High School Rod falling radially towards the center of a mass

[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.

 

[PAllen]

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.

My only justification is that if scenario changes sufficiently small amount going to the frame of the rod at rest, you have balanced tension forces at the ends of the rods (they better be balanced or the rod isn't going to remain at rest), and Fm going to zero . As I've said, this is dicey part - because it isn't a Galilean transform, while we must end up with balanced tidal tension forces, the curve of force in between may change non-linearly. So, all I have is intuition, that in the Newtonian realm, it won't change enough to make a difference. So really, no argument except intuition for this part.

 

[timmdeeg]

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.

Thanks. The main argument to my understanding at least regards the symmetry of forces around the force-free point. I think being in free fall in FRW-spacetime the rod is symmetric about its COM in case the latter is co-moving, whereas falling radially in Schwarzschild spacetime said symmetry doesn't exist.

 

[Dr_Zinj]

You've got the physics here.
Read Larry Niven's short story, "Neutron Star", for a good hypothetical experience of this situation.

 

[PeterDonis]

The main argument to my understanding at least regards the symmetry of forces around the force-free point.

If you are going to use "symmetry of forces", then a comparison to FRW spacetime is not workable, because there is no "force of gravity" in the Newtonian sense in FRW spacetime, even as an approximation. To make that sort of comparison, you need to look at geodesic deviation; what is symmetric (more precisely, isotropic--the same in all directions) in FRW spacetime is the deviation of neighboring comoving geodesics from a chosen comoving geodesic. But geodesic deviation is not the same thing even in Schwarzschild spacetime as "variation of the force of gravity with height". That is why I have said that I don't think intuition is enough to resolve this question; we would need to look at the detailed math.

 

[timmdeeg]

If you are going to use "symmetry of forces", then a comparison to FRW spacetime is not workable, because there is no "force of gravity" in the Newtonian sense in FRW spacetime, even as an approximation.

Instead of looking at geodesic deviation I was thinking of Birkhoff's theorem here. If the COM of the rod coincides with the origin, then presumably the forces are balanced (producing pressure in the center, because if one releases particles along the rod, they would fall towards the center).

To make that sort of comparison, you need to look at geodesic deviation; what is symmetric (more precisely, isotropic--the same in all directions) in FRW spacetime is the deviation of neighboring comoving geodesics from a chosen comoving geodesic. But geodesic deviation is not the same thing even in Schwarzschild spacetime as "variation of the force of gravity with height". That is why I have said that I don't think intuition is enough to resolve this question; we would need to look at the detailed math.

Perhaps my intuition is misleading. Regarding the math, would it make sense to integrate the force from both ends of the rod towards its center until both quantities are equal at the searched r-coordinate?

Searching some literature it seems that the COM of orbiting extended bodies doesn't follow a geodesic. If of interest, I can provide the link.

 

[PeterDonis]

Instead of looking at geodesic deviation I was thinking of Birkhoff's theorem here.

Birkhoff's Theorem doesn't say anything about "symmetry of forces". It says that a spherically symmetric vacuum spacetime must be isometric to Schwarzschild spacetime. But that is perfectly consistent with the standard GR view that gravity is not a "force", it is a manifestation of spacetime curvature.

If you are thinking of FRW spacetime, Birkhoff's theorem is irrelevant because the spacetime is not vacuum. You might be confusing it with the shell theorem, which says that if a spacetime is spherically symmetric outside a certain radius, what is outside that radius has no effect on the "gravitational field" inside that radius. That theorem applies whether the spacetime is vacuum or not.

If the COM of the rod coincides with the origin, then presumably the forces are balanced (producing pressure in the center, because if one releases particles along the rod, they would fall towards the center).

It is true that, if you release particles along the rod that are at rest relative to the rod, those particles will (slowly) fall towards the center. But that is not what produces pressure in FRW spacetime; the pressure is put in by hand in the equation of state you assume, it is not an output of the model. A matter-dominated FRW universe has zero pressure--more precisely, the term "matter-dominated" in this context means we are assuming an equation of state with zero pressure. But the statement above about what happens to particles released is still true in such a spacetime.

Also, if you release particles along the rod that are at rest relative to the rod, those particles will not be "comoving" particles. A "comoving" particle at one end of the rod will be moving outward slightly relative to the COM of the rod. Such a particle, in a matter-dominated FRW universe, will slowly decelerate relative to the COM of the rod. (If the universe is dark energy-dominated, it will slowly accelerate, that is, move away faster.)

Regarding the math, would it make sense to integrate the force from both ends of the rod towards its center until both quantities are equal at the searched r-coordinate?

Once again, as far as GR is concerned, gravity is not a force. The GR approach is to look at geodesic deviation, as I said. Geodesic deviation has nothing to do with "force"; it's geometry.

Searching some literature it seems that the COM of orbiting extended bodies doesn't follow a geodesic.

