Math of relativistic space travel

[Ike47]

Hi, Graal. I'm still waiting and hoping to get answers from Janus about calculating exhaust velocity and about how to determine when coasting should start in an accelerating-coasting-decelerating scenario.

I want to be able to vary the inputs to see the effect on flight time (both ship referenced and Earth referenced), but the initial set of parameters I came up with, assuming (100% efficient) fusion of hydrogen, and using the statistics of the Saturn V rocket as a starting point were:

M_T = 2.3 x 10⁶kg

MR = 4.6

Total energy: 1.134 x 10²¹ J

(This last based on 2.6 x 10⁴ m³ of liquid hydrogen, with a mass of 1.8 x 10⁶ kg, and an energy output of 6.3 X 10¹⁴ J kg^-1.)

If I could convert that energy to an exhaust velocity, I could solve the equations Janus suggested. Then I'd just have to be able to determine when the ship starts coasting, since its coasting speed will depend on the ideal time to shut down the engines.

As for interstellar drag, DaveC, I think he's referring to the particles in 'empty' interstellar space, which I believe are thought to be in the vicinity of one atom per cubic meter. While that amount of contact with a ship is negligible at ordinary speeds, if the ship is at relativistic speed, then the mass of the those particles becomes effectively great enough to slow down the ship by drag... if they don't destroy it instead. :)

 

[qraal]

Can you elaborate on this "interstellar drag"?

Same as regular drag, but it depends on how the starship handles it. If using a magnetic field to deflect ions, then that will create drag over the whole effective area of the field's influence (which I'm not sure exactly how to calculate, though I have a rough idea.)

The other question is whether any kind of aerodynamic reduction of the drag is feasible when handling a very high flux of interstellar ions, atoms and dust. Otherwise the incident flux on the frontal cross-section of a vehicle will be very high.

To illustrate, the Interstellar Medium (ISM) is typically quoted at being about 3 Jupiter masses per cubic light-year, and is composed largely of individual hydrogen atoms. It works out at, on average, about 1 million protons per cubic metre (it's lower than average near the Sun's part of the Galaxy) or ρ = 1.67x10^-21 kg/mγ. Crashing through the protons and absorbing their energy directly means the frontal incident flux is ~1/2ρv³ per square metre. That's a good estimate at low speeds < 0.5 c. Once you're traveling at relativistic speeds the γ factor appears in the equations and it gets a bit more elaborate mathematically. First the kinetic energy of the incident protons becomes m_pc²(γ-1), and the volume traveled through by each square metre of frontal area per second is (vγ) m³ (in the ship's reference frame.) Thus the incident energy is now E = ρc²(γ-1)(vγ) J/s which means it gets really, really hot in a hurry.

 

[qraal]

Hi Ike47

It's Qraal, which sounds a bit like my real name.

I worked out the equations for a boost-coast-brake flightplan, both classical and relativistic. Just let me know if you want them still.

As for your exhaust velocity computations it's usually best to use 'c' units and work out the energy in terms of mass ratios from before and after the reaction. For example, straight hydrogen fusion like the Sun does, has a net energy production of 0.007 relative to the mass of the original hydrogens. That means the resulting helium's energy-to-mass ratio is 0.007/(1-0.007) = 0.00705 and the resulting gamma-factor is that plus 1 i.e. 1.00705. Invert it, square it and subtract it from 1, then take the square root to get the velocity of the exhaust in units of 'c' (lightspeed.)

i.e. v = (1-γ^-2)^0.5

which in this case gives v = 0.118 c.

But - a big 'but' - that's an ideal case using a non-ideal reaction. Proton-fusion is slow and hard to start. A proton at 15.7 million K and 215 billion bars of pressure (i.e. inside the middle of the Sun) has a 50% probability of reacting with another proton in 9 billion years. Slow. Not good for rockets. Proton-proton reactions don't speed up much until about 10 billion degrees or more, a temperature we just can't contain at sufficient density.

A good place to find fusion reaction details is here...

http://www.ibiblio.org/lunar/school/InterStellar/Explorer_Class/Bussard_Fusion_systems.HTML" (forgive the bad spelling)

If you want to be especially precise you can look up all the reactions in detail around the web and use precise isotope masses to work things out. However in the real world fusion engines will get less than ideal performance.

Hi, Graal. I'm still waiting and hoping to get answers from Janus about calculating exhaust velocity and about how to determine when coasting should start in an accelerating-coasting-decelerating scenario. [snipped]

If I could convert that energy to an exhaust velocity, I could solve the equations Janus suggested. Then I'd just have to be able to determine when the ship starts coasting, since its coasting speed will depend on the ideal time to shut down the engines.

 

[Ike47]

Thanks, Qraal, and sorry for misreading your name.

First, just to respond about interstellar drag, your numbers are right, but probably overpessimistic. (My number, 1 atom m^-3, applies to intergalactic space, e.g., Smoot and Davidson, Wrinkles in Time, p. 158.) I say overpessimistic, in that the 1 million protons m^-3 is an average, and by avoiding dust clouds, nebulae, etc., one should manage with about a few thousand atoms or protons m^-3, at least according to the only partially documented Wikipedia page on interstellar medium.

