The thread thread: Strangeness of the expanding space paradigm

[chronon]

Given a infinitely long thread which is set to be uniformly expanding, then it must break in an inifnite number of places. If it is perfectly uniform then it will break everywhere, and the individual molecules will continue to expand uniformly. However, if it is not perfectly uniform then it will break at weak spots, and the resulting finite threads will each contract due to tension, but will move apart from each other - a situation similar to that of the galaxies in the universe.

 

[Zanket]

Do you understand why the molecules of a thread in equilibrium in expanding space look like:

*..*...*...*...*...*..*

?

(* is a molecule, . is empty space)

No, I don't understand why there would be differing distances between adjacent molecules, when the cosmic expansion is uniform. The thread could extend indefinitely beyond the ends displayed, making them arbitrary, so there is no reason for the differing distances.

How does assuming that "the breakage point of a large-scale thing is arbitrary" lead you to the conclusion that "a small-scale thing must break"?

Because, if the breakpoint of a large-scale thing is arbitrary, then scale is not a factor in its breakage. Then things of all scales must break. No information about the length of the thread could be gleaned by any experiment at an arbitrary breakpoint.

Notice that you're line of reasoning also "proves" that if you only have a pair of molecules (and nothing else), the pair of molecules will break.

Yes. I show that the paradigm implies that, hence it is inconsistent when it also says otherwise.

But you suggested you already worked through the pair of molecules example and understand they don't.

What I did was show my understanding of what the paradigm implies. I did not say that it made sense to me. I said that the paradigm implies that ** becomes *__* becomes _**_. That is silly, because why can it not end up **__ or __**? The paradigm calls for the particles to always end up at an arbitrary location in the expanded space. Silliness is an indication of an inconsistency, which I expose in the paradox.

 

[Zanket]

I still do not see a "smoking gun" for the paradigm being internally inconsistent (which appears to be your concern, but at this point I am not convinced that it is).

You should be convinced unless you can show where the paradox goes astray. Or at least hold the consistency of the paradigm in doubt.

We can however compute the forces that are required for the particles to maintain a constant distance from each other - these forces are the tidal forces I was talking about.

Keep in mind that the paradigm works as well in flat space having zero tidal force.

This is done via GR, not via your "expanding space paradigm" however.

As chronon pointed out above, and I confirmed by books, the paradigm is part and parcel of GR.

 

[Zanket]

Given a infinitely long thread which is set to be uniformly expanding, then it must break in an inifnite number of places. If it is perfectly uniform then it will break everywhere, and the individual molecules will continue to expand uniformly. However, if it is not perfectly uniform then it will break at weak spots, and the resulting finite threads will each contract due to tension, but will move apart from each other - a situation similar to that of the galaxies in the universe.

That’s the best logic against the paradox yet I think. But if the thread breaks anywhere then the paradox holds. We can imagine such a thread passing through our galaxy. Let it break within our galaxy—any breakpoint is as good as another. The distance between the pieces must expand, the pieces are not being pulled or pushed, so there is no explanation as to why our galaxy does not expand too. That finding does not change when the thread breaks in more than one place. Neither does it change when the thread is nonuniform, because the locations of the weak spots are still arbitrary.

 

[Hurkyl]

No, I don't understand why there would be differing distances between adjacent molecules, when the cosmic expansion is uniform.

Then you should work through it. Draw a free body diagram, and determine the tension required between neighboring molecules for the system to remain in equilibrium.

Because, if the breakpoint of a large-scale thing is arbitrary, then scale is not a factor in its breakage.

First off, IT IS NOT ARBITRARY. How many times do we have to say that? Once you've exactly specified the initial conditions, the point(s) where the string breaks can be determined.

Secondly, why would it follow that scale is not a factor?

Thirdly, if you are correct, then you should be able to make a direct proof that a two-particle system will break.

I said that the paradigm implies that ** becomes *__* becomes _**_.

Then you're wrong.

(1) ** cannot exist -- it is nonsensical. There must be separation between the particles.
(2) Secondly, why would you think expansion would make the particles actually separate and then come back together?