This is true, but the effect is negligible in weak fields, i.e., in any field where the Newtonian approximation is going to work. Certainly it's negligible in the solar system; for example, all of the planets move on geodesics to within our precision of measurement.

 

[timmdeeg]

Birkhoff's Theorem doesn't say anything about "symmetry of forces". It says that a spherically symmetric vacuum spacetime must be isometric to Schwarzschild spacetime. But that is perfectly consistent with the standard GR view that gravity is not a "force", it is a manifestation of spacetime curvature.

If you are thinking of FRW spacetime, Birkhoff's theorem is irrelevant because the spacetime is not vacuum. You might be confusing it with the shell theorem, which says that if a spacetime is spherically symmetric outside a certain radius, what is outside that radius has no effect on the "gravitational field" inside that radius. That theorem applies whether the spacetime is vacuum or not.

Thanks for correcting. It seems that I misunderstood Peacock in http://arxiv.org/pdf/0809.4573v1.pdf , saying (page 2)
"in an isotropic universe, Bikhoff’s theorem assures us that we can neglect the effect of all matter at distances greater than that of the test particle, and all that counts is the mass between the particle and us."
If I understand it correctly now, he applies Birkhoff's theorem to expanding space-time, assuming the Schwarzschild vacuum solution though.

It is true that, if you release particles along the rod that are at rest relative to the rod, those particles will (slowly) fall towards the center. But that is not what produces pressure in FRW spacetime; the pressure is put in by hand in the equation of state you assume, it is not an output of the model. A matter-dominated FRW universe has zero pressure--more precisely, the term "matter-dominated" in this context means we are assuming an equation of state with zero pressure.

Yes, the released particles can't exert pressure, I didn't want to say that. Instead I was thinking that the directions of their motion indicate the creation of pressure in the rod (not tension as discussed before concerning the radial free infall of the rod) with the maximum in its geometrical center, because in this case the situation is symmetric per se.
Unfortunately "the pressure is put in by hand in the equation of state" doesn't say much to me. Could one simply understand the pressure being due to the summarized "weight" of the parts of the rod? In equation (4) Peacock shows the acceleration as a function of M and r. Would it be possible in principle to calculate the pressure at the origin from this?

Once again, as far as GR is concerned, gravity is not a force. The GR approach is to look at geodesic deviation, as I said. Geodesic deviation has nothing to do with "force"; it's geometry.

I'm aware of this. Could one say gravity is the source of tidal forces which extended bodies are experiencing? Those forces have been discussed earlier in this thread as being at least related to the unresolved question where on the rod one expects the "locus of no acceleration".

 

[PAllen]

Thanks for correcting. It seems that I misunderstood Peacock in http://arxiv.org/pdf/0809.4573v1.pdf , saying (page 2)
"in an isotropic universe, Bikhoff’s theorem assures us that we can neglect the effect of all matter at distances greater than that of the test particle, and all that counts is the mass between the particle and us."
If I understand it correctly now, he applies Birkhoff's theorem to expanding space-time, assuming the Schwarzschild vacuum solution though.
.

Welcome to the world of published errors. That paper not only misspells but misapplies Birkhoff's theorem. It is that simple. Birkhoff's theorem is purely about the uniqueness of the Schwarzschild metric in a region that is vacuum and has spherical symmetry. As Peter says, what they are really applying is the GR version of the shell theorem (due to Newton for Newtonian gravity). There are derivations of the shell theorem that use Birkhoff as part of the reasoning, but it is just wrong to equate this with Birkhoff's theorem.

[Note, the physical statements made in this section are correct, it is just a terminology error. The rest of the paper, at a glance, is a reasonable advocacy of one interpretation of cosmology in GR; there are people who argue in favor of different interpretations; they all make the same physical prediction. Personally, I am mostly in agreement with the point of view of this paper.]

 

[PeterDonis]

It seems that I misunderstood Peacock

You read him correctly, but as PAllen said, Peacock himself is mistaken; he should have said the shell theorem, not Birkhoff's Theorem.

If I understand it correctly now, he applies Birkhoff's theorem to expanding space-time, assuming the Schwarzschild vacuum solution though.

No. The Schwarzschild vacuum solution has no reasonable interpretation as "expanding space-time". Peacock is using the FRW solution; he never uses the Schwarzschild solution at all in this paper.

I was thinking that the directions of their motion indicate the creation of pressure in the rod

No, they don't. If there is pressure or tension in the rod, the individual pieces of the rod are not moving on geodesics; they are not in free fall. But the description you gave only applies to geodesic, i.e., free-fall, motions.

"the pressure is put in by hand in the equation of state" doesn't say much to me. Could one simply understand the pressure being due to the summarized "weight" of the parts of the rod?

No. The pressure is a property of the "cosmological fluid", i.e., of a highly idealized model of all of the matter and energy in the universe as a continuous substance. The whole rod scenario is using a different model to start with; the pressure that appears in the Friedmann equations for the universe as a whole isn't even a meaningful concept in the rod scenario.

