B Rod falling radially towards the center of a mass

[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.