When a sphere of radius $r$ is placed in the path of a parallel beam of light of intensity $I$, the force exerted by the beam on the sphere is given by:

$$F=\frac{\pi r^2 I}{c}$$

I derived the above result by assuming the sphere to be perfectly reflecting. However, it turns out that the force exerted by a light beam of same intensity on a perfectly absorbing sphere of same radius is also given by the same formula. Further, even if the sphere partially reflects and partially absorbs the incident photons, the force exerted on it by the beam remains the same. I understood the final case (partially absorbing and reflecting) by imagining it to be a combination of the first two cases - totally reflecting and totally absorbing.

In short, the force exerted by the light beam on a sphere depends only on the area obstructed by the object, here it's just the area of the biggest circle in a sphere ($\pi r^2$). I understood the mathematics behind this result. But, this seems to be counter-intuitive for me because, the change in momentum in case of total reflection is twice that of the case when the light beam is totally absorbed. The force exerted on the object is nothing by the rate of change in momentum and therefore the force on the object which totally reflects is more compared to totally absorbing or partially absorbing objects.

However, in case of spheres placed in the beam, the force on it remains the same irrespective of the amount of light absorbed or reflected. What is the intuitive reason behind this fact? Also, is this a property of only spherical objects or are there even more examples for this?

  • 1
    $\begingroup$ I have deleted my answer because it was wrong. This simple video youtube.com/watch?v=OVbqppCV1pg shows how reflection and absorption work on a given geometry using classical light. It is a function of power of the beam where he shows that reflection transmits twice the momentum for the simple geometry, and for complicated geometries it is a sum of reflected and absorbed. $\endgroup$
    – anna v
    Mar 15, 2020 at 6:45
  • $\begingroup$ to see the complexity see youtube.com/watch?v=BGb6nGSnDow $\endgroup$
    – anna v
    Mar 15, 2020 at 6:54

4 Answers 4


Consider a photon that strikes the center and is reflected straight back on itself. That photon gives the sphere twice its momentum.

Consider a photon that strike the edge at a glancing angle and is only slightly deflected. It hardly affects the sphere at all. The momentum change is about $0$.

If you integrate over the sphere, you get an average momentum change in between the two extremes.

If a photon is absorbed, it doesn't matter what the angle of the surface it. It gives all its momentum to the sphere.

You have shown that the average value for reflection comes out that same as the uniform value for absorption.

For other geometries, consider a cone where the surface is at 45 degrees. Light everywhere would reflect at 90 degrees. This would impart the same momentum as being absorbed.

This would apply to a flat disk at 45 degrees too.

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    $\begingroup$ It certainly doesn't apply to a flat disk. $\endgroup$
    – mmesser314
    Mar 15, 2020 at 5:40
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    $\begingroup$ The net momentum transfer must be zero. A wedge with 2*45 degrees would also qualify, no lateral or spin component may remain. Most important, each reflection with less than 90 degrees must be paired up with one of more than 90. Now, prove that an integral over the surface of a sphere satisfies this condition... $\endgroup$
    – Bobby J
    Mar 15, 2020 at 18:00
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    $\begingroup$ Excellent qualitative answer. But "you have shown" is a bit too optimistic, isn't it? Shouldn't you integrate over the whole cylinder of incoming photons? For example, aren't there more photons around the edge than around the center? $\endgroup$ Mar 16, 2020 at 2:10
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    $\begingroup$ @EricDuminil - You are right. But since Guru Vishnu came up with an answer, he must have done that integral. I did not repeat it. It sounds reasonable, and he wasn't asking if his math was right. $\endgroup$
    – mmesser314
    Mar 16, 2020 at 3:10
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    $\begingroup$ I feel the need to comment that a photon that interacts with particles at the approximate center of the sphere is not "reflected straight back on itself". Photons interact with matter by being absorbed by it, typically by exciting an electron into a higher energy level, and then subsequently that energy is released when the electron de-excites and a new but statistically identical photon is created with a momentum with the highest probability of being the same but with opposite direction as the incident photon, but this is not a given as other momentums are possible although unlikely. $\endgroup$ Mar 17, 2020 at 10:38

