Is momentum conserved as a body falls through a gravitational field?

[Herbascious J]

TL;DR Summary

How is momentum treated in General Relativity as an object falls through a gravitational field?

If one stands on a large planetary body, like the moon, and throws a large object, like a rock straight up, the object will leave with some velocity, slow down to a stop, and then come back down with the same velocity once it returns to its origin. In Newtonian mechanics, the understanding is that the object's momentum is slowly converted into potential energy as it comes to a stand still at the top of it's trajectory, and then as the object's velocity reverses, it gains speed and momentum until it ends up back at its origin, with the same speed and velocity with which it was projected only in the opposite direction.

In General Relativity, I understand that objects following geodesics through a gravitational field, like a comet for example, are not being accelerated in the way we think of in Newtonian mechanics, but instead the objects follow a geodesic of space-time. My question is, when these objects move through a geodesic like this, are they maintaining their momentum just like an object floating in free space? The acceleration is like an illusion?

So in the case of a rock being thrown, does it maintain a constant momentum, but only that it is following a geodesic so that it only appears to have its momentum convert to potential energy and then reverse? The idea behind this question is to ask if the rock has the same momentum at the top of its arc/trajectory when it comes to a stand still as it did when it was initially thrown and at a high velocity? This is counter intuitive because it seems like it comes to rest momentarily, only higher in the gravitational field, and so it shouldn't really have any momentum at that moment. Does GR indicate that the objects momentum is unchanging during this exchange, and that the dynamics are simply due to the curvature of space time as the objects preserves its momentum in free fall so to speak?

 

[PeterDonis]

In Newtonian mechanics, the understanding is that the object's momentum is slowly converted into potential energy as it comes to a stand still at the top of it's trajectory

Not quite. The object's kinetic energy is converted into gravitational potential energy as it rises. Momentum is not energy and doesn't get inter-converted to energy.

The Newtonian view of momentum for this case is that the momentum of the object by itself is not conserved; only the momentum of the whole system, object plus Earth, is conserved. As the object rises, the Earth moves by a very tiny amount in the opposite direction, so that momentum as a whole is conserved.

For this particular case, the GR analysis is the same as the Newtonian analysis.

In General Relativity, I understand that objects following geodesics through a gravitational field, like a comet for example, are not being accelerated

They don't have any proper acceleration; an accelerometer attached to the object would read zero. But this is also true in Newtonian mechanics; it's just that Newtonian mechanics does not attach the same meaning to it as GR does.

The acceleration that Newtonian mechanics cares about is coordinate acceleration; in this case, coordinate acceleration in the frame in which the Earth is at rest. Or more precisely, the frame in which the center of mass of the system of Earth plus object is at rest. This acceleration is present in GR as well; it's just that GR does not attach the same meaning to it as Newtonian mechanics does.

when these objects move through a geodesic like this, are they maintaining their momentum just like an object floating in free space?

Momentum relative to what? Momentum is frame dependent. In the object's rest frame, of course it is maintaining its momentum: its momentum is zero in that frame, always.

The real question is whether there is any invariant, any quantity that is not frame-dependent, which is conserved in this scenario. There is no such quantity that has an interpretation as momentum. There is a conserved energy, called "energy at infinity", which can be interpreted as kinetic energy + potential energy in the Earth's rest frame; and there is a conserved angular momentum. But there is no conserved linear momentum.

 

[Dale]

the object's momentum is slowly converted into potential energy

Momentum and energy are each separately conserved. They cannot be converted into each other. You cannot even convert different components of momentum into each other.

Potential energy can be transferred to kinetic energy. The z component of momentum in a projectile can be transferred to the z component of momentum in the earth. But energy cannot be converted to momentum.

The conservation of momentum ultimately comes from the symmetry of the laws of physics under spatial translation. In GR momentum is conserved in any spacetime with that symmetry.

 

[Herbascious J]

Not quite. The object's kinetic energy is converted into gravitational potential energy as it rises. Momentum is not energy and doesn't get inter-converted to energy.

The Newtonian view of momentum for this case is that the momentum of the object by itself is not conserved; only the momentum of the whole system, object plus Earth, is conserved. As the object rises, the Earth moves by a very tiny amount in the opposite direction, so that momentum as a whole is conserved.

For this particular case, the GR analysis is the same as the Newtonian analysis.


They don't have any proper acceleration; an accelerometer attached to the object would read zero. But this is also true in Newtonian mechanics; it's just that Newtonian mechanics does not attach the same meaning to it as GR does.

The acceleration that Newtonian mechanics cares about is coordinate acceleration; in this case, coordinate acceleration in the frame in which the Earth is at rest. Or more precisely, the frame in which the center of mass of the system of Earth plus object is at rest. This acceleration is present in GR as well; it's just that GR does not attach the same meaning to it as Newtonian mechanics does.