(3) Let me try to interpret your diagrams as something sensible. Suppose you placed a piece of dust at each of the two molecules, then, the diagram

..*...*...

becomes, after a time delay,

.:..*...*..:.

where the dust particles are denoted with a colon. (Of course, in both diagrams, empty space continues in both directions) (I've purposely drawn the diagrams to have an offset, to emphasize that absolute position is irrelevant)

It's should be obvious why this happens: the left thread molecule is attracting the right thread particle, but not the right dust particle, and vice versa. (Intermolecular forces)

 

[pervect]

You should be convinced unless you can show where the paradox goes astray. Or at least hold the consistency of the paradigm in doubt.

Your arguments to me don't seem to have anything to do with the paradigm - basically, you are (as other posters mentioned) getting off on some weird sidetrack about infinite strings.

Your argument is something like "An infinite string can't break, because in order to break it would have to break at some specific point - but an infinite string must break. Therfore relativity is wrong".

It just doesn't pass muster. I think the problem lies with your infinite strings, not relativity.

I'd need to see a clearcut argument that assumed "the paradigm" and arrived at a contradiction to see a paradox. So far I don't see that. Your conclusion is not coming directly from your premisies, your conslusion is coming from your premises plus some additional assumptions (which may or may not be true - they seem extremely dubious to me, liike your assumptions about how an infinite string must behave).

Keep in mind that the paradigm works as well in flat space having zero tidal force.

What makes you think that a flat space-time (we are already assuming space is flat) with zero tidal force that has the property that it also expands exists?

[edit]
At this point I'm still investigating whether or not I think such a situation can exist, and if it does, whether or not it can be sustained.
[end edit]

As chronon pointed out above, and I confirmed by books, the paradigm is part and parcel of GR.

If there's a real problem, it should show up in the actual math of GR. So far, the only thing that's shown up on my end are a few calculation errors :-(, but the integrity of GR looks intact.

 

[Zanket]

Then you should work through it. Draw a free body diagram, and determine the tension required between neighboring molecules for the system to remain in equilibrium.

That seems unnecessary at this point. I gave a paradox above, post 68. To challenge it, please pick one statement (at a time) and tell me where it’s wrong. You haven’t done that AFAIK. Why should I do further analysis until that happens?

First off, IT IS NOT ARBITRARY.

It must be. Otherwise you could tell me where an infinitely long thread breaks. Right?

Secondly, why would it follow that scale is not a factor?

If the large-scale thing breaks at an arbitrary spot, then it can break anywhere. It breaks because space expands between adjacent particles. The paradigm implies that no force is strong enough to prevent the breakage. Then expanding space can separate adjacent particles anywhere, regardless of other forces. Then scale is irrelevant.

For example, let there be two parallel threads, one that is infinite and one that is two particles long:

________________**_________________
<<******************************>>

Let expanding space break the long thread right where the short thread is (I can choose that spot because the long thread breaks at an arbitrary spot):

________________*_*_________________
<<*************_*****************>>

Both threads broke and scale was irrelevant. Since no force was strong enough to prevent the long thread from breaking, no force can prevent the short thread from breaking too. And this could happen anywhere in the universe.

Thirdly, if you are correct, then you should be able to make a direct proof that a two-particle system will break.

Not sure what you mean by a "direct" proof, but the above comment shows that a two-particle system will break.

(1) ** cannot exist -- it is nonsensical. There must be separation between the particles.

By ** I mean two particles that are bound together by some binding force.

(2) Secondly, why would you think expansion would make the particles actually separate and then come back together?

Sorry, I meant that the paradigm implies that expanding space tries to make ** into *__* but gravity makes it _**_

(3) Let me try to interpret your diagrams as something sensible. Suppose you placed a piece of dust at each of the two molecules, then, the diagram

..*...*...

becomes, after a time delay,

.:..*...*..:.

where the dust particles are denoted with a colon. (Of course, in both diagrams, empty space continues in both directions) (I've purposely drawn the diagrams to have an offset, to emphasize that absolute position is irrelevant)

It's should be obvious why this happens: the left thread molecule is attracting the right thread particle, but not the right dust particle. (Intermolecular forces)

I don’t get why there are no dust particles in the top diagram. But instead of discussing intermolecular forces at this point, can you just pick a statement in the paradox to refute? Maybe you’re doing that but I don’t see how. The diagrams I gave are just extra info; they aren’t mentioned in the paradox. As long as all the statements in the paradox hold, the details of the paradigm are superfluous.

 

[Zanket]

Your arguments to me don't seem to have anything to do with the paradigm - basically, you are (as other posters mentioned) getting off on some weird sidetrack about infinite strings.

Not a sidetrack. Infinitely long threads are used in the paradox given in post 68.

