# The aging effect with regard to high speed travel

1. Nov 24, 2013

### geordief

Hi
Suppose an astronaut is accelerated to the speed of light (OK to one part in a million less than the actual speed of light ) and makes a journey to the event horizon of a black hole that is ,at the outset of the journey and hopefully also when he reaches it) 100 light year distant.

At his destination he slingshots back in direction of the Earth and so arrives home after 200 years.

How much time has elapsed in the spacecraft on the astonauts watch?

What would be the corresponding answer if the speed attained by the astronaut was 50 %, 90% or 100 %(theoretically) of the speed of light?

Many thanks.

2. Nov 24, 2013

### Mentz114

This is a non-trivial calculation. The graphs are from this paper. As you can see the proper time elapsed on the orbit is less than the elapsed value of coordinate time.

Uros Kostic, Analytical time-like geodesics in Schwarzschild space-time.
General Relativity andGravitation, 2012.
Preprint :http://arxiv.org/pdf/1201.5611v1.pdf

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3. Nov 24, 2013

### geordief

I am sorry to lay my ignorance bare but could it be that the information you have provided is more relevant to black holes themselves rather than the "simple" calculation of objective time lapse versus subjective time lapse in 2 inertial frames moving (should it be "accelerating -I am not sure?) wrt each other?

I agree it was myself who introduced to black hole to my question but I only did it so as to provided a plausible method of effecting the return lap of the journey.

I am afraid the mathematics of the analysis you provided seems to be beyond me and I not sure if the general thrust is of much help to what I was asking (unless I was just too dim/lazy to see).

Is it possible for me to reframe the question as having the astronaut simply decelerate to arrive at the furthest point (100 light yrs hence) ,turn around ("instanteously") and return home without need for a black hole?

4. Nov 24, 2013

### Mentz114

It does not have to be a black hole, a star or planet will do. You need to specify the closest approach distance to get the exact value.

Sure, that can be done with a simple space time (Minkowski diagram). ghwells will be along to show you how. (Go George.)

5. Nov 24, 2013

### phinds

Regardless of what the spaceship does, the time elapsed in it flows at one second per second. Your "200 years" is presumably from the point of view of the spaceship, so the time elapsed is 200 years.

EDIT: OOPS I thoughtless conflated light years w/ years. Sometimes, I get like that

Last edited: Nov 24, 2013
6. Nov 24, 2013

### ghwellsjr

Your question is answered very easily if you stipulate that the astronaut travels at a constant speed, β as a fraction of the speed of light, but different directions during his trip. The ratio of Coordinate Time in the Earth's rest frame to his accumulated Proper Time is equal to gamma. Since gamma is:

1/√(1-β2)

So if you just plug in your values of speed and divide gamma into 200 years, you will get how much the astronaut ages during his trip.

For β=0.999999, gamma=707.1 and the astronaut ages 0.283 years.
For β=0.9, gamma=2.294 and the astronaut ages 87.18 years.
For β=0.5, gamma=1.1547 and the astronaut ages 173.2 years.

Theoretically, a speed of β=1 is impossible since the calculation of gamma requires a division by 0.

EDIT: I was assuming that the trip took 200 years of Earth time no matter what the distance. If you want the distance to be 100 light-years, then you have to calculate how long in the Earth frame it takes to traverse the round trip at any given speed and use that for the Earth accumulated time. It's still very easy.

Last edited: Nov 24, 2013
7. Nov 24, 2013

### TumblingDice

Possible, and also desirable. It's fine to set unrealistic conditions as long as you keep their influence in mind. For example, if you're more interested in the largest impact that time dilation could have at relativistic levels, you could define conditions as 'immediately accelerate to X% of c at origin, and immediately change direction at other end.'. Of course that's impossible, but it allows focus on maximum dilation.

Or you might say acceleration to c occurs during first half each 'leg' and deceleration during second half, then repeated on journey home. A little more calc, but straightforward.

Trying to be 'plausible' isn't possible with most questions like these. Choose a thought experiment that focuses on the specific you want to examine closer.

8. Nov 24, 2013

### ghwellsjr

I don't have a screen big enough to show what happens at 0.999999c but here are two for 0.9c and 0.5c. They are not exactly set to 100 light-years as my software requires integer numbers for the Proper Times.

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9. Nov 24, 2013

### Bill_K

Is there in fact such an orbit? Can you have a Schwarzschild orbit in which a particle comes in from infinity, reaches a finite perihelion, makes a complete U-turn and goes back out in the same direction it came?

If so, what would be the perihelion distance? (Must be greater than 3M, I think.)

