Atomic Rockets


Mass ratios are the bane of atomic rocket designers. No matter how potent the drive is, you are going to have several kilograms of propellant for each kilogram of rocket. This puts severe limits on the sorts of missions a rocket can perform before the ever-hungry engine has to be fed again.

This is because all rockets utilize Newton's Third Law of action and reaction. You throw something backwards (the propellant) and in reaction the rocket moves forward. This is why rockets are called "reaction drives."

Naturally, the thought occurs that if you can figure out how to make a spacecraft move without using propellant, all the problems with mass ratio vanish. You'd have a "reactionless drive."

Which would be great, were it not for the unfortunate fact that it would violate the law of conservation of momentum.

Now, it is true that Newton's third law has some rare occasions where it does not apply (certain situations with magnetically coupled particles and gravitational forces acting between objects moving very rapidly), but the law of conservation of momentum is a genuine iron-clad rock-solid no-exception law. In a closed system the total quantity of momentum cannot change. It has been verified to within one part in 1e15, and no exception has ever been found.

Which means in a closed system, a reactionless drive is impossible, since it would change the total quantity of momentum. And even if you hand-waved one into existence for your SF novel, you've still got problems.

(Note that it is possible to avoid that law with an open system, with something like a solar sail, a spacecraft launched by a mass driver based on an asteroid, pellet-stream propulsion, or a Starwisp. In these cases, the propulsion system is external to the spacecraft, so the system is open and the law does not apply.)

However, a little thing like violating a law of physics isn't going to stop the crack-pots. Face it, the second law of thermodynamics hasn't stopped all the people attempting to create perpetual motion machines of the first kind.

Yes, before you all email me, I have heard about Roger Shawyer's EmDrive. It too violates the law of conservation of momentum, and the inventor's experiments have not been replicated.

Dean Drive

Oh, the Dean Machine, the Dean Machine,
You put it right in a submarine, And it flies so high that it can't be seen --
The wonderful, wonderful Dean Machine!

Damon Knight

The fun started in 1960 when the John W. Campbell (the father of the Golden Age of Science Fiction) decided to make some excitement by giving free publicity to Norman Dean and his infamous "Dean Drive". It allegedly could convert rotary motion into linear motion, i.e., it was a reactionless drive. U.S. Patent 2,886,976. "Just think," Campbell said, "stick one of these in a submarine and you have instant spaceship!"

Campbell was miffed that mainstream scientists were not even interested in looking at the drive. But in this case, the scientists were acting properly. Faced with the fact that the Dean Drive obviously violated the law of conservation of momentum, well, extraordinary claims require extraordinary proof. A box vibrating on a pan balance that makes the beam scale look like it had lost an ounce or two is not anywhere near convincing enough.

Interest in the Dean Drive faded away as Dean refused to let anybody examine the gadget, with the notable exception of John W. Campbell and G. Harry Stine. At least without forking over some money first. After Dean died, Stine made a brief resurgence of interest in the 1980's, but it died too, and later so did Stine. A close examination of the patent reveals that the device is actually a complicated ratchet pulling itself along a metal tape, not a reactionless drive.

Physicist Milton Rothman notes that Dean Drive apologists wave their hands and talk about the strange relationship between force and changing acceleration as a justification for the drive, but all they are doing is revealing the depths of their ignorance about basic physics.

Conservation of Momentum

I had thought that one could hand-wave a reactionless drive but control it with some kind of limit on the damage. Specifically I thought that one could figure the kilowatt equivalent of the momentum change created by such a drive, and use that as the required power.

The experts at quickly educated me as to how naive I was.

The underlying problem is that breaking the law of conservation of momentum shatters the entire mathematical framework. The specific problem is that you will get different values for the kinetic energy expended depending upon the reference frame of the observer.

Isaac Kuo said:

There are basically two approaches you can use:

1. There is a special frame of reference. In this case, the "reactionless" drive is really pushing against an infinitely massive special frame of reference.

(ed note: which means you've just destroyed Einstein's Relativity, with all the collateral science damage that implies)


2. There is no special frame of reference. In this case, the only way to sort of preserve conservation of energy is to limit drive efficiency to that of a photon drive. This is not a very useful drive, though, since it has the same (low) performance as a photon drive.

(ed note: the photon drive, where one lousy Newton of thrust takes three hundred freaking megawatts!!)

Isaac Kuo

Dr. John Schilling said:

There is the complication that "energy of the thrust" is as meaningless a phrase as, e.g., "mass of the time".

Thrust is a force, not an energy. Force, multiplied by distance, gives an energy. A force of one pound, applied as an object moves over a distance of one foot, equates to (unsurprisingly) one foot-pound of energy. The same force, over a greater or lesser distance, comes to proportionately more or less energy.