Your argument is something like "An infinite string can't break, because in order to break it would have to break at some specific point - but an infinite string must break. Therfore relativity is wrong".

No, it goes like it does in post 68.

It just doesn't pass muster. I think the problem lies with your infinite strings, not relativity.

There’s no problem with an infinitely long thread. They can exist in principle.

I'd need to see a clearcut argument that assumed "the paradigm" and arrived at a contradiction to see a paradox. So far I don't see that. Your conclusion is not coming directly from your premisies, your conslusion is coming from your premises plus some additional assumptions (which may or may not be true - they seem extremely dubious to me, liike your assumptions about how an infinite string must behave).

See post 68. Please pick one statement at a time and tell me exactly why it is wrong.

What makes you think that a flat space-time (we are already assuming space is flat) with zero tidal force that has the property that it also expands exists?

Whether it exists in nature is irrelevant. I’m arguing the paradigm only, not nature. If you mean “exists in theory,” then it is given by GR, which, when its cosmological constant is zero, says that space must either be expanding (according to the expanding space paradigm) or contracting. (Also I may have misled you. Instead of “flat spacetime having zero tidal force” I should have said “flat spacetime, which has zero tidal force”.)

If there's a real problem, it should show up in the actual math of GR. So far, the only thing that's shown up on my end are a few calculation errors :-(, but the integrity of GR looks intact.

Then you should be able to refute the paradox directly, by choosing a statement and telling me what is wrong with it.

 

[Hurkyl]

It must be. Otherwise you could tell me where an infinitely long thread breaks. Right?

I could, once you told me the initial conditions. (And if I had experience doing numerical computations in GR)

There are a vast number of initial conditions that have the qualitative description "There's an infinitely long thread". Obviously, I cannot tell you the outcome until I know the initial conditions.


Let's try an analogy:

The sum of two numbers is arbitrary. Therefore, arithmetic is silly and paradoxical. If you disagree with me, then tell me what the sum of two numbers is!


I'll respond to the rest when I get back home tonight.

 

[pervect]

Whether it exists in nature is irrelevant. I’m arguing the paradigm only, not nature.

It's quite releavant, actually.

All you can succeed in doing is showing that the flat-space frw metric

ds^2 = a(t)^2*(dx^2+dy^2+dz^2) - dt^2

with a(t) = k*t which would represent a uniform tidal-force less expansion is unphysical.

This appears to be a true statement. You have assumed that something non-physical exists, and found that it is non-physical. So I guess your baby paradigm actually has succeeded in demonstrating something, rather surprising for something with so little numerical content.

http://rocinante.colorado.edu/~pja/astr3830/lecture35.pdf

goes through the various cases. Matter dominated universes have a(t) = t^2/3. Energy dominated universe have a(t)=t^.5. So neither of them can have a(t)=kt.
.
One can have an absence of tidal forces for one bare instant of time (an instant in the cosmological frame) when a universe is in the process of "switching over" from a matter-dominated expansion to a cosmological constant dominated expansion. Here q switches from a value of .5 (for the matter-dominated flat universe) to q=-1 (for the cosmological constant dominated flat universe), so there is one instant where q=0. q=0 implies no tidal forces.

This case is not sufficient to make your argument work. Your argument requires that there was NEVER any deviation from a(t) = kt, otherwise there are ways to distinguish points on the string via their past history (when a(t) was not equal to kt). Your notion that all points on the infinite string are the same fails in a cosmology that it not static, a cosmology that is evolving. The only flat-space cosmologies that have q=0 are evolving ones, which only have q=0 for an instant.

You might try to ressurect your argument for a non-flat space cosmology, but I don't think it's going to actually work there, either. At the moment I'm not feeling like doing any more calculating, though, so I'm not going to try and calculate the tidal forces / Riemann for the non-flat space case.

 

[chronon]

Since all points on the threads are equal,

This is where the thread differs from the universe. Different points of the universe look different.

 

[pervect]

One other thing I ought to note about the connection between tidal forces and Riemann, because I managed to confuse even myself for a bit when looking over my past derivations (not a good sign for a clear derivation!).

In general, given a unit dispalcement vector $$\hat{\xi}$$, the tidal force, also vector, is given in component language by

$$R^{\hat{a}}{}_{\hat{t}\hat{\xi}\hat{t}}$$

'a' here is what parametizes the four components of the resulting vector.

