# Prove that acceleration and gravity are really the same?

1. Jul 5, 2014

Is there a way by which we can prove that acceleration and gravity are really the same?

2. Jul 5, 2014

### elegysix

Gravity is a force which causes acceleration. Gravity is not the same as acceleration.

If gravity were acceleration, then the radius of earth would have to be expanding with an acceleration of 9.8 m/s^2, and we would have been very confused when we measured it from higher altitudes.

Last edited: Jul 6, 2014
3. Jul 6, 2014

So how does the curvature of space-time gives birth to gravity?

4. Jul 6, 2014

5. Jul 6, 2014

### MikeGomez

Not quite. Due to the equivalence principle (in the absence of tidal forces) acceleration due to gravity is exactly the same as acceleration due to any other force. The source of the earth’s gravity is the energy of its components, as per the stress-energy tensor...

http://en.wikipedia.org/wiki/Stress–energy_tensor

6. Jul 6, 2014

### MikeGomez

I'm pretty sure that gravity works just fine without curvature.

7. Jul 6, 2014

### elegysix

Yes acceleration is acceleration, however the way in which gravity accelerates objects is a function of distance (inverse square).
So if we try to say that acceleration due to gravity was actually caused by earth's radius expanding with an acceleration of 9.8m/s^2, we would have no explanation for why the acceleration at higher altitudes was less than at the surface.

8. Jul 6, 2014

### A.T.

The radius is constant. But a reference frame at rest on the surface has a proper acceleration of 9.8m/s^2 upwards. Any accelerometer will confirm this:
http://en.wikipedia.org/wiki/Proper_acceleration

The Equivalence principle states a local equivalence. The proper acceleration of local hovering reference frames varies with position:
http://en.wikipedia.org/wiki/Equivalence_principle

9. Jul 6, 2014

### elegysix

No argument there. I know you know I know the radius of earth is constant.

10. Jul 6, 2014

### MikeGomez

Gravity does not accelerate objects by the inverse square law. The inverse square law is due to the 3D geometrical arrangement of mass. A body emitting gravity is like a body emitting radiation. Radiation also falls off as the inverse square, if originating from a sphere. The reason there is less measured radiation at a further distance is because there is the same amount of radiation being measured for a larger area.

Its the same situation for gravity. It falls off as the inverse square if originated from a sphere, as 1/r if originating from a cylinder or line, and it does not fall off at all if originated from an infinite plane.

In the case of radiation, even though the radiation falls off as the inverse square law, we can think of it as composed as individual photons. The amount of photons falls off as the inverse square, but the individual photons remain the same (except in extreme cases where there are relativistic effects like red/blue shift).

11. Jul 6, 2014

### WannabeNewton

The intensity falls off as inverse square but the radiation field itself in the far field zone goes like $1/r$.

12. Jul 6, 2014

### MikeGomez

You edited that! Long live Jimmy Page....

13. Jul 6, 2014

### WannabeNewton

Yes indeed

14. Jul 6, 2014

### elegysix

Yes, you are correct. I was assuming the gravitational acceleration due to an approximately spherical object such as earth, since that is what my thought experiment entailed.

I suppose to make my statement more general, for the case of a body undergoing constant linear acceleration the derivative of acceleration with respect to position is zero. However, for the acceleration due to earth's gravity, the derivative with respect to position is not zero. This causes a difference in acceleration based on position. In the thought experiment I posted earlier, this would be the source of the confusion I mentioned.

15. Jul 6, 2014

### Staff: Mentor

Can you elaborate on this? What's the difference between the intensity and the "radiation field itself"?

16. Jul 6, 2014

### WannabeNewton

The radiation field is just the electric field (and magnetic field of course) of the electromagnetic wave and goes like $|E(\vec{x},t)| \sim \frac{1}{r}$; it is called the radiation field because in the far field zone of a system of isolated charges it dominates over the non-radiative electromagnetic fields which go like $\frac{1}{r^2}$ and therefore represents radiation propagating to infinity. The average intensity of the radiation field is the average power per unit area and goes like $I \sim |\langle E(\vec{x},t) \rangle_t |^2 \sim \frac{1}{r^2}$.

17. Jul 9, 2014

What does 'body emitting gravity' mean???

18. Jul 9, 2014

How?

19. Jul 9, 2014

Sir, can you please elaborate the statement. It seems interesting:thumbs:

20. Jul 9, 2014

### A.T.

I think Mike means without intrinsic space-time curvature. That curvature is related to tidal-effects, not the gravitational coordinate acceleration itself.

Here a nice diagram by DrGreg (flat = no intrinsic curvature):

And here an animation that explains gravity in a space-time without intrinsic curvature: