# Constant Acceleration seen in an inertial frame

1. Dec 5, 2007

### lightarrow

I'm sorry, this topic has certainly already been covered, but I didn't find what I need.

I'm trying to compute x(t) in an inertial frame if a rocket has a constant acceleration "a" as measured with accelerometers inside of it.

I made these (clearly wrong) computations:

In a co-moving frame which speed v (with respect the stationary frame) equals the instantaneous speed of the rocket, the rocket is seen to acquire a speed du during the interval of time dt'. The variation of the speed, as seen from the stationary inertial frame is:

dv = (v + du)/(1 + v*du/c^2) - v = [1 - (v/c)^2]du

The interval of time in the inertial frame is:

dt = Sqrt[1 - (v/c)^2]dt' <-- this is the mistake

so: dv/dt = [1 - (v/c)^2]du/Sqrt[1 - (v/c)^2]dt' =

= Sqrt[1 - (v/c)^2]du/dt' = Sqrt[1 - (v/c)^2]*a

Integrating I have v(t) proportional to sin(a*t) which is a clearly wrong result.

Can you help me?
Thanks.

Last edited: Dec 5, 2007
2. Dec 5, 2007

### Staff: Mentor

Have you been introduced to four-vectors yet? If so, I think the way to go is to use the four-acceleration ( http://en.wikipedia.org/wiki/Four-acceleration ). The norm of the four-acceleration is the proper acceleration which is the acceleration measured by the accelerometers on board.

Warning: I haven't worked out your problem and I know that I am biased towards using four-vectors just about everywhere, even where they actually make the problem harder.

3. Dec 5, 2007

### Chris Hillman

Link to FAQ, suggest trigonometric approach

See this Physics FAQ entry.

If you want a slick approach, note that the world line of an object experiencing constant acceleration (constant magnitude and direction) is a hyperbola asymptotic to two null lines, the Minkowskian analog of a circle. This suggests (correctly) that problems involving constant acceleration can be solved by developing the Minkowski analog of ordinary trigonometry.

Last edited: Dec 5, 2007
4. Dec 5, 2007

### lightarrow

It was exactly this that I wanted to find as result; it seems so trivial (and probably is) in the books, but I still haven't grasped it.

Last edited: Dec 5, 2007
5. Dec 5, 2007

### robphy

Indeed, the "uniformly [4-]accelerated worldline [in Minkowski spacetime]" is
the "curve of constant-[Minkowskian worldline-]curvature".

[Bonus:
Can you guess the curve of a "uniformly accelerated worldline in Galilean spacetime"?
..that is, the "curve of constant-[Galilean worldline-]curvature".]

6. Dec 5, 2007

### pervect

Staff Emeritus
Since you're getting (proportioanl to) sin, and the correct answer is (proportioanl to) sinh, start looking very closely for sign errors. (d^2/dt^2) sin(t) = -sin(t), (d^2/dt^2) sinh(t) = sinh(t).

Last edited: Dec 5, 2007
7. Dec 5, 2007

### lightarrow

Circle. In the previous case hyperbola because the interval is constant there. So the curvature is proportional to the acceleration? And the interval is the analogous of the radius so the curvature is inversely proportional to it?

8. Dec 5, 2007

### lightarrow

I've found the mistake: actually dt = dt'/Sqrt[1 - (v/c)^2] so:

dv/dt = a[1 - (v/c)^2]^(3/2) and integrating:

v(t) = at/Sqrt[1 + (at/c)^2] -->

x(t) = (c^2/a)Sqrt[1 + (at/c)^2] -->

--> x^2 - (ct)^2 = (c^2/a)^2

Last edited: Dec 5, 2007
9. Dec 5, 2007

### robphy

It's not the circle.
Actually, I don't think it's a hyperbola "because the interval is constant there."

Acceleration is primarily about worldline-curvature not intervals.

In case it's unclear, the Galilean spacetime is the spacetime structure underlying much of introductory physics.

10. Dec 6, 2007

### jcsd

We can say it's not a circle because topologically the worldine of any object in Gallilean spacetime can never be a circle. But as dt is invariant under a Gallilean transformation there's no useful metric that can be imposed on Gallilean spacetime, so talking about curvature of worldlines is redundant?

11. Dec 6, 2007

### lightarrow

In the Galilean spacetime a constant acceleration is described by a parabola, but how do I compute the curvature? Why is it constant in it?

12. Dec 6, 2007

### robphy

Yes, the parabola (on a distance-vs-time graph or spacetime diagram).
Here is a poster of mine (from an AAPT conference) that describes it (page 16)