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.

 

[A.T.]

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.

Huh? I was just pointing out that your equivalency seems backwards, or at least can be easily misunderstood, and asking for clarification.

"Accelerating due to gravity" (coordinate-acceleration in free fall) is equivalent to a "non accelerating chest" (rocket in empty space with engines off).

"Hovering against gravity" (or standing on the Earth) is equivalent to an "accelerating chest" (rocket in empty space with engines on).

 

[stevendaryl]

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?

Right, if tidal forces are negligible, then accelerating due to gravity doesn't feel any different from floating in gravity-free space.

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.

No, it doesn't say that. It says that there is no experiment that can tell whether you are STATIONARY in a gravitational field, or accelerating in gravity-free space.

Here are 4 situations:

1. Being stationary (that is, NOT accelerating) in a gravitational field (for example, you're standing on a platform that is stationary on a planet).
2. Accelerating due to a rocket in gravity-free space (for example, you're standing on a platform that is on board an accelerating rocket).
3. Falling (that is, accelerating downward) due to gravity.
4. Floating inertially (that is, not accelerating) in gravity-free space.

The principle of equivalence says that situation 1, which does not involve acceleration, is equivalent to case 2, which does involve acceleration. It says that situation 3, which involves acceleration, is equivalent to case 4, which does not.

It seemed to me that you were saying that cases 2 and 3 were equivalent, that acceleration due to gravity is equivalent to acceleration due to a rocket. That's absolutely not the case.

Maybe the difficulty is with the definition of "acceleration". If you define acceleration to be the second time derivative of the position (coordinate acceleration), then cases 2 and 3 involve acceleration, but cases 1 and 4 do not. If you define acceleration to be measured relative to freefall (proper acceleration), then cases 1 and 2 involve acceleration, and cases 3 and 4 do not.

On the other hand, the use of proper acceleration only makes sense once you assume the equivalence of cases 3 and 4.

 

[stevendaryl]

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.

You're being a little thin-skinned. A.T. didn't do anything to indicate that he has a grudge against you. He just disagreed with you.

 

[MikeGomez]

...
Here are 4 situations:

1. Being stationary (that is, NOT accelerating) in a gravitational field (for example, you're standing on a platform that is stationary on a planet).
2. Accelerating due to a rocket in gravity-free space (for example, you're standing on a platform that is on board an accelerating rocket).
3. Falling (that is, accelerating downward) due to gravity.
4. Floating inertially (that is, not accelerating) in gravity-free space.

The principle of equivalence says that situation 1, which does not involve acceleration, is equivalent to case 2, which does involve acceleration. It says that situation 3, which involves acceleration, is equivalent to case 4, which does not.

It seemed to me that you were saying that cases 2 and 3 were equivalent, that acceleration due to gravity is equivalent to acceleration due to a rocket. That's absolutely not the case.

Maybe the difficulty is with the definition of "acceleration". If you define acceleration to be the second time derivative of the position (coordinate acceleration), then cases 2 and 3 involve acceleration, but cases 1 and 4 do not.
...

Thank you for taking the time to sort this out.

Yes, I do consider cases 2 & 3 to be acceleration by its definition of being the second time derivative of position. It seems reasonable (for me) to consider at least the possibility that these two situations can be related at a very fundamental level, because they are both physical processes which cause changes in momentum.

On the other hand, to say that case #3 (freefall) is not equivalent to case #2 (proper acceleration) because #3 is not acceleration, or that #3 should be called “coordinate” acceleration, is puzzling to me.

It’s not that I am totally against the idea, it’s just that I would like a reasonable explanation. The example of an accelerometer not measuring acceleration in freefall is complete rubbish. The apparatus is used in this example to take a measurement for which it was not designed. It’s like using carbon-14 dating on a live specimen.

 

[stevendaryl]

On the other hand, to say that case #3 (freefall) is not equivalent to case #2 (proper acceleration) because #3 is not acceleration, or that #3 should be called “coordinate” acceleration, is puzzling to me.

Well, that's what's so important about the equivalence principle, that it relates a situation involving acceleration to a situation that doesn't involve acceleration.

It’s not that I am totally against the idea, it’s just that I would like a reasonable explanation. The example of an accelerometer not measuring acceleration in freefall is complete rubbish.

