# Times circling the Earth

1. Jun 4, 2015

### Nick666

So after watching this, circling around the Earth 7 times per second, close to the speed of light, one week of travel results in 100 years on Earth. Simplifying, 1 second on the machine versus 5200 seconds on earth. So I have to ask, how many times did the machine travel around the Earth ? 5200*7 ?

Last edited: Jun 4, 2015
2. Jun 4, 2015

3. Jun 4, 2015

### Nick666

It just seems to me the obvious answer would be 5200*7. But then I try to imagine the perspective of the train rider, and the circling of the Earth 36400 times versus its measly one measured second, something doesnt seem right.

4. Jun 4, 2015

### Simon Bridge

It's a tad tricky because the train rider is not in an inertial frame.
But basically, everyone will see the same number of circuits... you can imagine the train going back and forth instead of in a circle and use the usual twin's paradox analysis.

5. Jun 4, 2015

### Nick666

I understand the usual twin paradox...but there's something thats bugging me.

Say the train sends a photon to a mirror in space thats ~150000 km away. Now shouldnt the train receive back the photon after 5200*7 rotations from its perspective, yet we on earth see the train receiving it after 7 rotations ?

6. Jun 4, 2015

### A.T.

150000 km as measured in which frame?

7. Jun 4, 2015

### Nick666

Ermm... we on earth measure it. Half the distance to the moon or something like that.

8. Jun 4, 2015

### A.T.

The train will measure a different distance to the mirror due to length contraction, which will vary depending which direction the train is currently moving. Also, in the non-inertial train frame light doesn't travel along straight lines at constant speed c. So there is no reason to assume that the train will "receive back the photon after 5200*7 rotations from its perspective". Just like with the Twin Paradox the fallacy is assuming rules of inertial frames for a non-inertial frame.

9. Jun 4, 2015

### Nick666

But I can imagine that mirror being a sphere-mirror of ~150000km in circumference, and sending the photon to whatever direction, wouldnt that mean that the length contraction happens in all directions ?

10. Jun 4, 2015

### Staff: Mentor

No. You're trying to say that all points on the mirror are a distance of $75000/\pi$ kilometers from the center of the earth... but an observer on the orbiter will find that the distance from center of earth to the mirror along a line parallel to the motion of the orbiter will be contracted while the distance along a line perpendicular to the motion of the orbiter will not. Thus, the sphericalness of the mirror is itself a frame-dependent thing.

(This shouldn't be surprising. Even in ordinary straight-line motion If I use light and mirrors, or radar, to determine the shape of an object moving relative to me, if it is spherical in one frame it won't be spherical in others).

11. Jun 4, 2015

### Nick666

I understand the sphericalness of the sphere being frame-dependent.

But if the distance along a line perpendicular to the motion will not contract, doesnt that mean that it will take 5200*7 rotations for the train to receive the photon while we see the train as receiving the photon after 7 rotations ? Cause thats what I understood from A.T. , "it shrinks so it will receive the photon way before 5200*7 rotations" . But you say it doesnt shrink in that specific direction.

12. Jun 4, 2015

### A.T.

You are ignoring the rest of my post:
- The length contraction, which will vary depending which direction the train is currently moving.
- In the non-inertial train frame light doesn't travel along straight lines at constant speed c.

13. Jun 4, 2015

### Staff: Mentor

Remember, the train is constantly changing direction so the direction of shrinkage is also constantly changing and the light is not travelling in straight line relative to the train observer. You've constructed a fairly complicated setup here so it's hard to see the answer clearly, but it really all comes down to what A.T. said above - you're applying inertial-frame simplifications to non-inertial movement, and that never works.

14. Jun 4, 2015

### Nick666

So what you're saying is that the speed of light is less or higher than c from some points of view of the non-inertial train ?

15. Jun 4, 2015

### A.T.

In non-inertial frames the time for a light round trip of length d can be different from d/c.

16. Jun 4, 2015

### Nick666

Yes, but can it be higher or lower than d/c ?

17. Jun 4, 2015

### Staff: Mentor

Either. In this context, $d$ is a coordinate distance not a proper distance, and it doesn't have much physical significance.

If you want to work this problem out properly, you will have to stop trying to understand it in terms of time dilation and length contraction; these are simplifications of the more general machinery of the Lorentz transformations between momentarily comoving inertial frames and they won't work here. Instead, pick a coordinate system, any coordinate system that works for you; describe the worldlines of the orbiter, the earth-bound observer, and some interesting light rays in that coordinate system; and calculate the proper time on the orbiter's and earthbound observer's worldlines between the points of intersection of these worldlines and the paths of the light rays.

18. Jun 4, 2015

### pervect

Staff Emeritus
The coordinate speed of light can be different from c in non-inertial frames - the depends on your coordinate choices.

Coordinate speeds are not usually regarded as "physical" however. I should explain more, probably, but I'm running out of time, so I'll leave it at that for now.

[add] I guess I should say that the issues involve the fact that differences between coordinates do not always reflect either proper distances or proper times. But this then leaves wanting a clear explanation of what is meant by proper distance and proper time.

19. Jun 4, 2015

### Simon Bridge

The light fired from a train going in a circle is a common misdirect on "einstein was wrong" websites. Its also a common exercize in introductory GR courses.

To understand the issues... it helps to be more careful. Lets get rid of the distracting parts of the description like the exact numbers and specific locations.

An observer O sees another observer T moving at a constant relativistic speed v on a stationary circular track radius r. At t=0 on O and Ts clocks, O and T are right next to each other.
At that time O notices a pulse of light travel radially outward to a mirror at distance d from the track (which is d+r from the center of the track)... where it is reflected directly back again. Meantime the train completes an integer n number of circuits of the track, arriving in time to receive the returning pulse.
(A circuit is completed each time T passes by O.)

O figures that $2d/c=2n\pi r/v$ since the time to go n times around the track is also the time for light to get to the mirror and back.

Is this the sort of setup you are thinking of?

To work out what T figures, you have to change to the situation where T is stationary.
This means the track and O are following an odd path around T, and so is the mirror. Thus a circuit is completed each time O passes by T.
The track is no longer a circle, and keeps changing shape while it moves. The path the light takes is no longer a straight line.

Any calculation anyone wants to do that ignores any of that is wrong.

The question is... how many circuits does T count compared with O?

That sound good to you?

20. Jun 5, 2015

### Nick666

Allright, I get it now. I never thought of imagining the T(rain) stopped and the rest of the setup "circling around" . Though I have to say this example I cooked up in my head all by my self, no anti-relativity sites, dont care about those.

Thank you all for the answers.