augeri@regina.DEC (03/06/85)
There has been several attempts to derive the momentum imparted to a light
sail as a result of impinging photons. I have not yet seen the correct
derivation, so here is my contribution.
From Einstein's work we know that the energy of the photon is given by:
E = hf = hc/l (1) where E = energy, h = Planck's constant,
f = frequency, l = wavelength (I use l for
lambda), c = speed of light, and f = c/l
From this we can derive the formula for the momentum of the photon as:
p[p] = E/c = hf/c = h/l (2) where p[p] is the momentum of a photon and
the other variables are the same as above
Expressing that momentum must be conserved we get:
momentum before collision = momentum after collision
If we assume that the initial momentum of the sail is 0, then for the
system we are discussing we have:
photon momentum before collision = photon momentum after collision
+ sail momentum after collision
Just rearranging this equation we get:
sail momentum after collision = photon momentum before collision
- photon momentum after collision
Using some variables to describe this equation we get:
p[s,a] = p[p,b] - p[p,a] (3) where the bracketed quantities represent
the momentum of the sail after the
collision, p[s,a], etc
Substituting the relation shown in equation 2 into equation 3 we get:
p[s,a] = h/l - h/l' (4) where l is the photon wavelength before the
collision and l' is the photon wavelength
after the collision
Compton's effect is given (without proof) as:
dl = l' - l (5a) where dl is the change in wavelength, h
= (h/(mc))(1 - cos(z)) (5b) and c are as above, m is the rest mass of
the sail, and z is the angle of incidence
of the photon (I use parentheses liberally
to keep the proper precedence)
Assuming that the angle of incidence of the photon is perpendicular to the
sail surface (this just simplifies the formula, since the cos(90) = 0) we
get:
dl = h/(mc) (6)
Rearranging equation 5a we get:
l' = l + dl (7)
Note that we see from equations 6 and 7 that the wavelength of the photon
after the collision (l') is longer than the wavelength before the collision
(l), therefore, the reflected radiation is red-shifted.
Substituting equation 7 into equation 4 we get:
p[s,a] = h/l - h/(l + dl) (8)
Rearranging equation 8 (left as an exercise) we get:
p[s,a] = (h/l)(dl/(l + dl)) (9)
Substituting for dl from equation 6 into equation 9, we get for photons
hitting the sail at an angle of 90 degrees:
p[s,a] = (h/l)(h/(mc))/(l + h/(mc)) (10)
The parentheses in equation 10 obscure the result (oh, for a universal
terminal that could display equations the way they should be!). Anyway,
by rearranging equation 10 (again, left as an exercise), we get what I
consider to be a more useful form of the equation:
p[s,a] = (h/l)/(mc/(h/l) + 1) (11)
There are many ways that equation 11 can be written, but the reason that I
think this form of the equation is more useful is that the values we know
are the momentum of the incident photons and the mass of the light sail,
and what we want to know is expressed directly in terms of those values.
Therefore, recalling that the momentum of the photon before the collision
is p[p,b] = h/l (see equations 3 and 4), we get for photons hitting the
sail at an angle of 90 degrees:
p[s,a] = p[p,b]/(mc/p[p,b] + 1) (12)
Finally, for any angle z we get:
p[s,a] = p[p,b]/(mc/(p[p,b](1 - cos(z))) + 1) (13)
I hope that this clears up the matter.
Mike Augeri (DEC, Maynard Mass)
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