Rod falling radially towards the center of a mass

[timmdeeg]

This example might be not very interesting, but perhaps helps to improve my understanding.

Is a radially falling rod in free fall?
Which frequency shift of light emitted at the ends of the rod would observers at rest at the respective other ends measure?

Here are some assumptions. Please correct accordingly.

What can one tell about the velocity of the ends of the rod, measured locally relative to freely falling test particles? Presumably the upper end falls faster and the lower end slower than the resp. particle. If true, there should be a point on the rod which is in rest with the particle at a moment.
In my opinion this would mean that no point on the rod is in free fall, means that no point follows a geodesic.

What can one say about the position of this very point? The gravitational acceleration is larger at the lower end of the rod. Can one conclude from this that this point is in the lower half and moves towards the lower end during the fall? Given the radial coordinates of the ends of the rod how would one calculate (in principle) the position of this point?

Two observers in radial free fall see themselves redshifted. So, I expect that the ends of the rod are less redshifted, maybe even blueshifted. I found this calculation

https://www.physicsforums.com/threads/black-hole.104577/#post-861282 #5

very interesting, but suspect that to calculate of the frequency shift along the rod is much more difficult.

Any help is very appreciated.

 

[pervect]

This example might be not very interesting, but perhaps helps to improve my understanding.

Is a radially falling rod in free fall?
Which frequency shift of light emitted at the ends of the rod would observers at rest at the respective other ends measure?

Here are some assumptions. Please correct accordingly.

What can one tell about the velocity of the ends of the rod, measured locally relative to freely falling test particles? Presumably the upper end falls faster and the lower end slower than the resp. particle. If true, there should be a point on the rod which is in rest with the particle at a moment.
In my opinion this would mean that no point on the rod is in free fall, means that no point follows a geodesic.

I'm not sure of your logic here. To summarize, though, I'll say that while I believe it's correct to note that the ends of the rod are not following geodesics, I disgaree with the conclusion that no point on the rod follows a geodesic.

Basically, if we consider the Newtonian case of a pair of freely falling radial test particles, we note that there is a tidal force that causes the pair of particles to accelerate away from each other, to separate.

GR tells us the same thing, in a rather different and probably unfamiliar language, involving geodesics, the separation vector between geodesics, and the curvature of space-time.

What does this tell us about the rigid rod case? Well, in the rigid rode case, internal forces of the rod try to keep the ends of the rod the same distance apart. The internal forces of the rod oppose the natural tendency of the particles at the end of the rod to separate as they would if they were both in free fall. So we conclude that the falling rod is under tension.

This tension causes the ends of the rod to depart from geodesic motion. At the center of the rod, though, the tension doesn't have any net effect on the trajectory. The effect of the tension is to "pull" both ends of the rod away from geodesic motion, while the center falls a geodesic path.

For a more technical exposition, read about the "Geodesic Deviation Equation". Unfortunately , the thread is labelled B - beginner level, and most of the detailed discussions of the geodesic deviation equation would be A (advanced - graduate level), and even the simpler ones s would be I (intermediate) level.

 

[timmdeeg]

This tension causes the ends of the rod to depart from geodesic motion. At the center of the rod, though, the tension doesn't have any net effect on the trajectory. The effect of the tension is to "pull" both ends of the rod away from geodesic motion, while the center falls a geodesic path.

For a more technical exposition, read about the "Geodesic Deviation Equation". Unfortunately , the thread is labelled B - beginner level, and most of the detailed discussions of the geodesic deviation equation would be A (advanced - graduate level), and even the simpler ones s would be I (intermediate) level.

Is it the center of the rod which experiences no forces? I mean this situation is not symmetric as it would be in a FRW universe. Because in Schwarzschild spacetime the local tidal forces are increasing with decreasing r-coordinate I expected an asymmetric location of the force-free point on the rod. And moreover that this point is moving relative to the rod for the same reason. Is this reasoning wrong?

In case it makes sense, can someone change the labeling of the thread?

 

[PAllen]

There is an interesting theorem on this that I have only seen in Synge's 1960 general relativity book. Unlike many theorems focusing on finding limit of a body's motion as it approaches a pointlike body, this talks about a feature of an arbitrary body. The body is not assumed to be small and no limitation on gravitational radiation is needed. The theorem is as follows:

Given a world tube of arbitrary stress energy, outside of which there is vacuum (thus no EM field, for example), and for which the weak energy condition holds (Synge doesn't use that term, but his criterion is identical - maybe the term wasn't in wide use yet in 1960), and given that the Einstein field equations are satisfied everywhere, there exists a curve within the world tube which is a locus of no acceleration. [edit: there is also a requirement that body not have 'extreme rotation' according to a specific criterion.]

