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Question about General Relativity.

Edit: I've gotten some really good input to help me better phrase my confusion.

I think the heart of my problem is understanding the difference between geodesic motion and unaccelerated motion.

Suppose there's an apple hanging in a tree. It feels the force of gravity which is cancelled out by the pull of the tree on the stem. The net result is no motion.

I was thinking that if free falling, the apple would follow a geodesic and that geodesic motion is unaccelerated motion. I was thinking I could use the fact that the motion was geodesic to ignore the question "not accelerating with respect to what?". I am now thinking this is one of my mistakes.

So is there any meaning to saying that motion along a geodesic is not accelerated without answering the question "accelerating with respect to what?"

Now consider the apple in the tree with its stem holding it up. The motion is now non-geodesic, and hence accelerating(maybe?). So the apple is accelerating when it is still and not accelerating when it is falling.

What is the expected read out of an accelerometer if one were attached to the apple in the case of the stationary apple, vs. the falling apple?

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    $\begingroup$ Acceleration is change in velocity. A freely falling object is accelerating. $\endgroup$ – Jasper Nov 9 '18 at 21:25
  • $\begingroup$ Objects in free fall don't accelerate though, right? $\endgroup$ – R. Romero Nov 9 '18 at 21:33
  • $\begingroup$ Without picking a reference system there is no such thing like "falling" or "accelerating". Your problem is that you are switching reference systems in the above examples. $\endgroup$ – Nobody-Knows-I-am-a-Dog Nov 9 '18 at 21:34
  • $\begingroup$ A freely falling apple is accelerating wrt to the apples still attached to the tree. A freely falling apple is not accelerating wrt to a second apple which had been cut loose from the tree at the same time as itself. $\endgroup$ – Nobody-Knows-I-am-a-Dog Nov 9 '18 at 21:36
  • $\begingroup$ If you watch an apple falling from a tree, does it change its velocity on its way? It surely does! $\endgroup$ – Jasper Nov 9 '18 at 21:38
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The best way to approach acceleration in special and general relativity is to use the four-acceleration. This is a four-vector so its norm is a scalar invariant that we call the proper acceleration. The physical significance of the proper acceleration is that is the acceleration felt by an observer in their rest frame.

So for example as I sit at my desk typing this I feel an acceleration of $1g$ so I know that my proper acceleration must be $1g$. In my case this acceleration is the result of the ground beneath me exerting an upwards force of $mg$ on me. The same applies to your apple, though in that case the proper acceleration of $1g$ is due to the upwards force $mg$ applied by the branch from which the apple hangs (which i guess comes ultimately from the ground the tree is rooted in).

The four acceleration is given by the equation:

$$ A^\alpha = \frac{\mathrm d^2x^\alpha}{\mathrm d\tau^2} + \Gamma^\alpha{}_{\mu\nu}U^\mu U^\nu \tag{1} $$

where $x^\alpha$ is the position and $U^\alpha$ is the four velocity. Look at the two terms in this equation. The first term looks like a simple double derivative of position with time (though it's proper time not coordinate time). The second term arises from the curvature. The objects $\Gamma^\alpha{}_{\mu\nu}$ are the Christoffel symbols and they describe the curvature (in my coordinates).

Consider the apple you used as an example. In the apple's rest frame it is stationary so $d^2x^\alpha/d\tau^2 = 0$. That means the four acceleration of the apple is given only by the curvature term:

$$ A^\alpha_{\,\textrm{apple}} = \Gamma^\alpha{}_{\mu\nu}U^\mu U^\nu $$

So the reason the apple has a non-zero four-acceleration, and therefore a non-zero proper acceleration, is due to the curvature of spacetime. In flat spacetime the Christoffel coefficients $\Gamma^\alpha{}_{\mu\nu}$ are all zero and therefore the four-acceleration must be zero. That's why the apple would be weightless far from any masses where spacetime was flat.

And our equation (1) for the four-acceleration neatly explains geodesic motion and what it means to be weightless. If you are weightless your proper acceleration must be zero and therefore your four-acceleration must be zero. If we take equation (1) and set $A^\alpha=0$ then we get:

$$ \frac{\mathrm d^2x^\alpha}{\mathrm d\tau^2} = -\Gamma^\alpha{}_{\mu\nu}U^\mu U^\nu \tag{2} $$

and this is the geodesic equation i.e. the trajectory $\mathbf{x}(\tau)$ that solves this equation is the one that describes free motion. When the stem of the apple breaks and it starts falling its motion is described by the geodesic equation.

If you're interested I go into more detail on this in my answer to How does "curved space" explain gravitational attraction? There is also a related discussion in If gravity isn't a force, then how are forces balanced in the real world?

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  • $\begingroup$ I think that answers my question! $A^\alpha$ is 0 traveling along a geodesic. So the proper acceleration is 0. So that's the since in which it's not accelerating when its moving. It experiences a proper acceleration when it is held up by the tree, but isn't moving, at least not relative to the branch. Further, this analysis can take place with a minimum of external reference. I didn't know how to ask the question, but I think this is exactly what i was looking for. Thanks! $\endgroup$ – R. Romero Nov 14 '18 at 15:08
  • $\begingroup$ "The apple would be weightless far from any masses where spacetime was flat." I guess this needs some qualification in view of the constantly accelerating frame in flat spacetime: apples have weight in such a frame. $\endgroup$ – Andrew Steane Nov 14 '18 at 21:41
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We cannot talk of an object to be "not accelerating". We can only talk of an object to be "not accelerating with respect to a specific reference system in which we measure position and time". One often avoids such long sentences to make life easier. However such attempts to make life easier may lead to misunderstandings.

