Problem in understanding how upwards acceleration works (General Relativity)

[Mike_bb]

Hello, PF!

As is known, if body stays on the ground then body is accelerated upwards by the ground.
For example, free-falling observer see that body on the ground move to him with acceleration $9.81m/s^2$

I have a problem in understanding how it is possible.

If the body is being accelerated by the ground then this is only to prevent body to fall down. But how is it possible that body accelerates (and move) relative to free-falling observer?

Thanks.

 

[Dale]

But how is it possible that body accelerates (and move) relative to free-falling observer?

Can you explain your confusion here? What would make you think it is not possible?

You can drop an object and visually see that relative to the object the ground is moving.

 

[Mike_bb]

Can you explain your confusion here? What would make you think it is not possible?

Free-falling motion assumes that falling observer is at rest or moving inertially. But body on the ground doesn't lose contact with the support. Nevertheless, body on the ground moves with upward acceleration. I can't understand how.

 

[Roberto Pavani]

body on the ground moves with upward acceleration.

You have to specify the reference frame to define motion, that's the key.

Here perhaps a too simple explanation:

Three stones: A and B rest on the ground, C falls.
A: "B is at rest relative to me, we both feel an acceleration ($F/m = g$), and C falls toward us."
C: "I feel nothing ($F/m = 0$), and I see A and B accelerate toward me."

 

[A.T.]

But how is it possible that body accelerates (and move) relative to free-falling observer?

This isn't specific to General Relativity, but also applies to Newtonian mechanics, which you should properly understand first, before tackling Relativity.

 

[Mike_bb]

This isn't specific to General Relativity, but also applies to Newtonian mechanics, which you should properly understand first, before tackling Relativity.

Why? I can't provide link but I saw that this is example from GR.

 

[Halc]

But how is it possible that body accelerates (and move) relative to free-falling observer?

This isn't specific to General Relativity, but also applies to Newtonian mechanics, which you should properly understand first, before tackling Relativity.

There are important differences between the two, but A.T. is correct in pointing out that the concept of reference frames (inertial or not) and of acceleration are common between Newton and GR.

It is possible that a body accelerates relative to a free falling observer simply by that body having a force applied to it. So the free falling observer can light a bottle rocket and off it goes (good trick in vacuum, but I didn't say vacuum).

Differences: Under Newtonian mechanics, gravity is exerting a continuous force on everything near Earth. So a house is stationary because the downward force of gravity is perfectly cancelled by an equal and opposite force applied upward by the ground. The guy jumping off the cliff accelerates due to gravity since nothing counters the force (at first).

Under GR, there is no gravitational force, so the guy jumping off the cliff has no forces applied to him (again, at first), so he's in free fall. No acceleration and no proper acceleration. The house on the other hand has an unbalanced force applied upward to it, so it (and the observer on the ground) all accelerate upwards. They have positive proper acceleration, as can be measured by any accelerometer.

As for reference frames, those two guys each using reference frames where they are stationary and the other guy is moving. Under Newton, the free falling guy is using an accelerating reference frame and the guy in the house is in using an inertial one. Under GR it's the other way around, with the freefalling guy being inertial and the guy on the ground using an accelerating frame.

The bold part answers your question in post 3

 

[Mike_bb]

You have to specify the reference frame to define motion, that's the key.

Here perhaps a too simple explanation:

Three stones: A and B rest on the ground, C falls.
A: "B is at rest relative to me, we both feel an acceleration ($F/m = g$), and C falls toward us."
C: "I feel nothing ($F/m = 0$), and I see A and B accelerate toward me."

Good. Is this explanation work if C is at rest on the height?

 

[Sagittarius A-Star]

For example, free-falling observer see that body on the ground move to him with acceleration $9.81m/s^2$

Yes. To be more precise: For the free-falling observer, a nearby body on the ground has locally a coordinate-acceleration relative to the falling (=inertial) frame of approximately $9.81 \ m/s^2$ (for $v<<c$).

An observer in a locally free falling elevator-cabin can throw a ball an notices, that the ball moves with constant velocity along a straight line, as discribed in Newton's first law. Therefore, the falling frame must be inertial. It has no proper acceletaion.

 

[Roberto Pavani]

I never specified the height where the stones are, A can be on the ground, while B and C on the rooftop with C falling from there. In that case your question is just a relabeling of B <-> C

 

[Mike_bb]

I never specified the height where the stones are, A can be on the ground, while B and C on the rooftop with C falling from there. In that case your question is just a relabeling of B <-> C

I mean that stone C has velocity = 0.

