A very technical post on matter-antimatter propulsion

There has been a ton of traffic based on a previous post that I wrote about an imagined trip to Gliese 581g.  A number of sources (Discover, slashdot, even io9 to some degree) seemed to suggest that I was optimistic about such a trip.  To the contrary.  The whole point of my calculation was to crunch the numbers to give some idea of the barriers we’d need to overcome to get to another planet.

I showed that constant acceleration/deceleration at g allow us to reach relativistic speeds quickly, meaning that the trip could be completed in a fraction of a human lifetime.  I also showed, however, that under the absolutely most optimistic conditions (full conversion of fuel into energy, harnessing the total incident solar energy of the sun on the earth), it would take 3 million years to fuel up.  Nevermind that I’m assuming 100% conversion to matter-antimatter pairs (which is very tough).  I thought the 3 million year lower bound was daunting enough.   That last bit was not stressed very much.

But at any rate, there was a fair amount of interest in understanding where the 530:1 fuel to cargo ratio came from, and so I figured I’d provide a bit of a mathematical treat.

Warning, this is technical.  As in super-technical.  I’m assuming you know calculus, a bit about special relativity and Lorentz transformations, and are solid on classical mechanics.  Don’t read this if you just want the numerical answer.

Now, on with our derivation.

Imagine you are in a ship of mass, m (including the ship itself as well as your remaining fuel supplies), which is moving along at some speed, v.  At some instant, you liberate a small amount of your matter-antimatter mix, of mass dm << m.  The matter-antimatter mix is immediately converted to radiation (energy) which is emitted out of the back of the back of the ship.

Since:

   dE=dm c^2

We’ll call the “primed” frame the one that’s moving along at speed v (and is initially moving with the ship).  In that frame, the momentum of the radiation is:

   dp=-dm c

which from conservation of momentum means that the ship is now moving at a speed

   u'=frac{dm}{m}c

(in the primed frame).  Of course, in the “unprimed frame” (earth, for example), the ship is now moving:

   u=frac{u'+v}{1+frac{u'v}{c^2}}

by the normal Lorentz transform of velocity.  However, we know that u' << c (because the dm << m), so this can be computed as:

   u=(v+u')left(1-frac{u'v}{c^2}right)simeq v+u'left(1-frac{v^2}{c^2}right)

to first order in u’ (which is the correct thing to do when you’re talking about an infinitesimal thrust).  Thus:

   du=frac{u'}{gamma^2}

where gamma is defined in the normal way:

   gammaequivfrac{1}{sqrt{1-v^2/c^2}}

Since du is the amount that the ship speeds up in the rest frame, we can combine everything to get:

   dv=-frac{dm}{m}frac{c}{gamma^2}

The minus sign comes from the fact that we’re losing mass.  Or, to put this in integral form:

   -int_i^f frac{dm}{m}=frac{1}{c}int_0^v gamma^2 dv

This is perfectly general.  If all you specify is the maximum speed, we can compute how much fuel you’ll need to burn (since presumably your final mass is just your ship plus cargo).

Writing it out:

   ln left(frac{m_i}{m_f}right)=tanh^{-1}left(frac{v}{c}right)

Plug in the peak speed of the ship (which you can compute using constant force on fixed mass relations — I’ll do another technical post on this if there’s interest) as v=0.99623 c (the peak value of the ship), and you get:

   ln left(frac{m_i}{m_f}right)=3.136

Of course, you have to decelerate as well, and so that doubles the RHS.  So:

   ln left(frac{m_i}{m_f}right)=6.273

or

   frac{m_i}{m_f}=530

I would not recommend explaining this ad nauseum at most cocktail parties.

-Dave

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