal sail.(18)

I don't understand how the square shape of the sail has anything to do with the pressure, and I don't understand why it would go to zero near 60°. Reference (18) is given as Space Sailing, Jerome Wright (1992), Gordon and Breach Science Publishers, Appendix B.

I've pasted that expression into an internet search and I see that it is repeated in many places, with similarly worded text, but so far I have not found any explanation for why it is supposed to better represent the performance of a solar sail.

Question: What is the physics behind this approximate expression? Why would the pressure drop from maximum to zero at about 60°? Would a square shape have anything to do with it at all?

Beyond 60° the expression goes negative; are we supposed to clip it at zero, rather than there being an attractive force due to negative photon pressure, or does this represent a known physical phenomenon? Under some conditions light can be used as an attractive force as in the case of a laser trap, but that usually involves either a strong gradient, or tuning to a specific resonance. In this case the light is uniform and the interaction is non-resonant.

In the plot below, the simple $cos^2(\theta)$ behavior is shown by the thick solid (blue) curve as a function of incident angle $\theta$, and the more complex expansion is shown by the thinner solid (green) curve. Zero is indicated by the dashed (black) line.

enter image description here

  • $\begingroup$ The physical principles section of the WP page mentions billow, and it's clear how a billowed sail receiving incident pressure from the side would develop force in the "wrong" direction (though not why the sail would remain billowed in that case). 60 degrees also seems like a very early crossover, implying a very deep billow. $\endgroup$ Commented Jul 17, 2017 at 14:14
  • $\begingroup$ @RussellBorogove are you saying this equation is meant to apply to a generically curved (billowed) surface, and it does not apply to a flat surface? In the case of specular reflection the force at any point will always be in the direction of the local surface normal (not true of absorption or some kinds of diffuse reflection though). I'm not sure what "wrong direction" means here. $\endgroup$
    – uhoh
    Commented Jul 17, 2017 at 15:02
  • 1
    $\begingroup$ The graph in the question is wrong. The actual graph should look like this: desmos.com/calculator/stxdjtz5a5 . Note that the three numerical coefficients sum to 1. The two expressions are each normalized to 1 at theta=0. $\endgroup$
    – user687
    Commented Jul 17, 2017 at 18:13

1 Answer 1


I doubt that we are going to answer this without finding a copy of the book by Wright or contacting him. The following is therefore not a complete answer, but it's too long to fit in a comment. The WP article's history shows that Wright was one of the article's main early authors, and that he wrote the skimpy text introducing the equation. Wright has a web site, where he has a very brief presentation of some information that is presumably given in more detail in his book.

The ideal function can be written as $$f(\theta)=\frac{1}{2}+\frac{1}{2}\cos2\theta,$$ while the numerical expression given in the WP article is of the form $$g(\theta)=A_0+A_2\cos2\theta+A_4\cos4\theta,$$ where they seem to have chosen the normalization such that $A_0+A_2+A_4=1$.

On one page on Wright's web site, he has the following somewhat more detailed description of how $g$ was obtained: "A ship can have multiple sails, each with different orientations. Each sail has some curvature in it. Using finite element analysis, the local pressure on parts of sails can be integrated to determine the total force and moments acting on the ship. A resulting approximation of the force versus its angle theta, valid for a square sail, is: [...]"

It is theoretically possible to have $g<0$ without any exotic physics. For example, one could imagine a system that accepted light coming in at an oblique angle, sent it through fiber optic cables, and threw it back out along the normal to the surface, on the dark side of the surface. However, it's not clear to me at all how anything like this can happen with a solar sail. In particular, it seems very difficult to explain the fact that $g$ attains a large negative value at $\theta=90$. At this angle, a non-billowing sail will intercept zero sunglight, so there should be zero force. Even if the sail is billowing, the amount of sunlight intercepted should be very small. Furthermore, at 90 degrees you would expect the component of the force along the sail's average normal to be zero by symmetry. The only way I can think of to break this symmetry would be if the sail had unequal absorbance and emittance on its two sides. Even so, this result seems very implausible.

  • $\begingroup$ As far as "I doubt that we are going to answer this without finding a copy..." it looks like you are converging on the answer anyway :) I'll check back later. $\endgroup$
    – uhoh
    Commented Jul 17, 2017 at 19:11
  • $\begingroup$ OK I think it's clear from the short paragraph in the link that this equation does not represent the pressure on a single flat reflecting surface, but is the result of modeling a complete spacecraft with a complex shape including various un-flat surfaces, such as this one shown on the site i.sstatic.net/nJpiv.jpg Since we can't see anything further, the equation is pretty much useless to other readers of the Wikipedia article who come looking for helpful equations and explanations and should probably be removed. Thanks for your answer and for tracking that down! $\endgroup$
    – uhoh
    Commented Jul 18, 2017 at 5:48

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