It's not complete rubbish, since it's true. If the variation of the gravitational field with location is small enough (no significant tidal forces), then an accelerometer cannot measure acceleration due to gravity.

Here's a simple model for an accelerometer: You have a cubic box. In the middle of the box is a massive metal ball. It is held in place in the center of the box by 6 identical springs connecting the ball to each of the 6 sides of the box.

When you accelerate, the ball will move from the center. By measuring the lengths of the 6 springs, you can compute what the acceleration vector is.

This accelerometer will certainly not show any acceleration if the box is falling in a gravitational field.

The apparatus is used in this example to take a measurement for which it was not designed. It’s like using carbon-14 dating on a live specimen.

Well, whether it was designed for freefall or not, it's a fact about accelerometers that situations 1 and 2 (stationary on a planet vs. accelerating in a rocket) show acceleration and situations 3 and 4 (freefall vs. inertial motion) show no acceleration. That's what the principle of equivalence says, although it generalizes this result to all experiments you might make in a small room.

 

[stevendaryl]

It’s not that I am totally against the idea, it’s just that I would like a reasonable explanation.

An explanation for why the equivalence principle is true? Well, from the point of view of Newtonian gravity, you can explain it this way:

The only time that you "feel" acceleration is when different parts of your body are accelerated in different ways. If you are on a rocket standing on the floor, the floor is pushing up on your feet, but the rest of your body has no (external) forces acting on it. So the stress that you feel is your feet being accelerated relative to the rest of your body. Your body will compress slightly until the internal forces in your body are in equilibrium.

In contrast, gravity accelerates all parts of your body equally. So your feet are not accelerating relative to the rest of your body, and your body isn't compressed. So you feel no acceleration.
​

Einstein's explanation for the equivalence principle is different. In his theory of GR, he explains that freefall feels like inertial motion because it IS inertial motion, only inertial motion in curved spacetime rather than flat spacetime.

 

[MikeGomez]

Here's a simple model for an accelerometer: You have a cubic box. In the middle of the box is a massive metal ball. It is held in place in the center of the box by 6 identical springs connecting the ball to each of the 6 sides of the box.

When you accelerate, the ball will move from the center. By measuring the lengths of the 6 springs, you can compute what the acceleration vector is.

Thank you. I know how an accelerometer works.

An explanation for why the equivalence principle is true?

Seriously? No. I wasn't asking for an explanation of the equivalence principle. I was asking for a better explanation (than the falling accelerometer) of why case #3 should not be called acceleration, but must be called "coordinate" acceleration.

So you honestly don't see falacy in the accelerometer example?

gravity accelerates all parts of your body equally

That's exactly why accelerometers doesn't measure the gravitational acceleration in case #3.

It's not complete rubbish

I'm still not convinced.

Anyways, thanks for your time. I don't see this going anywhere productive, so I'll just concede that case #3 is definitely not acceleration, and that great care should be taken to make the distinction that it is "coordinate" acceleration.

 

[stevendaryl]

Thank you. I know how an accelerometer works.

Then I don't understand your objection to the claim that an accelerometer can't distinguish between being stationary on a planet vsl accelerating on a rocket, or between being in freefall in a gravitational field vs. traveling inertially. It definitely CAN distinguish between accelerating due to gravity and accelerating due to any other force.

Seriously? No. I wasn't asking for an explanation of the equivalence principle. I was asking for a better explanation (than the falling accelerometer) of why case #3 should not be called acceleration, but must be called "coordinate" acceleration.

Because an accelerometer measures no acceleration. That's a coordinate-independent definition of acceleration. In contrast, coordinate acceleration depends on having a coordinate system to measure acceleration relative to. That's why it's called "coordinate acceleration".

So you honestly don't see fallacy in the accelerometer example?

No, I have no idea what you are talking about when you say there is a fallacy involved.

I'm still not convinced.

Convinced of what?

 

[LitleBang]

Wouldn't it be nice if the standard model people would show why gravity(graviton) appear to be identical with inertia?

 

[Dale]

I was asking for a better explanation (than the falling accelerometer) of why case #3 should not be called acceleration, but must be called "coordinate" acceleration.

So you honestly don't see falacy in the accelerometer example?

