plutherus

Can someone who's better at math than me help me out here?

How do I calculate the derivative of v with respect to t in the formula:

v=sqr(c² – t₀²c²/t²)

And once I get that, I'll need the second derivative as well. Calculus was a very long time ago and I'm afraid I'm rather rusty...

Why I would want to know this:

So, we have a spaceship, the Bradbury
Let's say it masses 10x10⁸ kg, (about ten times the mass
of the USS Ronald Reagan.) This is a big ship, capable of carrying 50,000 colonists.

TheBradbury is equipped with the latest antimatter engines, which produce a combined thrust
of 98x10⁸ kg m/s², just enough to accelerate the ship at 9.8m/s², so the colonists are at a nice
comfortable 1G for their trip.

It sets out for Rigel Kentaurus, which according to Starry Night is 4.395 light years away (or about 9.5 x 10¹⁵ meters).

So, the question is, how long will it take to get there? Now, in purely Newtonian physics, this would be easy. The formula for distance
traveled is:
d = h₀ + v₀t + ½ at²

We can call the starting position (Earth orbit) zero. Initial velocity is also 0 (well, technically, it would have to be about 25,000mph, but
that's close enough to zero to not really matter). So, since h₀ = v₀=0, we have:
d = ½ at²

Or:
t= sqr(2d/a)

So:

t =sqr(2 x 9.5x10¹⁵m/9.8m/s²) = 44031529
seconds, or about 1.4 years.

Now, obviously, the Bradbury can't make a 4.4 light year journey in only 1.4 years without exceeding the speed of light. (3x10⁸ m/s² – it's not just a good idea, it's the law!) We have a speed limit. And, you can't just accelerate up to it and stop either: as our colonists approach the speed of light (with respect to Earth), the mass of the ship increases. Since F=ma, then, as mass increases, since we have a constant force, the acceleration would have to decrease.

Actually, it's really the other way around: time itself decreases the closer to the speed of light you seem to be, so the change in velocity (or anything else) over time would have to decrease. So, since F=ma can also be written a=F/m. If a decreases, and F is constant, then m has to increase, which we know to be true from experimental evidence.

Now, here's where I need help, as my calculus is rather rusty.

I think we can still use our same Newtonian distance formula here by adding in the next derivative, the change in acceleration over time, known as jerk, so we have:

d = h₀ + v₀t + ½ at² + 1/3jt³

So, what's the change in acceleration? That would depend on the change in time.

The formula for time dilation is:

t = t₀ / sqr(1-v²/c²)
Where t is the rate of time on the Bradbury as seen from Earth, t₀is
Earth time, v is the perceived rate of the space ship from Earth, and
c, of course, is the speed of light.

Solving for v, we get:

v=sqr(c² – t₀²c²/t²)

So, acceleration is dv/dt = the derivation of all that stuff on the right.

And jerk is the derivative of acceleration. So the question becomes: what is the
second (and the first, come to think of it, because we'll need that,
too) derivative of sqr(c² – ₀²c²/t²)?

If I could calculate that, I could then plug those numbers in to
determine how long, earth time, it takes for a spaceship at 1G
acceleration to travel from here to Rigel.

The second question is, what about those on the Bradbury?
How long do they take to reach Rigel? I believe the answer to that
is correctly 1.4 years. As they accelerate, time slows down for
them. They don't perceive it this way, of course, what they see is
time slowing down for everyone else, and distance between Rigel and
Earth shrinking So,
strangely, they would perceive Rigel, and Earth, both rushing
towards them the further they get from either one. But that's a
calculation for another day. Besides, we still have to slow down and
re-enter the Earth/Rigel reference frame if we are intending to land.
(Assuming that Earth and Rigel are moving at roughly 0 relative to
each other.)