In equation (4) Peacock shows the acceleration as a function of M and r. Would it be possible in principle to calculate the pressure at the origin from this?

No. Notice that pressure never appears at all in this article. He only talks about density, because that's all you can get from arguments of the sort he is making. The only way to get the pressure is, as I said, to assume an equation of state. The reasoning in this article can't tell you anything about the equation of state.

Could one say gravity is the source of tidal forces which extended bodies are experiencing?

No. Spacetime curvature (which you could call "gravity", but a better term would be "tidal gravity" since it is more precise) determines what states of motion are geodesic, i.e., free fall. But in an extended body whose parts are experiencing forces, i.e., having nonzero proper acceleration, the forces are due to the non-gravitational interactions between the parts of the body. To know the details of those forces, you need a model of the material properties of the body, i.e., of how the non-gravitational forces between the parts are produced.

Those forces have been discussed earlier in this thread as being at least related to the unresolved question where on the rod one expects the "locus of no acceleration".

Once again, gravity is not a force in GR, and the "forces" being talked about previously were gravitational "forces" (in the Newtonian interpretation). These are not the same as forces that produce nonzero proper acceleration, as described just above. This is why I keep insisting on using GR terminology, in which gravity is not a force--because it avoids the confusion you have gotten into here.

 

[timmdeeg]

Welcome to the world of published errors. That paper not only misspells but misapplies Birkhoff's theorem.
...
[Note, the physical statements made in this section are correct, it is just a terminology error. The rest of the paper, at a glance, is a reasonable advocacy of one interpretation of cosmology in GR; there are people who argue in favor of different interpretations; they all make the same physical prediction. Personally, I am mostly in agreement with the point of view of this paper.]

Thanks, that's good to know. - To me its fascinating to interpret the cosmological redshift as shown in equation (16).

 

[PeterDonis]

To me its fascinating to interpret the cosmological redshift as shown in equation (16).

I actually don't like this interpretation, because it applies the concept of "gravitational potential" to a non-stationary spacetime, for which that concept doesn't really make sense. But as PAllen said, it's a matter of interpretation, not physics; all of the interpretations make the same physical predictions.

 

[timmdeeg]

No, they don't. If there is pressure or tension in the rod, the individual pieces of the rod are not moving on geodesics; they are not in free fall. But the description you gave only applies to geodesic, i.e., free-fall, motions.

No. The description I gave was related to the "released particles" which are in free fall then.

No. The pressure is a property of the "cosmological fluid", i.e., of a highly idealized model of all of the matter and energy in the universe as a continuous substance. The whole rod scenario is using a different model to start with; the pressure that appears in the Friedmann equations for the universe as a whole isn't even a meaningful concept in the rod scenario.

Mentioning "pressure" I had the rod scenario in my mind, not the "cosmological fluid".

No. Notice that pressure never appears at all in this article. He only talks about density, because that's all you can get from arguments of the sort he is making. The only way to get the pressure is, as I said, to assume an equation of state. The reasoning in this article can't tell you anything about the equation of state.

Ok thanks. Unfortunately I have no notion how one would proceed to obtain said equation of state. Perhaps a crude explanation isn't possible using simple wording, right?

No. Spacetime curvature (which you could call "gravity", but a better term would be "tidal gravity" since it is more precise) determines what states of motion are geodesic, i.e., free fall. But in an extended body whose parts are experiencing forces, i.e., having nonzero proper acceleration, the forces are due to the non-gravitational interactions between the parts of the body. To know the details of those forces, you need a model of the material properties of the body, i.e., of how the non-gravitational forces between the parts are produced.

Would it be sufficient here to assume Born rigidity to have a model?

Once again, gravity is not a force in GR, and the "forces" being talked about previously were gravitational "forces" (in the Newtonian interpretation). These are not the same as forces that produce nonzero proper acceleration, as described just above. This is why I keep insisting on using GR terminology, in which gravity is not a force--because it avoids the confusion you have gotten into here.

Ok, thanks for clarifying.

 

[PeterDonis]

The description I gave was related to the "released particles" which are in free fall then.

Yes, exactly. Which means there can't be any pressure. Pressure would make the particles not be in free fall.

I have no notion how one would proceed to obtain said equation of state.

The usual way is, as I have said, to assume one. Typical ones used in cosmology are:

Matter: $p = 0$ (i.e., no pressure). This would apply to both ordinary visible matter and dark matter.

Radiation (and relativistic matter): $p = \frac{1}{3} \rho$ (where $\rho$ is the energy density). This is negligible now since there is no significant relativistic matter in the universe, the main radiation energy density in the present universe is the CMBR, and its energy density is several orders of magnitude smaller than that of matter. But in the early universe this equation of state is very important.

