Prove that acceleration and gravity are really the same?

[Aditya Vishwak]

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

 

[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.

 

[Aditya Vishwak]

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

 

[A.T.]

Have you read about the Equivalence principle already?
http://en.wikipedia.org/wiki/Equivalence_principle

https://www.youtube.com/watch?v=2MquzTW5nq0

 

[MikeGomez]

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.

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

 

[MikeGomez]

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

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

 

[elegysix]

acceleration due to gravity is exactly the same as acceleration due to any other force.

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.

 

[A.T.]

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...

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

...we would have no explanation for why the acceleration at higher altitudes was less than at the surface.

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

 

[elegysix]

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

 

[MikeGomez]

Yes acceleration is acceleration, however the way in which gravity accelerates objects is a function of distance (inverse square).

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).

 

[WannabeNewton]

Radiation also falls off as the inverse square, if originating from a sphere.

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

 

[MikeGomez]

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

You edited that! Long live Jimmy Page...

 

[WannabeNewton]

Long live Jimmy Page...

Yes indeed

 

[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.

 

[Drakkith]

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

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

 

[WannabeNewton]

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

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 $I \sim |\langle E(\vec{x},t) \rangle_t |^2 \sim \frac{1}{r^2}$.

 

[Aditya Vishwak]

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).

What does 'body emitting gravity' mean?

 

[Aditya Vishwak]

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

How?

 

[Aditya Vishwak]

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.

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

 

[A.T.]

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

How?

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):

• Two inertial particles, at rest relative to each other, in flat spacetime (i.e. no gravity), shown with inertial coordinates. Drawn as a red distance-time graph on a flat piece of paper with blue gridlines.
  
• B₁. The same particles in the same flat spacetime, but shown with non-inertial coordinates. Drawn as the same distance-time graph on an identical flat piece of paper except it has different gridlines.
  
  B₂. Take the flat piece of paper depicted in B₁, cut out the grid with some scissors, and wrap it round a cone. Nothing within the intrinsic geometry of the paper has changed by doing this, so B₂ shows exactly the same thing as B₁, just presented in a different way, showing how the red lines could be perceived as looking "curved" against a "straight" grid.
  
• Two free-falling particles, initially at rest relative to each other, in curved spacetime (i.e. with gravity), shown with non-inertial coordinates. This cannot be drawn to scale on a flat piece of paper; you have to draw it on a curved surface instead. Note how C looks rather similar to B₂. This is the equivalence principle in action: if you zoomed in very close to B₂ and C, you wouldn't notice any difference between them.

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

https://www.youtube.com/watch?v=DdC0QN6f3G4

 

[elegysix]

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

$F_g=-GMm/r^2$, for $r = r_{earth}, \ F_{g}=mg \ or \ a=9.8m/s^2$.

$for \ r > r_{earth}, \ F_g < F_g(r_{earth})$

if the radius of Earth was increasing with an acceleration of 9.8m/s^2, it would not matter where we measured it from. we would come up with the same value.

If we believed the Earth's radius was accelerating like that, then when we measured it on top of a mountain, we would have to conclude that the mountain is sinking into the earth, because $a(r=r_{mountain})<a(r=r_{earth})$ This would lead to a contradiction, because it would not actually be sinking. and so we would be confused

 

[A.T.]

if the radius of Earth was increasing with an acceleration of 9.8m/s^2...

Nobody suggested that the radius of Earth is increasing, so you are arguing against straw man. The surface of the Earth has a proper acceleration of 9.8m/s^2 away from the center. But in curved space-time that doesn't imply actual movement away from the center.

 

[elegysix]

From the OP's original question, I got the impression he was thinking of simple acceleration (constant, linear) which is why I introduced the straw man to begin with. It does a good enough job highlighting the difference between simple acceleration and acceleration due to a gravitational field. I'd have stopped talking about it already, but the OP asked me to elaborate

 

[A.T.]

I find proper acceleration or curved space-time to be more confusing than helpful in this discussion.

I mentioned these concepts only to address your misleading talk about an increasing Earth radius. But they are in fact related to the Equivalence principle, which is the OP asks about.

 

[LitleBang]

A possible mechanical explanation for Gravity.