 

[Roberto Pavani]

V = 0 doesn't matter.
Throw C upward: at the top $v=0$ for an instant, but C ir's still free-falling all the time (accelerometer = 0).

This video can provide a rough idea:

 

[Mike_bb]

V = 0 doesn't matter.
Throw C upward: at the top $v=0$ for an instant, but C ir's still free-falling all the time (accelerometer = 0).

This video can provide a rough idea:

Ok. Thanks.

If C is free-falling why can't we say that Earth moves with acceleration $9.81m/s^2$ relative to stone C?

 

[Roberto Pavani]

Yes,

If C is free-falling

we can

say that Earth moves with acceleration 9.81m/s2 relative to stone C

Note: acceleration is $\approx 9.81 \frac m { s^2 }$ because $g \propto \frac 1 { R^2 }$

 

[PeterDonis]

Free-falling motion assumes that falling observer is at rest

No.

moving inertially.

If this means "an accelerometer attached to the observer reads zero", then yes. And that's the only thing free-falling motion means.

 

[A.T.]

Why? I can't provide link but I saw that this is example from GR.

But your question regarding that example had nothing to do with GR. It rather indicates that you don't understand the basics, while trying to learn GR from some links you can't even find.

 

[Dale]

Free-falling motion assumes that falling observer is at rest or moving inertially. But body on the ground doesn't lose contact with the support. Nevertheless, body on the ground moves with upward acceleration. I can't understand how.

Free-falling motion means that an accelerometer attached to the body reads 0. It has nothing to do with any observers. Every observer will agree that the accelerometer reads 0 so every observer will agree that the free-falling body is inertial.

An accelerometer attached to a body on the ground reads 9.8 m/s^2 upwards. That also has nothing to do with any observer. Every observer will agree that the accelerometer reads 9.8 upwards, so every observer will agree that the body on the ground is non-inertial.

This is proper acceleration, the acceleration measured by an attached accelerometer. This is a relativistic four-vector, and all frames and all observers agree on this vector, even if they use a different basis to describe it.

 

[Mike_bb]

Every observer will agree that the accelerometer reads 9.8 upwards, so every observer will agree that the body on the ground is non-inertial.

Yes. But I want to provide such example:

Let we have water, the motorboat and the boat. Water(blue) and the motorboat (grey) are non-inertial frames. The boat (green) has constant velocity $V$ (inertial frame).
Water flow moves with acceleration $a$ and motorboat moves with the same acceleration. The motorboat is at rest relative to observer on the shore (by analogy with the non-inertial still body on the surface of the Earth). But relative to the man on the boat (inertial frame) the motorboat is still. As is mentioned above, non-inertial frame move with upwards acceleration to free-falling observer. But my example show that non-inertial frame is still relative to inertial frame.

How is it possible that non-inertial body on the surface of the Earth move with upwards acceleration to free-falling observer if non-inertial body is still relative to the surface?

Where did I wrong? Thanks.

 

[Dale]

The boat (green) has constant velocity V (inertial frame).

Just having constant V doesn't make it inertial. The boat is non-inertial because an accelerometer it carries reads an acceleration of 1 g upwards.

 

[A.T.]

The motorboat is at rest relative to observer on the shore (by analogy with the non-inertial still body on the surface of the Earth)

I assume horizontal here is meant to be analogus to vertical in the free fall case? But it doesn't match:
- An accelerometer on the motorboat will measure zero horizontal proper acceleration.
- An accelerometer on the body on the surface of the Earth doesn't measure zero vertical acceleration.

How is it possible that non-inertial body on the surface of the Earth move with upwards acceleration to free-falling observer if non-inertial body is still relative to the surface?

The surface and the body are both moving with upwards coordinate acceleration g relative to a free-falling frame of reference, so where is the problem? This is true for both: Newton and GR.

The only GR specific part is which frame is considered inertial, but you seem stuck at basic kinematics.

 

[Mike_bb]

A.T., No. I have no problem with kinematics.

The surface and the body are both moving with upwards coordinate acceleration g relative to a free-falling frame of reference, so where is the problem?

Problem: surface and the body are still. They don't lose contact with support.

 

[A.T.]

A.T., No. I have no problem with kinematics.

Yes, you obviously have:

Problem: surface and the body are still.

Not in the free-falling frame of reference that you were talking about:

How is it possible that non-inertial body on the surface of the Earth move with upwards acceleration to free-falling observer if non-inertial body is still relative to the surface?

