Position of the problem

It was proposed to use a new form of energy, for example the nuclear energy to propel spaceships manned to - outside of the solar system. The problem is clearly that nearest to the known stars (α of the Centaur) is at 4 light-years, i.e. about 40  000 billion km. To launch a crew in a reasonable voyage from one duration, and this under bearable conditions of acceleration over the long life, one encounters an insurmountable difficulty.

However, on paper, one can think of the fact that when one reaches speeds close to that of the light, time passes more slowly than for the people who were not accelerated. One can try to exploit this fact to shorten voyages at distances which appear inaccessible.

Study of a concrete case

A proposal is made in the sub-section quoted on nuclear energy to in one way or another use nuclear energy with an aim of obtaining this kind speed, by thus shortening the time of voyage of a factor 100 compared to the land ones.

The goal of this article is to calculate the consequences of such a company, on the basis of same restricted Relativité used to obtain this effect.

One will not go into the details of the technique to implement for realization, but one will restrict oneself to calculate energy to be used in the most favorable case of output, and the loss of mass thus caused. One will suppose thereafter the reader familiar with the fundamental formulas of restricted relativity, if not, one will advise to him to jump to conclusions.

Basic assumptions

One will limit oneself to a hypothetical system whose output is the best possible one as regards acceleration for a loss of energy given.

If one considers the vessel of current mass M in his instantaneous system inertial, and that one ejects in a small time clean a certain quantity of matter or radiation, it will acquire by reaction an impulse dp opposed to that of what is emitted, and thus it will be accelerated of a = \ frac {dp} {M D \ tau} .

It also will lose an energy of opposed to that of what is emitted. However, in all the cases, one has |   dP  |   ≤  of . Let us take the most favorable case, that of the equality. One has then a = - \ frac {of} {M C D \ tau} , or since E = Mc ² :
a = - \ frac {C DM} {Mandelevium \ tau} .

To pass to the total formulas, and either differentials, we must hold account owing to the fact that the inertial system of the vessel changes all the time. Most convenient the additive parameter \ varphi of the group of Lorentz. It is usual to standardize it by noting speed not-relativist like c \ varphi, so that the factor of Lorentz \ gamma = \ cosh \ varphi and acceleration in the inertial system is a = \ frac {C \; D \ varphi} {D \ tau} .

One arrives finally at the relation: \ frac {C \; D \ varphi} {D \ tau} = \ frac {C \; DM} {M \; D \ tau} , or while simplifying: \ frac {D \ varphi} {D \ tau} = - \ frac {DM} {M \; D \ tau}

This equation is integrated by M = e^ {- \ varphi} = e^ {- \ tau a/c} .

Numerical application

The average relationship between terrestrial time T and clean time τ aimed being of 100, very arbitrarily, so that the time passed in years in the voyage is that past in centuries on ground, it is necessary to calculate the value of this average ratio.

The instantaneous report/ratio \ frac {dt} {D \ tau} = \ cosh \ varphi = \ cosh \ left (\ frac {\ tau has} {C} \ right) . While integrating on τ , one finds t = \ frac {C} {has} \ sinh \ left (\ frac {\ tau has} {C} \ right) , from where \ frac {T} {\ tau} = \ frac {C} {has \; \ tau} \ sinh \ left (\ tau \ frac {has} {C} \ right) = \ frac {\ sinh \ varphi} {\ varphi} = 100.

By numerical resolution, one finds \varphi \simeq 7,284. There is thus for acceleration outward journey only a report/ratio of masses between the departure and the change of sign of the acceleration of \ frac {M_0} {M} \ simeq e^ {7,284} \ simeq 1457.

For the complete mission, acceleration-décération, return ticket, it is necessary to divide the initial mass 4 times of continuation by this factor, that is to say by a total factor unrealistic of 4,5·1012 .

The distance covered X , by supposing an acceleration with which we are familiar, that is to say G , would be given by \ frac {dx} {D \ tau} = C \ sinh \ left (\ tau \ frac {G} {C} \ right) , is of x = \ frac {c^ {2}} {G} (\ cosh \ varphi - 1) = 727,4 \; c^ {2} /g, or 727,4  c/g time-light, is approximately 700 light-years, since \ frac {C} {G} \ simeq 0,97 \; \ mbox {year} .

Therefore, under the conditions indicated, one could go at approximately 1400 light-years, but it would be necessary for that a clean time of 2 \ times \ tau = 2 \ times \ frac {C \ varphi} {G} \ simeq 2 \ times 7,284 \ simeq 14,5 \; \ mbox {years} , one-way ticket, is 29 years on the whole and 29 centuries for the land ones which await the return of the mission.

Conclusions

The made assumptions are clearly intolerable. The loss in mass due to the consumption of energy, by supposing an output of 100%, leads to completely absurd figures. One could think of limiting this effect by decreasing the range of the mission, for example while limiting oneself to a relativistic parameter of 1 only, instead of more than 7. Always by keeping an acceleration equal to that we support on the Earth, G , and an output of 100%, one would then lead to divide the mass of the vessel by a factor e^ {4} \ simeq 55 for the whole of the mission. The time of mission would be 4 years on board vessel and 6,5 years on Earth, but obviously, the range of the mission would be only one light-year.

While waiting for even the prospect for suitable outputs compared to the maximum assumption of 100%, one sees that the voyage using the relativistic contraction of time is far from being seriously possible To fix the ideas, we give below a range of parameters of voyage within the framework above, always with an output of 100%:

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