# GR and constant acceleration

1. Nov 15, 2012

### DiracPool

I just watched a video whereby the man says that you could accelerate at a constant one g for 10 years and not reach the speed of light due to GR effects. I did a non-GR calculation and found that you would hit c at about one year. Could someone tell me specifically how GR affects this rate? Here's the vid, see the last 5 min.

Last edited by a moderator: Sep 25, 2014
2. Nov 15, 2012

### CJ2116

I don't think that you need GR to analyze this; SR is more than adequate as it can handle constant acceleration. For starters, you can't travel at the speed of light!

Here is a link (From the Problem Book in Relativity and Gravitation) with a similar problem and solution. Type in page 14 for the problem and page 169 for the solution.

This should give you some idea of what's going here.

3. Nov 15, 2012

### Staff: Mentor

As long as tidal gravity is not significant you can use SR only, there is no need to use GR in this scenario.

The first thing to understand is the difference between coordinate acceleration and proper acceleration. Proper acceleration is the amount of acceleration that you "feel", it is measured by an accelerometer. Coordinate acceleration is just the second time derivative of your position.

At any point in time there is a special reference frame called the momentarily comoving inertial frame. In that frame for that instant the proper acceleration equals the coordinate acceleration.

If you work out the math you find that curves of constant proper acceleration asymptotically approach c, but never reach or exceed it. As they get close to c the coordinate acceleration goes to 0.

4. Nov 15, 2012

### pervect

Staff Emeritus
The Relataivistic Rocket (Sci.physics.faq) has the equations of motion.

To get some insight, consider the Newtonian formula

$v(\tau+d\tau) = v(\tau) + g*d\tau$

and replace the addition of velocities with the relativistic velocity addition formula

$$v_1 + v_2 = \frac{v_1+v_2}{1+v_1\,v_2/c^2}$$

giving

$$v( \tau+d\tau ) = \frac{v(\tau) + g d\tau}{1+ (g/c^2)\,v(\tau)\,d\tau} \approx v(\tau) + \frac{g d\tau}{1-v^2(\tau)/c^2 }$$

(The approximate answer can be derived with a taylor series, among other methods, using calculus).

You can use a spreadsheet or calculus to find $v(\tau)$ this way, and compare it to the exact known results in the sci.physics.faq.
Not that $\tau$ here is proper time, what the spaceship's clock measures.

If you are familiar with 4 velocites and 4-accelerations, MTW's textbook "Gravitation" has a more formal derivation.

It turns out that the 4-velocity is $[\cosh g\tau, \sinh g\tau]$
while the 4-accleration is its derivative $[g \sinh g\tau, g \cosh g\tau]$
and the 4-position its integral $[(1/g) \sinh g\tau, (1/g) \sinh g\tau + K]$

Once you know that the magnitude of the 4 velocity must be -1 (with MTW's sign convention), and the 4-acceleration must be perpendicular to the 4-velocity, plus the fact that the 4-acceleratio is just the derivative of the 4-velocity with respet to proper time $\tau$ its pretty easy to solve the equations of motion. Realizing why all of these are true will require some familiarity with 4-vectors and their application to relativity.

This sort of motion is also known as "hyperbolic motion", See the wiki

Last edited: Nov 15, 2012
5. Nov 16, 2012

### DrStupid

You result is correct in a single frame of reference. The result in the video is correct, if velocity v and acceleration a are measured in different frames of reference - v in the original inertial rest frame and a in inertial frames moving with v.