Momentum relative to what? Momentum is frame dependent. In the object's rest frame, of course it is maintaining its momentum: its momentum is zero in that frame, always.

The real question is whether there is any invariant, any quantity that is not frame-dependent, which is conserved in this scenario. There is no such quantity that has an interpretation as momentum. There is a conserved energy, called "energy at infinity", which can be interpreted as kinetic energy + potential energy in the Earth's rest frame; and there is a conserved angular momentum. But there is no conserved linear momentum.

This is very helpful thank you. So, If I am on the earth and I watch a comet move around the earth in an arc, then does GR say that the momentum of the object, relative to me and the earth, changed and that it was accelerated? I'm imagining the object is accelerated along a geodesic, but I'm not confident in that interpretation.

 

[PeroK]

This is very helpful thank you. So, If I am on the earth and I watch a comet move around the earth in an arc, then does GR say that the momentum of the object, relative to me and the earth, changed and that it was accelerated? I'm imagining the object is accelerated along a geodesic, but I'm not confident in that interpretation.

It seems you are trying to understand GR in the context and terminology of Newtonian physics. This is unlikely to lead to an understanding of GR.

 

[Ibix]

does GR say that the momentum of the object, relative to me and the earth, changed and that it was accelerated?

In general, it says that "momentum relative to me" is meaningless unless you are co-located with the comet.

In an asymptotically flat static spacetime (as you have here if you treat the comet gravity as negligible) then you can compare the comet's momenta at infinity before and after the interaction and they are different. However, this is treating the Earth as an immovable object and we violate momentum conservation in the equivalent Newtonian model, so you can't read much into it.

I'm imagining the object is accelerated along a geodesic, but I'm not confident in that interpretation.

"Accelerated along a geodesic" is pretty much a direct self contradiction. Geodesics are the paths you follow when you don't accelerate.

 

[PeterDonis]

If I am on the earth and I watch a comet move around the earth in an arc, then does GR say that the momentum of the object, relative to me and the earth, changed

To the extent you can even define such a thing (which, as @Ibix has pointed out, is problematic since you and the comet are not co-located), yes.

and that it was accelerated?

No. First, as @Ibix points out, an object in geodesic motion is not accelerated in GR. Second, acceleration in GR has nothing to do with change of momentum, since proper acceleration is not frame-dependent while change of momentum is.

 

[PeterDonis]

momenta at infinity

There is no such thing as "momentum at infinity". The only "at infinity" quantities which have meaningful definitions in the spacetime of our solar system are energy and angular momentum. (Even these are approximate because the spacetime of our solar system is not exactly stationary--which is what gives a meaningful definition of energy at infinity--or axisymmetric--which is what gives a meaningful definition of angular momentum.)

 

[Ibix]

There is no such thing as "momentum at infinity".

You can write down the four momentum $p^a(\tau)$ of a body in an open orbit and consider the limits as $\tau\rightarrow\pm\infty$. You can then transport the momentum vector from the outbound leg to the inbound leg, staying in the nearly flat region, and the result must tend to path independent due to the asymptotic flatness.

 

[PeterDonis]

You can write down the four momentum $p^a(\tau)$ of a body in an open orbit and consider the limits as $\tau\rightarrow\pm\infty$. You can then transport the momentum vector from the outbound leg to the inbound leg, staying in the nearly flat region, and the result must tend to path independent due to the asymptotic flatness.

Ah, I see, you are doing the same sort of analysis as for bending of light by the sun--how much is the body's orbit "bent" by the gravity of the central mass.

 

[cianfa72]

Potential energy can be transferred to kinetic energy.

I believe that the potential energy is actually a property of the "rock + Earth" physical system. To me doesn't make much sense to "attach" potential energy to the system "rock" alone.

 

[PeterDonis]

I believe that the potential energy is actually a property of the "rock + Earth" physical system.

It is, but since the Earth is many, many orders of magnitude more massive than the rock, it is a very good approximation to treat the Earth as stationary, in which case we can assign a potential energy to the rock as a test object moving in the spacetime geometry around the Earth.

 

[Dale]

I believe that the potential energy is actually a property of the "rock + Earth" physical system. To me doesn't make much sense to "attach" potential energy to the system "rock" alone.

I agree. That doesn’t change anything I said.

 

[cianfa72]

It is, but since the Earth is many, many orders of magnitude more massive than the rock, it is a very good approximation to treat the Earth as stationary, in which case we can assign a potential energy to the rock as a test object moving in the spacetime geometry around the Earth.

So the model is: "system rock" in an external gravitational field (spacetime geometry around the Earth). This way the field acts on the "system rock" but the rock doesn't act on the field that is supposed to be assigned.

