# A new veiw on gravity

1. Jan 4, 2008

### Raisin-toe

I had an interesting thought about what gravity really is, and I had to ask someone to see if they would get the same spark.

I have been reading "E=mc^2" by David Bodanis, and after I had read a part about how the space-time continuum flexes around a star in the sky, I had this Idea that maybe gravity is a force created from the warped space time continuum. I pictured how the space time continuum curves around our earth for example, then someone is walking along the earth, but what holds him down to the earth, is the warped space-time. Being that time would move slightly faster where his head is than where his feet are, he is forced to stay on the ground. I think you could compare the warped space-time continuum, to rubber bands being stretched around the earth. I wish I could even understand this idea more fully. It's hard to explain exactly what is in my mind.

2. Jan 4, 2008

### matticus

so is this set theory, logic probability or statistics?

3. Jan 4, 2008

### Raisin-toe

Theory.

but, It looks like I'm not the first to notice this affect after all. I was just reading another forum where it mentions that Einstein also explained this same affect.

4. Jan 4, 2008

### DaveC426913

Unless I'm completely bonkers and am misreading you, that is exactly what Einstein's Theory of General Relativity is all about.

5. Jan 4, 2008

### Raisin-toe

Oh . . .

I always thought that the theory of relativity was just the space time continuum, in relation with light.

6. Jan 4, 2008

7. Jan 4, 2008

### DaveC426913

The Theory of Special Relativity is known for its posits about how light behaves.

The Theory of General Relativity deals more with gravity.

Wiki is your friend. It will bring you up to speed faster than we can.

8. Jan 5, 2008

### HallsofIvy

First, this physics, not mathematics so I am going to move it. Second, "gravity caused by warped space time continuum" is not new. Third I do not understand where you get "Being that time would move slightly faster where his head is than where his feet are, he is forced to stay on the ground." Why does time moving faster where his head is "force" him to stay on the ground?

9. Jan 5, 2008

### gel

Doesn't this combined with the strong equivalence principle imply exactly what the OP suggested? Someone standing in an accelerating spaceship would also observe time running very slightly faster at his head than at his feet, and would feel an effective force similar to gravity.

10. Jan 5, 2008

### Gokul43201

Staff Emeritus
The equivalence principle presupposes the existence of a gravitational field. To use this to demonstrate the existence of one is circular.

11. Jan 6, 2008

### HallsofIvy

Both of those statements are true. I just don't see how one implies the other.

12. Jan 6, 2008

### gel

Assuming that physics in space-time is locally Lorentz invariant, there should be a local coordinate frame in which special relativity holds (ignoring tidal effects...). The only way to do this is to use a frame which is accelerating downwards, in order to cancel out the difference in the rate at which time is flowing at the observer's head and feet. Then, an object moving at constant velocity in this accelerating frame would actually be accelerating downwards in a frame fixed to the surface of the Earth.
I wasn't trying to argue that this difference in the rate at which time flows at the observer's head and feet is the "real cause" for the gravitational force he feels, just that either one does follow logically from the other according to the assumptions of General relativity.

13. Jan 6, 2008

### Raisin-toe

I think I said this wrong. Time should move faster at his feet, and slower at his head, correct?

14. Jan 6, 2008

### gel

Last edited by a moderator: Apr 23, 2017
15. Jan 6, 2008

### Raisin-toe

Thanks for that explanation, that is what I viewed in my mined. But if something could hold perfectly still, would it float as if it were in space?

Also, everything that the object was made up of would have to be perfectly still.

Last edited: Jan 6, 2008
16. Jan 6, 2008

### gel

No. That doesn't follow from what I was saying, and would violate the equivalence principle.

17. Jan 6, 2008

### Raisin-toe

What would violate the equivalence principle?

18. Jan 6, 2008

### gel

For one thing, I was arguing that the inertial frame under which special relativity is locally valid is one which is accelerating downwards. Any object that is not subject to external non-gravitational forces would move with constant velocity, and would not accelerate when viewed in such a frame. So, switching to the frame fixed to the surface of the earth, all objects would accelerate downwards at the same rate. It doesn't matter whether it or any of its component parts are holding perfectly still.

Furthermore, the definition of the strong equivalence principle as given in Wikipedia (http://en.wikipedia.org/wiki/Equivalence_principle#The_strong_equivalence_principle") explicitly states that the gravitational motion of a body does not depend on its constitution.

Last edited by a moderator: Apr 23, 2017
19. Jan 6, 2008

### OmCheeto

hmmmm.... So If I took a 1 meter rod and held it vertically, would it weigh more than if I held it horizontally? And if I flattened the rod into a sheet, would it weigh even less? Or am I totally not following this thread at all?

20. Jan 6, 2008

### Raisin-toe

Actually, I think you have a good point. If you were holding a rod up vertically, it would weigh less, because the intensity of the warped time frame would be more where it is nearest to the earth, than the intensity of the time frame up higher into the atmosphere.

I may be able to explain that better: The time frame nearest to the earth would have a smaller radius than the time frame that is farther up into the atmosphere.

Because of the Theory of General Relativity, The rod would weigh slightly less if it were standing upright, and weigh more if it were laying horizontally, I do believe.

Just as a space ship leaves the earth, and is nearly released of the gravitational pull, your pole would also be farther out of the gravitational pull, and would weigh less.

Last edited: Jan 6, 2008
21. Jan 7, 2008

### paw

No it doesn't. Gravity acts on an irregular mass as if through it's center of mass (center of gravity if you perfer). The only factors that matter are the mass of the rod and the distance of the center of mass from the Earth's center of mass.

Assuming you rotate the rod through it's com then it weighs the same whether vertical or horizontal. If you don't rotate through the com them the rod may weigh more or less depending on the movement of the com (the difference isn't likely detectable for any rod of resonable length).

In a very mis-informed way I think you are describing tidal stabilization. A very long rod in orbit around the Earth with one end pointing to the Earth's com will remain stable in that orientation because the gravitational 'pull' on the end closer to the Earth's com is higher than on the opposite end. This does NOT make the rod weigh less than a horizontal rod for the reasons outlined above. It does make an incremental 'piece' of the rod at one end 'weigh' more than an equivalent 'piece' at the other end. This has nothing to do with gravitational time dilation.

No spaceship is 'released', or in any way leaves, the gravitational field of the Earth. This is a common misconception. Gravitational fields only tend toward zero in the limit as distance approaches infinity. A space craft may reach a distance where the effect of Earth's gravity in undetectable but it never truly leaves it.

22. Jan 7, 2008

### pervect

Staff Emeritus
I think you can get from one to the other, if in addition you assume the principle of extremal aging. But I don't think you can get there without an additional assumption of this sort.

Last edited: Jan 7, 2008
23. Jan 7, 2008

### Raisin-toe

I know, that is why I said it is nearly released.

I guess I should have worded that better.

24. Jan 7, 2008

### Raisin-toe

I think I'm starting to understand what you mean.

Is this first picture how you are viewing the difference in how the rod is sitting? I am viewing it like the rod in the second picture.

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25. Jan 7, 2008

### paw

Yes, that's what I was saying with regard to the position of the center of mass. In the first view the rod is rotated around its center of mass and it will 'weigh' exactly the same in either orientation.

In the second view the rod is rotated and also the center of mass is at a different gravitational potential (ie the com is closer to the Earth's com). In this case the horizontal rod will 'weigh' more than the vertical rod. I must point out though that it would have to be a big difference in order to measure it and it has nothing to do with the orientation of the rod. It is due to the distance of the com from the Earth ONLY. You would get the same result with two vertical rods at different distances from the Earth.