This statement follows directly from the geodesic deviation equation, but it requires distinghishing between coordinate forms of the Riemann and coordinate basis forms of the Riemann - i.e. the "hats" - which are very important, and which I've sometimes glossed over and not typed.

However, in this particular case, the hats don't matter- which is one reason why I glossed over them, not wanting to type them all the time - but this could get one into trouble in a different situation. The hatted and non-hatted components will only differ by a scale factor. When the scale factors are either uniform in all directions (true in this case), or when the tidal force has no shear components (also true in this case), there is no problem. Problems will arise when both of these conditions are not met.

 

[Hurkyl]

I don’t get why there are no dust particles in the top diagram.

They were supposed to start where the thread particles were. Sorry, I thought that was clear from the text.

If the large-scale thing breaks at an arbitrary spot, then it can break anywhere.

As I mentioned in my last post, it breaks in specific places and times that depend on the initial conditions. It doesn't break just anywhere.

It breaks because space expands between adjacent particles.

AND the net force on each individual particle is zero.

Draw a free body diagram:


<---*--->

The left force is the attraction to the next particle to the left. The right force is the attraction to the next particle to the right. I've assumed the distances are equal either way, so we have that the net force on this particle is zero.

Since the net force is zero (and will remain zero, if we continue on the assumption that the spaces between particles are perfectly equal and expansion is perfectly uniform¹), the particle will follow a geodesic, and be carried away by the expansion of space, and will eventually break.


But in the two particle case, look at the free-body diagram on the left particle:

*--->

The arrow is the attraction to the particle on the right. We see that the forces are not balanced, so the particle will not follow a geodesic.


If we draw your picture correctly, this happens:

*   *   *   *   *   *
        *   *

*     *     *     *     *     *
             *   *

As you see, the thread particles separate (they will remain comoving with dust particles!) because there is no net force on each particle. But, in the two-particle case, the intermolecular forces cause it not to follow a geodesic.

What happens if we start with four particles equally spaced? Something like this:

      *   *   *   *
   . *   *     *   * .
.    *  .*     *.  *    .

The periods are dust particles, and they start where the thread particles are.

Initially, the interior particles have no net force, so they're carried along with the expansion of the universe. However, the end particles are tugged inwards, so they're not carried along.

Then, the intermolecular forces are imbalanced. The inner-right particle is more strongly attracted leftwards than rightwards, so we see that in the text frame, the inner-right particle is now located to the left of the inner-right dust particle.


Anyways, as we see, the effect of expanding space on the short threads is quickly balanced by intermolecular forces -- they do not travel in the same manner as the particles in the infinite thread... and the fact the infinite thread breaks does not suggest the finite thread breaks.

That was a large gap in your argument anyways -- you never argued that the particles in the short thread remained comoving with those in the long thread.

By ** I mean two particles that are bound together by some binding force.

I know, but I think that neglecting the intermolecular space is one of your biggest problems.


1: However, this is sort of like assumping a pencil will stand on its point, because the forces pulling it in the various directions are perfectly balanced.

 

[Zanket]

I could, once you told me the initial conditions. (And if I had experience doing numerical computations in GR)

There are a vast number of initial conditions that have the qualitative description "There's an infinitely long thread". Obviously, I cannot tell you the outcome until I know the initial conditions.

The ability to do GR computations is unnecessary here. By “infinitely long thread,” I mean a thread that stretches to infinity in opposite directions from any given point on it. No more initial conditions than that are necessary in the paradox, for the paradigm is clear that on the largest scales no force can withstand the cosmic expansion. Then the thread must break according to GR.

The sum of two numbers is arbitrary. Therefore, arithmetic is silly and paradoxical. If you disagree with me, then tell me what the sum of two numbers is!

The sum of two numbers is a specific value among an infinity of possible values, the opposite of arbitrary. The sum of two variables is arbitrary. If the thread broke at its center, say, that would be a specific point of the thread. But an infinitely long thread has no center or other identifiable location relative to itself, so wherever it breaks is an arbitrary location; that is, the breakpoint cannot be specified relative to itself. That means that it can break at any point along itself, and since the thread can cross any point in the universe, any point in the universe is a potential breakpoint that no force can withstand.

 

[Zanket]

It's quite releavant, actually.

GR is a theory about nature, so in an argument regarding its consistency one might mention things that exist in nature as we suppose they do, and one might take as givens what GR says about nature in some respects, but that would not be debating how nature really is, which is irrelevant. An inconsistent theory is invalid regardless of how nature really is.

You have assumed that something non-physical exists, and found that it is non-physical.

Huh?

One can have an absence of tidal forces for one bare instant of time...

Again, no talk of tidal forces is necessary. The paradigm works in flat spacetime, which has zero tidal force. Why not make your argument simpler and stick to flat spacetime?

You might try to ressurect your argument for a non-flat space cosmology, but I don't think it's going to actually work there, either.

I didn't grasp much of the rest of your post. I didn't see any mention of a statement of the paradox. Again I ask that you pick a statement from post 68 and tell me what is wrong with it. Otherwise I have to assume you’re blowing hot air, no offense.

 

[Zanket]

This is where the thread differs from the universe. Different points of the universe look different.

All I mean by “all points on the threads are equal” is that no point on the thread can be specifically identified relative to the thread itself; e.g. there is no midpoint of the thread itself.

 

[Zanket]

As I mentioned in my last post, it breaks in specific places and times that depend on the initial conditions. It doesn't break just anywhere.

I will examine the rest of your post and respond later, but in the interest of time, since my claim that it breaks anywhere along the thread (that is, at an arbitrary point) is the crux of my argument (in the paradox in post 68), I suggest we focus on this disagreement, which I did in my post to you above. If you can prove me wrong on this, then the paradox fails.

 

[pervect]

An inconsistent theory is invalid regardless of how nature really is.

You're apparently missing the point. Maybe you can try taking a break and re-reading my post in a bit, because at this point I'm not sure I can explain things any more clearly, though I'll try. Possibly the URL will help too (it may not help unless you accept the facts that the tidal forces exist, are proportional to the decleration parameter q, and are important to your problem). Basically, if you would take those three ideas seriously and get rid of whatever mental block is preventing you from thinking about them, I think you'd probably see what we are trying to say.

It's not just a question that our particular universe isn't the way your scneario demands, it's that the particular scenario you are envisioning (a flat universe expanding uniformly without acceleration or deceleration) is impossible to create within the framework of GR.

Again, no talk of tidal forces is necessary. The paradigm works in flat spacetime, which has zero tidal force. Why not make your argument simpler and stick to flat spacetime?

You are missing the point, alas. Tidal forces are key to the whole problem. You have created a "paradox" by ignoring the solution to the paradox, which is the tidal forces.

Tidal forces are not something that I am just adding in arbitrarily. They aren't just a nuisance. They are a necessary part of the theory, and are needed to explain why some things hold together (things hold together because their inter-molecular bonds are stronger than the tidal forces) and why other things rip apart or are crushed (one or the other happens to objects that are too weak - and the bigger the object is, the stronger it has to be to prevent the tidal forces from stretching/crushing it).

In fact, it takes an exceptional and highly unusual set of conditions for the tidal forces to be made equal to zero. These exceptional and highly special conditions can only occur for an instant of time, in a universe that is not static (a universe where q changes as a function of cosmological time). The instant occurs in the cosmological coordinate system, which is a flat piece of space because of the flat-space model we are using.

I didn't grasp much of the rest of your post. I didn't see any mention of a statement of the paradox.

I'd suggest trying to read it again - as I said in the post, your assumptions would make some sense if a static universe existed that expanded uniformly with no tidal forces ever existing. But this can't happen.

Let's consider the simplest case first. We demand the universe be static, so that it's properties (such as q and H) don't change with time.

Then the universe has a non-zero q, and tidal forces exist. This is the solution we started out with and trying to explain to you. The tidal forces prevent any infinite string from ever forming (it will either be crushed or stretched out of existence).

However, as you argued your idea, the question came up as to whether or not the tidal forces could ever be zero. The answer is that they can be zero, but only for an instant, in a universe where q is evolving as a function of time.

Because the universe isn't static, though, you cannot appeal to the symmetry of the situation to say that all points on the string are the same. All points are not the same, because the universe itself is evolving with time.

 

[Hurkyl]

I suggest we focus on this disagreement

It's just like the case of adding two numbers -- once I tell you what the two numbers are, you're able to tell me the sum. Once you tell me what your infinite thread is (i.e. the initial space-time geometry, the location & motion of the thread particles, and the nature of the binding force), I can tell you every place and time where the thread breaks.


Your latest posts seem to suggest you want the problem to be perfectly symmetric under a particular translation. Then, the thread will break everywhere.

