Why do things in free fall accelerate?

[Z3kr0m]

TL;DR

Why do things in free fall accelerate?

Hello, I have a problem to understand acceleration in GR, objects in free fall move along a geodesic, they are in inertial motion. But observer on Earth can clearly see that falling thing accelerates. What causes the acceleration, when there is no gravititional force? Thanks for answers.

 

[Ibix]

Free fall means that they aren't accelerating in any physically meaningful sense. Things that are not in free fall (e.g., you) are accelerating, which is why you have weight - which feels just like being pressed back into your seat in an accelerating car.

Regarding something you drop, then, it's moving in free fall. You are accelerating upwards. If you want to pretend that you are moving inertially you need to invoke a fictitious force, like centrifugal and Coriolis forces, to explain why the ball isn't moving with constant velocity. We call this particular fictitious force "Newtonian gravity".

This is the point of Einstein's elevator thought experiment. There's no difference (locally, anyway) between neing on the surface of the Earth and dropping something, and being in an accelerating (rocket powered) elevator far from any gravity and dropping that same object.

 

[Orodruin]

It is the observer that is not in free fall that is accelerating. Typically because the floor is pushing up on that observer. This leads to relative acceleration - but again it is the observer that is accelerating, not the object in free fall.

 

[Z3kr0m]

So the observer is being accelerated due to floor pushing him upwards? It looks like force to me

 

[Z3kr0m]

If you want to pretend that you are moving inertially you need to invoke a fictitious force, like centrifugal and Coriolis forces, to explain why the ball isn't moving with constant velocity. We call this particular fictitious force "Newtonian gravity".

So the fictitious force which relatively accelerates falling objects is just upwards acceleration from the ground?

 

[Orodruin]

So the observer is being accelerated due to floor pushing him upwards? It looks like force to me

That is a force on the observer, not on the object in free fall.

 

[Ibix]

So the observer is being accelerated due to floor pushing him upwards? It looks like force to me

As Orodruin says, that's a real force on the observer. If you want to use the observer's frame as a rest frame you need another force to balance the real upwards force so he doesn't accelerate. This is the fictitious "force of gravity", which is also what "accelerates the ball downwards".

 

[A.T.]

Hello, I have a problem to understand acceleration in GR, objects in free fall move along a geodesic, they are in inertial motion. But observer on Earth can clearly see that falling thing accelerates. What causes the acceleration, when there is no gravititional force? Thanks for answers.

This might help (see also links in the video description on youtube):

 

[Z3kr0m]

That is a force on the observer, not on the object in free fall.

But usn't general relativity based on the fact, that force doesn't exist? The effecte are just the curvature of spacetime.

 

[Orodruin]

But usn't general relativity based on the fact, that force doesn't exist? The effecte are just the curvature of spacetime.

No. Forces still exist. What is the effect of the curvature of spacetime is gravity. The force on the observer from the floor is not a gravitational force. It is a contact force originating from electromagnetic interactions.

 

[Z3kr0m]

No. Forces still exist. What is the effect of the curvature of spacetime is gravity. The force on the observer from the floor is not a gravitational force. It is a contact force originating from electromagnetic interactions.

So that is the force which stands behind the relative acceleration? Anyway, thank you very much, I understand the concepts a bit more I think. :)

 

[Ibix]

But usn't general relativity based on the fact, that force doesn't exist? The effecte are just the curvature of spacetime.

No. Gravity is not a force in general relativity, but other forces still exist.

I think there have been efforts to make the other forces disappear in a similar way, but not so effectively (e.g. Kaluza-Klein). Also, it's probably true that people tend to write down a Lagrangian and solve it without explicitly working with forces. But the concept is still available hidden in the maths if you want to use it - just not for gravity.

 

[A.T.]

So that is the force which stands behind the relative acceleration?

There is no force needed to have that relative acceleration (a.k.a coordinate acceleration). It just a consequence of the chosen reference frame.

 

[Ibix]

There is no force needed to have that relative acceleration (a.k.a coordinate acceleration). It just a consequence of the chosen reference frame.