Example: You are sitting in a sports car and push the gas pedal. The sports car accelerates with respect to the reference system of the road. Also you accelerate with respect to the reference system of the road. However you do not accelerate with respect to the driver's seat. That is why you stay in the drivers seat (you are not accelerated wrt the driver's seat).

In order to describe the apple example correctly, we thus have to choose a reference system first. The description we will make will depend on the reference system and on the physical theory we use.

Let's use the point of view taken by general relativity, since it is a good theory for understanding gravitation. Here, gravitation is not considered a force but part of the surrounding geometry.

Now let us choose the reference system of an apple in free fall. This falling apple feels no force and it is not accelerating WRT to the chosen reference system. The falling apple will look up to its brothers and sisters which are still attached to the stem. The apple will say: "My brothers and sisters are pushed upward by the stem and it is due to this force that they are accelerating upward wrt to myself". In GR a free falling reference system is an inertial system. So the observation also is consistent with the statement that "force leads to acceleration" only applies in inertial systems.

Now let us choose the reference system of the apples still attached to the tree. These apples feel an upward pulling force from the stem. They are astonished that they are not accelerating wrt to the chosen reference system. However they understand that the law "force leads to acceleration" only applies in inertial systems. So they see no inconsistency, since they are not in an inertial system.

Let's now use the point of view taken by special relativity. Here, the notion of an inertial system is different and that of gravitation as well. Gravitation is now a force and a freely falling reference system no longer is an inertial system. The explanation still is consistent inside of the theory.

First let us choose the reference system of the apple in free fall. The apple is acted on by the gravitational force. However the apple is not accelerating. This is fine. In SR a freely falling system is not an inertial system and so we do not expect "no acceleration = no force". The apple is feeling no force. This is fine as well, since the gravitational force is compensated by the inertial force which is due to observations being made in an non-inertial system.

Now let us choose the reference system of the apples still attached to the tree. This is no inertial system either (rotation of earth around its axis, rotation of earth around the sun). However, for our purposes we can neglect these effects, since they are small. We have an inertial system here, more or less. The apples sitting in the tree will argue: We are acted on by gravity and this is compensated by the forces of the stem. So we are not acted upon by a net force. We feel the tension between the gravity and the stem in our necks, however, but net force is zero. Since we are in an inertial system, the concept applies that no net force means no acceleration wrt this system. This is fine. The apple falling down is pulled by gravity and is accelerating wrt to this inertial system. This is fine as well.

The confusing thing is: It depends 1) on the chosen reference system and 2) on the chosen theory, how we have to reason. The arguments always look different and the concepts as well. For example, in general relativity, gravity is not considered a force. Still, in the chosen framework we can argue consistently.

If we are familiar with the concept of a geodesic in space-time, we can even extend the reasoning to this setting. The freely falling apple is moving on a geodesic and feels no force. The apples on the stem are pulled by the stem from the geodesic and feel a force.

Edit: Since you edited the question - let me edit the answer as well. :-)

An accelerometer attached to the apple on the tree shows a readout of 1g. An accelerometer attached to the falling apple shows a readout of 0g (neglecting air flow, assuming vacuum, neglecting field inhomogeneity). Fun experiment: Go for a skydive (and you will experience this yourself).

Staying in GR and using geodesic: We can construct a generic reference system when speaking geodesics. This is locally inertial in the GR sense. The free falling apple is unaccelerated wrt to this geodesic reference system (and stays on the geodesic), it feels no force. The stem apple is accelerated wrt to this geodesic reference system (and therefore deviates from the geodesic in its movement), it feels the pull of the stem (and in this model no gravity, since this is assumed to be in the geometry).

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  • $\begingroup$ Awesome and detailed answer! I'm going to take my time wrapping my head around it. I think that final sentence gets at the heart of my confusion. I think i have it in my head that Geodesic motion = unaccelerated motion. But unaccelerated with respected to what? I see from these responses that that doesnt' make sense. Then if I See moving along geodesics as unaccelerated motion, then the apple stuck in the tree is not moving along a geodesic and is there for accelerating. Again probably an incorrect statement. Hence my part about the apple accelerating when its not moving. $\endgroup$ – R. Romero Nov 9 '18 at 22:31
  • $\begingroup$ The best answer yet, all the other answers so far are wrong since the question bears the general relativity tag. The only thing that's not correct is the claim that special relativity makes assumptions about gravity, which it clearly doesn't. Nevertheless, +1. $\endgroup$ – Yukterez Nov 9 '18 at 22:36
  • $\begingroup$ @СимонТыран Before you put down other answers, you should realize the GR tag was recently added. There was no specification earlier $\endgroup$ – Aaron Stevens Nov 9 '18 at 22:39
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    $\begingroup$ Well that's a lesson learned to check edits then. $\endgroup$ – Gradient137 Nov 9 '18 at 22:46
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    $\begingroup$ No you don't have them in SR nor under Newton, that's why you need GR for gravity, at least in the regime where this effects become noteable. That's why I said "if you treat gravity as a force, you don't get them". $\endgroup$ – Yukterez Nov 9 '18 at 22:57

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