Macroscopic objects in free-fall

[cianfa72]

TL;DR Summary

Analysis of macroscopic objects in free-fall in gravitational field

Hi,
very basic question. Take an object like a rock or the Earth itself. If we consider their internal constituents, there will be electromagnetic forces acting between them (Newton's 3th law pairs).

From a global perspective if the rock is free from external non-gravitational forces, then it will be in free-fall in gravitational field (i.e. spacetime geometry).

What about the aforementioned internal forces ? If we look at the center of mass (as object’s representative), then the electromagnetic internal forces driving the object’s costituents away from their geodesic paths will not enter into account, I believe.

 

[phinds]

You clearly, and correctly, state that absent any external forces, a macro object will be in a geodesic freefall path in whatever gravitational field it exists in. THEN, you ask if there is some way internal forces could throw then entire object off that path. Do you see the contradiction?

 

[cianfa72]

Do you see the contradiction?

Yes, I see. I was trying to better understand the reason behind it.

Thus, each object’s costituent path through spacetime is not geodesic due to electromagnetic forces acting on it from other constituents (i.e. they have non-zero path curvature). Yet, the motion of object's center of mass is still geodesic (i.e. in free-fall).

 

[phinds]

I have often found that a way to clarify things is to think of extremes. In this case, think of a fairly thin-walled sphere with some thickness and mass, with a heavy rubber highly elastic ball inside. You shake the sphere vigorously, getting the ball slamming back and forth between opposite sides.

Now you let it go in free fall. The ball keeps bouncing and you think, well, heck, it's obviously going to move the sphere off to one side when it hits that side, so the COM of the sphere clearly moves. Yes, it does, BUT ... the ball is also going move (rebound) thus keep the COM of the entire macro object unmoved (well, still following the geodesic).

 

[PeterDonis]

each object’s costituent path through spacetime is not geodesic due to electromagnetic forces acting on it from other constituents (i.e. they have non-zero path curvature). Yet, the motion of object's center of mass is still geodesic (i.e. in free-fall).

Yes, that's correct.

 

[Hill]

I have often found that a way to clarify things is to think of extremes. In this case, think of a fairly thin-walled sphere with some thickness and mass, with a heavy rubber highly elastic ball inside. You shake the sphere vigorously, getting the ball slamming back and forth between opposite sides.

Now you let it go in free fall. The ball keeps bouncing and you think, well, heck, it's obviously going to move the sphere off to one side when it hits that side, so the COM of the sphere clearly moves. Yes, it does, BUT ... the ball is also going move (rebound) thus keep the COM of the entire macro object unmoved (well, still following the geodesic).

How does the system behave if instead of a massive ball it is a massless light bouncing inside the sphere?

 

[phinds]

How does the system behave if instead of a massive ball it is a massless light bouncing inside the sphere?

Why would it be any different?.

 

[Hill]

Why would it be any different?.

Where is a center of mass of the system sphere+light?

 

[PeterDonis]

Where is a center of mass of the system sphere+light?

Assuming the system is spherically symmetric (i.e., the light inside the sphere is distributed in a spherically symmetric way), the center of mass would be at the geometric center of the sphere.

You seem to be under a misconception that a system containing parts with zero rest mass, like light, cannot have a center of mass. That is not the case.

A better general term would be "center of energy", or even, if we want to take relativistic effects into account, "center of energy-momentum".

 

[Hill]

You seem to be under a misconception

I am not and I do not.

A better general term would be "center of energy", or even, if we want to take relativistic effects into account, "center of energy-momentum".

This is what I aim to.

if we want to take relativistic effects into account

This thread is in the Relativity forum.

 

[PeterDonis]

I am not and I do not.

Then I am puzzled as to why you highlighted the words "center of mass" in your post #8. Or indeed why you felt it necessary to ask the question you asked in that post at all, since, as @phinds implied in post #7, the answer is the same for any spherically symmetric distribution, no matter what it is composed of.

 

[jbriggs444]

If tidal gravity is significant, it is not clear that the center of mass will coincide with the center of gravity.

 

[PeterDonis]

If tidal gravity is significant, it is not clear that the center of mass will coincide with the center of gravity.

How are you defining "center of gravity" in this case?

 

[PeterDonis]

If tidal gravity is significant

In this case, the composition of the body, given that the distribution of energy and momentum is the same, still will not affect the location of the center of mass.

 

[cianfa72]

In this case, the composition of the body, given that the distribution of energy and momentum is the same, still will not affect the location of the center of mass.

You mean the distribution of energy and momentum is the same since spherically symmetric distribution for the "sphere + light" system is assumed.