I think this is potentially confusing. In this particular scenario, with a physical person releasing a body into free fall, you do need a real force on the person. Otherwise the person is in free fall alongside the ball, like an astronaut.

However, it's true that you can have a force-free scenario. You don't actually need a real person to be able to describe what a real person would see. A ball that is free falling in a gravitational field will have apparent acceleration ("coordinate acceleration") relative to a coordinate system that would regard observers standing on a non-rotating planet as "at rest", regardless of whether or not there are any such observers.

 

[PeroK]

So that is the force which stands behind the relative acceleration? Anyway, thank you very much, I understand the concepts a bit more I think. :)

Just to add something. In general, in both classical physics and in GR, an object can accelerate (in your reference frame) for one of two reasons:

1) There is a force on the object.

2) There is a force on you and you are accelerating!

Now, if you are in an accelerating car or train or aeroplane, it doesn't seem strange that the objects on the ground are "accelerating backwards" relative to you (in your reference frame).

The thing with GR is to understand that, standing on the surface of the Earth you are subject to an upward force from the ground and you are, therefore, accelerating upwards. Objects in free fall do not have a force on them, so they accelerate downwards in your reference frame. This is analogous to the example above.

The other point is that gravity is the curvature of spacetime about the Earth, due to the Earth's mass, and means that the natural path (geodesic) for any object is towards the centre of the Earth. You cannot follow that path because the ground prevents you, but the falling object does follow that path until it hits the ground.

 

[Z3kr0m]

Thank you very much for your answers!

 

[jbriggs444]

1) There is a force on the object.

2) There is a force on you and you are accelerating!

For objects that are close to one another, those are the two possibilities. For objects at a distance there is a third possibility:

3) There is no force on either of you, but space-time is curved. A natural coordinate system in which you are unaccelerated and at rest is one against which the other object is judged to be accelerating.

The situation described in the original post appears to involve an observer and an object that are adjacent. So #3 does not apply. It would apply if an observer in free fall at the north pole were to ask why an object in free fall at the south pole has an upward relative acceleration.

 

[Dale]

So that is the force which stands behind the relative acceleration?

Not all relative acceleration is due to force. In GR you simply attach an accelerometer to an object or an observer. If the accelerometer reads 0 then the object is not subject to a net real force. If the accelerometer reads non-zero then the object is subject to a real net force.

For an observer on the ground and a ball in the air, the observer’s accelerometer reads g in the upwards direction while the ball’s accelerometer reads 0. The only real force on the observer is the upwards contact force from the ground so that is the force causing relative acceleration.

Now, consider two objects in different orbits. They each have accelerometer readings of 0, so they have no real forces on them. However, they have relative acceleration. This relative acceleration is not due to a force, it is due to the curvature of spacetime. Curvature essentially represents tidal gravity effects.

 

[PeroK]

For objects that are close to one another, those are the two possibilities. For objects at a distance there is a third possibility:

3) There is no force on either of you, but space-time is curved. A natural coordinate system in which you are unaccelerated and at rest is one against which the other object is judged to be accelerating.

The situation described in the original post appears to involve an observer and an object that are adjacent. So #3 does not apply. It would apply if an observer in free fall at the north pole were to ask why an object in free fall at the south pole has an upward relative acceleration.

Another good example is objects in free fall at different heights, so that there is acceleration relative to each other and different accelerations relative to the Earthbound observer.

 

[Z3kr0m]

Another good example is objects in free fall at different heights, so that there is acceleration relative to each other and different accelerations relative to the Earthbound observer.

Last question, what causes the magnitude of the upward acceleration? You have larger acceleration on bigger objects, is it becouse of the bigger spacetime curvation?

 

[PeroK]

Last question, what causes the magnitude of the upward acceleration? You have larger acceleration on bigger objects, is it becouse of the bigger spacetime curvation?

The magnitude of the acceleration relates to how much you are deviating from a geodesis. Strictly speaking, the spacetime curvature determines tidal forces.