I didn't understand this at first so I thought I would write up a quantitative explanation. Suppose a light beam with intensity $I$ is incident from the right on the surface of a sphere and consider what happens to light that hits an an angle $\theta$ from the leading point of the sphere:
fig 1
The incident light carries momentum $$\Delta p_x=-\frac Ic\cos\theta\,d^2A$$ In the $x$-direction, where $d^2A=R^2\sin\theta\,d\theta\,d\phi$ is the areal element on which the light is incident and $c$ is the speed of light. The factor of $\cos\theta$ is there because the areal element is inclined at angle $\theta$ to the incident beam. As can be seen in the figure above, the reflected light carries momentum $$\Delta p_x^{\prime}=\frac Ic\cos2\theta\cos\theta\,d^2A$$ Thus the change in momentum of the sphere is $$\Delta p_x-\Delta p_x^{\prime}=-\frac Ic(1+\cos2\theta)\cos\theta\,d^2A$$ Adding this up over the front face of the sphere we get $$\begin{align}F_x&=\int_0^{2\pi}\int_0^{\pi/2}-\frac Ic(1+\cos2\theta)R^2\cos\theta\sin\theta\,d\theta\,d\phi\\ &=-\pi R^2\frac Ic\int_0^{\pi/2}\left[1+(\cos2\theta)\right]\sin2\theta\,d\theta\\ &=-\pi R^2\frac Ic\left[-\frac12\cos2\theta+\left(-\frac14\cos^22\theta\right)\right]_0^{\pi/2}\\ &=-\pi R^2\frac Ic\left[1-(0)\right]\end{align}$$ So with or without reflection I am getting $$F=\frac{\pi R^2I}c$$ To the left as the force on the sphere due to the light beam.

As to intuition: if the above picture were for a cylinder, reflection would increase the force by $33\%$ compared to absorption. I had thought that it might have been related to the fact that a slice of a sphere of radius $R$ of thickness $h$ always has $A=2\pi Rh$ of the original surface, no matter where the slice was taken, but looking at the above derivation we can see that it's a completely different phenomenon.


Let’s consider the two cases in detail. To make it specific, consider light as coming along the Z axis, aligned with the $\theta = 0$ of a spherical coordinate system.

Now consider two points:

First, small bit of area $dA$ at the pole ($\theta = 0$), with $I\,dA$ light impinging on it. If it’s absorbed, that’s $dp = {I \over c}\,dA$ on momentum transferred. If it’s reflected back, it’ll be twice that.

Now consider an area at the limb, near but not quite on the edge ($\theta = \pi/2$). Because the $dA$ there is inclined, it only projects $dA \,\cos\theta$ into the light, so $I\,dA\,\cos\theta$ impinges. If it’s completed absorbed, that’s $dp = {I \over c}\,dA\,\cos\theta$.

But if it’s reflected from there, it’s being reflected by just glancing off the surface and mostly going forward. If you do the trigonometry you’ll find that instead of the increase by a factor of 2 seen above, a reflective surface here provides momentum transfer reduced by a factor of $1 - \cos{2 \theta}$. At the limb, that’s zero.

Different shapes will have different distributions of surface inclinations: a flat perpendicular disk looks more like the pole case (more force if reflective), a long thin needle or cone like the limb case (more force if absorptive). For a sphere, the surface distribution is just right for the two to conceal out when averaged over the surface.

  • $\begingroup$ +1 for some quantitativeness while still being intuitive enough. However it should be emphasized that these two cases are in no way sufficient to actually prove that the outgoing light carries no net momentum. $\endgroup$ Mar 16, 2020 at 9:04
  • $\begingroup$ @leftaroundabout I’m not sure what you mean. The he outgoing light definitely does carry momentum in the plate and needle cases. By subtraction, it doesn’t for the sphere. $\endgroup$ Mar 16, 2020 at 13:24

While this doesn't directly answer the question I would like to add that the rigorous force exerted by a light beam on a sphere needs to be computed by solving Maxwell's equations. This solution is called Lorenz-Mie Theory and corresponding software can be found here. In the rigorous solution the force is a function of the sphere size, permittivity, and the shape of the incident beam, which can be exploited e.g. in optical tweezers.


Here is a current reference on calculating optical forces on spherical particles.


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