"Locus of no acceleration" is defined by covariant derivative of 4-velocity [edit: of an element of the body] vanishing along the curve. One might say this is just a geodesic [edit: but it isn't because it is not the 4-velocity of curve whose covariant derivative vanishes]. A distinction from similar theorems is that there is no concept of 'geodesic of background metric' or about motion of the body as a whole. In particular, not only is it not expected that any particular material element of the body follow this path, but it can be established that this is impossible (in the general case), because this locus need not be timelike. What it does say, is given any foliation of the world tube, every slice contains a momentarily free falling element, and they are connected.

I would say this notion lends support to the intuition in the OP's last post. In particular, that a technical definition corresponding to what he calls a 'force free point', may move within the body as it falls.

 

[PeterDonis]

There is an interesting theorem on this that I have only seen in Synge's 1960 general relativity book.

Can you give a specific reference from the book?

 

[PAllen]

Can you give a specific reference from the book?

Sure. I had a vague memory about this, so I looked it up and re-read the section (twice, I'm rusty). It is Chapter IV, section 7, "Note on the motion of an isolated body". Starting on p. 194 of my printing.

 

[PeterDonis]

Chapter IV, section 7

Got it, thanks!

"Locus of no acceleration" is defined by covariant derivative of 4-velocity vanishing along the curve. One might say this is just a geodesic.

I don't think you can say that, because I don't see anything in Synge's proof that requires what he calls the "4-velocity" field to be tangent to the curve that is the locus of no acceleration. In fact, if the locus is not timelike, obviously the 4-velocity, which is timelike, can't be tangent to it.

In fact, I don't even see anything in Synge's proof that requires the locus to be a smooth curve; it looks to me like it could have arbitrary "kinks" in it where its tangent vector is discontinuous.

What it does say, is given any foliation fo the world tube, every slice contains a momentarily free falling element, and they are connected.

I don't think you can say this either, because at points where the locus is not timelike--i.e., the tangent vector to the locus curve is not timelike--its tangent vector obviously can't be the 4-velocity of any element of matter in the object. So it's entirely possible for there to be slices in the foliation where no element of the object is in free fall, because the tangent vector to the locus in that slice is not timelike.

 

[PAllen]

I don't think you can say that, because I don't see anything in Synge's proof that requires what he calls the "4-velocity" field to be tangent to the curve that is the locus of no acceleration. In fact, if the locus is not timelike, obviously the 4-velocity, which is timelike, can't be tangent to it.

In fact, I don't even see anything in Synge's proof that requires the locus to be a smooth curve; it looks to me like it could have arbitrary "kinks" in it where its tangent vector is discontinuous.

I agree with you here. The locus only need be connected.

I don't think you can say this either, because at points where the locus is not timelike--i.e., the tangent vector to the locus curve is not timelike--its tangent vector obviously can't be the 4-velocity of any element of matter in the object. So it's entirely possible for there to be slices in the foliation where no element of the object is in free fall, because the tangent vector to the locus in that slice is not timelike.

I don't agree with you here. The points in the locus are points where some an element of the body (changing from slice to slice, in general) has covariant derivative of that element's 4-velocity vanishing. They are a locus of 'no acceleration' elements of the body.

 

[pervect]

Is it the center of the rod which experiences no forces? I mean this situation is not symmetric as it would be in a FRW universe. Because in Schwarzschild spacetime the local tidal forces are increasing with decreasing r-coordinate I expected an asymmetric location of the force-free point on the rod. And moreover that this point is moving relative to the rod for the same reason. Is this reasoning wrong?

In case it makes sense, can someone change the labeling of the thread?

It's only the exact center in the appropriate approximation where the rod is "short". But as you make the rod shorter and shorter, this "short rod" approximation becomes better and better. This can be formulated as an appropriate limit, but considering that this is a B-level thread, I didn't even try to make that part rigorous.

 

[timmdeeg]

It's only the exact center in the appropriate approximation where the rod is "short". But as you make the rod shorter and shorter, this "short rod" approximation becomes better and better. This can be formulated as an appropriate limit, but considering that this is a B-level thread, I didn't even try to make that part rigorous.

Being just a layman, interested in physics though, I 'v chosen B-level. I can follow the math shown in the link #1, but that's my limit.

It seems challenging to write down the $r$-coordinate of the "locus of no acceleration" as a function of the $r$-coordinates of the ends of the rod and of time.