10. Nov 24, 2013

### yuiop

The black hole part complicates things considerably because we then have to deal with gravitational time dilation and continuously changing velocity on top of the velocity related time dilation. Also, there is the technicality that a trip to an event horizon is a one way trip because you cannot leave the event horizon.
That is much better as it simplifies everything considerably. In flat spacetme you can ignore any acceleration as far as time dilation is concerned, it is only the velocity that counts. By making the acceleration instantaneous, we can ignore the infinitesimal time of the turnaround event and treat the whole problem as one of constant speed. (See calculations by ghwellsjr for the details.)

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11. Nov 24, 2013

### Myslius

"Is there in fact such an orbit?"

There is. Photon sphere (where photons orbits around BH is 1.5 Schwarzschild radius). Depending on speed, U-turn distance should be greater than 1.5 Schwarzschild's radius.

12. Nov 24, 2013

### yuiop

It does seem that you are correct that the minimum perihelion must be greater than 3M, because
using the equations listed here the minimum perihelion asymptotically approaches 3M as the angular momentum L gets arbitrarily large.

Here is a plot of the perihelion radius for angular momentum varying from 1 to 200 with M=1.

(Change the sign in front of the square root to + to plot the radius of the apogee of the precessing orbit.)

Not sure about infinity, but for a given initial finite radius (r2) it is always possible to reach a finite perihelion (r1) where 3M<r1<r2 and return to the same spot even when taking precession into account. (The apogee will be be slightly greater than this crossing point).

Last edited: Nov 24, 2013
13. Nov 24, 2013

### Mentz114

Yes. See post#2. There is also a trajectory in which there is a complete orbit before escaping.

14. Nov 24, 2013

### yuiop

Nice link and interesting trajectories! Thanks for posting it Mentz. Missed it the first time.

15. Nov 24, 2013

### Bill_K

I guess I need leading by the hand. I looked in the ref quoted in #2 and couldn't find either a discussion or an example of what I mean. The closest he comes is orbit A in Fig. 7, which is a crossover, not a U-turn. Maybe I should assume that if you can have one, you can have the other, but that doesn't answer the question of what its perihelion distance would be.

16. Nov 24, 2013

### Mentz114

I think there is a hyperboloid orbit in fig 4 ( type A). Like the Newtonian hyperbola if the perihelion >> 2m.

The Schwarzschild geodesics take a simple form in terms of the conserved E and L but the parameter space is almost chaotic so finding the right values is very tricky. This is the geodesic with $\theta=\pi/2$

\begin{align}\label{uvel} \vec{u} &= \frac{E\ r}{r-2m}\ \partial_t + \frac{\sqrt{\left( 2m-r\right) {L}^{2}+(E^2-1){r}^{3}+2m{r}^{2}}}{{r}^{\frac{3}{2}}} \ \partial_r + \frac{L}{{r}^{2}}\ \partial_\phi \end{align}

The trick is to reparameterize E and L with other contstants, like the perihelion distance and initial positions and so on.

17. Nov 24, 2013

### yuiop

Just to add to that, it states on page 5 that:

type A: scattering orbits with both endpoints at infinity. Scattering orbits can
never extend below r = 3M.

type B: plunging orbits with one end at infinity and the other behind the
horizon,
type C: near orbits with both ends behind the horizon of the black hole.
type D: bound orbits. Highly eccentric orbits can never reach below r = 4M
while circular orbits can never reach below r = 6M.

The link I posted earlier appears to use a simplification where the effect of the radial velocity on the proper time is ignored and this becomes significant for highly eccentric orbits.

18. Nov 25, 2013

### Mentz114

Just to round off, I looked up my solution for the type A orbit. This 4-velocity goes to $\vec{\partial}_t$ as $r\rightarrow \infty$ and the radial component is zero when $r=R$, with $r>R$ always. So $R$ is the closest approach distance.

$\vec{u}=\frac{r}{r-2\,m}\ \vec{\partial}_t \pm\ \frac{\sqrt{r-R}\,\sqrt{\left( 2\,m\,r-4\,{m}^{2}\right) \,{R}^{4}-4\,{m}^{2}\,r\,{R}^{3}}}{{r}^{\frac{3}{2}}\,{R}^{\frac{3}{2}}\,\sqrt{R-2\,m}} \ \vec{\partial}_r + \frac{\sqrt{2}\,\sqrt{m}\,R}{{r}^{2}\,\sqrt{R-2\,m}}\ \vec{\partial}_\phi$

This is the 4-velocity I posted earlier with the constants L and E replaced by R so as to satisfy the two constraints above.