I also fail to see the fallacy. It is simply a matter of definition. By definition "coordinate acceleration" is the second derivative of position wrt time. By definition "proper acceleration" is the second covariant derivative along the worldline (physically the acceleration as measured by a correctly functioning accelerometer). There are cases where the two definitions coincide and cases where they do not coincide, nothing fallacious about that.

You can certainly use the unqualified term "acceleration" how you choose, but if the case under discussion is one where they do not coincide then you invite confusion and ambiguity.

 

[MikeGomez]

No, I have no idea what you are talking about when you say there is a fallacy involved.

Here’s a picture.

Accelerometer A is at rest in the lab frame produces no reading (no compression or tension in the springs.

Accelerometer B is being accelerated from the outside of the box (as designed) and therefore gives a reading.

Accelerometer C is being accelerated with the same force as Accelerometer B, but in a manner in which the apparatus was not designed for. I have opened the lid of the box, and I have accelerated it evenly so that there is no tension/compression in the springs. I couldn’t draw it very well, but you can imagine, at least in principle, that every unit of mass is accelerated.

I believe that at a fundamental level, every unit of mass in accelerometer 'C' gains momentum in precisely the same way as it would by gravity in free fall. I could be mistaken, but that is how I currently understand it. So when I say "Accelerating due to gravity" (coordinate-acceleration in free fall) is equivalent to an "accelerating chest", that is my reasoning.

On the other hand, what would you think if I were to tell you that accelerometer C is not accelerated because it shows no reading? Doesn’t that sound silly? So why should that same situation (the misuse of the apparatus) suddenly start making sense simply because the accelerometer is in freefall?

Hopefully that explains why I believe the accelerometer in freefall explation is bunk. If not, oh well.

 

[MikeGomez]

...
By definition "coordinate acceleration" is the second derivative of position wrt time. By definition "proper acceleration" is the second covariant derivative along the worldline
(physically the acceleration as measured by a correctly functioning accelerometer). There are cases where the two definitions coincide and cases where they do not coincide, nothing fallacious about that.

O.k. Thank you for that. I guess in my example, case 'C' is the case where they coincide?

 

[stevendaryl]

Here’s a picture.

Accelerometer A is at rest in the lab frame produces no reading (no compression or tension in the springs.

That would be true if your lab were floating in space. Here on Earth, an accelerometer would measure compression. It would indicate upward acceleration. Of course, since g is roughly constant on the surface of the Earth, a practical accelerometer would be calibrate to subtract off this effect, but it is certainly measurable.

Accelerometer B is being accelerated from the outside of the box (as designed) and therefore gives a reading.

Accelerometer C is being accelerated with the same force as Accelerometer B, but in a manner in which the apparatus was not designed for. I have opened the lid of the box, and I have accelerated it evenly so that there is no tension/compression in the springs. I couldn’t draw it very well, but you can imagine, at least in principle, that every unit of mass is accelerated.

I believe that at a fundamental level, every unit of mass in accelerometer 'C' gains momentum in precisely the same way as it would by gravity in free fall. I could be mistaken, but that is how I currently understand it. So when I say "Accelerating due to gravity" (coordinate-acceleration in free fall) is equivalent to an "accelerating chest", that is my reasoning.

Okay, I would agree that accelerating due to gravity is equivalent to an accelerating chest in which it has been carefully arranged so that every single particle in the box accelerates in exactly the same manner. But that's not the normal way of accelerating things. If you accelerate something by pushing on one end, that end will accelerate more than the other end. If you accelerate something by pulling on one end, that end will accelerate more. If you accelerate a charged particle by an electric field, then oppositely charged particles will be accelerated in the opposite direction.

There is no way to accelerate all parts of an object equivalently OTHER than using gravity. You might be able to do it approximately, though.

On the other hand, what would you think if I were to tell you that accelerometer C is not accelerated because it shows no reading? Doesn’t that sound silly?

Frankly, exactly the opposite. It seems silly to insist that something is accelerating, even though there is no way to measure it. Your case C is a very special case. It's not something that can be arranged except through very special forces that treat all phenomena in the same way. (And it really has to be ALL phenomena; even the path of a light beam is accelerated by gravity. You can't arrange that by any other means, as far as I know).

So why should that same situation (the misuse of the apparatus) suddenly start making sense simply because the accelerometer is in freefall?

