# Why Do We Fall Back to Earth?

1. May 7, 2012

### Chevron

I have almost no formal training in physics, and I am only a junior in high school, so this may be a stupid question. But, anyways, I recently watched some documentaries on gravity and general/special relativity. I understand why objects orbit in accordance with GR (the orbiting object is moving in a straight line, but its path is curved due to the curvature), but I don't understand why a person would come back to Earth after jumping due to the space-time curvature described in GR. If anyone could clarify, it'd be very helpful and much appreciated!

2. May 7, 2012

### clamtrox

It's an excellent question! In general relativity, things which are freely falling are moving "straight". But the surface of the Earth is not! So in GR picture, it's the surface of the Earth which is accelerating upwards while you are falling freely.

3. May 7, 2012

### A.T.

I recommend this introductory visualization, which explains exactly the vertical free fall case you are asking about:
http://www.relativitet.se/spacetime1.html

4. May 7, 2012

### Staff: Mentor

This picture won't work, because things fall towards the center of the Earth from all directions. The Earth can't be moving "up" towards me in the United States and someone in Australia at the same time.

Also, while the surface of the Earth is curved, its curvature is very different from the curvature of spacetime around Earth that GR uses to account for the Earth's gravity (or even just the spatial components of that curvature).

The site A.T. linked to gives a better visualization of the "shape" of spacetime around a gravitating body (more precisely, the shape of particular projections of that spacetime).

5. May 7, 2012

### Chevron

Thanks! That really cleared up a lot for me.

6. May 7, 2012

### A.T.

He didn't say moving upwards, just accelerating upwards. That is not the same (Even in Newtonian mechanics: In circular motion you are accelerating inwards without moving inwards). Clamtrox is correct: The Earth's surface is undergoing proper acceleration away from the Earth's center.

I don't think he meant the curved surface of the Earth, but rather its curved world-lines:

From: http://www.relativitet.se/spacetime1.html

Unlike the green & red world-lines, the blue 0.0m world-line is not straight. The straight going toy car would not follow such a line. This is easier to see when you approximate the trumpet with a cone, which can be unrolled onto flat. Then you see that the worldliness of constant spatial radial coordinate are bend upwards, so objects which follow them (like the Earth's surface) are accelerating upwards:

http://www.physics.ucla.edu/demoweb/demomanual/modern_physics/principal_of_equivalence_and_general_relativity/curved_time.gif
From: http://www.physics.ucla.edu/demoweb..._and_general_relativity/curved_spacetime.html

7. May 7, 2012

### Staff: Mentor

It's possible that I misread what he was trying to say, and that he did mean this. But bear in mind that viewing the worldlines of individual points on the Earth's surface as accelerating upwards in this sense is a local viewpoint: an individual point on the surface of the Earth is accelerating upwards only with respect to the inertial frames in its immediate vicinity. It is *not* accelerating upwards with respect to inertial frames elsewhere on the Earth. If you are accelerating upwards at the North Pole, you are accelerating sideways with respect to the equator and downwards with respect to the South Pole.

8. May 7, 2012

### julian

In GR the straightest line is the one that regesters the maximal proper time of a clock taken along for the ride (for a given fixed time as registerd by a clock on Earth). Now two things are in conflict, the higher up you go the faster your clock runs, but the faster you go to get higher up the more time you loose cus of time dialation. The best compromise is given by newtonians law - parabolic curves (that is if we are in a weak grav field, small velocities, and small masses) you get out Newtonian gravity. It's in the 3rd volume of the Feynman lectures.

9. May 7, 2012

### A.T.

The proper acceleration of points on the Earth's surface is frame invariant. The problem is that in curved space time inertial frames exist only locally.
By "upwards" we usually mean "away from the Earth's center". The proper acceleration of points on the Earth's surface is always away from the Earths's center.

