# Hyperbolic path in Minkowski space

1. Aug 27, 2013

### luitzen

The path described by a constantly accelerating particle is given by:

$x=c\sqrt{c^2/a'^2+t^2}$

where a prime denotes an observer traveling with the particle and a letter without a prime a resting observer.

If we leave the $c^2/a'^2$ out it reduces to x=ct, which makes sense. The distance traveled by a photon is simply it's speed times the travel time. If we think of a constantly accelerating particle, a particle which would first need to accelerate to a speed c (or at least close to it), we would expect that it would travel a shorter distance in the same amount of time, since it would first need to accelerate and hence travel at a smaller average speed. However, when you look at the equation above, you can see the travel time gets a little bonus, $c^2/a'^2$:

$x=c\left[t+f\left(c^2/a'^2\right)\right]$

According to the equation, the slower one accelerates, the larger distance one will be able to travel. To me that does not make any sense at all, so I wondered whether I made any mistakes and decided to verify the answer on the internet. However, without blinking an eye, every single source I find online gives me the same answer.

So where do I go wrong in all this?

Sources:
http://en.wikipedia.org/wiki/Hyperbolic_motion_(relativity [Broken])
http://www.physik.uni-leipzig.de/~schiller/ed10/Uniform relativistic acceleration.pdf

Last edited by a moderator: May 6, 2017
2. Aug 27, 2013

### The_Duck

Careful: where is the particle at t = 0?

3. Aug 27, 2013

### luitzen

$x\left(t=0\right)=c^2/a'$

So I should define a new axis x'' then?

$x''=x-c^2/a'=c\sqrt{c^2/a'^2+t^2}-c^2/a'=c\left(\sqrt{c^2/a'^2+t^2}-c/a'\right)$

If a' goes to infinity:

$x''=ct$

If a' goes to zero:

$x''=c\left(\sqrt{c^2/a'^2}-c/a'\right)=0$

Let's say a'=0.5c/yr, t=5 yr:

$x''=c\left(\sqrt{4+25}-2\right)=2.39ly$

t=infinity:

$x''=c\left(\sqrt{c^2/a'^2+t^2}-c/a'\right)=c\left(\sqrt{t^2}-c/a'\right)=c\left(t-c/a'\right)=ct$

This makes all sense to me. Thank you.

4. Aug 27, 2013

### yuiop

This source http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken] gives the distance travelled as:

$$d =\frac{c^2}{a}\left(\sqrt{1+\left(\frac{at}{c}\right)^2}-1\right)$$

d in this case increases with increasing proper acceleration a.

For your example in your second post, when a=0.5c/yr, t=5 yr, d=3.385 lyr.

Last edited by a moderator: May 6, 2017
5. Aug 27, 2013

### luitzen

It's the same, but I made a calculation error. I did $\sqrt{25+4}-3$ instead of $\sqrt{25+4}-2$

I prefer my version though, since it's in the form of c times a time.

In my formula it's also much more obvious that d increases with increasing a'.

6. Aug 27, 2013

### yuiop

Here is how to get from the equation given in the Baez link I posted, to the equations in your links.

Starting with:

$$d =\frac{c^2}{a}\left(\sqrt{1+\left(\frac{at}{c}\right)^2}-1\right)$$

which can be rewritten as:

$$d =\sqrt{\frac{c^4}{a^2}+c^2t^2}-\frac{c^2}{a}$$

where d=0 when time t=0.

In the Rindler coordinate system (X,T), the distance from the origin is c^2/a at time T=0 and the proper acceleration (α) of a given particle is proportional to c^2/X also at T=0. This means that the above equation can be written in terms of X as:

$$X = d+\frac{c^2}{\alpha} = \sqrt{\frac{c^4}{\alpha^2}+c^2T^2}$$

which is essentially your equation at the start of this thread. Note that α is constrained to be the value of c^2/X at time T=0.

P.S. I see I missed your post while posting this and that you had already figured this out. I was thrown by the fact you got a different result but that was just a numerical error.

Last edited by a moderator: May 6, 2017
7. Sep 18, 2013

### qraal

If we rearrange that equation in terms of the co-ordinate time we get this:

$$t =\sqrt{\left(\frac{2d}{a}\right)+\left(\frac{d}{c}\right)^2}$$

...so the co-ordinate time can be thought of as having two time components - the Newtonian time to travel a displacement d at a constant acceleration a, and the light travel time. I'm not sure if that makes sense, but that's how it looks to me from the equations.

Last edited by a moderator: May 6, 2017
8. Sep 18, 2013

### luitzen

Thanks, that makes total sense.