We can develop an algorithm that can be executed by a computer to find the force exerted by a laser beam on a dielectric sphere because of refraction. We initially switch back to the coordinate system in which the laser comes to a focus at the origin.
Assume that for any photon in the laser beam 0 £ q0 £ qmax and 0 £ f0 £ 2p. The path of an incident photon can be described by the equations z = (1/tanq0)r, f = f0.
The magnitude of the momentum of an incident photon is p = n1hn/c = n1h/l. The components of its momentum vector are pz = -pcosq0, px = -psinq0cosf0, and py = -psinq0sinf0.
Assume the center of a sphere of radius R is located at r0 = (x0,y0,z0). Let us calculate the change in the photon’s momentum if its path crosses this sphere. We assume that the sphere can absorb this change in momentum with negligible change in energy, i.e. that it is "infinitely heavy".
The algorithm consists of a series of rotations and a translation. If we let the z-axis of the coordinate system denote the optic axis, then a skew ray entering a sphere in one coordinate system can become a meridional ray in rotated and translated coordinate system. We can calculate the momentum transferred to the sphere because of refraction of the meridional ray as outlined above. Then we reverse the order of the rotations of the coordinate system to find the momentum transfer in the original coordinate system.
First we find the coordinates of the point where the photon enters the sphere. To do this, we rotate our coordinate system ccw about the z-axis by an angle f0.
The photon is now incident in the x’z’-plane. It has momentum components
pz’ = -pcosq0, px’ = -psinq0.
The center of the sphere is located at r0’ = (x0’, y0’,z0’), where
.
The photon will contact the sphere at a point located in the x’-z’ plane. The equation for the photon’s trajectory is z’ = (1/tanq0)x’. The equation for the surface of the sphere is (x’-x0’)2 + (y’-y0’)2 + (z’-z0’)2 = R2. The surface of the sphere intersects the x’z’-plane at points which satisfy the equation (x’-x0’)2 + (z’-z0’)2 = R2-y0’2. This is the equation of a circle. If R2 < y0’2 then there exists no intersection and the photon’s momentum is unchanged.
Assume that the surface of the sphere intersects the x’z’-plane. If the line giving the photon’s trajectory passes through the circle the we can find the point where the photon enters the sphere by inserting z’ = (1/tanq0)x’ into the equation of the circle. This yields
(z’tanq0-x0’)2 + (z’-z0’)2 = R2-y0’2,
or Az’2 + Bz’ + C = 0, with
A = (1 + tan2(q0)) = 1/ cos2(q0),
B = -2(x0’tanq0 + z0’),
C = x0’2 + y0’2 + z0’2
- R2.
The constant A is equal to or larger than 1. Therefore the larger value of z’ is given by
.
A solution exists if B2-4AC is positive. Then x1 = tanq0z1.
Assume now that in the primed coordinate system the photon enters the sphere at the point (x1, 0, z1). We will now shift the center of this coordinate system to the origin of the sphere. This shift does not change the orientation of the axes, and the momentum components of the photon are the same before and after the shift. The coordinates of the point where the photon enters the sphere become
(x2, y2, z2) = (x1-x0’, -y0’, z1-z0’).
The vector pointing from the center of the sphere to (x2, y2, z2) makes an angle q2 with the new z-axis and an angle f2 with the new x-axis, where
,
.
We now rotate this new coordinate system ccw about the z-axis through an angle f2. This moves the intersection point into the x-z plane of the rotated coordinate system, which we will label with superscript 1. The components of the photons momentum in this coordinate system become
.
To move the intersection point onto the z-axis, we will now rotate the coordinate system ccw about the y-axis by an angle q2. We will denote the new coordinate system with a double prime. The components of the photon’s momentum in this new coordinate system become
.
Let tanf3 = py’’/px’’.
One more rotation ccw about the z’’-axis through an angle f3
yields the situation shown in the figure below with
.
In this figure qe = 2q3’ - q3, where q3’ is found from Snell’s law, n1sinq3 = n2sinq3’. The momentum components of the photon leaving the sphere in the double primed coordinate system are
pzf’’= -pcosqe, pxf’ ’= psinqecosf3, pyf’’ = psinqesinf3,
with
and 
.
To find the momentum vector of the photon in the original unprimed coordinate system the rotations of the coordinate axes have to be reversed.
,
,
with
,
.
Finally,
.
The momentum transferred to the sphere is P =( Px,Py,Pz) = (px-pxf, py-pyf, pz-pzf).
To find the total momentum transferred to the sphere by the laser beam we have to choose an intensity profile for the laser beam. The simplest assumption is a uniform angular distribution of the photons in a beam with circular cross section, i.e. every solid angle element DcosqDf contains the same number of photons up to the maximum angle qmax. We can now calculate P for many photons in the distributions and find the average P per refracted photon. The intensity of the laser beam yields the number of photons per second n. The force on the sphere is given by F = nP.