 

[Mike_bb]

Yes, you obviously have:

Not in the free-falling frame of reference that you were talking about:

I've read about Einstein's lift experiment and I understood it. But I can't imagine how it is possible that body doesn't lose contact with support but accelerate, although experiment works indeed.

 

[A.T.]

But I can't imagine how it is possible ...

1) Learn about kinematics in different reference frames
2) Learn to express yourself precisely and to differentiate proper acceleration and coordinate acceleration

Otherwise this thread is just as pointless as your previous threads.

 

[PeterDonis]

I've read about Einstein's lift experiment and I understood it.

Did you? Do you understand that if you are standing inside Einstein's lift and drop a rock, it accelerates downward, relative to you? Just like it would if you were standing on the surface of the Earth and dropped a rock?

I can't imagine how it is possible that body doesn't lose contact with support but accelerate

If you're standing inside Einstein's lift, you're not accelerating (in the sense of coordinate acceleration) relative to the lift. But if you drop a rock, it accelerates (in the sense of coordinate acceleration) downward relative to you.

Just as if you are standing on the surface of the Earth, you're not accelerating (in the sense of coordinate acceleration) relative to the Earth. But if you drop a rock, it accelerates (in the sense of coordinate acceleration) downward relative to you.

In both cases, if you use a frame in which the dropped rock, not you, is at rest, you (and the lift/surface of the Earth in your vicinity) will be accelerating upward (in the sense of coordinate acceleration) relative to the rock.

And in both cases, an accelerometer attached to the rock reads zero, and an accelerometer attached to you reads a nonzero upward acceleration.

If you don't understand all of the above, you don't understand Einstein's lift thought experiment, since all of the above points are part of what that thought experiment is intended to illustrate.

 

[Dale]

Let we have water, the motorboat and the boat. Water(blue) and the motorboat (grey) are non-inertial frames. The boat (green) has constant velocity $V$ (inertial frame).

The boat is not an inertial frame since an attached accelerometer will not read 0. Nothing here is inertial, the water, motorboat, and boat are all non-inertial.

Water flow moves with acceleration $a$ and motorboat moves with the same acceleration.

You mean that they move with a horizontal acceleration $a$ and their vertical acceleration $g$.

relative to the man on the boat (inertial frame) the motorboat is still.

The boat is not inertial, and the motorboat is not still relative to the man on the boat.

But my example show that non-inertial frame is still relative to inertial frame.

Your example is wrong in several places, including specifically the claim that the non-inertial frame is still relative to the inertial frame.

How is it possible that non-inertial body on the surface of the Earth move with upwards acceleration to free-falling observer if non-inertial body is still relative to the surface?

Moving and accelerating are two different things. In all frames a body on the surface of the earth is accelerating (proper acceleration) upwards. In the ground's frame that constant upwards acceleration produces no motion. In the free-falling frame that constant upwards acceleration produces motion.

Proper acceleration is frame-invariant, all frames agree.

Motion is frame-dependent, different frames disagree.

But I can't imagine how it is possible that body doesn't lose contact with support but accelerate, although experiment works indeed.

There is no need to imagine it. You can measure it directly. Every smart phone has an accelerometer to measure proper acceleration. Look at it and see. You are accelerating upwards without losing contact with the ground. And in fact, if you do lose contact with the ground then you stop accelerating. It is precisely the contact with the ground that provides the force for the acceleration.

 

[Mike_bb]

Dale,
Thanks.

"In the free-falling frame that constant upwards acceleration produces motion."

Could you elaborate how does constant upwards acceleration produce motion?

 

[Dale]

Dale,
Thanks.


Could you elaborate how does constant upwards acceleration produce motion?

In the free falling frame if the ground is at rest at $t=0$ then at $t=1$ it is moving upwards at $v=9.8 \mathrm{\ m/s}$ and at $t=2$ it is moving upwards at $v=19.6 \mathrm{\ m/s}$.

 

[Mike_bb]

Dale,

I've found explanation why the body doesn't lose contact with the support and accelerates upwards. Infinitesimally small deformations play key role in the process of the acceleration. These deformations cause atoms to accelerate synchronically and the body accelerates.

I think that since deformations are continuous then we get that body is accelerated upwards. We can consider motion as sum of infinitesimally small displacement that is caused by infinitesimally small deformations.

 

[Dale]

These deformations cause atoms to accelerate synchronically and the body accelerates

There certainly are deformations (Hooke’s law), and they are finite rather than infinitesimal. However, if the deformations are the explanation for the acceleration then you are left with a pretty big problem:

You can put one accelerometer on rubber and another on steel. Both accelerometers read the same acceleration, but the deformation of the rubber is much more since it is less stiff.