Last edited: May 7, 2012
10. May 7, 2012

### Staff: Mentor

The geometric object, the proper acceleration 4-vector, is, yes. Its components in particular coordinates will change if the coordinates are changed.

Yes, which makes "upwards" in the sense we've been using the term a local term.

The real question is which usage clamtrox meant. When I first read his post, the picture which came to my mind was of two people, one standing on the Earth's surface and the other falling past him, and the falling one saying that he was the one "standing still" and the other was "moving upwards" past him. Then I pictured two other people on the opposite side of the Earth having the same discussion. The problem with the falling people's claims as they stand is that they are going to fall past each other (assume for simplicity that there is a long tunnel through the Earth that they will both fall into), so they are obviously *not* "standing still"--but the ones standing on the Earth's surface are at rest relative to each other and stay that way. Given that, to say that both observers on the Earth's surface are "accelerating upwards" does not seem to be the natural way that a lay person would describe the situation.

11. May 7, 2012

### the_emi_guy

First, Chevron it is really great that you are interested in this while still in high school.
A key concept is that things move in a certain way when no forces are applied to them, they do not necessarily "stand still". In Newtonian mechanics objects move in straight lines with constant velocity in the absence of force. In GR a free falling man is following a "straightest possible path" in spacetime (geodesic), analogous to the straight line constant velocity in Newtonian, because there are no forces acting on this man. The man standing on the earth *does* have a force acting on him, and he can feel it on the bottoms of his feet, and this is why this second man is not traveling along the spacetime geodesic. A man free falling on the opposite side of the earth is also following his geodesic, which will be different because he is in a different location. These two men may collide at the center of the earth, this would be analogous to two marbles moving in straight lines colliding in Newtonian.

Imagine redoing this experiment in far away space, not near any planet. The man that was standing on earth instead has rocket motors mounted on the bottoms of his shoes accelerating him at 9.8m/s^2 while he passes another man floating is space. Einstein's idea is that this is equivalent to the gravitation situation. The floating man is just doing what masses do when they are not subject to any forces (and this is true to his senses, free falling to earth feels just like floating is empty space). Its also clear that a marble and a truck would appear to "fall" at the same speed relative to the rocket-shoe man. Of course the marble and truck remain side by side because it is really the rocket-shoe man that is accelerating due to force.

We would laugh at the rocket-shoe man if he conjured up a strange force that was causing the marble and the truck to accelerate identically while ignoring the force that he could feel on his own feet. This is what we are doing in Newtonian gravitation.
Cheers.

Last edited: May 8, 2012
12. May 8, 2012

### A.T.

Again, the issue is not movement but acceleration. Movement is relative, but they will agree what their accelerometers measure.

Why not? When two ships (engines off) pass each other in empty space, they both can rightfully consider themselves to be at rest.

It may seem counter intuitive, but it is similar for two people on opposite sides of a merry-go-round. The are both accelerating inwards, towards each other, but still are at rest relative to each other and stay that way.

13. May 8, 2012

### clamtrox

What's the shape of that cone, exactly? That looks like a great way of visualizing gravity, but the cone is kind of pulled out of a hat there.

14. May 8, 2012

### A.T.

The math for those embedding diagrams is in the papers here:

http://www.relativitet.se/Webarticles/2005AJP-Jonsson73p248.pdf
http://www.relativitet.se/Webarticles/2001GRG-Jonsson33p1207.pdf
http://www.relativitet.se/Webtheses/lic.pdf
http://www.relativitet.se/Webtheses/tes.pdf

Note that there is a free parameter in those embeddings, which determines how tight you roll the space time diagram (how much time per circumference). This affects how the embeddings look like but has no physical meaning.

Also note that some of the diagrams use coordinate time as the circumferential coordinate, and some proper time. The latter variant is described in chapter 6 of:
http://www.relativitet.se/Webtheses/lic.pdf
and made into an interactive diagram here:

15. May 8, 2012

### clamtrox

Ooh, I like that way of visualizing a lot (at least based on a quick glance). So from what I gather, the contour reaches a plateau at center, which corresponds to spatial infinity. The event horizon is reached when the cone slope is 45 degrees (in natural units, c=1), at which point the embedding scheme breaks down (or you'd need to draw a different cone for the region inside event horizon). Is that about right?