 

[Zanket]

AND the net [attractive] force on each individual particle is zero.

Draw a free body diagram:


<---*--->

There is a subtle flaw with your reasoning here. The net force on a floating particle is always zero; otherwise it would feel an acceleration and would not be floating. Regardless how it is typically done, I would draw the diagram like this:

--->*<---

The left force is the particle’s attraction of other particles inward from the left. The right force is the particle’s attraction of other particles inward from the right. So we have that the net force on this particle is zero.

Then in the two particle case, the diagram of the left particle is:

--->*<---

We see that the forces are balanced, so the particle will follow a geodesic. And intuitively this is true: Two particles free-falling toward each other follow geodesics. Neither particle feels a pull toward the other. And this remains true for any number of particles, such as for the particles of any length of a floating thread.

Anyways, as we see, the effect of expanding space on the short threads is quickly balanced by intermolecular forces -- they do not travel in the same manner as the particles in the infinite thread... and the fact the infinite thread breaks does not suggest the finite thread breaks.

Since the net force on a floating particle is always zero, for both attractive forces and cosmic expansion, the length of the thread is irrelevant as to whether it breaks. No experiment on any given pair of adjacent particles could indicate the length of the thread.

It's just like the case of adding two numbers -- once I tell you what the two numbers are, you're able to tell me the sum. Once you tell me what your infinite thread is (i.e. the initial space-time geometry, the location & motion of the thread particles, and the nature of the binding force), I can tell you every place and time where the thread breaks.

I think these were assumable or irrelevant, but here are some conditions: An empty flat universe, expanding of course. The thread’s curvature of surrounding spacetime is negligible. Location and motion of the thread are indeterminate due to lack of reference points (and would be relative anyway, hence meaningless). The thread is floating. From any given point on it, the thread extends to infinity in opposite directions. Not sure what you mean by “nature of the binding force”; I guess the nature of whatever binding forces keep a thread together normally.

Your latest posts seem to suggest you want the problem to be perfectly symmetric under a particular translation. Then, the thread will break everywhere.

Perfectly symmetric, flat universe, whatever’s the simplest scenario. Agreed in that case it will break everywhere, according to that part of the paradigm.

 

[Zanket]

It's not just a question that our particular universe isn't the way your scneario demands, it's that the particular scenario you are envisioning (a flat universe expanding uniformly without acceleration or deceleration) is impossible to create within the framework of GR.

“Expanding uniformly” means expanding uniformly throughout the universe at any given cosmic time, not throughout time.

Then the universe has a non-zero q, and tidal forces exist. This is the solution we started out with and trying to explain to you. The tidal forces prevent any infinite string from ever forming (it will either be crushed or stretched out of existence).

We’ll have to agree to disagree on that. A flat universe has zero tidal force on a large scale. The thread contributes a tidal force, but we can assume it’s negligible except at small scales. One of my books says, paraphrasing, “a tidal force is an indicator of spacetime curvature.” And for curvature it says, “a property of spacetime evidenced by a tidal force.” If no tidal force, then no curvature and flat spacetime. If flat spacetime, then no tidal force and no curvature. An always-flat universe is possible in principle, so I can use it in a thought experiment and then there is no tidal force on large scales to consider.

Because the universe isn't static, though, you cannot appeal to the symmetry of the situation to say that all points on the string are the same. All points are not the same, because the universe itself is evolving with time.

All points are the same at any given cosmic time; that’s implied.

 

[Chronos]

I perceive a subtle flaw in your reasoning. You are inserting a simultaneous reference frame for all points in your conceptual 'infinite thread'. That's a back door version of any number of common SR paradoxes. But, a nice try.

 

[chronon]

the particular scenario you are envisioning (a flat universe expanding uniformly without acceleration or deceleration) is impossible to create within the framework of GR.

A flat universe has zero tidal force on a large scale. The thread contributes a tidal force, but we can assume it’s negligible except at small scales. One of my books says, paraphrasing, “a tidal force is an indicator of spacetime curvature.” And for curvature it says, “a property of spacetime evidenced by a tidal force.”

You need to be careful about what it is that is flat. A flat spacetime indeed has no tidal force, but a flat space (with the time coordinate being proper time) will mean a curved spacetime. See http://www.chronon.org/Articles/milne_cosmology.html

 

[chronon]

All I mean by “all points on the threads are equal” is that no point on the thread can be specifically identified relative to the thread itself; e.g. there is no midpoint of the thread itself.