Dark energy: $p = - \rho$. This is the dominant component of the present universe, and the pressure being negative corresponds to accelerating expansion. (More precisely, $p < - \frac{1}{3} \rho$ corresponds to accelerating expansion.) A similar equation of state is used in inflationary cosmology, with a much larger $\rho$ and therefore a much faster acceleration of the expansion.

All of these equations of state are really special cases of the general perfect fluid equation of state, whose general form is $p = w \rho$, where $w$ can take various values. As you can see, "matter" corresponds to $w = 0$, "radiation" corresponds to $w = 1/3$, and "dark energy" corresponds to $w = - 1$.

More here:

https://en.wikipedia.org/wiki/Equation_of_state_(cosmology)

Would it be sufficient here to assume Born rigidity to have a model?

That is one way of obtaining a model, but it's not always possible; some states of motion of extended objects are not compatible with Born rigid motion at all. The gory details are in something called the Herglotz-Noether theorem.

Also, specifying Born rigid motion doesn't tell you how that motion is physically produced; it doesn't tell you the internal forces between the parts, it just tells you that, whatever those forces are, they are capable of producing Born rigid motion. So Born rigidity by itself doesn't give you things like the pressure and stress in the object.

 

[PAllen]

I actually don't like this interpretation, because it applies the concept of "gravitational potential" to a non-stationary spacetime, for which that concept doesn't really make sense. But as PAllen said, it's a matter of interpretation, not physics; all of the interpretations make the same physical predictions.

That's why I said 'mostly agree'. The parts I like are 'expanding universe' versus 'expanding space'; that what determines the solution are initial conditions; and that there is no value to making statements about superluminal motion; and that cosmological redshift is a generalization of Doppler. The specific factoring they do, I also don't think is meaningful. I agree that in non-stationary cases, factoring redshift into 'motion' and 'gravitational' effects is not meaningful or well defined. All you can say is that the result is influenced both by curvature and motion.

 

[pervect]

A few comments I wanted to make, primarily on the issues of "force in General Relativity". The most correct statement is probably that forces don't exist in GR, but while correct, this may not be very helpful. We can say this a lot, but if it's not understood why forces don't exist, it doesn't matter how many times we repeat it :(. Another totally correct, but not necessarily helpful, observation is to say that Riemann curvature tensor exists in the GR description, and that this tensor describes the curvature of space time, and can be regarded as the source of gravity. But it doesn't have much intuitive significance, either.

A more intuitive description of the Riemann curvature tensor might be based on the Geodesic deviation equation. A particularly important special case of the geodesic deviation tells us how the separation between two particles, both of which are following geodesics, changes with time.

Let's consider what happens in flat space-time, i.e. special relativity, without gravity, first. If we have two particles separated by some distance D, and they both follow geodesics in space-time, they both move in straight lines. If we imagine a pair of particles that are initially not in relative motion, and both particles are following geodesics in the flat space-time (which are straight lines), the distance D remains constant. Thus we can write $D(t) = d_0$ where $d_0$ is some constant, the initial separation.

Now, when we change to curved space-time, this is no longer true. There is a very simple formula for what happens to the separation. We can still imagine two particles that start out at relative rest, though the operational details of how to define "relative rest" rigorously via observation would be rather long and perhaps even a bit tricky. I'll settle for just saying that the concept of relative rest makes sense, at least for sufficiently nearby particles. In this special case, we can write the following: $D(t) = d_0 + \frac{1}{2} a t^2$. Here a is the relative acceleration between geodesics.

We can flesh out these ideas by drawing space-time diagrams on some curved surfaces - for instance the surface of a sphere, where we know geodesics are great circles - rather than on flat pieces of graph paper, but that would be too long and require too much work for my short post. It might be helpful, though, to do some actual computations on the apparent "relative acceleration" between two different great circles on a sphere, even though I'm not going to go through and do it myself in this post.

The point I want to make is that a pair of geodesics on a curved surface that start out initially parallel to each other (with an initial relative rate of separation of zero) don't keep a constant distance, or a constant relative velocity. Their relative velocity changes with time, causing an apparent relative acceleration between them, even though both lines are "straight" in the sense of being geodesics.

This relative acceleration between geodesics is a property of the local curvature of space-time, and (with some suitable formalism of tensor notation that I won't get into at all) can be described by something that is called the Riemann curvature tensor.

In the GR picture, there isn't any actual "force" involved, at least not in the case of geodesic motion. Instead, all we actually have are particles in natural motion, particles which are undisturbed by any forces whatsoever. If we separate these particles by an idealized rigid bar, though, rather than letting them move naturally, both particles follow a paths that are unnatural. In this case, then, there are actual, real forces in the bar which cause both particles to deviate from natural motion. We can describe the force in the bar which opposes the natural motion and makes the particles unnaturally keep a constant distance apart (rather than following their natural tendency to accelerate towards or away from each other) as a tension or a compression in the bar.