 

[PeterDonis]

So the model is: "system rock" in an external gravitational field (spacetime geometry around the Earth). This way the field acts on the "system rock" but the rock doesn't act on the field that is supposed to be assigned.

In the approximation I described, yes. But in that approximation, momentum is not conserved. If you want a model that conserves momentum, you have to include the (tiny) motion of the Earth in your model, and then you can no longer assign a potential energy to the rock alone.

 

[cianfa72]

If you want a model that conserves momentum, you have to include the (tiny) motion of the Earth in your model, and then you can no longer assign a potential energy to the rock alone.

But it might be possibile to assign a momentum to the field in order to "balance" it manteining the conservationofmomentum ?

 

[PeterDonis]

But should be possibile to assign a momentum to the field in order to "balance" it ?

In the model where you include the motion of the Earth, there is no need to try to assign any momentum to the "gravitational field"; the sum of the momentum of the Earth and the momentum of the rock is conserved.

If you were actually asking about the approximation where the Earth is treated as stationary, no, you can't assign momentum to the "gravitational field"; you just have to accept that the rock's momentum is not conserved, and use other conserved quantities, such as kinetic plus potential energy, to analyze the scenario.

 

[Ibix]

If you were actually asking about the approximation where the Earth is treated as stationary, no, you can't assign momentum to the "gravitational field"; you just have to accept that the rock's momentum is not conserved, and use other conserved quantities, such as kinetic plus potential energy, to analyze the scenario.

Just to add to this, note that this is exactly the same as the Newtonian case where we simply fix the Earth at $r=0$ and then stick out fingers in our ears and whistle real loud. We can't conserve momentum here because momentum leaks out of whatever we nailed the Earth to, but KE-vs-PE works fine.

This model is more defensible in the Newtonian case because the two body problem is the same as the one body problem with the gravitational field of a reduced mass fixed at the origin. In GR there's quite a lot of qualitative difference between a one and two body problem (albeit often negligible), not just a small error term in the potential strength.

 

[PeterDonis]

whatever we nailed the Earth to

 

[cianfa72]

In the model where you include the motion of the Earth, there is no need to try to assign any momentum to the "gravitational field"; the sum of the momentum of the Earth and the momentum of the rock is conserved.

Sorry, in this model "Earth + rock" system is the gravitational field part of the system or is it just the "mediator" of the interactions between system's constituents (i.e. Earth and rock) ?

 

[cianfa72]

We can't conserve momentum here because momentum leaks out of whatever we nailed the Earth to, but KE-vs-PE works fine.

Sorry, not sure to get your point.

 

[PeterDonis]

Sorry, in this model "Earth + rock" system is the gravitational field part of the system or is it just the "mediator" of the interactions between system's constituents (i.e. Earth and rock) ?

What's the difference?

 

[PeterDonis]

Sorry, not sure to get your point.

He's just saying that the GR and Newtonian models in the approximation described both have the property I described in the post he quoted, of not conserving momentum.

 

[cianfa72]

What's the difference?

I was asking that from a logical point of view. The difference is between "Earth + rock" system vs "Earth + rock + field" system.

 

[PeterDonis]

I was asking that from a logical point of view.

We're not doing logic here, we're doing physics.

The difference is between "Earth + rock" system vs "Earth + rock + field" system.

Again, what's the difference? The physical difference?

 

[pervect]

This is very helpful thank you. So, If I am on the earth and I watch a comet move around the earth in an arc, then does GR say that the momentum of the object, relative to me and the earth, changed and that it was accelerated? I'm imagining the object is accelerated along a geodesic, but I'm not confident in that interpretation.

Momentum in GR is complicated, as is energy in GR. There are several concepts of energy, and momentum, in GR, not just one, that can be applied in different situations. If this seems unsatisfying to you, I would agree, though that's probably a matter of personal opinion.

There is a formulation of energy-momentum in GR that has a "yes" answer to the question of energy-momentum conservation in the situation you describe, which is the ADM concept. However, you would need an additional assumption, called "asymptotic flatness", for this formulation to be valid.

This is an advanced topic in GR, if (as seems likely) you are just starting out, you almost surely don't have the required background. It is, however discussed in Wald's text, "General Relativity". I am not intimately familiar with this topic personally, by the way, mostly I've glossed over this section and nodded my head before moving onto something simpler.

I was going to discuss some of the background you might need to attempt reading Wald, but I thought better of it.

On a less technical level, I will point out that if one has two heavy bodies, which deflect each other, you will generate gravitational waves, which carry energy and momentum. This is usually neglected, because the amount of energy and momentum carried away in this manner is usually negligible, and because it's _very_ complicated to perform such an analysis. Usually, this sort of analysis is only done in extreme situations like binary inspirals where it is actually required. I'm mildly familiar with some approximations that describe how much energy is carried away by gravitational radiation within an order of magnitude or so, but I'm not familiar with any approximations that would apply to the momentum. One might use the former to put a limit on the later.