Don't you mean that the thread has no weak points. If it does have weak points then it will break at those points, just as the universe 'breaks' in intergalactic space, where, due to the large distance, gravity is not strong enough to keep the galaxies together.

 

[Hurkyl]

Regardless how it is typically done, I would draw the diagram like this:

--->*<---

The left force is the particle’s attraction of other particles inward from the left. The right force is the particle’s attraction of other particles inward from the right. So we have that the net force on this particle is zero.

Ack. You even have the concept of a free-body diagram wrong! (Unless they teach it differently wherever you're from) -- a free-body diagram, by definition, pictures the forces acting on a particle.

Then in the two particle case, the diagram of the left particle is:

--->*<---

We see that the forces are balanced, so the particle will follow a geodesic.

Why do you have two arrows?


And in the case of 4 particles, (which you didn't draw), I imagine you'd also say:

--->*<---

for the internal particles. Why are they the same length?


Though, I think your comments on these two diagrams have brought to light another mistake you've been making:

"The net force on a floating particle is always zero; otherwise it would feel an acceleration and would not be floating."
"And intuitively this is true: Two particles free-falling toward each other follow geodesics. Neither particle feels a pull toward the other. And this remains true for any number of particles, such as for the particles of any length of a floating thread."
"Since the net force on a floating particle is always zero"

You seem to have been assuming from the outset that each individual molecule experiences no net force, and that's bad.

Location and motion of the thread are indeterminate due to lack of reference points (and would be relative anyway, hence meaningless).

Wrong. Not only does each individual molecule of the thread provide something that could be used as a reference point, but you can always coordinatize space-time, so that every event in space time can be uniquely specified by coordinates.

You should have learned this in high school geometry. You can describe a problem by using a coordinate chart. (There are, of course, many coordinate charts you can use, but the point is you can always do it)

There's even another way you could do it! The problem remains identical if there are a few dust particles scattered about, and you could use the dust particles as reference points.

Not sure what you mean by “nature of the binding force”; I guess the nature of whatever binding forces keep a thread together normally.

I mean that, to turn the numerical crank, one would need an exact relationship between distance and the force experienced.

All points are the same at any given cosmic time

I would like to nitpick, just in case this is another source of your problems. I know you didn't choose the original phrasing, though... all points are not the same, though the conditions at those points might be.

 

[Zanket]

I perceive a subtle flaw in your reasoning. You are inserting a simultaneous reference frame for all points in your conceptual 'infinite thread'. That's a back door version of any number of common SR paradoxes. But, a nice try.

Please elaborate. How is it any more a simultaneous reference frame than the concept of cosmic time?

In any case, the only purpose of the infinitely long thread, as I said above, is to help the intuition realize that it will break at an arbitrary spot, rather than, say, at the midpoint. I thought you had agreed that since the expansion is uniform, that a finite-length thread will also break at an arbitrary spot. Then an infinitely long thread should not be needed to convince you of the validity of the rest of the paradox.

And it doesn't really matter if a finite-length thread breaks always at the midpoint, say, for in that case the midpoint of any really long thread can still be put anywhere.

On thing that is definitely true, though, is that--regardless of validity--the paradox is far too complicated to be convincing.

 

[Zanket]

You need to be careful about what it is that is flat. A flat spacetime indeed has no tidal force, but a flat space (with the time coordinate being proper time) will mean a curved spacetime. See http://www.chronon.org/Articles/milne_cosmology.html

As you say in the link, "The critical density space is flat if you use the cosmological time coordinate". That's the time coordinate that can be used.

 

[Zanket]

Don't you mean that the thread has no weak points.

I mean that it has no identifiable point relative to itself, e.g. a midpoint.

If it does have weak points then it will break at those points, just as the universe 'breaks' in intergalactic space, where, due to the large distance, gravity is not strong enough to keep the galaxies together.

Note that the universe had to break in intergalactic space before there was intergalactic space, i.e. before the galaxies had broken. But I know what you mean.

 

[pervect]

Since the net force is zero (and will remain zero, if we continue on the assumption that the spaces between particles are perfectly equal and expansion is perfectly uniform¹), the particle will follow a geodesic, and be carried away by the expansion of space, and will eventually break.