Given the $r$-coordinate of the "locus of no acceleration" at a distinct moment, would the velocities $dr/dt$ as measured by shell observers at $r_u$ and $r_l$ (upper and lower end) follow from this? In other words can the known free-fall velocity of the locus at r_locus be transported to the ends of the rod? And if correct and if one includes the gravitational shift, would the knowledge of these velocities allow to calculate the frequency shifts mentioned in #1? Sorry that I express myself very vague, its all purely speculative assumptions.

 

[PeterDonis]

The locus only need be connected.

Actually, I'm not even sure that is the case. Synge's proof uses the fixed point theorem to say that there must be some point in each spacelike slice of a foliation where $DV^i = 0$. But this is a non-constructive proof--it doesn't tell you where in each slice that point is. So it doesn't tell you whether the set of such points is connected. Synge does say "these points form a curve", but if by "curve" he means a connected set of points, I don't see how that actually follows from his proof.

The points in the locus are points where some an element of the body (changing from slice to slice, in general) has covariant derivative of that element's 4-velocity vanishing.

Hm, yes, I was not distinguishing between the tangent vector to the locus itself (assuming that even exists--see above) and the vector field $V^i$ at a given point of the locus.

 

[PAllen]

Actually, I'm not even sure that is the case. Synge's proof uses the fixed point theorem to say that there must be some point in each spacelike slice of a foliation where $DV^i = 0$. But this is a non-constructive proof--it doesn't tell you where in each slice that point is. So it doesn't tell you whether the set of such points is connected. Synge does say "these points form a curve", but if by "curve" he means a connected set of points, I don't see how that actually follows from his proof.

My guess is that for textbook (rather than research paper) he doesn't prove this but just implies it would follow from smoothness assumptions for the stress energy tensor. I assume he believes the 'curve' statement is provable.

 

[Ibix]

I was wondering what discontinuous would mean in this context. I can see that it would happen if the falling object broke into two parts which could go their own ways. Loosely, one center of mass became two which were not co-located with the first. But I don't immediately see what physical situation is being described by a single non-continuous locus. Or is that why PAllen was talking about smoothness assumptions ruling out discontinuities?

 

[PAllen]

I was wondering what discontinuous would mean in this context. I can see that it would happen if the falling object broke into two parts which could go their own ways. Loosely, one center of mass became two which were not co-located with the first. But I don't immediately see what physical situation is being described by a single non-continuous locus. Or is that why PAllen was talking about smoothness assumptions ruling out discontinuities?

Yes, that is why I think Synge claimed the locus was a curve, and I assume he know what he was talking about (though it would be nice if justified this in the text).

 

[PeterDonis]

I was wondering what discontinuous would mean in this context.

It would not imply discontinuity of the worldline of any piece of the object, since the locus of no acceleration is not a worldline of any piece of the object. What I meant was that there is nothing in what is given in Synge's book, that I can see, that rules out the possibility that the points where $DV^i = 0$ in neighboring spacelike slices are not neighboring points, where "neighboring" is meant in the topological sense. However, I suspect PAllen is right and there probably is a way of showing that this is ruled out when we make appropriate assumptions about smoothness of the stress-energy tensor.

I don't immediately see what physical situation is being described by a single non-continuous locus.

The locus of no acceleration is not describing a physical "thing". It's just an abstract set of points that happen to share a particular property. It doesn't describe the worldline of anything. So I don't think it describes a "physical situation", whether the set of points is continuous or not.

 

[pervect]

Being just a layman, interested in physics though, I 'v chosen B-level. I can follow the math shown in the link #1, but that's my limit.

It seems challenging to write down the $r$-coordinate of the "locus of no acceleration" as a function of the $r$-coordinates of the ends of the rod and of time.

Given the $r$-coordinate of the "locus of no acceleration" at a distinct moment, would the velocities $dr/dt$ as measured by shell observers at $r_u$ and $r_l$ (upper and lower end) follow from this? In other words can the known free-fall velocity of the locus at r_locus be transported to the ends of the rod? And if correct and if one includes the gravitational shift, would the knowledge of these velocities allow to calculate the frequency shifts mentioned in #1? Sorry that I express myself very vague, its all purely speculative assumptions.

I'm not sure what link #1 is - the first link in your post? Anyway, I'll expound a bit more if you're comfortable with more math.

Let's consider the Newtonian case. Suppose we have a rigid rod. Then the force on a particle with mass m at distance r is $\frac{GmM}{^2}$ The total force on the rod will be the $F_{tot} = \int \frac{GM}{r^2}dm$ where we integrate $dm = \rho dV$ over the volume V of the rod, $\rho$ being the density, dV being the volume element.

