1998/09/29 ; 2002/08/07 ; last change 2003/08/31 ; (concept)

Travel to very far spots in the physical Universe .

1. This note is a reaction on a Science Fiction story (D. MOFFITT 1986 Second Genesis) about a very far travel of duration 500 years . In the present note I show that the travel indeed can be accomplished in a lifetime WITHOUT violating the laws of physics .
2. This note has been unofficially verified by the theoretical physicist prof. Blote from Leiden, in fact using the method of so-called rapidities.

- The physical universe contains very many stars and constellations of stars. The farthest visible object (by telescopes) is at a distance of many billions of lightyears. Can we reach such an object? The answer is perhaps surprising. According to the laws of physics known at present a human being can reach in its lifetime an object at a distance of ten billion lightyears, though with extreme difficulty.

- This note discusses only the theoretical possibilities. It will probably never be realized in practice because of many giant problems.
- In short, the possibility of the travel lies in the so-called 'clock-paradox', the retardation of time at very large velocities, as described by the Theory of Special Relativity. Because of this the time for a space-traveler goes slower than for someone on earth. Remind: in this text 'time' will refer to the time as feeled and observed by the space-travellers !
- Roughly one can realize a time-retardation of nearly a factor three each year (i.e. compared to the time in the previous year). After 23 years this amounts to a factor ten-billion (10 to the power 10). In a year you do .. tenbillion lightyears-as-seen-from-earth . In a minute you cross a complete constellation of stars.

- The most huge problem is the amount of fuel needed for the spacecraft. In order to reach the speed one has to emit each year at least 65% from the total mass of the spacecraft, probably even 85%, the next year again 65% or 85% from the remaining mass, etc. therefore we need to travel with a spacecraft with mass-at-the-start as heavy as a whole planet like Venus or Jupiter.
- Another very serious problem is how to evade the many meteorites etc. !
- Also, when you want to see something on that far spot in the Universe, visit some planet or something, then you need to lower your huge speed again to zero.. that takes again 23 years. And the fuel for that? Hopefully pick up during flight.
- We will fly at a continuous accelleration of 1g, i.e. the gravity feels as on earth. In total then the travel will last 46 years, but.. if you can endure permanently a double gravity.. then the travel can last about 24 years.
- Then you are perhaps able to return to earth again, in 50 years total .. Then you can examine how our "milky-way" constellation has become after 20 billion years. It is uncertain though whether we still can find our earth then ..


- Here comes the calculation of the travel-time that is rougly needed. The calculation is based on the following formulae of the Theory of Special Relativity. See e.g. the Encyclopedia Brittannica.
- Use of symbols: x*y means x times y , x^y means x to-the-power y , x&<&<1 means x near-to-zero , c means the speed of light .
- Beneath one should read:
- object A is the earth
- object B is the space-vessel
- object C is the space-vessel shortly later .
1. When v= speed of object B as seen from object A
- - and w= speed of object C as seen from object B
- - and v'= resulting speed of object C as seen from object A
- -then v'= (v+w)/(1+v*w/c*c) .
2. When t= time between event 1 and 2 at object B as seen from object A
- - and t'= time between event 1 and 2 at object B as seen from object B
- -then t'= t*sqrt(1-v*v/c*c) .
3. E = m*c*c .

- We write y= 1-v/c , y'= 1-v'/c and z= w/c . So y and z are speeds relative to c , and when v approaches c then y approaches zero: y&<&<1. And we will always suppose z&<&<1 .
- Then v= (1-y)*c and w= z*c and so v'= (v+w)/(1+v*w/c*c) =
((1-y)*c+z*c)/(1+(1-y)*c*z*c/c*c) = (1-y+z)*c/(1+z-y*z) ,
so y'= 1-v'/c = 1- (1-y+z)*c/(1+z-y*z)*c =
((1+z-y*z)-(1-y+z))/(1+z-y*z) = (y-y*z)/(1+z-y*z) .
So y'= y* (1-z)/(1+z-y*z) .
Using logarithms with base the number e (=2.7182) :
: ln(y')= ln(y)+ ln(1-z)-ln(1+z-y*z) , and because z&<&<&<1 : ln(y')= ln(y) -z -(z-y*z) = ln(y) -2z +y*z .

- When the space-vessel is gaining speed with accelleration 1g, then each second the speed increases with 9.8 meter per second , so in 306000 seconds with 3000 km/sec . 306000 seconds is 306000/(24*60*60) days = 306000/86400 days = 3.55 day.
- Then w= 1/100 *c so z = 0.01 .
- The calculations are made by hand, so you donot need to use a computer.

- If y &< y0 then ln(y')=ln(y) -2z +y*z &< ln(y) -2z +y0*z .
- Always y &< 1 so to start we put y0=1 .
- Then after 50*3.55 days = 177 days: ln(y') &< 50*(-0.02+0.01) = -0.5 so y' &< e^-0.5 =0.6 .
Now we can put y0 = 0.6 , and then after again 35*3.55 days = 124 days: ln(y') &< -0.5 -35*(0.02-0.006) = -1 so y' &< e^-1 = 0.36 .
Now y0=0.36 and after 107 days ln(y')&< -1 -30*(0.02-0.0036) = -1.5
Now y0=0.22 and after 100 days ln(y')&< -1.5 -28*(0.02-0.0036) = -2
Now y0=0.13 and after 95 days ln(y')&< -2 -27*(0.02-0.0013) = -2.5
Now y0=0.08 and after 185 days ln(y')&< -2.5 -52*(0.02-0.0008) = -3.5
Now y0=0.03 and after 360 days ln(y')&< -3.5 -102*(0.02-0.0003) = -5.5
, y0=0.004 and after 3560 days ln(y')&< -5.5 -1003*(0.02-0.00004) = -25.5
, y0=0 and after still 3550+178 days ln(y')&< -25.5 -1050*0.02 = -46.5
- So now y' = e^-46.5 = 0.6*10^-20 , and this in
(177+124+107+100+95+185+360+3560+3550+178) = 8436 days = 23 years.

- The time-retardation is then sqrt(1-v*v/c*c) = sqrt(1-(1-y)*(1-y)) =
sqrt(2y-y*y) = (because y&<&<1) sqrt(2y) = sqrt(2*0.6*10^-20) =
1.1* 10^-10 , that is about tenbillion times.


- Now we should say something about the fuel needed to gain the speed. Let m denote the mass of the vessel, seen from the space-vessel itself. When one wants to increase the speed with 0.01*c then, in order to keep the gravity-center the same (law of coservation of momentum), one must shoot-off a mass of 0.01m with the speed of light: impuls 0.01*c*m .
- The energy needed is the E= 0.01*m*c*c so equal to 1% of the mass of the space-vessel and this only when one should succeed to convert the whole mass of 1% in energy , with efficiency 100% ..
- If you want to have still 10 tons (10000 kilo) after 23 years, then, since 23 years = 2300*3.55 day, you must begin with 1.01^2300 *10 ton = e^23 *10 ton = 10^11 tons.
- In order to attain a complete transition of mass into energy, it is probably needed to take with you half of the mass in the form of antimatter (consisting of antiprotons and antineutrons).
- Also it is probably not possible because of basic laws of nature, to obtain an efficiency of more than 50%. In that case the mass that you need each time doubles so the total load is used-up twice as fast. So then you need to start with e^46 = 10^20 *10 ton.. that is 10000000^3 ton, that is a "planet" of 10000 km diameter like the planet "Venus" , plus an equal amount of antimatter.
- The driving of the vessel is indeed a nearly unsurmountable problem.

Leo Horowitz - home