Now, while one might see how we can do this, it might not immediately be obvious why it's absolutely necessary to do things in this manner. The reason for this is a bit subtle, and has to do with the concept of fictitious forces. The first inclination might be to replace effects of curved space-time with some fictitious forces that mimic its effects. In Newtonian physics, these fictitious forces are just the traditional Newtonian "force of gravity". Other than acting instantaneously at a distance, these fictitious forces act enough like real forces that we can make this approach work.

The problem with this approach is that in special relativity, fictitious forces become more complicated and unwieldy to manipulate. The clearest example of this is the notion of gravitational time dilation - if we imagine an accelerating, rigid, space-ship, we find that not all clocks on the space-ship appear to run at the same rate. This necessitates a broadening of the idea of "fictitious forces". The fictitious forces become what's known as Christoffel symbols, a rather more complicated mathimatical entity, not a the simple vector quantity that we can use in Newtonian mechanics. We need a more complex formalism that can affect how we measure the rate of clocks. This formalism exists, and has a name (Christoffel symbols) - but knowing this doesn't help too much if it's not familiar, it's not particularly intuitive without the right mathematical background.

The point is, that the approach we used to use for gravity, which explained it via a fictitious force, just doesn't work anymore because the Newtonian concept of fictitious forces isn't strong enough to allow us to make statements about how clocks tick. Further more we know (though I haven't explained the details in this post) that in an accelerating rigid space-ship, or in Einstein's Elevator, we do see these effects on clocks due to the "artificial gravity" of the accelerating space-ship. So, we realize we need something more powerful than the notion of fictitious forces. The simplest approach winds up to be to just forget about the "fictitious forces", and think about things in terms of natural motion, instead, which is the approach I've tried to advocate in this post.

 

[timmdeeg]

The usual way is, as I have said, to assume one. Typical ones used in cosmology are:

Matter: $p = 0$ (i.e., no pressure). This would apply to both ordinary visible matter and dark matter.

Radiation (and relativistic matter): $p = \frac{1}{3} \rho$ (where $\rho$ is the energy density). This is negligible now since there is no significant relativistic matter in the universe, the main radiation energy density in the present universe is the CMBR, and its energy density is several orders of magnitude smaller than that of matter. But in the early universe this equation of state is very important.

Dark energy: $p = - \rho$. This is the dominant component of the present universe, and the pressure being negative corresponds to accelerating expansion. (More precisely, $p < - \frac{1}{3} \rho$ corresponds to accelerating expansion.) A similar equation of state is used in inflationary cosmology, with a much larger $\rho$ and therefore a much faster acceleration of the expansion.

All of these equations of state are really special cases of the general perfect fluid equation of state, whose general form is $p = w \rho$, where $w$ can take various values. As you can see, "matter" corresponds to $w = 0$, "radiation" corresponds to $w = 1/3$, and "dark energy" corresponds to $w = - 1$.

Thanks for explaining, I'm sorry about this misunderstanding though. According to my question here

Yes, the released particles can't exert pressure, I didn't want to say that. Instead I was thinking that the directions of their motion indicate the creation of pressure in the rod (not tension as discussed before concerning the radial free infall of the rod) with the maximum in its geometrical center, because in this case the situation is symmetric per se.
Unfortunately "the pressure is put in by hand in the equation of state" doesn't say much to me. Could one simply understand the pressure being due to the summarized "weight" of the parts of the rod?

I was erroneously assuming that you mentioned the "equation of state" in the context of the rod (describing the "internal forces") and was astonished about that. It is clear that the cosmological fluid doesn't exert pressure.

That is one way of obtaining a model, but it's not always possible; some states of motion of extended objects are not compatible with Born rigid motion at all. The gory details are in something called the Herglotz-Noether theorem.

Also, specifying Born rigid motion doesn't tell you how that motion is physically produced; it doesn't tell you the internal forces between the parts, it just tells you that, whatever those forces are, they are capable of producing Born rigid motion. So Born rigidity by itself doesn't give you things like the pressure and stress in the object.

So, assuming Born rigidity would it be possible to do a rough calculation (or at least a rough estimate) to find out on which point of the rod the internal forces are balanced? The answer to this question seems clear in FRW- but not in Schwarzschild-spacetime. Are these forces a function of the speed of the rod relative to a freely falling particles measured locally? I think this speed increases from zero if one moves along the rod starting at the balanced point and is probably greater at the bottom than at the top (in contrast to the FRW case). It seems however you disagree that such considerations will guide the intuition (of PAllen and myself) in the right direction. In short, is it hopeless to come to an definite conclusion without to "work through the math"?

By the way, here
http://users.ugent.be/~nvdbergh/workshop/info/Filipe Costa.pdf, slide 2
Costa mentions
"- no real (i.e., extended) bodies move along geodesics (only idealized point monopole) particles
- in general, no point in the bodies moves along a geodesic (not even the center of mass)"

But this might not regard the case we are discussing here, because you mentioned already "This is true, but the effect is negligible in weak fields".

