Relativistic rocket equation(s) WITH initial velocities?

As we all know:

t = sqrt [(d/c)[sup]2[/sup] + (2*d/a)]

Where:

• t is time elapsed in a fixed observer’s frame of reference,
• d is the distance travelled from a fixed observer’s frame of reference, and
• a is the acceleration of the spacecraft from the spacecraft’s frame of reference

What I’d like to know is, what if the spacecraft starts out with some initial velocity? E.g., suppose my spacecraft can cruise at 9.8 m/s[sup]2[/sup] in its own frame of reference indefinitely, but only after it’s been boosted up to 20% of the speed of light by some outside source?

What does the equation work out to if, in addition to knowing d and a, there’s also an initial velocity v[sub]0[/sub] to contend with?

There are at least three easy ways to find this. First, of course you can just redo the integration that you used to find the original equation, but with a nonzero initial velocity.

The other two ways are even easier. Because the acceleration in the spacecraft’s frame is the same in both cases, the curve you’re looking for will of course look the same as the stationary-start curve, except that the origin will be shifted to some other point on the curve. That is, let’s write the spacecraft’s world line in the form t=f(d) as you did. Pick a point (d[sub]0[/sub],t[sub]0[/sub]=f(d[sub]0[/sub])) on this curve, and shift the origin to this point, giving the translated curve
t=f(d+d[sub]0[/sub])-f(d[sub]0[/sub]).
Here you are trying to set an initial velocity, so you just pick d[sub]0[/sub] so that f’(d[sub]0[/sub])=1/v[sub]0[/sub].

Another way is to do a Lorentz transform. Because the curve is defined to have constant acceleration a in the spacecraft’s frame, this property is independent of which inertial frame we are using. So if we just consider this same curve from the point of view of an observer moving with constant velocity -v[sub]0[/sub], we will get the desired curve.

Combining the last two methods, you can see that this hyperbolic constant-acceleration world line is essentially invariant under Lorentz boosts; all that a boost does is translate the curve. (In fact, if you shift the origin so that the hyperbola is centered at the origin, it doesn’t even translate under boosts.)

Er … that’s gonna be a problem for me, seeing as that I didn’t integrate to find the original equation. Somebody else did the derivation, and didn’t show his steps. :o

Oh … and I should mention that I’ve … er … never done a Lorentz transform, either. My knowledge of relativistic mechanics comes from high school physics. The lower-division physics courses I took in college never touched on Relativity.

Well, then, door #2 is the way for you. All that it needs is high-school algebra.

If you’re interested in it, special relativity is really not that hard. I could write down the Lorentz transform for method #3 and you could get the same answer in a few lines of algebra, but that wouldn’t really teach you anything. The usual introductory approach is probably a better way to go, where you think about moving clocks and meter-sticks (time dilation and length contraction, two special cases of the Lorentz transform) and understand some of the fundamental concepts before writing down the general transformation.

Good references include Einstein (Relativity - The Special And General Theory) and Taylor and Wheeler (Spacetime Physics; this seems to be out of print). There are undoubtedly lots of good web references now, too (but lots of cranks, too). You might be able to find online lectures somewhere. MIT’s relativity course has lecture notes, but no lectures.

I mean, the acceleration A observed by a stationary observer at any moment, assuming the spacecraft is accelerating at a uniform acceleration a in its own reference frame, would be:

``````

9.8 m/s[sup]2[/sup]
A = ---------------
SQRT (1 - v[sup]2[/sup]/c[sup]2[/sup])

``````

But don’t forget, A is dv/dt. This sure sounds like a differential equation to me!

Does it make any difference what the initial velocity is? From the rocket’s frame of reference it is still motionless.

“Door #2” was referring to this method, the second one in my list:

There’s no differential equation to solve there, since it’s already been solved for you: the answer was t=f(d). OK, I guess you have to take the derivative of f(d) to find the velocity so that you can figure out the new origin, I forgot about that. So that’s not quite high-school algebra; but it’s still simple high-school differential calculus.