The rod as a whole accelerates rigidly, so that it has a single value of acceleration a of $a = F_{tot}/m = \int \frac{GM}{r^2}dm / \int dm$. Now, if we want to find the exact value of this average force and thus the acceleration, we need to perform the above integral. But this is really overkill. You can do it if you really want to. I don't really want to. The easy approach is to just say that we can approximate the force with a Taylor series around $r = r_0$ and write the force on the mass element dm as:

$$F(r) = \frac{GM}{r^2} dm \approx \left[ \frac{GM}{r_0^2} + \frac{dF}{dr} (r-r_0) \right] dm = \left[ \frac{GM}{r_0}^2 - \frac{2GM}{r_0^3} (r - r_0) \right] dm$$

Here we've just evaluted the first derivative of $dF/dr$ to get the first term in the Taylor series, $- \frac{2GM}{r_0^3}$

So we break down the total force into an average force, $GM/r_0^2$, and a tidal force, $-(2GM/r_0^3) (r-r_0)$

The tidal force is just the derivative of the force law, and we are just doing a first-order Taylor series approximation for the force. But if our rod is short, this will tell us what we need to know.

And what we need to know is basically this. The end of the rod closer to the end of the central mass will be tugged towards it harder than the further end of the rod. We can view this as an average force, that acts on every particle in the rod, and a tidal force, that stretches the rod. At least approximately.

This is the Newtonian case - the GR analysis is more complicated and unfamiliar. But thinking about the Newtonian case gives us some insight as to what happens in the GR case, at least in the simple example of an infalling radial rod. If we imagine a bunch of freely falling particles, they'd all follow geodesics. And we can compute the second derivative of the distance between geodesics by the geodesic deviation formula.

One reference on this that's somehwat simple is http://math.ucr.edu/home/baez/gr/geodesic.deviation.html

But to understand the way GR views things, we need to know about tensors in general, parallel transport and the Riemann curvature tensor. And probably the notion of vectors as partial derivatives. If we're familiar with all of these things, the resulting equation is fairly simple. The equation is:

$A^a = R^a{}_{bcd} v^b w^c v^d$, where all the quantities are tensors, A being the 4-acceleration, R being the Riemann curvature tensor, v being the 4-velocity, and w being a separation vector.

I suspect the GR formalism may not make a lot of sense to the OP not knowing their background. IT's A-level stuff, not B-level or even I-level. But I've included a brief sketch of it, it just in case it does. The GR formula tells us how the freely falling particles would accelerate with no other forces on it. Because GR and Newtonian mechanics give the same predictions in the weak field, we know that the freely particles will separate in GR, just as they did in Newtonian physics. But we attribute this separation not to "forces", but to the curvature of space-time, in the form of the Riemann curvature tensor, rather than a force. We also know that the rod is "rigid", so it doesn't separate. We haven't even discussed the mathematics of what "rigid" means in the context of relativity. The applicable notion would be Born rigidity, which says that every pair of particles in the rod maintains a constant separation as the rod falls. It takes quite a bit of work to show that this is even possible - a crucial and non-intuitive consideration that's required to make this possible is that our rod isn't rotating (or rather, isn't changing it's state of rotation - but not rotating before and after means it's not changing it's rotational state). As it turns out, Born rigidity is possible, and the mechanism that makes a physical rod rigid is rather similar to what happens in our Newtonian cases. There are internal forces in the rod, these forces act to keep the spearation between different parts of the rod constant, and these forces can be described as a tension in the falling rod. Furthermore, we can say that the magnitude of these forces is roughly equal and opposite to the rate at which the particles in the rod would accelerate away from each other, if they were not bound by the mechanical forces that hold the rod together. In other words, the particles in the rod would separate, if left to their natural motion. The tension in the rod fights this tendency to separate, and holds the rod together.

 

[timmdeeg]

Thanks for explaining the Newtonian case.

The equation is:

$A^a = R^a{}_{bcd} v^b w^c v^d$, where all the quantities are tensors, A being the 4-acceleration, R being the Riemann curvature tensor, v being the 4-velocity, and w being a separation vector.

I suspect the GR formalism may not make a lot of sense to the OP not knowing their background.

That's right, but hopefully it is useful to others who read your explanation.

There are internal forces in the rod, these forces act to keep the separation between different parts of the rod constant, and these forces can be described as a tension in the falling rod.Furthermore, we can say that the magnitude of these forces is roughly equal and opposite to the rate at which the particles in the rod would accelerate away from each other, if they were not bound by the mechanical forces that hold the rod together. In other words, the particles in the rod would separate, if left to their natural motion.

