# Deriving the displacement of an object

1. Jul 18, 2014

### Solibelus

Hi, I'm trying to derive the displacement of an object with proper acceleration (constant) α.
t is our laboratory time.

The goal:
$$x(t) = {1 \over α} [ \sqrt{1+α^2t^2} -1 ]$$

How to approach this? :(

Note for anyone who's wondering: no, this is not homework...

2. Jul 18, 2014

### WannabeNewton

Do you know how to obtain the parametric equations $x(\tau)$ and $t(\tau)$ for a uniformly accelerating object of magnitude $\alpha$, where $\tau$ is the proper time of a comoving clock? If so then your result is immediate, you just have to use a bit of algebra.

If not then your first step would be to calculate $(t(\tau),x(\tau))$. Use $a^{\mu}a_{\mu} = \alpha^2$ and $u^{\mu}u_{\mu} = -1$ to get the differential equations for $x(\tau)$ and $t(\tau)$ and solve them.

3. Jul 18, 2014

### pervect

Staff Emeritus
The easiest approach is the textbook approach using 4-velocities, or proper velocities.
You'll find the following worked out in MTW's "Gravitation", in tensor notation. I've re-written it to use non-tensor notation. I've also chosen to use subscripts in places where MTW uses superscripts for readability.

Let u be the 4-velocity (proper velocity) of the object, and let $\tau$ be proper time. We wish to solve for $u(\tau)$, we will take care of the conversion to find x(t) later.

Then by definition we can write the components of the 4-velocity $u_0$ and $u_1$ as:

$u_0 = dt/d\tau \quad u_1 = dx/d\tau$

where $x(\tau)$ is the displacement in terms of proper time.

The proper acceleration a is just the rate of change of the proper velocity with respect to proper time, i.e

$a_0 = \frac{d }{d\tau} u_0$ and $a_1 = \frac{d }{d\tau} u_1$

The 4-magnitude of the proper velocity is a constant. With the sign conventions being used, it's -1. Thus ${u_1}^2 - {u_0}^2 = -1$.

The 4-magnitude of the proper velocity is a constant. With the sign convetions being used it's just $\alpha$ (no minus sign). Thus

$a_1^2 - a_0^2 = \alpha^2$

Because the magnitude of the proper velocity is always constant, the proper acceleration is always perpendicular to the proper velocity

$u_1\, a_1 - u_0 \,a_0 = 0$

If you wish, you can derive this mathematically and formally from the fact that $u_1 = \sqrt{u_0^2 - 1}$ using the chain rule.

The simple way to proceed is to note that we have

${u_1}^2 - {u_0}^2 = -1 \quad u_1\,a_1 - u_0\,a_0 = 0 \quad a_1^2- a_0^2= \alpha^2$

where the last equation comes from the fact that the magnitude of the proper acceleration is constant and equal to $alpha$.

We can write down the solution $a_0 = \alpha \, u_1 \quad a_1 = \alpha \, u_0$ which we can easily verify by inspection.

This is equivalent to the differential equation:

$\frac{d}{d\tau} u_0 = \alpha \, u_1 \quad \frac{d}{d\tau} u_1 = \alpha \, u_0$ Combining this with the fact that $u_1^2 - u_0^2 = -1$ we immediately write down the solution

$u_0 = \cosh (\alpha \, \tau) \quad u_1 = \sinh (\alpha \, \tau)$

We integrate to find:

$t = \frac{1}{\alpha} \sinh(\alpha \tau) \quad x = \frac{1}{\alpha} \cosh (\alpha \, \tau)$
This doesn't have x=0 at t=$\tau$=0, which is just a matter of an integration constant. But we'll address this issue later, we'll just note at this point we can add a constant value to x and we will still have a solution, the proper velocity won't be affected by a choice of origin for x.

To solve your problem, we don't need to write down the proper acceleration, but it's a good check. We see that it's magnitude is constant as desired.

$a_0 = \alpha \, \sinh(\alpha \, \tau) \quad a_1 = \alpha \, \cosh (\alpha \, \tau)$

To find x as a function of t (rather than x and t as a function of tau) we start with
$t = \frac{1}{\alpha} \sinh(\alpha \tau) \quad x = \frac{1}{\alpha} \cosh (\alpha \, \tau)$

We note that
$x^2 - t^2 = \frac{1}{\alpha^2}$

Solving for x, we have

$x = \sqrt{t^2 + \frac{1}{\alpha^2}}$

The solution won't be affected if we add a constant to x to make it equal to 0 when t=0 as we mentioned previously. Adding this offset, we have

$x = \sqrt{t^2 + \frac{1}{\alpha^2}} - \frac{1}{\alpha}$

which is equal to (though formatted differently) than the expression you wished to find.

Last edited: Jul 18, 2014