 

[PeterDonis]

assuming Born rigidity would it be possible to do a rough calculation (or at least a rough estimate) to find out on which point of the rod the internal forces are balanced?

Meaning, which point of the rod feels no acceleration and is in free fall? I'm not sure, because, as I said, Born rigidity by itself doesn't tell you anything about the internal forces.

Are these forces a function of the speed of the rod relative to a freely falling particles measured locally?

There is no unique "speed" of the rod relative to freely falling particles; at any point on the rod you can always find a freely falling particle that is momentarily at rest relative to that point.

is it hopeless to come to an definite conclusion without to "work through the math"?

I think so.

you mentioned already "This is true, but the effect is negligible in weak fields".

Yes. What Costa is saying is true as a matter of exact theory, but in practice the effects are too small to measure in, for example, the solar system. In fact, AFAIK they are too small to measure even for binary pulsar systems, in which the fields are considerably stronger than they are in the solar system--models in which both pulsars in a binary move on geodesics match observations to within our current measurement accuracy.

 

[timmdeeg]

Thanks a lot for this excellent overview.

...
In the GR picture, there isn't any actual "force" involved, at least not in the case of geodesic motion. Instead, all we actually have are particles in natural motion, particles which are undisturbed by any forces whatsoever. If we separate these particles by an idealized rigid bar, though, rather than letting them move naturally, both particles follow a paths that are unnatural. In this case, then, there are actual, real forces in the bar which cause both particles to deviate from natural motion. We can describe the force in the bar which opposes the natural motion and makes the particles unnaturally keep a constant distance apart (rather than following their natural tendency to accelerate towards or away from each other) as a tension or a compression in the bar.

I understand so far.

Now, while one might see how we can do this, it might not immediately be obvious why it's absolutely necessary to do things in this manner. The reason for this is a bit subtle, and has to do with the concept of fictitious forces. The first inclination might be to replace effects of curved space-time with some fictitious forces that mimic its effects. In Newtonian physics, these fictitious forces are just the traditional Newtonian "force of gravity". Other than acting instantaneously at a distance, these fictitious forces act enough like real forces that we can make this approach work.

How would you describe these fictitious forces in relation to tidal forces, if there is any? Is there any difference between these two kinds of forces at all in case the rigid bar, which prevents the two particles from geodesic deviation, is very short?

The fictitious forces become what's known as Christoffel symbols, a rather more complicated mathimatical entity, not a the simple vector quantity that we can use in Newtonian mechanics. We need a more complex formalism that can affect how we measure the rate of clocks. This formalism exists, and has a name (Christoffel symbols) - but knowing this doesn't help too much if it's not familiar, it's not particularly intuitive without the right mathematical background.

Yes, no chance, I don't have this background.

So, we realize we need something more powerful than the notion of fictitious forces. The simplest approach winds up to be to just forget about the "fictitious forces", and think about things in terms of natural motion, instead, which is the approach I've tried to advocate in this post.

I must have missed something. To me this conclusion is surprising, because natural motion means no forces. If there are "fictitious forces", thinking of the two particles separated by a rigid bar, then we have no natural motion. As I understood your approach, it leads from natural motion to fictitious forces. Why then should we forget about them?

May I ask what your intuition tells you regarding the position of the "locus of no acceleration" on the rod?

If I see it correctly, the r-coordinate of the COM of the rod in radial free fall is not the arithmetic mean of the r-coordinates of top and the bottom. Thus the difference top between COM is smaller than the difference between COM and bottom. If so, shouldn't one expect larger "fictitious forces" in the lower part of the rod? Somewhere they change sign and cancel each other, but where?

 

[PAllen]

Yes. What Costa is saying is true as a matter of exact theory, but in practice the effects are too small to measure in, for example, the solar system. In fact, AFAIK they are too small to measure even for binary pulsar systems, in which the fields are considerably stronger than they are in the solar system--models in which both pulsars in a binary move on geodesics match observations to within our current measurement accuracy.

What do you mean by the comment on binary pulsars? Unless one is smaller by a factor of some thousands that the other, there is no geodesic of a background geometry to speak of, and it is two body problem for which there is no exact solution. Are you referring to the 'effective one body' approximation methods? I wouldn't describe those in terms of the bodies following geodesics.

 

[pervect]

Thanks a lot for this excellent overview.

I understand so far.

How would you describe these fictitious forces in relation to tidal forces, if there is any? Is there any difference between these two kinds of forces at all in case the rigid bar, which prevents the two particles from geodesic deviation, is very short?

In the case of Newtonian gravity, you have a universal notion of time shared by all observers, and you can regard tidal force at some time t as being the the spatial gradient of this fictitious force. It's easiest to illustrate the concept with only one spatial dimension, which we'll call x. Letting f be the fictitious force (due to gravity) and x be the position of the bar, you can write that the tidal force is $\partial f/\partial x$. To perform the partial derivative, you need to specify what you're holding constant, in this case t - we've eliminated y and z from the problem.

