Defense Intelligence Reference Document Negative Mass Propulsion

Defense Intelligence Agency · 43 pages · text from the file's own layer

This Defense Intelligence Reference Document from the Defense Intelligence Agency is dated 3 January 2011. It was produced under the Advanced Aerospace Weapon System Applications (AAWSA) Program as part of a series of advanced technology reports. The report asks whether negative mass exists and whether it could be used for propulsion. It concludes that one route might be an ultra-light form of matter that could have gathered at the Moon's center. Reaching it would take a tunnel dug with thermonuclear shape charges.

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“The Times”16 pages

p. 7 …sign, the total energy and momentum of the combined system remains zero for all times, leaving…
p. 8 …of special relativity, being the expression of a Minkowskian space-time structure, there is an older…
p. 10 …With lxl = (1/r, where OJ= 2tr/T, with T the time of revolution, and r…
p. 11 …time before their fields propagating with the velocity of light would reach the earth. Adding the…
p. 16 …Even though its time-averaged velocity is always less than the velocity of light, this means…
p. 18 …0~ m· =12r ~6X1Q-13g, ( 45) larger by a factor 3.6x10 11 times the mass…
p. 22 …by saying that the zero-point vacuum energy "generates" the Minkowski space-time. 7. Dynamic Interpretation…
p. 23 …the rod is /, and if the rod rests in the distinguished reference system, the time needed…
p. 24 …c -v Slll-1// -VCOSlf/ (67) with the time t' for a to and fro signal…
p. 25 …If tR is the absolute reflection time of the light signal at clock B, one has…
p. 26 …using a rod to measure the distance and the time it takes a light signal in…
p. 34 …The time needed to bring back the universe to its original low entropy state, is the…
p. 37 …times more dense than water 16. Suppose that in the center of the moon the accumulation…
p. 38 …the heat diffusion equation given by (109) where Xis the heat diffusion coefficient, the diffusion time…
p. 39 UNCLASSIFIED/ ,'P81l 81'1'11!111it 1!1!11!! 811LY The time needed for the liquid…
p. 42 …This means that a small, continuous removal of the heat through the injection and circulation of…

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If rs is the horizontal radius up to which the rocks at a certain depth have to be
shattered (cylindrical case), one finds by equating p(r) in (115) and (117)
(116)
The total energy required to shatter the rocks to make a tunnel from the center at the
moon where r = 0, to its surface where r = R, is then given by summing up over the
slices with radius rs and thickness dr
(117)
where as in eqn (110) s,. ~1010 erg/cm3 is the cohesive binding energy of the rocks. For
(119) one can write
'
E=llc 2 Rt:, (p,,,,)p 0 )"" f(1-x 2 )"
0
dx. (118)
0
With the help of Euler's betafunction one has
'( ')'N (! !!J .,J;;f 1-(xr dx=(ll2)B -,- ec-.
0 2 9 2 (119)
Hence,
(120)
finds that E;:;:2xl0 24 erg:::;,5xl04 kiloton=50Megaton.
It must be emphasized that this energy must be quite nonuniformly released along the
tunnel shaft. Nuclear fusion explosions below a yield of 10 kiloton become
uneconomical with only a fraction of the energy in the fissionable material (needed to
36
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Report, from the dia collection. The PDF is mirrored here; the original link is under it. 43 pages are in the text index: search them above, or from the library's search.