Saturday, January 06, 2007


Newton says that objects accelerate as long as force is applied.

Does this apply when object's speed increases beyond c, speed of light?

Einstein, says the speed of light, c, is absolute; thus, no object can attain a speed greater than c. So, these two giants of science appear to disagree. (NOTE: Thought Experiment (TE) shows agreement in subsequent text.) Einstein also says that all observers measure same value for c regardless of observer's own speed. This also seems to conflict with Newtonian concept of speed differential which leads us to believe that faster observers would observe a slower c than slower observers.

To resolve this paradox, TE suggests the remainder concept which resembles the complement concept of Probability Theory.
See associated table for tabulated results..

In the search for practical interstellar travel, Thought Experiment (TE) considers a notional spaceship using a particle accelerator to produce g-force propulsion. The accelerator speeds fuel particles to significant portions of light speed; thus, enormous momentum from a very high speed of a small exhaust mass transfers to a small velocity increase of the spaceship. The propulsion process is designed to produce velocity increase equal to g, acceleration due to Earth surface gravity.

Thought experiment further assumes technology advances sufficient to produce a well controlled, consistent flow of exhaust particles at a set speed. Such exhaust particle flows would provide sufficient momentum such that a relatively small mass of fuel could accelerate a larger spaceship at g-force to simulate gravity and gain enormous speeds. Such speeds would require spaceship to g-force decelerate for same duration as acceleration.

QUESTION: How does Probability Theory's
"Complement" Concept

relate to interstellar g-force velocities????
TE proposes the "remainder" concept, an analog of the complement. COMPLEMENT: For any spin of the Roulette Wheel (1st spin, 2nd spin, 100th spin, etc.), the odds of spinning 00 is always 2.6% and the complement (i.e. "odds of not spinning 00") is always 97.4%.
LIGHT SPEED REMAINDER ANALOG: For any day of g-force acceleration (1st day, 2nd day, 100th day, etc.), vessel always attains velocity of .283%c; remainder (remaining velocity til c, light speed) is always 99.717% c.
To illustrate probability, consider an American roulette wheel which has:
  • 18 red slots
  • 18 black slots
  • two green slots (0 and 00)
for a total of 38 slots. Thus, the probability of rolling a double zero is 1/38.

P(00 ¦ 1 spin) = 1/38 = .026

COMPLEMENT: If the event of interest is spinning a 00; then, the event's complement is spinning any other slot (i.e., not spinning a 00).
To compute the probability of NOT rolling a double zero on first roll, use the complement concept as shown:
P(~00¦1 spin)=37/38=.974=1 - P(00¦1 spin)
P(x) is the probability of event x occurring; P(~x) is the probability of x not occurring (complement of x):
P(~x) = 1 – P(x)

EXAMPLE: Use the complement's probability to determine probability of rolling at least one double zero (00) in three rolls of the wheel?
First, determine the probability of a double zero on the first roll. This is easily calculated by dividing number of possible occurrences by total number of possible events.
P(00 ¦ 1 spin) = 1/38 = .026
Next, use the complement to determine probability of NOT rolling a double zero on first roll:
P(~00 ¦ 1 roll) = 37/38 = .974 = 1 - P(00 ¦ 1 roll)
Use that value to determine probability of NOT rolling 00 during three successive rolls:
P(~00 ¦3 rolls)=P(~00 ¦1 spin)×P(~00 ¦1 spin)×P(~00 ¦1 spin)
P(~00 ¦ 3 spins) = P(~00 ¦ 1 spin)3 = .924
Finally, use the complement once again to determine likelihood of at least one double zero during three rolls of the American roulette wheel.
P(00 ¦ 3 spins) = 1 - P(~00 ¦ 1 spin)3 = .076

What about 10 or any number (n) of spins?? What are the chances of getting at least one double zero?
P(00 ¦ 10 spins) = 1 - .97410 = 1 - .768 = .332
P(00 ¦ n spins) = 1 - P(~00 ¦ 1 spin)n
Thought Experiment (TE) assumes following:
For convenience, assume following values:

G-force acceleration uses same value as free falling object near Earth's surface.
9.807 m/sec2
Velocity increases for
each second of g-force flight.
v = g × t
time (t)velocity (v)
1 sec
9.807 m/sec
2 sec
19.613 m/sec
3 sec
29.420 m/sec
Convert seconds to days.
day  =86,400 sec
day2 =7.965×109 sec2
Convert meters to
Astronomical Units (AUs)
AU= 1.49598×1011 m
g= 9.8067m×AU