16. May 8, 2012

### A.T.

I will refer to the proper-time version, because in the paper there are different variants.
Not sure what you mean by "spatial infinity". The "plateau" at the mass center (or bulge radius maximum) means that (unlike on the surface) you can have a geodesic with constant spatial coordinate there. The toy car would go around that maximal circumference, without steering left or right. This corresponds to an object floating at the center.

This might be true for the coordiante-time variant, but I'm not sure. The cone slope depends on the embedding parameter, which can be chosen freely. IIRC for any choice of that parameter the embedding scheme breaks down at some point before a horizon forms. So you can adjust that parameter for different masses but then you loose comparability between pictures.

For the proper-time version the slope of the trumpet goes to 90° when you go towards a black hole. So the bulged cylinder breakes down into two disconnected trumpets. When something falls into the BH it will climb that 90° slope forever creating an infinite world-line. In space-propertime diagrams the length of the world-line corresponds to coordinate time of a distant observer, which will have to wait to infinity for the object to cross the horizon.

However, while the object climbs that slope forever, it goes less and less around the trumpet, so it experiences less and less propertime according to the distant observer. This corresponds to the fact that while it takes infinite distant-observer-time for the object to pass the horizon, the proper time experienced by the object is finite.

Last edited: May 8, 2012
17. May 8, 2012

### Staff: Mentor

Yes, but the two observers whose accelerometers show different readings (one reads zero, one reads 1 g) are in relative motion, and that fact is frame-independent, as is the magnitude of their relative velocity at the event when they pass each other. So the term "accelerating upward", from the viewpoint of the freely falling observer, does imply "motion upward" as well.

Considered in isolation, yes. But in my scenario, the two observers standing on the Earth's surface are at rest *relative to each other*, as they can verify by exchanging light signals and observing a constant round-trip travel time for them. The freely falling observers are not at rest relative to each other (or to the observers standing on the Earth's surface, of course). Again, these facts are frame-independent. So there is an invariant sense in which the observers standing on the surface are "at rest" while the freely falling ones are not.

Good point, the use of the term "accelerating" does seem more natural when viewed this way.

I should clarify that I'm not really trying to disagree here, I'm just trying to work out possible implications of taking a sort of "naive" viewpoint about the terms we use to describe how a freely falling worldline can be "straight" and an accelerated one can be "curved".

18. May 9, 2012

### A.T.

I wouldn't describe being "at rest relative to something" as being "at rest in an frame invariant sense". It seems like quite the opposite of what "frame invariant" usually means. What you are doing is basically defining "at rest" as "at rest in a frame where the gravitational field is static." That is intuitive, practical and the embedding diagrams are also based on that frame.

However it is important to draw the distinction between rest vs. motion on one hand and inertial vs. proper accelerated on the other.

19. May 9, 2012

### Staff: Mentor

The fact that two objects are at rest *relative to each other* is frame invariant. That would be true even for a pair of objects that were at rest relative to each other but were *not* at rest relative to the static gravitational field. For example, a pair of objects held at a constant proper distance apart by a short massless rod, while the assembly as a whole is freely falling. The fact that these two objects are at rest relative to each other would also be frame invariant.

What distinguishes the pair of objects at rest relative to the static field is that they can remain at rest relative to each other indefinitely while experiencing a constant proper acceleration. The pair of objects separated by the short rod while the assembly as a whole falls freely would experience changing proper accelerations as they fell.

I agree.

20. May 10, 2012

### A.T.

Yes, that is my point. Being at rest relative to each other and accelerating in different directions is not a contradiction. So it is perfectly valid to see the Earth's surface as accelerating outwards.