augeri@regina.DEC (03/13/85)
Lew Mammel claims that my derivation of the momentum transfer to a light
sail "is all wet". That's fine with me, but in my opinion, if he wants to
make the claim, he should support it with better evidence than he did with
his "derivation". I did not find his derivation very enlightening and I
would like to see a more detailed analysis of his derivation.
Further, he claimed that the Compton effect cannot be used to explain the
momentum transfer. That may be true, but I am convinced that there is a
lengthening of the wavelength of the reflected photon, and that the amount
of lengthening is a function of the amount of momentum transferred to the
sail. My conviction is based on the fact that energy and momentum must
be conserved in the transfer. Thus, if any momentum is transferred to the
sail, the momentum and energy of the photon must be less after the
collision than before the collision. The energy of the photon is:
E[p] = hc/l (1)
and its momentum is:
p[p] = E[p]/c = h/l (2)
where E[p] and p[p] are used to indicate the energy and momentum of the
photon respectively, h is Planck's constant, c is the velocity of the
photon (also a constant), and l is the wavelength of the photon. Since h
and c are constants, the only other quantity left to change is the
wavelength. For the energy to decrease, the wavelength must increase. If
the wavelength increases, the visible portion of the electromagnetic
radiation will appear to be shifted toward the red end of the spectrum. I
don't believe that there is any way around this conclusion, regardless of
how you choose your frame of reference. The only question is, how much
does the wavelength change? I thought that the Compton effect gave a
reasonable accounting of that change. If that relationship is invalid,
there must be another. All I ask is that you give us the relationship.
If we go back to my original equation (3) we have:
p[s,a] = p[p,b] - p[p,a] (3)
where p[s,a] is the momentum of the sail after the collision, etc.
The only difference between equation (3) above and equation (12) that was
derived in my original message is that I had eliminated the photon momentum
after the collision ( p[p,a] ) through the use of the Compton relationship.
Give me a valid relationship and equation (3) still holds.
As for Lew's comment about my derivation not taking into account the "speed
of the lightsail", I think there is a misunderstanding here. I assumed
that the initial momentum of the sail was 0 (zero) in order to simplify the
equations that had to be rewritten several times. If you want to give the
sail some initial momentum, you can just add it to equation (3) as follows:
p[s,a] = p[s,b] + p[p,b] - p[p,a] (3a)
The only problem that I have with my original derivation is that I did not
include a discussion of two extreme cases that we can consider in the
momentum transfer. The first case is where the photon is totally absorbed
by the sail material (corresponding to a perfectly inelastic collision),
and the second case is where the photon is totally reflected by the sail
material (corresponding to a perfectly elastic collision). I have some
comment to make on the second case, but let's start with the first case.
Using equation (2), we can rewrite equation (3) as:
p[s,a] = E[p,b]/c - E[p,a]/c (4)
where E[p,b] and E[p,a] are the energies of the photon before and after the
collision.
If the photon is totally absorbed, the energy of the photon after the
collision is E[p,a] = 0 (for example, say that the photon is absorbed by an
electron), thus all of the photon's original energy is transferred to the
sail. Therefore, equation (4) becomes:
p[s,a] = E[p,b]/c (5a)
= p[p,b] (5b)
If the photon is totally reflected, the energy of the photon after the
collision is E[p,a] = E[p,b], thus none of the photon's original energy is
transferred to the sail. Since momentum is a vector quantity we have for
the case of total reflection:
p[p,a] = - p[p,b] (6)
Therefore equation (4) becomes:
p[s,a] = p[p,b] - p[p,a] (7a)
= p[p,b] - ( - p[p,b] ) (7b)
= 2 * p[p,b] (7c)
The comment that I wanted to make about this case is that I believe that
equation (7c) is never completely valid since there is no such thing as
"total reflection". For any real sail material there will be a finite
amount of momentum transferred from the photon to the sail. Thus, in the
real universe, the momentum transfer will be somewhere between the limits
given by equations (5b) and (7c). Actually, I think the lower limit is
valid, that is, the photon can be completely absorbed.
Mike Augeri
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