Is it correct to argue that there is a distinct point on the rod which feels no force ("Locus of no acceleration", as mentioned by PAllen) and that this abstract point will move in the direction of the lower end of the rod (closer to the center of mass)? My reasoning is that during the fall the tension and thus the rate of relative acceleration of the particles which you mentioned is increasing faster at the lower end of the rod compared to the opposite end. Kindly correct if wrong.

 

[PAllen]

Is it correct to argue that there is a distinct point on the rod which feels no force ("Locus of no acceleration", as mentioned by PAllen) and that this abstract point will move in the direction of the lower end of the rod (closer to the center of mass)? My reasoning is that during the fall the tension and thus the rate of relative acceleration of the particles which you mentioned is increasing faster at the lower end of the rod compared to the opposite end. Kindly correct if wrong.

I think this is right. It certainly can't be solved based on Synge's theorem (because, as Peter noted, it is non-constructive existence proof). Solving this accurately in GR would be challenging exercise, but my intuition agrees with yours.

 

[PeterDonis]

Is it correct to argue that there is a distinct point on the rod which feels no force ("Locus of no acceleration", as mentioned by PAllen)

This part has to be true by Synge's theorem.

and that this abstract point will move in the direction of the lower end of the rod (closer to the center of mass)?

By "center of mass" I assume you mean "the large gravitating body that the rod is falling toward", correct? I mention this because the term "center of mass" has another common meaning which is also useful in this connection, as we'll see in a moment. For clarity, I will call the thing I think you were referring to here the "LGB" (for "large gravitating body") below.

Whether the locus of no acceleration "moves" depends on your choice of coordinates. In coordinates in which the LGB is at rest, yes, the locus of no acceleration, intuitively, should move towards the LGB--i.e., radially inward. As PAllen notes, this can't be shown just using Synge's theorem alone; but in this particular case, we can bring other tools to bear to (I think) show that our intuition is correct. Those tools are too technical for a "B" level thread, but the gist of them would be to switch to coordinates in which the center of mass of the rod in the usual sense of that term, which I'll call the "COM"--for a rod with constant mass per unit length, this would be the point at the rod's exact geometric center--is at rest. These coordinates are called "Fermi Normal Coordinates", and they are a very useful tool in GR. Synge's Chapter II goes into these coordinates in some detail. In these coordinates, in this particular scenario, the rod should be symmetric about its COM, which means that it should be possible to show, using the tools from Synge's Chapter II, that the worldline of the COM is a geodesic, and this is sufficient to show that the COM worldline is the locus of no acceleration for this particular case.

 

[timmdeeg]

PAllen, I thank you for this confirmation.

 

[pervect]

Thanks for explaining the Newtonian case.

That's right, but hopefully it is useful to others who read your explanation.


Is it correct to argue that there is a distinct point on the rod which feels no force ("Locus of no acceleration", as mentioned by PAllen) and that this abstract point will move in the direction of the lower end of the rod (closer to the center of mass)? My reasoning is that during the fall the tension and thus the rate of relative acceleration of the particles which you mentioned is increasing faster at the lower end of the rod compared to the opposite end. Kindly correct if wrong.

In the Newtonian case, it's easy to argue that there is a point that does not "feel a force" - at least once we define what we mean by "feeling a force". I'll interpret this as meaning that a point on the rod follows the same path through space-time as a free particle at the same location that is not connected to the rod would follow, i.e. that there is some point on the rod that follows a geodesic.

In the Newtonian case, we know that all points on the rod accelerate at some average acceleration, $a_{avg} = F_{tot} / m$, because that's the only way the rod can stay rigid. The difference between the acceleration of a free-falling particle $F/dm$ and the average acceleration $a_{avg}$ changes sign, and is continuous, therefore there is some point on the rod at which a free particle accelerates at the same rate as a particle on the rod, because a continuous fuction that is positive at one point and negative at another have at least one point at which it is zero.

I'm not aware of any argument simpler that Synge's theorem for the GR case. (And the only thing I know about Synge's theorem is what I read here - it looks like it applies, but it seems to have a lot of caveats). The Newtonian proof won't translate directly.

 

[timmdeeg]

Yes, with center of mass I meant the large gravitating body.

These coordinates are called "Fermi Normal Coordinates", and they are a very useful tool in GR. Synge's Chapter II goes into these coordinates in some detail. In these coordinates, in this particular scenario, the rod should be symmetric about its COM, which means that it should be possible to show, using the tools from Synge's Chapter II, that the worldline of the COM is a geodesic, and this is sufficient to show that the COM worldline is the locus of no acceleration for this particular case.