There's a good piece here:

http://www.einstein-online.info/spotlights/geometry_force.html

 

[Z3kr0m]

Yes, the curvature, which equates to the strength of the gravitational field, depends on the mass of the Earth and the distance from the centre of the Earth.

Which formula is used to calculate the acceleration? The geodesic equation?

 

[PeroK]

Which formula is used to calculate the acceleration? The geodesic equation?

You can get it directly from the metric. If you are at rest relative to the Earth, then you can calculate your "proper" acceleration, which requires a force to sustain.

 

[Dale]

Last question, what causes the magnitude of the upward acceleration? You have larger acceleration on bigger objects, is it becouse of the bigger spacetime curvation?

The magnitude of the acceleration is determined by the mass of the object and the strength of the force acting in the object: $a=F/m$ where a is the four-acceleration, F is the four-force, and m is the invariant mass. If the interaction is, for example, due to standing on an elastic floor, then you could use Hookes law to calculate F.

 

[Ibix]

The floor resists you falling. In stronger spacetime curvature you need more force to stay at the same altitude. Obviously a floor may not be able to provide sufficient force - one can sink in marshy ground on Earth.

 

[jbriggs444]

In stronger spacetime curvature you need more force to stay at the same altitude

This is the second time I've seen this picture presented in this thread. I believe it to be an incorrect over-simplification.

Spacetime curvature is essentially equivalent to tidal gravity -- the rate at which gravitational acceleration changes with position. If you want to compare an acceleration here against a state of rest over there, what matters is, roughly speaking, the integral of curvature over a path.

The local acceleration of gravity is independent of the local spacetime curvature.

My mistaken impression has been corrected.

 

[Nugatory]

If you want to compare an acceleration here against a state of rest over there, what matters is, roughly speaking, the integral of curvature over a path.

But here we are comparing the deviation between the geodesic worldline of an object in freefall and the non-geodesic worldline of an object hovering with constant Schwarzschild $r$ coordinate at the point where the two worldlines intersect. That's a measure of the force required to hold the hovering object on its worldline, and it is indeed greater in a more strongly curved spacetime.

 

[Ibix]

The local acceleration of gravity is independent of the local spacetime curvature.

The proper acceleration of an observer hovering at constant Schwarzschild $r$ (with $G=c=1$) is $$a=\frac{M}{r^2\sqrt{1-2M/r}}$$The Kretchmann curvature invariant, which is what I was thinking of as "the strength of the curvature", is $K=R_{ijkl}R^{ijkl}=48M^2/r^6$. You can eliminate $r$ between those two equations and see that the acceleration needed to hover behaves as $K^{1/3}$ for small $K$ and as $K^{1/4}$ for large $K$. So the proper acceleration needed to hover increases in more strongly curved spacetime.

Intuitively, something like this must be true. Else, why would you feel heavier closer to a gravitating mass?

 

[FactChecker]

GR-curved space-time implies that the geodesics are not the same as they would be without GR. Without GR, we would think that an object held stationary at a constant altitude was following a geodesic, $A_{nonGR}$. But we know that the true GR-geodesic, $B_{GR}$, curves away from that. The increasing separation between $A_{nonGR}$ and $B_{GR}$ appears as "acceleration" to one who does not take GR into account. So an object that is left to follow the true (GR) geodesic, $B_{GR}$, will appear to be accelerating. That fact requires no calculations at all. The details of how much "acceleration" there is and in which direction requires calculations.

 

[Dale]

The local acceleration of gravity is independent of the local spacetime curvature.

My mistaken impression has been corrected.

I don’t know. I am more or less onboard with your “mistaken” impression. At each point on the Earth the local spacetime is approximately flat and yet you are accelerating upwards. So locally I don’t think that curvature explains the acceleration. The acceleration is (IMO) explained by real forces. The floor pushes up on your feet, the ground pushes up on the floor, the bedrock pushes up on the ground ... Every part of the surface of the Earth is accelerating upward due to unbalanced real forces acting on it.

What curvature explains is why every point on that surface can accelerate away from each other without the distance changing.