# I Why doesn't the Earth's radius change while the surface accelerates?

1. Dec 28, 2016

### FallenApple

So if I drop an apple, we can consider the ground accelerating upwards and the apple still. And the radius of the earth doesn't change due to space time being curved.

I'm having a hard time conceptualizing this.

Is it analogous to a classically orbiting object? Classically, under simplfying assumptions, the Moon is accelerating towards the center of the Earth, yet it's radial position realitive to the center is unchanged. This is similar to the ground accelerating up, but radial position doesn't change. Except there are two differences, the apple meets the ground, and there is no rotation, so there is something wrong with this analogy.

2. Dec 28, 2016

### BvU

Same difference: angular momentum. Moon yes, apple no.

3. Dec 28, 2016

### Ibix

The idea that the ground is accelerating up at you is valid in a local sense: if you simply ignore everythin going on further afield than the immediate "Earth's surface is flat" kind of distance. It doesn't generalise well to a global view. This is a fairly common feature in general relativity. Things that are good explanations locally don't always make sense if you try to apply them to a global point of view.

In particular, what you are using here is the equivalence principle, which says that (over a sufficiently small patch of spacetime) one cannot distinguish "standing on the surface of the earth" from "accelerating in a rocket in flat spacetime". This is only valid over a small patch of spacetime, one where the tidal effects of gravity are negligible. The patch of spacetime covering the entire Earth does not meet this criterion. So it isn't necessarily valid to generalise the local "surface is accelerating" perspective to the global one.

4. Dec 28, 2016

### A.T.

Uniform circular motion shows that even in flat space time, you can accelerate towards a point with getting closer to it. But I would not use gravitational orbits as an a example, because that is not proper acceleration as the ground has:

- A mass swung on a string has proper acceleration towards the center, without getting closer to it.
- In curved space-time a mass can have proper acceleration away from the center, without getting further away from it.

That's because proper acceleration is equivalent to deviation from geodesic worldlines.

5. Dec 28, 2016

### Staff: Mentor

I wrote an insights article on exactly this topic:

https://www.physicsforums.com/insights/understanding-general-relativity-view-gravity-earth/

I would appreciate feedback if you find it useful or not.

6. Dec 28, 2016

### FallenApple

Ah I see. Because uniform circular motion is free fall. The accelerometer would read 0.

Hmm this is the confusing part. So if I put a mass on the ground, in GR, since gravity is ficticious, there is only one force on the mass, the normal force caused by E&M. Force equals to instantaneous change in momentum over time, or whatever the GR analogue is. So a vector position vector in spacetime, (t,x,y,z), would be changed because there is a net force. Similarly, a patch on the ground would have a net force on it, causing its path in spacetime to change, causing a collision with the path in spacetime of the apple.

7. Dec 28, 2016

### FallenApple

Thanks, its very helpful and interesting.

I have a question on the lines of latitude part.

What do you mean by the top line constantly turning to north while the bottom line constantly turns to south? Do you mean that if I consider two caps cut out by the two lines into circles, a test particle on the boundaries of each cap would be accelerating towards there own centers(being the poles) just like if I topologically transform the two caps into circles, the test particles would be accelerating towards their own respective points on different circles?

8. Dec 28, 2016

### Staff: Mentor

A circular orbit under the influence of gravity is free fall and an accelerometer will read zero. Uniform circular movement in general is not free fall, and an accelerometer will register accelerations - that's how a centrifuge works.

9. Dec 28, 2016

### Staff: Mentor

It may be easier to see this for the latitude lines near the poles. Consider the 89.999 degree latitude line. The one 1 meter from the pole. Clearly, if you want to walk on that latitude line you have to turn in a very tight circle. To stay on that line you have to constantly turn towards the pole.

The same thing happens for latitude lines further away from the poles, just the turning is not as tight.

10. Dec 29, 2016

### A.T.

Uniform circular motion on a string is not free fall. You register (inwards) radial proper acceleration without radial movement, just like the ground does (outwards). It's not an analogy that explains GR, just a classical example showing that this aspect of GR isn't that weird.

Look at the last part of the video below, where space-time is a cone (time 1:10). The green worldline is like the one of the ground, and requires a net force upwards. On the opposite side of the Earth the cone is mirrored (also gets wider towards the center). So the green worldlines on it has a net force in the opposite direction, but the distance between the green worldlines (Earth's diameter) is constant.

Here is the global diagram, which shows how those local cone pieces connect to a continuous but curved space-time:

See also Chapter 2 of this thesis:

http://www.relativitet.se/Webtheses/tes.pdf

11. Dec 31, 2016

### Battlemage!

I found it pretty insightful, especially the difference between GR and Newton on a free falling object, and the use of accelerometers for this, and it has a tangible and easy to understand explanation for why inertial frames must be localized in GR.