So, if you imagine you describe 1-d Newtonian space-time by t,x if you hold t constant, you can find the fictitious force from the tidal force by taking $\int (\partial f / \partial x) \, dx$ up to some constant of integration. By setting the boundary condition that there is no fictitious force on an object that's far away from the source of gravity, you can set this integration constant as well, and find the fictitious forces from the tidal forces by taking an integral.

So - that describes the relation between the fictitious forces and the tidal forces in a 1-d Newtonian space-time. Which is what I think you were asking (except for the 1d part) I thought about discussing the 3d aspects, but I'll save that for another post, and only if you actually want to know about it and ask - it gets involved and possibly off-track.

When you attempt to duplicate this argument in a manner compatible with relativity, you run into the obstacle that there is no such thing as "universal time" t. However, two observers that are sufficiently close together and moving at low velocities relative to another do share a common notion of time. The upshot of this is that the notion of the tidal force survives in relativity, but the notion of integrating the tidal force into a fictitious force doesn't really work as well.

There's one other wrinkle. You ask if geodesic deviation is exactly the same as tidal force. At least one textbook, MTW's "Gravitation", would suggest they are the same, and under most circumstances it's an excellent approximation. However, at sufficiently high accelerations, you might have difficulties For instance, if you mount an accelerometer on the bow and stern of a rigid accelerating spaceship in the flat space-time of SR, the two accelerometers won't read the same value, because the clocks at the bow and stern will tick at different rates. If you define tidal force by the difference in proper accelerations divided by the separation of the bow and stern, then you would have a very tiny tidal force by this definition in this case, while the geodesic deviation is zero. Possibly the authors of MTW had some way around this difficulty, but they didn't specify it in enough detail to know what it might be. So some caution is warranted.

A more serious issue is that tidal force, measured with a bar and a pair off accelerometers, can be induced by rotation. So the identification between the two totally fails if the observer is rotating, and partially fails (though not very seriously) if the observer is accelerating.

I must have missed something. To me this conclusion is surprising, because natural motion means no forces. If there are "fictitious forces", thinking of the two particles separated by a rigid bar, then we have no natural motion. As I understood your approach, it leads from natural motion to fictitious forces. Why then should we forget about them?

One of the motivations of General relativity was to explain the identity of inertial mass with gravitational mass - something that was just a coincidence in Newtonian theory. The basic idea is that if we assume that inertia and gravity have the same origin, this unexplained equivalence becomes explained. The obvious way to unite the two is to describe gravity as an inertial force. But a better way to equate the two is to avoid using inertial forces for the reasons I discussed, and identify tidal gravity with geodesic deviation.

May I ask what your intuition tells you regarding the position of the "locus of no acceleration" on the rod?

In a nutshell, my first thought is that "the locus of no acceleration" isn't worded very well. What is the acceleration being measured relative to? So I would rephrase the question as "the locus of points with no relative acceleration to an observer in natural motion". My basic intuition is that under most static circumstances, the center of the bar wouldn't accelerate relative to an observer in natural motion. But the notion of a static bar is probably not quite right. If the bar is actually falling, it'll get closer to the object it's falling towards as time progresses. This means the stretching of the bar will change as it gets closer to the objet it's falling towards, meaning that the situation isn't exactly static. Being non-static, the bar will probably vibrate. If the bar is perfectly symmetrical, it should vibrate about its center. But if the bar isn't symmetrical - there point with no acceleration relative to a geodesic might vary periodically or semi-peridiocally with time due to the vibration.

Another thing that I worry about is gravitational radiation. If the bar isn't just composed of test particles, but has gravity of its own, the solution to Einstein's equations may result in gravitational radiation being emitted as the bar falls. The intuition is that the momentum carried away by the gravitational radiation will affect the motion of the bar, making it's center not follow a geodesic. But I haven't worked it out. Usually , though, this effect would be tiny.

If I see it correctly, the r-coordinate of the COM of the rod in radial free fall is not the arithmetic mean of the r-coordinates of top and the bottom. Thus the difference top between COM is smaller than the difference between COM and bottom. If so, shouldn't one expect larger "fictitious forces" in the lower part of the rod? Somewhere they change sign and cancel each other, but where?

That happens as well, but I think I've already discussed that. Basically my position is that's a second order effect, and when you take the limit of a short enough bar, you can make the importance of the effect as small as you like. Since my previous posts haven't seem to communicated this, I'd suggest trying it for yourself. Consider two 1 kg weights connected by a massless rigid, 10 meter bar, which is instantaneously at rest over the surface of an idealized Earth. Work out the total net force on the bar, the total mass of the bar (that's easy, it's 2kg), and the accleration (force/mass) of the assembly. Because the assembly is rigid, this gives you the acceleration of all points on the assembly. Compare this acceleraton to the acceleration of freely falling mass at the midpoint of the bar. Repeat for 1 meter bars, and .1 meter bars.