Now I'm a bit confused. Why is the rod "symmetric about its COM"? I think its lower end (closer to the LGB) "feels" a higher gravitational force (perhaps better a higher tension) than the opposite end. Therefor I was assuming "that this abstract point (locus of no acceleration) will move in the direction of the lower end of the rod", (post #17), i.e. will move relative to the rod, away from the COM. But then the worldline of the COM can't be a geodesic (perhaps unless we are talking about a very short rod). If the before said is correct, then presumably no part of the rod describes a geodesic.

 

[PAllen]

Whether the locus of no acceleration "moves" depends on your choice of coordinates. In coordinates in which the LGB is at rest, yes, the locus of no acceleration, intuitively, should move towards the LGB--i.e., radially inward. As PAllen notes, this can't be shown just using Synge's theorem alone; but in this particular case, we can bring other tools to bear to (I think) show that our intuition is correct. Those tools are too technical for a "B" level thread, but the gist of them would be to switch to coordinates in which the center of mass of the rod in the usual sense of that term, which I'll call the "COM"--for a rod with constant mass per unit length, this would be the point at the rod's exact geometric center--is at rest. These coordinates are called "Fermi Normal Coordinates", and they are a very useful tool in GR. Synge's Chapter II goes into these coordinates in some detail. In these coordinates, in this particular scenario, the rod should be symmetric about its COM, which means that it should be possible to show, using the tools from Synge's Chapter II, that the worldline of the COM is a geodesic, and this is sufficient to show that the COM worldline is the locus of no acceleration for this particular case.

Actually, I think the OP means something different, and I think they are probably right. Hypothesizing that Newtonian gravity is good enough for a non-extreme situation, the difference in force between the bottom and center of the rod should be slightly greater than between the center and top. Further, this differential increases as the rod gets closer to the LGB. To me (consistent with the OP), this suggests that in the rod COM FN coordinates, the locus of no acceleration will move toward the end of the rod closer to the LGB.

 

[PeterDonis]

The Newtonian proof won't translate directly.

Actually, Synge's proof is pretty much the same as the Newtonian one you gave in concept: find a function describing "acceleration" that goes from positive to negative as you go from one end of an object to the other, and the mean value theorem then tells you that function must be zero somewhere in between. The difference is just in the machinery Synge has to set up to define the function in question in GR, which is more complicated than it is in Newtonian physics.

o me (consistent with the OP), this suggests that in the rod COM FN coordinates, the locus of no acceleration will move toward the end of the rod closer to the LGB.

I don't think this is correct (though it might have been what the OP meant). In rod COM FN coordinates, the locus of no acceleration should be at the spatial origin for all time, and the rod as a whole should occupy a constant range of spatial coordinates $(-x, +x)$, where $x$ is the spatial coordinate along the rod's direction--at least as long as the rod's tensile strength is sufficient to keep it rigid against tidal gravity. The only thing that should change in these coordinates is the stress in the rod; at a given value of $x$ (except for $x = 0$, where the stress should always be zero), it should get larger in magnitude as the time coordinate $t$, which is the same as proper time for the rod COM (but not necessarily for other points on the rod), increases.

(By "stress" I mean the appropriate components of the stress-energy tensor; I haven't worked those out in detail, but intuitively I would expect $T^x{}_x$, the tension in the $x$ direction, and $T^t{}_t$, the energy density, to increase in magnitude in both directions away from the spatial origin; and I would expect $T^t{}_t$ to be positive everywhere and $T^x{}_x$ to be zero at $x = 0$, and to have the same sign as the $x$ coordinate itself elsewhere.)

 

[PAllen]

I don't think this is correct (though it might have been what the OP meant). In rod COM FN coordinates, the locus of no acceleration should be at the spatial origin for all time, and the rod as a whole should occupy a constant range of spatial coordinates $(-x, +x)$, where $x$ is the spatial coordinate along the rod's direction--at least as long as the rod's tensile strength is sufficient to keep it rigid against tidal gravity. The only thing that should change in these coordinates is the stress in the rod; at a given value of $x$ (except for $x = 0$, where the stress should always be zero), it should get larger in magnitude as the time coordinate $t$, which is the same as proper time for the rod COM (but not necessarily for other points on the rod), increases.