sec2 × day2


= g = .489AU

For flight durations of days and weeks, determine velocities
as AUs per day.
vt = g × t
time (t)velocity (v)
1 day
0.489 AU/dy
2 days
0.978 AU/dy
3 days
1.467 AU/dy
Newtonian: Determine g-force velocity by multiplying g by t.
Daily Increase in Velocity. After first day of g-force acceleration, velocity can be determined as follows:
v = g × t = 9.8065 m/sec2 × t
time (t)velocity (v)
1 sec9.8065 m/sec
86,400 sec847,294.56 m/sec = 847.3 km/sec
1 day847.294 km/sec × 86,400 sec/day = 73,296,201 km/day
1 day73,296,201km/day × AU/149,597,871km=.4899 AU/day
1 day.4899 AU/day × c/(173.145 AU/day) = 0.283% c
Daily Difference.  Velocity after first day's g-force acceleration can be expressed as a portion of light speed, c ( = 299,792 km/sec). TE calls this the "daily difference".
Δ = .00283 c = .283% c
With exponentials, the complement concept turns a very difficult probability calculation into a very simple one. Note following points:
  • For any particular roll, the odds of rolling a double zero remain 1/38 regardless of how many previous rolls were not double zero.
  • Though initial roll's complement is quite high (97.4%), prediction of repeated rolls will continue to whittle down the probability of NOT rolling a 00.
  • For an objective predictor who calculates probabilities prior to any gaming, they notice that as number of rolls increase, probability of 00 increases.
Of course, common sense tells us that more spins greatly increase the probability of getting at least one double zero; however, the odds remain very low for any particular spin producing a "00" (1 divided by 38).
Dest.Accel-time (tAcc)Max-vel (vMax)
2.0 days
.566% c
3.2 days
.906% c
4.5 days
1.274% c
Given(d/g) Δ × tAcc
For interplanetary destinations,

g-force velocities

closely approximate

linear Newtonian values.
Ship departs Earth and accelerates at g-force. After precisely one day of g-force flight, ship ejects Module-1 with observer. Since Module-1 has no propulsion system, it will maintain constant speed, v1, while g-force ship will continue increasing speed.

Earth bound observer measures c at 299,792,458 m/sec and measures Module One's velocity at .283%c.

NOTE:  Remaining velocity till light speed ("Remainder") is 99.717%c.
Second Day Scenario: After precisely two days of g-force flight, ship ejects Module Two (M2) with another observer. 
M1 observes M2 at .283%c; thus, a remainder of 99.717%c.
Earth observes M2 at .566%c; thus, a remainder of 99.434%c.
However, all observers measure same value for c, regardless of platform velocity. (Recall Einstein)
In fact, all observers (Earth, M1, M2, g-force ship) are at different speeds; yet, they all observe c at same value: 299,792,458 m/sec.
v = Δ × t
time (t)velocity (v)
1 day
.283% c
2 days
.566% c
3 days
.849% c
As modules eject at precise daily intervals throughout the entire flight, one day's duration of g-force flight will always increase ship's velocity .283%c.

This is true during start, middle or end of flight.

However, Earth bound observer will measure different velocities for different modules. First few modules appear to increase speed at a linear rate.
v = Δ × t
time (t)velocity (v)
100 days28.3% c
200 days56.6% c
300 days84.9% c
352 days99.616% c
353 days99.899% c
354 days100.182% c
Succeeding Day's Velocities can also be expressed as percentages of light speed (%c). Using Newton's linear method, we get values in table. Eventually, this violates Einstein's Special Relativity Theory by increasing speed above c. This paradox can be restated:
G-force velocities approximate
a linear relationship for short durations,
but this linearity cannot be sustained.

How can we resolve this paradox??
Linear model predicts g-force vehicle will exceed c.

Einstein says this can't happen.
Most scientists agree;
thus, TE considers another way
to compute g-force velocities.
Δ = .283% c
100 days
24.678% c
200 days
43.266% c
300 days
57.267% c
(1 - Δ)t
(1 - Rt)c
File:Nearby Stars (14ly Radius).svg
For interstellar destinations,

 g-force velocities

must use ingenious, 

Einsteinian values.
To account for relativity, use remainders and exponentials to compute interstellar velocities.
Daily Difference (Δ)
First day's g-force velocity.
Δ= 0.283%c = .489AU/day

Remainder (R)
Let R be the difference between
first day's final velocity
and c, light speed.
R= c (1 - Δ) = .99717 c
TE assumes subsequent remainders, Rt , can be computed with exponential, t (time), days of g-force acceleration.
c (1 -Δ)t
time (t)Remainder (Rt)
1 day
.99717 c
2 days
.99435 c
3 days
.99153 c
Daily Speeds.
Thus, compute daily g-force speeds as light speed minus that day's remainder.
vt = c - Rt
timex (t)Velocity (vt )
1 day
.283% c
2 days
.565% c
3 days
.847% c
time (t)NewtonianEinsteinian
1 day
.283% c
.283% c
2 days
.585% c
3 days
.847% c
t × Δ
c × (1 - Rt)
First few days of g-force
demonstrate approximate agreement
between Newtonian and Einsteinian models.
Day 200 Scenario.
Our notional g-force vessel keeps ejecting modules at precise daily intervals.  After exactly 199 and 200 days, modules M-199 and M-200 are respectively ejected.