I'm sorry that the answer to such a basic question turns out to be so complicated - but it is what it is. If you're curious about GR, I would not really want to scare you away from it, but I would suggest considering a simpler question first.

I'll give some general un-asked for advice for learning about GR. To parse the idea of gravity as curved space-time, one needs to understand curvature, and one needs to understand space-time, before one can paste the two concepts together and understand curved space-time. Curvature arises naturally in the concept of navigating on the globe. This is actually probably a slightly harder topic than learning about curvature in GR, if if you take it really seriously. It is still interesting because it is a less abstract situation in which we see the effects of curvature. I'll give an honorable mention to "sector models" as a way to dip one's toes into curvature without a high-level of commitment to the math of differential geometry. A search on PF should give some hits for "sector models".

Understanding space-time is also needed. Being able to visualize space-time as a space-time diagram would probably be the most basic start. For textbooks, I'd suggest Taylor and Wheeler's "Space-time physics" as an approach to introductory special relativity focused on the geometrical ideas of space-time. "The parable of the surveyor" is particularly interesting, I think. It poses the question - why do we consider a plane to be a two-dimensional space, and not two one-dimensonal spaces? The answer to this question goes a way to explaining why we regard space-time as a unified entitiy, in special relativity, not space as something sepearate from time.

While these to topics might be enough to understand curved-space time, it's not quite enough to understand Einstein's field equations, which can be thought of as the end goal. I'll defer giving any recommendations on this at the time being unless asked as this is already getting too long and I'm starting to ramble.

 

[cianfa72]

Again, what's the difference? The physical difference?

No, there is not any observable physical difference.

The difference is, I believe, that if the field (i.e. gravitational field/spacetime geometry in case at hand) can be conceived as a system's constituent then we may assign it physical properties (such as energy, momentum etc..) otherwise may not.

 

[cianfa72]

We can't conserve momentum here because momentum leaks out of whatever we nailed the Earth to, but KE-vs-PE works fine.

Therefore, let me say, whatever we nailed the Earth to (or the Earth itself if we assume it has infinite mass) acts a "source" of momentum: it can leak out momentum without changing its own momentum.

 

[Ibix]

Therefore, let me say, whatever we nailed the Earth to (or the Earth itself if we assume it has infinite mass) acts a "source" of momentum: it can leak out momentum without changing its own momentum.

In terms of the system under consideration, yes. Or you can (more realistically) say it's subject to a force external to the system we are considering. For example we could hold the Earth in place with a sufficiently powerful rocket.

 

[cianfa72]

Or you can (more realistically) say it's subject to a force external to the system we are considering. For example we could hold the Earth in place with a sufficiently powerful rocket.

Ok, this way the system "Earth + rock" is no longer isolated hence it doesn't conserve momentum.

 

[PeterDonis]

there is not any observable physical difference.

Then you should reconsider your apparent belief that there is any difference. In other words, you aren't describing two different ways the world could be. You are just describing the same physical configuration using two different strings of words.

The difference is, I believe, that if the field (i.e. gravitational field/spacetime geometry in case at hand) can be conceived as a system's constituent then we may assign it physical properties (such as energy, momentum etc..) otherwise may not.

Why would you think that? How can we magically give anything physical properties by drawing a "system" boundary around it and saying it's "part of the system" instead of not.

In other words, since you agree that "part of the system" vs. "not part of the system" makes no physical difference, that is telling you that "part of the system" is a human convention, not physics. It's no different from assigning a coordinate chart. You can't change physics by changing coordinate charts. Similarly, you can't change physics by changing how you draw the boundaries of "the system".

 

[PeterDonis]

Or you can (more realistically) say it's subject to a force external to the system we are considering. For example we could hold the Earth in place with a sufficiently powerful rocket.

The "more realistically" juxtaposed with the Earth example is very entertaining.

 

[cianfa72]

How can we magically give anything physical properties by drawing a "system" boundary around it and saying it's "part of the system" instead of not.

Of course that's true. I was reasoning along the lines of Feynman in lecture 10-5.

He claims that in order to check out the law of conservation of momentum for electric charges, it makes sense to assign a momentum to the electromagnetic field. Hence the system he actually considers is "charges + field" assigning a physical property (momentum) to the field itself.

 

[PeterDonis]

He claims that in order to check out the law of conservation of momentum for electric charges, it makes sense to assign a momentum to the electromagnetic field.

Yes. But what he describes does not work for gravity.

 

[cianfa72]

Yes. But what he describes does not work for gravity.

Why not ? Could you be more specific ?