The key pint is justifying that x=0 remains the point of vanishing stress. You haven't provided justification for this. I will provide again the argument which leads me to doubt this. Let's stick to the Newtonian case for now. The force as a function of radius from the LGB is non-linear. Therefore, shifting the force curve so it is +f at one end and -f at the other end will not place the force zero point at the mid point. It will, instead, be slightly below the mid point. I would claim that this force zero point is where you would expect zero stress. Let's stop here and see if you agree, and have an argument for why the stress zero point would not be at the force zero point, in the Newtonian example. (To see that the force zero point is below the mid point, all you need is the concavity of force curve; as I am graphing it, concave down means force zero point below the mid point.)

 

[PeterDonis]

The force as a function of radius from the LGB is non-linear. Therefore, shifting the force curve so it is +f at one end and -f at the other end will not place the force zero point at the mid point.

I see what you mean, but the counter-argument is that the coordinate transformation from LGB-centered global coordinates to rod COM Fermi Normal Coordinates is also nonlinear. My intuition is, heuristically, that the two nonlinearities cancel each other out, so to speak. But I think the only way to know for sure would be to actually work through the math; my intuition could well be wrong.

 

[PAllen]

I see what you mean, but the counter-argument is that the coordinate transformation from LGB-centered global coordinates to rod COM Fermi Normal Coordinates is also nonlinear. My intuition is, heuristically, that the two nonlinearities cancel each other out, so to speak. But I think the only way to know for sure would be to actually work through the math; my intuition could well be wrong.

I am not in a position (time wise) to do the math now, but the math I would propose is 'stay Newtonian'. If in weak gravity/slow speed, Newtonian math says the force balance point is not the mid point, it is hard for me to believe GR predicts something different. However, even here one notes that coordinates with the rod at rest are non-inertial coordinates in Newtonian physics, and the transform is not Galilean. The math still needs to be done, but it is much simpler than any GR analog.

 

[PeterDonis]

Newtonian math says the force balance point is not the mid point

Actually, on thinking this over further, I'm not even sure we know that this is true. Here is what pervect posted earlier along these lines:

In the Newtonian case, we know that all points on the rod accelerate at some average acceleration, $a_{avg} = F_{tot} / m$, because that's the only way the rod can stay rigid. The difference between the acceleration of a free-falling particle $F/dm$ and the average acceleration $a_{avg}$ changes sign, and is continuous, therefore there is some point on the rod at which a free particle accelerates at the same rate as a particle on the rod, because a continuous fuction that is positive at one point and negative at another have at least one point at which it is zero.

How do we calculate $F_{tot}$? Well, it's $GMm/r^2$--but what do we use for $r$? Is it the $r$ for the COM? If so, then since we are equating the point where $a = a_{avg} = F_{tot} / m$ with the point that follows a free particle trajectory, we are implicitly assuming that the rod's COM is that point.

If, OTOH, you say that the $r$ to use to calculate $F_{tot}$ is the $r$ of whatever point the "locus of no acceleration" turns out to be, then we can't get started, because that is supposed to be an output of the calculation, not an input. This doesn't necessarily mean the "force balance point" is the midpoint--to me it just means that intuition isn't a reliable guide either way.

The same issue arises if we try to do a GR calculation this way, because in order to get started, we have to know what the congruence of worldlines is that describes the rod. All we know at the outset is that it is a rigid congruence and that some "average" of it should equate to a free particle trajectory. But without some additional assumption--which, as above, amounts to putting into the calculation the answer we are supposed to be getting out of it--I don't see how the calculation can be done.

The only other way to proceed would be to numerically integrate starting with the rod at rest, in order to find the congruence of worldlines explicitly instead of having to assume it. But now we have to impose some constraint on the relative motion of the particles in the rod--in the simplest case, we would assume rigid motion, which means the proper distance between any two points on the rod would be constant with respect to the proper time of either of the points. But that still leaves a lot of room for different possible results. Again, I think all this just shows that intuition isn't a reliable guide either way for this problem.

 

[PAllen]

Peter, you credited Pervect's words to me (not that I disagree with them - all he is saying is the Newtonian analog of Synge's theorem, as you noted earlier; he has taken no position there about where the force free point is).

As for the quote that is due to me, you left the qualifier, which indicates (based on your earlier argument) that I am not sure what the answer is. The complete statement you snipped is: "If in weak gravity/slow speed, Newtonian math says the force balance point is not the mid point, it is hard for me to believe GR predicts something different." And the rest of the post indicates I am not sure.

Your answer to Pervect's words mixes in things I said into a an argument neither of us made.