Like all the previously ejected modules, M-199 observes velocity of the subsequent module, M-200, at .283%c with 99.717%c remaining until light speed.
The earth bound monitor observes all 200 modules.

 Subsequent modules attain faster and faster velocities

as they're ejected throughout the voyage.
time (t)NewtonianEinsteinian
100 days
28.3% c
24.678% c
200 days
43.266% c
300 days
57.267% c
t × Δ
c × (1 - Rt)
For greater durations,
values diverge even more between
the models of Newton and Einstein. 
 In fact, we eventually see impossible values for Newtonian method.

After 354 days, Newtonian method exceeds c, an impossible feat.

time (t)Newtonian
352 days99.616% c
63.123% c
353 days99.899% c
63.227% c
354 days100.182% c
63.331% c
GivenVt = g × t
Vt = c(1 - (1 - Δ)t)

Thought experiment assumes Einsteinian method

more accurately determines g-force velocities for longer durations.

Newtonian Method
closely approximates velocities for first few days of g-force flight.
Δ (Daily Difference of g-force velocity) remains a constant 0.489 AU/day ≈ 0.283% c. For convenience, thought experiment assumes following formula:
.489 AU/day × tDay = VFin = Δ × tDay = .283%c × tDay
works well for short interplanetary flights. Thus, 10 day interplanetary trip would reach velocity of 4.89 AU/day ≈ 3% c, well short of light speed. Relativity concerns prevent this linear increase to go on indefinitely, but it's a fairly good model for interplanetary travel with g-force propulsion systems. For much longer interstellar voyages, this formula will need adjustment to produce usable values.

Newtonian values don't prove true for much greater durations. (Example: After 354 days of g-force, above formula  predicts velocity exceeds c, an impossibility.)
Einsteinian Method
more closely approximates g-force speeds for interstellar flights.
Thus, TE assumes following equation:
Vt = c - c(1 - Δ)t

G-force interstellar flight durations take years (much longer than the few days required for interplanetary flights). The distances are too vast and the flight durations are too great; thus, we cannot get by with the convenient heuristic used for Volume 2, Interplanetary.

Following cells demonstrate how we can
leverage a simple probability concept to derive above equation.

Spin the Wheel
G-force to the Stars.
Probability of rolling a double zero on first roll is very low.G-force velocity after first day of flight is very low, a small percentage of c.
P (00) = 1/38 = .026 = 2.6%V (1 day g-force) = 847 km/sec ÷ 299,792 km/sec = .283% c = V1dy
Complement(00): P (~00) =1 - P(00) = 97.4% = C1SpinRemainder: c - V1dy = 99.717% = R1dy
If one knows the probability of an event's complement; then, easily determine the probability of the event:
P(E) = 1 - P(C)
If one knows the remainder of an g-force vessel's velocity; then, one knows the velocity of the vessel:
Vt = c - Rt
Spin 38 Roulette Wheels
Intuitively, it seems that at least one must show double zero.
NOT SO!!!!
After accelerating for one year (365.25 days of g-force),
Newtonian physics leads one to believe that velocity will exceed c, light speed.  NOT SO!!!
After the first few spins of the wheel, it looks like the probabilities are additive. But after many spins, this proves to not be true.After the first few days of g-force, the resultant velocities appear to add.  However, Einsteinian Relativity demonstrate that g-force velocities cannot continue to add indefinitely.
To observe results of 38 spins, one could record each spin in sequence. Alternatively, 38 observers could simultaneously spin 38 separate wheels and record the results.To observe one year of g-force velocities, an Earth bound observer can measure all 365 daily velocities of a g-force vessel.  Alternatively, one could  conveniently observe 365 markers with constant velocity achieved at time of ejection.
As number of spins increase, complement of double zero decreases; this is easily computed with exponentials.
C(00| n Spins) = .974n= Cn-Spins
As g-force flight duration increases, decreasing light speed remainder is easily computed with exponentials. EXAMPLE: After 1 year of g-force, Remainder (R) decreases to  .99717365.25= .3552.
V1Year = c - R365.25days = c (1 - .99717365.25) = 64.48%c
The complement of the complement gives the probability that at least one double zero will happen during a given quantity (n) of spins.The remainder of the remainder gives us the velocity of the vessel after a given quantity (t) of days with g-force propulsion. See associated 1G TABLE: Accelerate for 1 Year.
P(00| n Spins) = 1 - Cn-Spins
Vt = c - Rt = c - c(1 - Δ)t



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