[add]
You can use M=6*10^24 kg, r = 6.4*10^6 meters for the Earth - or whatever you like, actually. It's probably the most convenient if you keep the center of the bar at r=6.4*10^6 meters, so the ends of the bar would be at r+e, r-e, where e = 5 meters, .5 meters, and .05 meters.

Chose a suitable scale for acceleration where the ends of the bar for the 10 meter bar fall within the scale, and plot the free-fall accelerations of the end of the bar, the average acceleration of the entire bar (total force / total mass), and the acceleration of a freely falling mass at the midpoint (6*10^4) meters on this scale. See what happens as the length of the bar decreases from 10 meters to .1 meters.

[change]
Actually, if you want something visible discrpeancies, it'd be more informative to try a 1000 km bar (e=500000) for the long bar, and a 200km bar for the "short" bar.

 

[PeterDonis]

What is the acceleration being measured relative to?

"No acceleration" in this context means no proper acceleration--if an event on the worldline of a given piece of matter is also on the locus of no acceleration, that piece of matter has zero proper acceleration (i.e., is in free fall) at that event. However, it is quite possible, as far as Synge's theorem is concerned, that the worldline only meets the locus of no acceleration at one point--so that piece of matter would only be in free fall for an instant, so to speak.

 

[PeterDonis]

Are you referring to the 'effective one body' approximation methods?

Basically, yes. But on thinking it over, you're right that these can't really be viewed as involving geodesic motion--if for no other reason than that we know binary pulsar systems emit gravitational waves.

 

[timmdeeg]

Sorry for being late.

There is no unique "speed" of the rod relative to freely falling particles; at any point on the rod you can always find a freely falling particle that is momentarily at rest relative to that point.

Hmm, would you kindly check the following reasoning:

Lets assume a continues rain of freely falling particles starting at rest from infinity. The r-coodinate of the top the rod at a certain instant of time shall be $r_t$. Then a shell observer at $r_t$ measures the velocity of a particle $v_t=-\sqrt{r_s/r_t}$. At the same instant of time this shell observer records the top of the rod moving with a higher velocity $v_t-x$, because of the rigidity of the rod.
Somewhere down the rod at point $p$ a shell observer measures the same velocities of rod and particle. So, only here and only for an instant of time the particle is at rest with the rod.
Whereas the bottom moves with a slower velocity $v_b+y$ than the particle as measured by the shell observer at $r_b$, whereby $x<y$.

Supposed $r_p$ and thus $v_p$ is known, would it be possible to calculate $x$ and $y$ from that (I don't trust my naive notion)?

 

[timmdeeg]

Thanks for your detailed answer.

So, if you imagine you describe 1-d Newtonian space-time by t,x if you hold t constant, you can find the fictitious force from the tidal force by taking $\int (\partial f / \partial x) \, dx$ up to some constant of integration. By setting the boundary condition that there is no fictitious force on an object that's far away from the source of gravity, you can set this integration constant as well, and find the fictitious forces from the tidal forces by taking an integral.

Ok. Generally fictitious forces point opposite to the direction of acceleration, as I have read elsewhere. So, below a certain point on the rod, where these forces are balanced, they should point downwards and upwards above this point. But how would one perform the integration (neglecting that "that the notion of the tidal force survives in relativity, but the notion of integrating the tidal force into a fictitious force doesn't really work as well") if the sign of the force changes? Or would one integrate in two steps, below and above said point?
I'm not sure if I made myself clear, sorry if all this is very confusing.

In a nutshell, my first thought is that "the locus of no acceleration" isn't worded very well. What is the acceleration being measured relative to? So I would rephrase the question as "the locus of points with no relative acceleration to an observer in natural motion". My basic intuition is that under most static circumstances, the center of the bar wouldn't accelerate relative to an observer in natural motion.

Would the locus of no proper acceleration coincide with the locus of no acceleration relative to an observer in natural motion?

But the notion of a static bar is probably not quite right. If the bar is actually falling, it'll get closer to the object it's falling towards as time progresses. This means the stretching of the bar will change as it gets closer to the objet it's falling towards, meaning that the situation isn't exactly static. Being non-static, the bar will probably vibrate. If the bar is perfectly symmetrical, it should vibrate about its center. But if the bar isn't symmetrical - there point with no acceleration relative to a geodesic might vary periodically or semi-peridiocally with time due to the vibration.

Very interesting and surprising idea.

... Chose a suitable scale for acceleration where the ends of the bar for the 10 meter bar fall within the scale, and plot the free-fall accelerations of the end of the bar, the average acceleration of the entire bar (total force / total mass), and the acceleration of a freely falling mass at the midpoint (6*10^4) meters on this scale. See what happens as the length of the bar decreases from 10 meters to .1 meters.

[change]
Actually, if you want something visible discrpeancies, it'd be more informative to try a 1000 km bar (e=500000) for the long bar, and a 200km bar for the "short" bar.

I would love to do that but am simply overstreched.