My argument in standard coordinates centered on the LGB in Newtonian gravity is a pretty trivial piece of math, that sidesteps all the complications you raise:

We assume the rod is uniform density and rigid. There is some force of gravity F1 on an element at one end. There some force F2 on an element at the bottom. Then, purely due to concavity of F(r), the point where force on an element is (F1+F2)/2 is slightly below the midpoint (call this simple average of the end forces Fm). This much, I think is inarguable. Next I proposed (effectively) that in natural coordinates where the rod is stationary, you have F1' equal to the Fm - F1 in the +z direction, and F2' = F2 - Fm in the -z direction (toward the LGB), and Fm' = 0, at this point below the midpoint. However, in my post after this, I backed off and said I had no idea without (at least) accounting for the fact that rod stationary coordinates are non-inertial, and the transform to them is not Galilean.

 

[PeterDonis]

you credited Pervect's words to me

Oops, sorry! Fixed now.

The complete statement you snipped is: "If in weak gravity/slow speed, Newtonian math says the force balance point is not the mid point, it is hard for me to believe GR predicts something different." And the rest of the post indicates I am not sure.

I understand.

 

[pervect]

The only position I am take on what we've been calling the force-free point is that in the Newtonian case it approaches the center of the bar in the limit of a short bar of uniform density and cross section. A taylor series expansion of the specific force (force / unit mass) on a small segment of the bar with length dr and mass dm can be expressed as $f(r) \approx f(r_0) + f'(r_0) \Delta r + \frac{1}{2}f''(r_0) (\Delta r)^2 + ...$ where in this case we place $r_0$ at the center of the bar and let $r = r_0 + \Delta r$. Letting $\epsilon$ be the length of the bar, we can evaluate the total force of the bar as the sum of all the forces on each section of the bar, i.e. $\int_{\Delta r = -\frac{\epsilon}{2} ... \frac{\epsilon}{2}} f(r) \, \rho \, dr$. Using the Taylor series approximation for f(r), we can see that the quadratic terms in the approximation for f(r) become arbitrarily small compared to the lower order terms as $\Delta r$ approaces zero, leading to the conclusion that the total force on the bar is approximately equal to the specific force at the center of the bar multiplied by the total mass of the bar, making the center of the bar approximately the "force-free point" for short bars.

The GR case is a lot trickier, due to such issues as time dilation and the potential for gravitational radiation.

 

[PAllen]

The only position I am take on what we've been calling the force-free point is that in the Newtonian case it approaches the center of the bar in the limit of a short bar of uniform density and cross section.

This, I think is pretty clear. What is interesting is that already in Newtonian mechanics it is unclear for a 'long' uniform rigid rod falling radially.

 

[PAllen]

It seems that my (and the OPs) arguments gravitational balancing point being below the center of mass for a tall object in central gravity are well accepted. At least, answers like the following are accepted on physics stack exchange:

"Consider the Sears Tower. Its CG is about 1 millimeter below its CM. The reason why is because the base of the tower is closer to the center of the Earth than the top of the tower (by 442 m), and therefore receiving a slightly higher pull of gravity than the top of the tower. As a result, the CG is closer the the ground than the CM, because the part of the tower below the CM is being pulled by gravity (slightly) harder than the part of the tower above the CM"

So despite any complications, my strong intuition remains the same as the OP: that the force free point for this example is below the center (also the center of mass, by construction for a uniform density rigid rod), and that this force free point moves further from the center toward the Earth as the rod falls.

 

[PeterDonis]

It seems that my (and the OPs) arguments gravitational balancing point being below the center of mass for a tall object in central gravity are well accepted.

The CG here is not the same as the "locus of no acceleration". The Sears tower has no "locus of no acceleration", because no part of it is in free fall; it is, as a whole, at rest relative to the Earth, and is subjected to an external force (the force of the Earth's surface pushing up on it) to keep it in that state. We were talking about a rod that is subjected to no external force whatsoever; the only reason any part of it would not be in free fall would be because of internal forces resulting from tidal gravity. That is a different scenario.

 

[PeterDonis]

purely due to concavity of F(r), the point where force on an element is (F1+F2)/2 is slightly below the midpoint (call this simple average of the end forces Fm).

You are assuming that the force at the "locus of no acceleration" must be the simple average of the end forces. What justifies this assumption? I don't think we know, a priori, what the force at the "locus of no acceleration" should be. I think that's something a proper detailed model should tell us, not something we put in.

I should clarify also that the "force" in question here is the "force" of gravity, which even in Newtonian physics does not work exactly like non-gravitational forces, since it isn't felt by an object as weight or detectable by accelerometers. So intuitions that work for non-gravitational forces, even in Newtonian physics, won't necessarily work for gravity.

At this point, though, I think we simply have differing intuitions and we would have to look at a detailed model.