Question on feasibility of relativistic travel

1. Jan 1, 2005

Curious3141

I originally posted this as a reply to a post in the Michio Kaku forum, but it hasn't been replied to in a while. I would really like this cleared up, so please excuse the double post.

Someone else made the argument that accelerating even a human nostril hair beyond 0.9c would make its relativistic mass as much as the entire Universe. Hence it would be impossible, even in theory, to accelerate a ship to close to light speed, because we would just run out of mass in the Universe with which to fuel the craft.

But I believe this is wrong. Here's my thinking :

Let's say we have some sort of advanced fusion engine on board the spacecraft.

At a speed of 0.999c relative to the Earth, the spacecraft would appear to have the mass of a "gazillion" Universes or whatever. But the nuclear fuel woud go up in mass by the same factor and the extractable energy would still be related by E = mc^2.

From the rest frame of the ship, the remaining nuclear fuel would only have its rest mass, but the ship would be at rest mass too. As far as the ship itself is concerned, there is no great difficulty in accelerating itself further to 0.9999c, for example.

Is my logic totally off ? Is it always going to be impossible to bring a ship to near light speeds ?

Thanks.

Last edited: Jan 1, 2005
2. Jan 1, 2005

Chronos

You answered your own question. It would take a huge amount of energy to accelerate a macroscopic body anywhere near c. For example, to deliver a payload of 1 kilogram from earth to alpha centauri in five years would require the mass energy equivalent of a large [very large] asteroid. Math available upon request.

3. Jan 1, 2005

Curious3141

I'm afraid that doesn't answer my question. My question pertained to the fact that the fuel (e.g. nuclear fuel) used to propel the craft would go up by the same Lorentz factor as the rest of the ship. Ergo, from the perspective of us Earth observers, a much larger reaction mass is undergoing fusion to power the much more massive craft. So there really is no problem in generating energy to accelerate the craft.

From the perspective of the craft itself, it doesn't "feel" it's own relativistic mass, only its rest mass. The fuel remains at rest mass too. So from the frame of the ship, it doesn't find it any more difficult to accelerate either.

Where am I wrong ?

4. Jan 1, 2005

pervect

Staff Emeritus

Your extractible energy will not go up as you increase in velocity.

The easiest way to avoid the confusion you are suffering is simple - just don't use relativistic mass at all. You can calculate your speed at any point form the relativistic rocket equation

http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html

v = c * tanh(a/c * Tau)

where Tau is the time elapsed on the spaceship according to the spaceship's clocks (proper time). 'c' is of course the speed of light. Tanh is the hyperbolic tangent function (most good calculators can handle this nowadays). You don't need relativistic mass to use or derive this equation. In fact, "relativistic mass" is somewhat unpopular among phsyicicst nowadays (there are some holdouts who will vocally protest otherwise) - but see for instance

http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.html

This equation tells you how long (ship time) you must accelerate at a given constant acceleration a to reach a desired velocity.

If you know how much thrust a pound (or kilogram) of fuel delivers, you can work out the mass ratio of the rocket. This is known as the "specific impulse" of the rocket.

The equation for this is

accleration* time = specific impulse * ln(mass ratio)

here ln is the natural lograthim, specific impulse is a characteristic of the rocket engine (you can look it up on the www, though you may expericne some confusion with units - don't be afraid to ask for help on this if it comes up), and the mass ratio is the moss of the ship fully fueled to the mass of the ship empty.

So if you have some specific mass ratio, and you know the specific impulse, of your engine, you can caluclate accelration * time.

Then you can use the equation

v = c * tanh (accleleration * time / c)

to calculate the velocity that your rocket will reach.

The next comment is fairly simple. Fusion rockets are not great candiates for high speed interstellar missions. So analyzing a fusion rocket and saying it can't realiistically reach .9c is not the same as saying that .9 c cannot be reached. Light sails (driving by solar pumped lasers) are a reasonably good candidte for reaching high velocities. (Stopping them is tricky, though - but you didn't actually specify that the object had to stop!).

The actual energy contained in a nostril hair moving at .9c is not as large as your post might lead someone to believe it is. It is simply 2.3 times the total mass of the nostril hair. I'm not sure how much a nostril hair weighs, exactly, or I'd calculate it out in terms of equivalent megatons - I would guess that a nostril hair striking the earth at .9c would be on the order of a largish nuclear explosion in terms of total energy release - nothing very dramtic, really.

Finally, while it might be technically possible to get a nasal hair up to .9c, there's really not a lot of point to it. The best approach to interstellar travel is simply to be patient. Because human lives are currently short, everyone is in a big hurry to get where they are going. The best approach to interstellar travel will probably not be advances in physics, but in biology. Longer lifespans, ultra-efficient recycling systems, hibernation / suspended animation, human level artificial intelligence in a robotic ship, and/or humans in robotic bodies, are all alternatives to ultra high energy physics for interstellar travel. A trip may take a hunderd years, or a thousand - but this is not necessarily a problem with the right attitude and some long-term thinking.

5. Jan 1, 2005

Staff: Mentor

Ignore relativistic calculations altogether: the fusion reactor is always at rest with respect to the spaceship so its mass (relativistic or otherwise) is constant.

6. Jan 1, 2005

Curious3141

Thanks for the responses. Pervect, I read the site you quoted and it was quite clear, although I have to look up the derivations of those equations. My mistake was in assuming that from Earth's perspective, the apparent mass/energy on the ship would change. Of course, that isn't true because as Earth is inertial, that quantity has to remain constant by conservation, meaning that the total relativistic mass of (fuel + payload) + energy of emitted light from drive = total rest mass of (fuel + payload) at start. I get this now : no free lunch. ;)

BTW, that nostril hair example wasn't mine, I was quoting MonstersfromtheId's post in the Michio Kaku forum. I realised that the mass was many orders of magnitude lower than the mass of the Universe after I had posted it, but it was just a f'r instance, so I let it be.

Thanks again. :)

7. Jan 2, 2005

pervect

Staff Emeritus
There's a derivation of the relativistic rocket equation in "Gravitation" by Misner, Thorne , & Wheeler - it's a GR textbook, but they do some special relativity calculations too. It might be a bit difficult to follow their derivation,though, unless you are familiar with four-vectors.

My favorite derivation is based simply on the velocity addition formula.

Let v(t) be the ship's velocity as a function of time, t. Then by the relativistic velocity equation, v(t+dt) is the relativistic sum of v(t) and a*dt.

(Note that tau is usually used for proper time, but it was too hard to type!)

Thus we write

$$v(t+dt) = \frac{v(t)+a\,dt}{(1+(v(t)\,a\, dt) / c^2}$$

This isn't in a convenient form, so we multiply the numerator and denominator of the right hand side by (1-v*a*dt/c^2).

then we have

$$v(t+dt) = \frac{(v+a*dt)(1-v*a*dt/c^2)}{1-v^2*a^2*dt^2/c^4}$$

We can drop the second order terms in dt^2 and get

v(t+dt) = v(t) + a*dt - v^2*a*dt/c^2

Thus
$$\frac{v(t+dt)-v(t)}{dt} = a*(1-(v/c)^2)$$

Next we do a sanity check - if v=.9c, and a*dt = .0001c, does the formula give approximately the right answer? We predict .900019 using the above formula, the relativistic velocity addition of .9c + .0001c gives .9000189983, so it passes the test.

Now all we have is a simple differential equation dv/dt = a*(1-(v/c)^2), and rather than integrate it, we can just confirm that v(t) = c*tanh(a*t/c) satisfies this equation.

We get dv/dt = a*(1-tanh(a*t/c)^2) when we take the derivative of v(t), and if we substitute tanh(a*t/c) = v/c, we see that this is just a*(1-(v/c)^2) as desired.

8. Jan 2, 2005

Curious3141

Nice, thank you. :) A simple substitution of $\frac{v}{c}=\sin \theta$ and knowing the integral of $\sec \theta$ gave a direct solution.