Suppose we could make a sphere out of some mythical material that is totally non-elastic. Inside the sphere we place six balls made of the same material such that the balls almost touch the sphere but not quite. The sphere is located somewhere out in space where gravity is as close to zero as possible. We shake the sphere to start the balls bouncing around inside. Each ball has the same mass and inside of each ball is a clock that we can observe. Now we use the clock of each ball to measure it's momentum just before it collides with another ball and obviously we find that the momentum of each ball is identical. Next we move our sphere close to the edge of a black holes event horizon. If we calculate the momentum of a ball bouncing toward the EH we find that because it's clock slows down as compared to a ball bouncing away from the EH that it's momentum is greater. The total momentum of the balls will always be greater toward the EH. Instead of balls we could be talking about atoms. When we apply a force to an object we cause the object to accelerate making it's atoms act just as they would in the above discussion. Can someone show me why this does not imply that gravity is just an artifact of time dilation?

 

[A.T.]

We shake the sphere to start the balls bouncing around inside... and obviously we find that the momentum of each ball is identical.

Why should the momentum of each ball be identical, if they bounce around randomly?

Can someone show me why this does not imply that gravity is just an artifact of time dilation?

Gravity and gravitational time dilation are closely related in General Relativity.

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

 

[stevendaryl]

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

"acceleration due to gravity is exactly the same as acceleration due to any other force". I wouldn't put it that way. Acceleration due to gravity means falling, and that is very different from any other kind of acceleration in that it doesn't put any stress on the object that is accelerating (well, as you say, ignoring tidal forces). In contrast, if you are accelerating due to any force--say, a rope is pulling on your neck--you feel stresses due to the accelerating force.

What the equivalence principle says is that freefall--that is, accelerating solely due to gravity, and no other forces acting on you--is equivalent to inertial motion (drifting at constant velocity in the absence of any forces at all). Alternatively, the equivalence principle can be stated in terms of noninertial reference frames: The physics inside a room at rest in a gravitational field is the same as the physics inside a room that is on board an accelerating rocket. Once again, these equivalences are only true in the limit that the variation in gravity with location, or the variation in "g-forces" in a rocket, can be ignored.

 

[MikeGomez]

"... if you are accelerating due to any force--say, a rope is pulling on your neck--you feel stresses due to the accelerating force.

Geez. Nice anaolgy.

Acceleration due to gravity means falling, and that is very different from any other kind of acceleration in that it doesn't put any stress on the object that is accelerating

Acceleration due to gravity originates from energy. Tidal stresses are due to the spherical configuration of mass, and are not the source of gravity. Are you talking about non-tidal stresses?

What the equivalence principle says is that freefall--that is, accelerating solely due to gravity, and no other forces acting on you--is equivalent to inertial motion (drifting at constant velocity in the absence of any forces at all). Alternatively, the equivalence principle can be stated in terms of noninertial reference frames: The physics inside a room at rest in a gravitational field is the same as the physics inside a room that is on board an accelerating rocket. Once again, these equivalences are only true in the limit that the variation in gravity with location, or the variation in "g-forces" in a rocket, can be ignored.

The equivalence principle says that there is no experiment that you can perform which will tell you whether you are accelerating due to gravity or because you are in an accelerating chest. If there is no experiment that you can perform which can distinguish one situation from the other, then what sense is there to say that the two situations are different? If tomorrow a brilliant scientist demonstrates some new experimentally verifiable insight into gravity, then it will surely apply to acceleration, and vice versa. In my opinion to imagine otherwise is to defy the equivalence principle.

 

[A.T.]

The equivalence principle says that there is no experiment that you can perform which will tell you whether you are accelerating due to gravity or because you are in an accelerating chest.

"Accelerating due to gravity" (coordinate-acceleration in free fall) is not equivalent to an "accelerating chest" (rocket in space with engines on).

If you meant something else, they you should be more clear and explicitly distinguish between proper-acceleration and coordinate-acceleration.

 

[MikeGomez]

"Accelerating due to gravity" (coordinate-acceleration in free fall) is not equivalent to an "accelerating chest" (rocket in space with engines on).

If you meant something else, they you should be more clear and explicitly distinguish between proper-acceleration and coordinate-acceleration.

You seem to have some kind of a grudge against me for some reason. I will be happy to have a discussion regading coordinate acceleration versus proper acceleration with stevendaryl or someone else, but not with you.