Mostly I wanted to respond to your comments and offer about space travel formulae, however. Yes, I'd very much like to see the formulae you've calculated for acc-coast-dec space flight. Especially because I'd like to be able to vary the input data and see the effect on the time various destinations would require.

Thanks as well for the exhaust velocity formula. I quite understand that 100% energy conversion is impossible, but it provides a rough estimation. Plus, assuming unknown scientific and engineering advances in, say, the next 500-1000 years, perhaps 90-95% efficiency might be possible. Or not. :)

I didn't try to maximize all my data idealistically, btw. For example, I just used H to He, rather than the slightly more powerful H to Fe (which, from your formula, I calculate would create a v of 0.13 c). Nor did I assume a compressed gaseous state for the hydrogen, as in the Sun, but only liquid hydrogen. So I figured assuming 100% energy conversion efficiency wouldn't be too far off, hypothetically. Of course, the controlled fusion of liquid hydrogen within the confines of a rocket engine is fantasy at this point anyway, but that's why I'm talking about using this for science fiction, not serious speculation on the possibilities of interstellar flight.

Another complication I thought of is that efficient fusion of H might require a catalyst, as in the CN cycle of H to He. That would further worsen the vehicle's mass ratio. (This, and most of my information on fusion, comes from a book I've found very useful, Tayler's The Stars: Their Structure and Evolution, 1994 edition.)

In any case, with the formula you provided for exhaust velocity, I think I have everything I need except the formulas for acc-coast-dec (or boost-coast-brake). I'm guessing that these formulas would be similar to the ones Janus provided earlier in this thread, except that they would allow the calculation of the coasting beginning and ending points (time, distance), so the coasting velocity could be known. If you could provide those formulae (it sounds like you've already calculated them), I would really appreciate it!

 

[DaveC426913]

... by avoiding dust clouds, nebulae, etc.,

Have you considered the implications of trying to dodge obstacles with your ship in terms of trip-time - but more significantly - fuel?

It could easily double your fuel requirements. Say there were a mere two dust clouds blocking your path to your destination and the shortest route was between them. You'd have to accelerate your ship to relativistic velocities perpendicular to your path to do so. In other words, you've just added a whole trip within a trip fuel-wise.

It's entirely possible it would be more efficient to add a shield simply plow through the clouds.

 

[Ike47]

True, DaveC. But if your object was to explore the galaxy, or at least the relatively nearby stars in it (for example, all the stars and possible associated planets within, say, 50 parsecs of the Sun), I'd think you could find target stars from Earth with much less intervening matter than other target stars. You could try going to those first, and find paths that had low interstellar density from those stars to the stars that weren't good destinations from Earth directly.

In this manner, I would think you could get to most of the stars within any given region of the galaxy more efficiently than just 'plowing through' whatever density you might come upon. Of course, all this assumes you could measure the interstellar density at any given location reasonably accurately from Earth. To what degree that is, or may become true, I don't know.

 

[DaveC426913]

True, DaveC. But if your object was to explore the galaxy, or at least the relatively nearby stars in it (for example, all the stars and possible associated planets within, say, 50 parsecs of the Sun), I'd think you could find target stars from Earth with much less intervening matter than other target stars.

I would hope that the destination would be deciding factor, not the ease of the journey.

"Hey, this star has an Earth-like planet with an oxygen atmo!"
"Yeah, but look all thst dust in the way... let's go for that dead iceball instead."
"Uh, this is my entire life's work hangin' in the balance..."

 

[qraal]

Hi All

These equations are for boost-cruise-brake trajectories with constant acceleration, a; displacement, S; travel-time (planet reference frame) t, and speed, v, with γ = (1-(v/c)²)^-1.

Classical: a = c²/(vt - S)*((γ² - 1)/γ²)

Relativistic: a = c²/(vt - S)*((γ - 1)/γ)

...which is how I used the equations originally. I put the classical version in terms of γ (and v) to compare the two equations more readily - naturally it has no physical meaning in classical dynamics. A simpler version for the classical situation is:

t = v/a + S/v

which becomes...

a = v²/(vt - S)

...the two are related by the factor a_r/a_c = 2γ/(γ + 1), where a_r is the relativistic acceleration needed for the same v in time t, and a_c is the classical acceleration.

From the ship-board point of view the travel time, τ, is a bit more complicated by a cosh(γ) factor. Here's how it goes...

τ_a = 2c/a*arcosh(γ)

τ_c = (S - S_a)/vγ

thus τ = τ_a + τ_c

where τ_a is the total boost/brake time, and τ_c is the coast time. S_a is the total displacement while boosting/braking. In terms of S_a, a & c the γ factor is 1+aS_a/2c². From there you can infer the rest.

As for constant thrust dynamics, the relativistic equations are quite elaborate, but the difference between them and the classical result is minor. One problem with big mass-ratios and a constant thrust is that the average velocity ends up much less than the exhaust velocity over the acceleration track. That can be true for constant acceleration too, but it's worse for constant thrust. Better performance comes from constant acceleration and dropping off spent fuel-tanks and engines as you go, or using them for propellant (which you can do with lithium for example.)