I had been thinking about how much energy would be needed to actually accelerate a spaceship to a speed that is a significant fraction of the speed of light. Because of the huge energies involved, I resorted to matter-anti matter conversion as the fuel needed in the ship. I have made a formula for the proportion of the mass you would need in a rocket/spaceship for fuel, assuming that fuel is converted into energy through E = mc² and all this energy is completely converted into kinetic energy, i.e. it directly raises the kinetic energy of the spaceship and nothing get's wasted. This is an idealization, but it's about the basic idea.

Now I want to know what fraction of the mass you need to essentially change the gamma factor of your ship(as seen from outside) from one value to another. What makes this problem a little tricky is that the (rest)mass of the ship decreases as the speed increases. The way I tackled this problem was by using the following line of reasoning:

$$dE_{kin} = -c² dm$$ $$dE_{kin} = mc²d\gamma $$

Using seperation of variables I then get:

$$- \int_{m_{1}}^{m_{2}}\frac{dm}{m} = \gamma_{2}-\gamma_{1}$$


$$ \ln(\frac{m_{1}}{m_{2}})=\gamma_{2}-\gamma_{1}$$

To me this looks like a sound analysis, and even easier than using conservation of momentum as one would typically do for a classical problem of the same form. It also looks a lot like the classical counterpart in it's form.

What bothers me though, is the extreme amount of fuel relative to useable mass needed to reach a significant gamma factor, this goes up exponentially and is way beyond what I originally expected. If I would want to reach a gamma factor of about 1 Thousand, so one year for me would be like 1000 years for you and I could travel 1000 light years in only 1 year(from my POV) I would have to have:

$$ \frac{m_{1}}{m_{2}} = e^{1000} \approx 1.97*10^{434}$$

I expected a huge amount of mass was needed but never anything close to a value like this, especially considering how much energy you get by only converting a small amount of mass into energy. This is more than the number of atoms in the universe, and it certainly makes spacetravel between distant regions of space using time dilation completely impossible even for hypothetical type III civilisations. This gives me the feeling I made a mistake in my derivation above.

Is this derivation correct and does it really take such an amount of energy, or has something gone wrong in my reasoning above?


3 Answers 3


I think you have the wrong formula. I'm not sure I understand all the steps you took in deriving it.

You want a matter-antimatter engine. This essentially produces photons which you shoot behind your ship, in a process which preserves 4-momentum. Now, suppose you start from rest and accelerate. If $m_0$ is the initial mass of your ship, $m$ is the final mass and $E$ is the energy of the light emitted, by conservation of-4 momentum we have: $$m_0 c^2 = \gamma m c^2 + E$$ $$0 = \gamma m v -E/c$$ The first equation expresses conservation of energy, while the second one express conservation of momentum. Eliminating $E$ we obtain: $$\frac{m}{m_0}=\sqrt{\frac{1-\frac{v}{c}}{1+\frac{v}{c}}}$$ or using mass as a variable: $$v=c\frac{m_0^2-m^2}{m_0^2+m^2}$$ This is a lot more favourable than an exponential function.

You want to make $m/m_0$ as small as possible. The real constraint to travel, apart from the actual engine you have to build, is the amount of antimatter you need. Say you want to get to $v=0.9c$ and suppose that the mass of the rocket itself is a conservative $m\approx 1000 kg$. If you burn all the fuel, you'll need a mass of fuel equal to: $$m_f = \left(\sqrt{\frac{1+\frac{v}{c}}{1-\frac{v}{c}}}-1 \right) m \approx 4400 kg$$ Therefore you'll need roughly two tons of antimatter.

Notice however that this says nothing about the time it will take to reach that speed.

EDIT: Some people may be confused as to why other answers give different formulas. The more general formula for rocket motion is given by: $$v=c\tanh{\left(\frac{I}{c}\log{\frac{m_0}{m}}\right)}$$ which can also be formulated in terms of the rapidity $r=\tanh^{-1}(v/c)$. $I$ is known as the specific impulse. In the case of a photon drive, for which $I=c$, this reduces to the formula I give above. Notice that the general formula is harder to obtain.


You've made two mistakes here.

The first mistake is that you can't just convert mass energy into kinetic energy, since that pretty clearly violates conservation of momentum. The ratio of momentum delivered to mass of fuel is called specific impulse, denoted by $I$. Note that this has units of speed. The highest possible value for the specific impulse even in principle is $c$, achieved when the fuel can be converted into photons directly, in exactly the right direction.

It can be shown that the relativistic rocket equation is given by: $$ \Delta r= \frac{I}{c}\ln\frac{m_0}{m} $$ where $r $ is the rapidity.