 

[PeterDonis]

You mean the distribution of energy and momentum is the same

Yes.

since spherically symmetric distribution for the "sphere + light" system is assumed.

It is in the original case, but if tidal effects are present, the distribution won't be spherical.

 

[cianfa72]

It is in the original case, but if tidal effects are present, the distribution won't be spherical.

Yet, the Center of Mass (COM) of the system "sphere + light" will follow a geodesic path through spacetime (i.e. a path with zero curvature).

 

[hutchphd]

It is in the original case, but if tidal effects are present, the distribution won't be spherical.

I am confused by your objection. When you assume the cow to be spherical, it is a "done deal". We already assumed the cow to be spherical, knowing that this might not actually be troooo.

 

[PeterDonis]

the Center of Mass (COM) of the system "sphere + light" will follow a geodesic path through spacetime (i.e. a path with zero curvature).

Yes.

 

[PeterDonis]

When you assume the cow to be spherical, it is a "done deal".

No, when you assume the cow to be spherical, you are also (though you might not realize it) assuming that there are no non-negligible tidal effects. So if you want to consider the case where there are non-negligible tidal effects, you can't assume the cow to be spherical.

 

[hutchphd]

Perhhaps I don't understand the term "tidal effects" in this context?

 

[cianfa72]

When you assume the cow to be spherical, it is a "done deal". We already assumed the cow to be spherical

Sorry, what is the term cow ?

 

[Ibix]

Perhhaps I don't understand the term "tidal effects" in this context?

If spacetime curvature is small then the spacetime around our internally mirrored ball is flat (or near enough) and the mass distribution will be spherically symmetrical. If the curvature is larger spacetime is no longer near enough flat, and we don't expect the geometry of spacetime inside the ball to be isotropic. That's tidal gravity. The ball will be non-uniformly stressed and (even if its shape is negligibly affected) the light flight time across it will depend on the angular coordinate, so the mass distribution shifts (in general, anyway - you can probably find special cases).

 

[Ibix]

Sorry, what is the term cow ?

Google "spherical cow". It's an old joke about physicists' tendancies to build simple models where the maths is tractable but the model is somewhat divorced from reality.

 

[cianfa72]

Google "spherical cow". It's an old joke about physicists' tendencies to build simple models where the maths is tractable but the model is somewhat divorced from reality.

 

[hutchphd]

The ball will be non-uniformly stressed

This ain't no spherical cow where I come from.... Thanks for the explain.

 

[PeterDonis]

Perhhaps I don't understand the term "tidal effects" in this context?

@Ibix's explanation is correct. The only technical point I would make is that the tidal distortions of objects that he described are due to Weyl curvature.

 

[cianfa72]

The only technical point I would make is that the tidal distortions of objects that he described are due to Weyl curvature.

Such a Weyl curvature/tensor describes how the shape of a body is distorted due to spacetime curvature (i.e. geodesic deviation).

My point is: Can electromagnetic forces between body's constituents cope with those "tidal forces" inside the body ?

 

[Ibix]

My point is: Can electromagnetic forces between body's constituents cope with those "tidal forces" ?

Are you being torn apart by the non-zero tidal force on your body now? If not, then yes they can cope in at least some circumstances. However, extreme tidal forces can overwhelm them - a close orbit over a neutron star might be fatal to a human and you would certainly be killed this way short of the singularity in a black hole. If you search the forum for "ouch radius" (Wheeler's term) you will find some order-of-magnitude discussion.

 

[PeterDonis]

Such a Weyl curvature/tensor describes how the shape of a body is distorted due to spacetime curvature (i.e. geodesic deviation).

More precisely, it describes how the shape of the body is distorted in the absence of any internal forces, i.e., if the body's atoms follow the geodesic paths that the Weyl curvature dictates. Weyl curvature is geodesic deviation.

My point is: Can electromagnetic forces between body's constituents cope with those "tidal forces" inside the body ?

Internal forces inside the body will prevent the body's atoms from following the geodesic paths that the Weyl curvature dictates. Instead they will follow non-geodesic paths. But if the tidal effects are non-negligible, those non-geodesic paths will still result in a shape of the body that is different from what its shape would have been in flat spacetime in the absence of any geodesic deviation (i.e., spacetime curvature).

 

[cianfa72]

But if the tidal effects are non-negligible, those non-geodesic paths will still result in a shape of the body that is different from what its shape would have been in flat spacetime in the absence of any geodesic deviation (i.e., spacetime curvature).

However, as you highlighted before, even in this case the body's COM continues to follow a geodesic path in free-fall.

 

[PeterDonis]

as you highlighted before, even in this case the body's COM continues to follow a geodesic path in free-fall.

Yes.