Question: Is there some simpler way of finding the geodesic than using the geodesic equations? Unfortunately I'm finding the solution of the geodesic equations rather intractible, even for the simple (and unrealistic) case a(t) = Ht

For the metric

ds^2 = a(t)^2(dx^2+dy^2+dz^2) - dt^2

We parameterize our geodesic in terms of an affine parameter lambda
$$x(\lambda), t(\lambda)$$

and we let differentiaion with respect to lambda be represented by a "dot", so that
$$\dot{x} = \frac{dx}{d\lambda}, \dot{t} = \frac{dt}{d\lambda}$$

Then the geodesic equations for this metric are

$$\ddot{x} + f1(t)\dot{x}\dot{t} = 0$$
$$\ddot{t} + f2(t)(\dot{x}^2 +\dot{y}^2+\dot{z}^2) = 0$$

[the equations for y and z are similar to those for x, so I've omitted typing them out explicitly]

Here

f1(t) = 2*(da/dt)/a(t)
f2(t) = (da/dt)*a(t)

The trival solution x = constant is easy to find, but to solve the problem for a geodesic for a pair of particles maintaining a constant distance, we need the geodesic solutions for the case when $$\dot{x}$$ is nonzero - because two particles at a different x which both have $$\dot{x}=0$$ will have a non-zero relative velocity.

 

[Hurkyl]

Don't count me as an expert, and I hate ugly differential equations.


But I can do this one: using a(t) = Ht, we have:

x'' + (2/t) t' x' = 0
t'' + H² t (x')² = 0

The first equation is a linear first order differential equation in x'. The general solution is:

x'(λ) = C / t(λ)²

Substituting into the second gives:

t'' + H² t (C / t²)² = 0
t'' + H² C² t^-3 = 0

The form suggests a solution of the form t(λ) = D λⁿ. Plugging in and solving gives:

t(λ) = √(2 H C λ)

(And also the negative of this)

So, in the end, one family of solutions (I don't think the only one -- I should be missing a class of solutions to that second equation) is:

x(λ) = K + (log |λ|) / (2H)
t(λ) = √(2 H C λ)


(Of course, I've just looked at the x-t slice, assuming y' and z' were both zero. This should suggest the approach for the 4-D case)

 

[pervect]

So, in the end, one family of solutions (I don't think the only one -- I should be missing a class of solutions to that second equation) is:

x(λ) = K + (log |λ|) / (2H)
t(λ) = √(2 H C λ)


(Of course, I've just looked at the x-t slice, assuming y' and z' were both zero. This should suggest the approach for the 4-D case)

Cool - this partial solution should be sufficient to define the path of the geodesic, because the the exact nature of the affine parameterization is irrelevant to the path the particle actually follows. Hopefully I should be able to use it as a double check of my previous calculations.

 

[Chronos]

Hmm, this is wrong. I feel it in my bones. Will think about this a few days and comment again. Hurkyl is on the right track, I think.

 

[Hurkyl]

Well, the problem itself irritates me, and I meant to ask about it: it's not invariant under reparametrization.

My first thought was to choose a clever parametrization -- &lambda; = t. While the problem becomes much easier to solve, the only solutions were x = C.

 

[pervect]

You can multiply the affine parameter by a constant, and you can add a constant to it, but unfortunately I don't think you can do a general reparameterization :-(.

BTW, working on the problem some more in the morining, I found that Hurkyl's solution was apparently wrong in detail, though his obsevation that

x' = c/t^2

was the key to solving the problem. The following solution

$$k+{\frac {\ln \left( HC+\lambda \right) -\ln \left( HC-\lambda \right) }{2H}}$$
$$t = \sqrt {{H}^{2}{C}^{2}-{\lambda}^{2}}$$

appears to be correct according to my computer. And the general solution for t is just an affine reparameterization of the one above.

There is a singularity in the solution for x at the big bang (t=0), but it's interesting to note that this solution does not exist for t > HC. This is very interesting, and probably even important to the issue being discussed, though I can't clearly explain why the solution is limited in time at this point.

 

[Hurkyl]

I told you I hate differential equations! Even when I check my work, I'm still not confident in it. (And, as we see, with good reason!)

The quality of a path being a geodesic is a purely geometric fact -- it doesn't depend on a choice of coordinates. That is why I complain about the fact your differential equation is not invariant under a coordinate transform. If you have it correct, that must mean some assumption went into the derivation of those actual equations, and I'm curious what that was. (And wonder if some problems might be easier if you do not make that assumption!)