I wouldn't use the phrase "misuse of the apparatus". It's not a matter of misusing the apparatus, it's a matter of how to interpret its reading. You would like to interpret it as measuring coordinate acceleration. It doesn't measure that, except in special circumstances. It measures proper acceleration, which is acceleration relative to freefall.

That's true in general. Devices don't measure coordinate-dependent quantities. You have to do an act of interpretation to interpret their measurements as saying something about coordinates.

Hopefully that explains why I believe the accelerometer in freefall explation is bunk. If not, oh well.

The equivalence of freefall to inertial (acceleration-free) motion is just true (to the extent that variability of gravity with position can be ignored). And that equivalence was instrumental for Einstein to develop General Relativity.

So to me, it makes no sense to call something bunk when it is both truth and important.

But, anyway, regardless of whether you like it, that is what is meant by "the principle of equivalence". So your way of describing the principle of equivalence is not what "the principle of equivalence" means.

 

[stevendaryl]

O.k. Thank you for that. I guess in my example, case 'C' is the case where they coincide?

No. The two types of acceleration (coordinate vs. proper) coincide when there is no gravity. Your case C is a case where they DON'T coincide.

 

[DrGreg]

MikeGomez,

Acceleration has to be measured relative to something. So, to avoid all ambiguity, whenever anyone talks of "acceleration", they need to specify relative to what. Some of the confusion in this thread has been that the "what" hasn't always been explicitly mentioned and different people have been assuming different "what"s.

After you throw a ball in the air, you can say it has a (more-or-less) constant acceleration g downwards relative to the ground and the ground has an acceleration of zero relative to the ground. Or you can say that ground has a (more-or-less) constant acceleration g upwards relative to the ball and the ball has an acceleration of zero relative to the ball. They are both equally valid descriptions.

There is a special case when the reference object (that we measure relative to) is moving inertially, and is momentarily moving at the same speed as the object being measured. In that case we call this "proper acceleration". In Newtonian physics, the ground was taken to be inertial and so the falling ball would have non-zero "Newtonian proper acceleration". Einstein's revolutionary idea was to take the ball to be inertial and the ground to have a non-zero (relativistic) proper acceleration. That last sentence is really another way to describe GR's equivalence principle.

It turns out, using the GR definition of inertial, that accelerometers always measure (relativistic) proper acceleration whatever object you attach them to, whereas in Newtonian physics accelerometers fail to measure "Newtonian proper acceleration" when used in a gravitational field (unless you "calibrate out" the effects of gravity).


I put the phrase "Newtonian proper acceleration" in quotation marks because I don't think anyone actually uses the term "proper acceleration" in Newtonian physics.

 

[Dale]

O.k. Thank you for that. I guess in my example, case 'C' is the case where they coincide?

I am not 100% clear on your example C, but in simplified terms if a force is proportional to mass then it can be geometrized. That means that it can be lumped into the same category as the centrifugal and Coriolis forces in a rotating reference frame, and canceled out by a coordinate transform in the same way. The proper acceleration is given by a covariant derivative, so it cannot be canceled by a coordinate transform.

So, in your scenario C, if you intend the force acting on the accelerometer to be proportional to mass then you can find a coordinate system where it is locally canceled out. In this coordinate system proper acceleration and coordinate acceleration coincide locally. In other coordinate systems they would not coincide.

 

[A.T.]

The apparatus is used in this example to take a measurement for which it was not designed.

The point is that you can't design a local measurement of coordinate acceleration due to gravity. Maybe this video will help you:

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

 

[davenn]

Thanks AT

interesting video ... went on to watch a few of the others as well :)

Dave

 

[HALON]

I also found the video good value. It made me think of this question:

Body A at rest on planet X’s surface experiences gravity equivalently to body B undergoing acceleration not near any major sources of gravity in open space.

What is the formula used to convert B’s acceleration to X’s mass?

 

[Nugatory]

I also found the video good value. It made me think of this question:

Body A at rest on planet X’s surface experiences gravity equivalently to body B undergoing acceleration not near any major sources of gravity in open space.

What is the formula used to convert B’s acceleration to X’s mass?

There's no single answer, as the necessary mass also depends on the radius of the planet. You want the gravitational force from X on A to be sufficient to produce the acceleration that B is experiencing. Newton's law for gravitational force $F=Gm_1m_2/r^2$ and $F=ma$, plus a little algebra, will get you to the formula.