12. Dec 31, 2016

### Staff: Mentor

To understand the importance of curvature, consider two latitude lines on a sphere. Except for the equator, latitude lines are not geodesics (recall that on a sphere the geodesics are great circles), so to stay on a latitude line you cannot walk straight but must continually turn towards the nearest pole with the turning being tighter for latitude lines near the pole. So, for example, the 5° N line is constantly turning to the north and the 5° S line is constantly turning to the south. They are turning away from each other but maintaining constant distance. This is impossible on a flat surface, but possible in a curved surface.

13. Dec 31, 2016

### FallenApple

Ah, yes that makes sense. So near the poles, they are approximately circles.Turning on them, the acceleration vector has to point towards the center. But in general, as on transverses a latitude line, the acceleration vector cannot point into the sphere as we have to assume that there is no embedding, so considering the 2d surface as the space, the only way is for the "centripetal" acceleration to point towards a center, that is a pole.

14. Dec 31, 2016

### FallenApple

Oh so the apple on the tree is accelerating through spacetime because it has a force on it. The apple that is falling is not accelerating though spacetime. The apple falling is accelerating though space, but on spacetime, it isn't.

Also, if I have two point masses in a curved spacetime, would they fall towards each other without any forces whatsoever just because they are moving at constant "velocity" relative to each other on following geodesics that geometrically intersect?

15. Dec 31, 2016

### FallenApple

Makes sense. On a string, you can put a device to measure the tensile strength. On a orbit, it's not possible.

16. Dec 31, 2016

### Staff: Mentor

You're thinking about this part backwards. It's not the force (in this case, the tension in the string) that we need to measure, it's the acceleration. Once we have the acceleration, we can calculate the force from $F=ma$.

We measure the acceleration with an accelerometer, basically a mass suspended in all directions by stretched springs; any non-zero acceleration will cause the length of the springs to change. The accelerometer will register zero when in a gravitational orbit, but will register something non-zero if it's being twirled by a spring or being moved around by other non-gravitational forces.

17. Dec 31, 2016

### Staff: Mentor

Yes, that is correct. The surface of a sphere is a 2D manifold, and the curvature that we are interested in is entirely within that 2D surface, without any reference to any possible embedding space.

18. Jan 1, 2017

### A.T.

No. The disinction is between:
- Proper acceleration (what an accelerometer measures)
- Coordinate acceleration (what some observer measures as dv/dt)

The free falling apple has zero proper acceleration. Whether it has coordinate acceleration depends on the observer : for a ground fixed observer it has.

Yes, over larger regions that cone becomes "trumpet" where initially parallel geodesics intersect (objects initially at relative rest converge). See the other links I gave you.

19. Jan 3, 2017

### Peter Martin

I know that if two objects are in inertial motion either one can be legitimately considered at rest while all movement is attributed to the other. But this principle doesn't apply to accelerated motion.

The reason why the space twin is younger than his brother is that he was doing the "actual" moving because he accelerated several times during his journey while the Earth twin did not. Where accelerated motion is concerned -- as in your example -- the two objects (Earth and the falling apple) are not symmetrical, and therefore their motions cannot be interchanged. Only one of them is "really" moving.

20. Jan 3, 2017

### PeroK

Which one is "really" moving? The apple or the Earth?

21. Jan 3, 2017

### Staff: Mentor

That's not what's going on. The space twin is younger than the earth twin because the space twin's path through spacetime between the separation event and the reunion event was shorter than the earth twin's path between those two events.

The acceleration is something of a red herring - it just so happens that in flat spacetime there is no way of setting the two twins on different paths through spacetime without accelerating one or the other. However, there is a completely acceleration-free version of the experiment in which the space twin turns around by doing a hyperbolic orbit around a distant star. In this variant neither twin experiences any acceleration, but their paths through spacetime are still different lengths and the one that takes the longer path will age more.

There's a good explanation in the first chapter or so of MTW, and you could also look at the spacetime analyzis section of the twin paradox FAQ at http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html

Last edited: Jan 3, 2017
22. Jan 3, 2017

### Staff: Mentor

In GR it does. In SR it is somewhat arguable, because global inertial coordinates do have a "privileged" position in flat spacetime; but you can do SR in non-inertial coordinates in which an accelerated object is at rest (for example, Rindler coordinates for an object with constant proper acceleration) and get all of the same physical predictions. So the principle can indeed be applied to accelerated motion even in SR.

23. Jan 8, 2017

### Battlemage!

Is GR this but with tidal forces accounted for through curved spacetime?

24. Jan 9, 2017

### Staff: Mentor

I'm not sure what you mean by "this". Global inertial coordinates do not exist in a curved spacetime, so what I said about SR and global inertial coordinates isn't even meaningful in GR to begin with.

25. Jan 10, 2017

### Battlemage!

I won't learn tensors for about a year, but when I look at SR written in tensor format, it looks a lot like GR.

So I was wondering if the SR way of handling an acceleration could be modified by adding curvature of spacetime and if doing so would that basically be a formulation of GR.