So you set
1/v[sub]0[/sub] = f’(d[sub]0[/sub]) = (d[sub]0[/sub]/c[sup]2[/sup]+1/a)/sqrt((d[sub]0[/sub]/c)[sup]2[/sup]+2d[sub]0[/sub]/a)
(there, I took the derivative for you), solve for d[sub]0[/sub] in terms of v[sub]0[/sub] (a quadratic equation), and plug the result in for the translated curve above.

Right; this is the insight that allows method #2 (and #3) to work. The only difference in the trajectory is a change in origin–the shape of the trajectory is the same. If the rocket “starts” at velocity v[sub]0[/sub], all this means is that the origin is moved to the point along the original curve where the rocket had accelerated to v[sub]0[/sub] on its own.

You sure that d[sub]0[/sub]/c[sup]2[/sup] term shouldn’t be (d[sub]0[/sub]/c)[sup]2[/sup] ?

Yes. Dimensional analysis tells you that your way isn’t right, anyway. Also, we know that for large d[sub]0[/sub] the speed should approach c; my expression has a numerator ~d[sub]0[/sub]/c[sup]2[/sup] and a denominator ~d[sub]0[/sub]/c, so the expression f’(d[sub]0[/sub]) for the inverse velocity approaches 1/c, yay.

If you don’t trust me, [%2Cd]"]ask Mathematica](D - Wolfram|Alpha[Sqrt[(d%2Fc)^2%2B(2d%2Fa)).

If I let Wolfram Alpha solve for d[sub]0[/sub] for me, I get:

``````

+/- sqrt(a[sup]2[/sup]c[sup]8[/sup] - a[sup]2[/sup]c[sup]6[/sup]v[sub]0[/sub][sup]2[/sup]) - ac[sup]4[/sup] + ac[sup]2[/sup]v[sub]0[/sub][sup]2[/sup]
d[sub]0[/sub] = -------------------------------------
a[sup]2[/sup]c[sup]2[/sup] - a[sup]2[/sup]v[sub]0[/sub][sup]2[/sup]

``````

… which I now have to substitute in to my original equation (!)…

Well, get to it! Are we mice or mathematicians? (…as a professor of mine used to say…)

Actually, this is why we don’t use computer algebra systems blindly. The given answer is correct, but can be put into a much nicer form for our purposes.

First, let’s get rid of that annoying +/-; for v[sub]0[/sub]=0 we want d[sub]0[/sub]=0, which means we want the + sign. OK, Mathematica has helpfully cleared all the complex fractions; let’s put some of them back. Factor ac[sup]4[/sup] out of the numerator and a[sup]2[/sup]c[sup]2[/sup] out of the denominator. Now there are lots of instances of v[sub]0[/sub]/c, so define b[sub]0[/sub]=v[sub]0[/sub]/c. [This is a common notation in special relativity, except usually “beta” is used instead of b. This is the velocity in units of c.] Now you get
d[sub]0[/sub]=(c[sup]2[/sup]/a)(sqrt(1-b[sub]0[/sub][sup]2[/sup])-(1-b[sub]0[/sub][sup]2[/sup]))/(1-b[sub]0[/sub][sup]2[/sup]).
Obviously we should make something equal to 1-b[sub]0[/sub][sup]2[/sup] as well. Following the convention in relativity I will actually define g[sub]0[/sub]=1/sqrt(1-b[sub]0[/sub][sup]2[/sup]) [again, this is usually a gamma]. Now after a bit more cancellation you get
d[sub]0[/sub]=(c[sup]2[/sup]/a)
(g-1).
This isn’t so bad any more, right? If you like you can make things easier by rewriting your original equation as

Okay. So.

d[sub]0[/sub] = (c[sup]2[/sup]/a) * (γ[sub]0[/sub] - 1)