In MKS, by the way, that would be one Newton of force over one meter of distance equals one Joule of energy. If we assume constant force and motion, we can extend that to one Newton of force applied constantly at a velocity of one meter per second, equals a power of one watt.

The question is, velocity relative to what?

If it is a rocket, the relevant velocity is that of the rocket's own exhaust relative to the rocket itself. For an "intertialess thruster", the answer isn't clear and the power or energy associated with a given thrust will change widely depending on what reference frame you use to measure your velocity.

Which is one facet of the reason "inertialess thrusters" seem to be physically nonsensical. However, if you really need one for some SFnal purpose, you could try either

  1. the one universally invariant velocity in real physics. That being the velocity of light, giving you a figure of three hundred megawatts of power per Newton of thrust. A tad high for most purposes, I think, and functionally equivalent to saying your thruster is a photon drive or a (nearly-)massless-neutrino drive or a Dark Energy Rocket or whatever.
  2. the velocity of the spacecraft relative to some absolute reference frame. Either a cosmic absolute, or a local absolute tied e.g. to the nearest massive body or bodies in whatever manner is most convenient to the story. This is functionally equivalent to the old aetheric theories, and you can mine those for ideas.
Dr. John Schilling

Why doesn't this reference frame problem occur with an ordinary rocket? Isaac explains:

Because an ordinary rocket has a "reaction". The amount of kinetic energy added to the rocket by a rocket thrust depends upon what frame of reference you look at it. Indeed, there are plenty of frames of reference where the rocket thrust subtracts kinetic energy from the rocket! So you can't meaningfully talk about THE amount of kinetic energy added to the rocket. However, you CAN meaningfully talk about how much kinetic energy the rocket adds to the system because kinetic energy is also added to (or subtracted from) the rocket exhaust. No matter what frame of reference you use, the total amount of kinetic energy in the rocket plus the exhaust is increased by the same amount.

Isaac Kuo

In an ordinary rocket, both the kinetic energy of the rocket and the kinetic energy of the exhaust will change. Different observers will disagree about the absolute change of each, but will agree about the net change in kinetic energy, and so energy conservation can be enforced.

Example: A hundred-kilogram satellite ejects one gram of nitrogen through a cold-gas thruster at a velocity, relative to the spacecraft, of one hundred meters per second.

An observer at rest relative to the initial position of the spacecraft will see it accelerate to 0.001 meters per second, increasing its kinetic energy by 0.05 millijoules. The exhaust will be observed to accelerate to 99.9995 meters per second, with resulting kinetic energy of 4.99995 Joules. The total kinetic energy increase, provided by the expanding gas, comes to 5 Joules.

An observer zipping along in the opposite direction at 1,000,000 meters per second, will see both the spacecraft and the propellant as having had an initial velocity of 1,000,000 meters per second, and an initial kinetic energy of 50,000,000,000,000 Joules and 500,000,000 Joules, respectively. The spacecraft accelerates to 1,000,000.001 meters per second, giving it a new kinetic energy of 50,000,000,100,000 Joules - a gain of 100,000 Joules. Far cry from the .05 millijoules the stationary observer had thought the spacecraft acquired.

But the moving observer will have seen the slug of exhaust gas decelerate from 1,000,000 m/s to 999,900.0005 meters per second, with a new kinetic energy of 499,900,005 Joules. That's a loss of 99,995 Joules. So the net change in energy is, spacecraft +100,000.0, exhaust -99,995.0, or plus 5.0 Joules. Both observers agree on conservation of energy. And, for that matter, momentum.

If there were only the spacecraft involved, they'd be arguing about the missing hundred kilojoules.

Dr. John Schilling

If you're talking about a "true" reactionless drive, where energy is converted directly into momentum (or angular momentum), then there are lots of complications. Consider for instance that kinetic energy goes as the square of the speed:

K = (1/2) m v^2

So the power P you need to accelerate is dK/dt:

dK/dt = (1/2) m [2 v dv/dt] = m v a

As you can see, the power is a function of not only the acceleration that you want (which seems obvious), but also the speed at which you're currently traveling. The snag there is that your current speed is frame dependent. Consider that you're already doing your job and accelerating along nicely. At that point you pass someone who is already coasting at nearly the same speed you are. He sees you using much less power! Who's right?

The solution is that you either need to play by the rules of the game and use reaction drives (even if it's just reaction momentum, like a photon drive), or posit a special frame in violation of special relativity. With the special frame, now there's a "correct" frame where all the kinetic energy calculations are "official" and everyone agrees on them.

Erik Max Francis