With the best possible specific impulse, in terms of $\gamma $, and assuming it is initially at rest, this becomes $$\gamma=\cosh\ln\frac{m_0}{m}$$

Assuming the final $\gamma$ is large, this simplifies to $$m_0=2\gamma m $$

Your other mistake was in the differentials. The second one should be $dE=\gamma c^2 dm+mc^2 d\gamma $.


The problem with this approach is the assumption that "lost" rest mass of the rocket equals kinetic energy of the rocket. You can't make that assumption, not the least because the exhaust also has kinetic energy. You really do have to return to the fundamental approach, by conservation of momentum in the frame always momentarily comoving with the rocket. You can work with something like the approach you propose, but you must account for the kinetic energy of the expelled fuel and you must also account for conservation of momentum - this second equation tells you how much of the expelled rest mass winds up as kinetic energy of the rocket. John Baez tells the tale of a similar problem well on his website (search "Relativistic Rocket John Baez" if the link breaks). Look at the section for "How Much Fuel is Needed". He's assuming the optimal condition where the exhaust is expelled at light speed and perfect conversion of fuel to light is achieved. Your situation is a little more general (non lightspeed, non zero rest mass exhaust).

Let's work in the Rocket's proper frame. In the momentarily comoving inertial frame, the rocket converts a small amount of fuel of rest mass $\delta\,m$ into total energy of the exhaust - beware: the rest mass of the exhaust $\delta\,m_e$ is different from $m$ because some of the fuel contributes to the total energy represented by the exhaust's momentum. The exhaust's total energy is $\sqrt{(\delta\,m_e)^2+(\delta\,p)^2}$ (I'm working with $c=1$ so that That is, $\delta\,m^2 = \delta\,m_e^2 + \delta\,p_e^2$ (I'm working with $c=1$). In John Baez's calculation, he's assuming a light beam as exhaust, so that $\delta\,m = \delta\,p_e$ because the exhaust has no rest mass. In general, we have a relationship $\delta\,p_e = \kappa_e\,\delta\,m$, where the $\kappa_e\leq 1$ depends on the particle mix that is thrown out. For a simplified rocket, we assume a uniform fuel burning process, so that $\kappa_e$ is assumed constant and completely characterizes the burning process for the purposes of this problem. Again, $\kappa_e=1$ for John Baez's case with lightspeed, rest-massless exhaust.

As in Newtonian mechanics, we simply apply conservation of momentum to the momentarily comoving frame but we must replace increment in velocity with increment in rapidity. This is because rapidities of collinear Lorentz transformations add linearly and we want to calculate the total change in motion relative to our beginning frame as a composition of infinitessimal boosts (I discuss this idea more here) as a simple summation (integral). So now we have the differential equation for the rocket's rapidity relative to its beginning frame:

$$m\,\mathrm{d}\eta = -\kappa_e \mathrm{d} m$$

so that $\eta = -\kappa_e\,\log\left(\frac{m}{m_0}\right)$. You can then convert this into an equation for $\gamma$ factors through $\cosh\eta = \gamma$. For large $\gamma$, and for the perfectly efficient case of lightspeed exhaust ($\kappa_e=1$), we have $m_0\approx 2\,\gamma\, m$; that is, you need a rocket that will convert $\frac{1\,999}{2\,000}$ of its mass to exhaust and have a payload of $5\times 10^{-4}$ of its beginning mass.

One can, of course, stick with velocities relative to the beginning frame but the integral is going to be much more complicated to account for time dilation on the way.

Unfortunately if you now add the constraint that the proper acceleration has to be reasonable (i.e. survivable by humans, for example), you're going to take an awfully long proper time to reach high $\gamma$ and travel significant distances. To get to Andromeda galaxy at $1\,g$ acceleration takes of the order of $3\,000$ years.

Also be sure to read Section 6.2 of Misner Thorne and Wheeler "Gravitation" for background. A key observation useful in these calculations is that the four velocity of any object always has unit Minkowski norm.

  • $\begingroup$ Could you expand a bit on the calculation needed to calculate that number of 3000 years? In classical mechanics you simply divide the final speed by the acceleration, but here the continiously varying gamma factor of time dilation in the dynamic force equation makes it a lot harder. $\endgroup$
    – Dirkboss
    Nov 25, 2017 at 0:43
  • 1
    $\begingroup$ @Dirkboss Have a look at John Baez's page. Yes, the integration is a bit more involved, but not too much. There's a tabulation of the results of the calculation for trips to different places. $\endgroup$ Nov 25, 2017 at 2:14

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