Therefore,

t(d+d[sub]0[/sub]) - t(d[sub]0[/sub]) = t(d + (c[sup]2[/sup]/a) * (γ[sub]0[/sub] - 1)) - t((c[sup]2[/sup]/a) * (γ[sub]0[/sub] - 1))
= (c[sup]2[/sup]/a) * sqrt(((a* (d + (c[sup]2[/sup]/a) * (γ[sub]0[/sub] - 1))/c[sup]2[/sup])+1)[sup]2[/sup]-1) - (c[sup]2[/sup]/a) * sqrt(((a* (c[sup]2[/sup]/a) * (γ[sub]0[/sub] - 1)/c[sup]2[/sup])+1)[sup]2[/sup]-1)
= (c[sup]2[/sup]/a) * (sqrt(((a*d + c[sup]2[/sup] * (γ[sub]0[/sub] - 1))/c[sup]2[/sup])+1)[sup]2[/sup]-1) - sqrt(γ[sub]0[/sub][sup]2[/sup]-1))

Does that look right to you?

Yes, except for two minor details. First, that is actually an expression for ct, not t (typo in my previous post), so divide the whole thing by c. Second, you can simplify the stuff inside the (…)[sup]2[/sup] by cancelling factors of c[sup]2[/sup] to get
c
t = (c[sup]2[/sup]/a) * {sqrt[(a*d/c[sup]2[/sup]+γ[sub]0[/sub])[sup]2[/sup]-1] - sqrt[γ[sub]0[/sub][sup]2[/sup]-1]}.
Finally, the last part would probably be simplified to
sqrt[γ[sub]0[/sub][sup]2[/sup]-1] = β[sub]0[/sub]*γ[sub]0[/sub]
(more numerically stable for calculations when β[sub]0[/sub]<<1).

So, that means

t = (c/a) * (sqrt[(a*d/c[sup]2[/sup]+γ[sub]0[/sub])[sup]2[/sup]-1] - sqrt[γ[sub]0[/sub][sup]2[/sup]-1])

Yep.

Okay, I’ve unfortunately run into a tougher problem.

My Bussard ram powered space ship needs to accelerate to a certain speed, say 0.4 c, from an initial “start up” velocity where the ram starts to work, which it attains at 0.2 c. But, the maximum acceleration my ram engine can produce (in the space ship’s reference frame) is proportional to its current velocity with respect to the interstellar medium, because the faster it’s going the more nuclear fuel it can scoop up:

a = v / (6,000,000 sec)

So at 0.2 c, it can accelerate at 10 m/s^2, while at 0.4 c it can accelerate at 20 m/s^2. (My crew are in suspended animation and can withstand a continuous 2 G’s of acceleration indefinitely with no ill effects.)

Solving The Above as a differential equation is pretty easy. It’s just

dv/dt = (1 / 6,000,000) * v.

BUT, if we bring relativity into the picture, we need to calculate what dv/dt is from the reference frame back on Earth. There:

dv/dt = (1 / 6,000,000) * v) / SQRT (1 - v^2/c^2)

I plugged this equation into Wolfram Alpha and asked it to solve it, but it came back with a hideously complicated equation with v in it multiple times (inside square roots, no less). Solving for v in terms of t was impossible for Wolfram Alpha in that case.

Is there some way to make this … well … solvable?

I can’t grok any of this in the slightest, but I thought I’d just say I love this place. And it gives this nerdtastic thread a free bump.

Unfortunately, your ramscoop is also limited by drag from the interplanetary medium, which scales up faster than your increased rate of fuel-gathering. Assuming fusion-powered engines, continual use of the scoop, and 100% efficiency, this ends up limiting you to a top speed of .12c, relative to the ISM. You could in principle get faster, but that’d require you to not use all of the fuel you gather, but instead store a surplus in big tanks, then turn off the scoop and run off of the saved fuel. I haven’t yet done the calculations for whether this would let you shave off any net time from your trip.

Let’s assume I’ve come up with some kind of “scramjet” technology that allows me to scoop up the interstellar medium, and burn it in proton fusion, without slowing it down – so that it induces negligible drag.