pervect
Staff Emeritus
Hurkyl said:
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
Gold Member
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
Staff Emeritus
Gold Member
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
Staff Emeritus
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.

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Hurkyl
Staff Emeritus
Gold Member
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!)

Hurkyl
Staff Emeritus
Gold Member
Actually, I'll have to admit error in that! I checked Spivak and Wikipedia, and both of them have the parametrization as part of the definition of a geodesic.

pervect
Staff Emeritus
I'm certainly not immune to making mistakes. I believe the geodesic equation I wrote is accurate for the metric specified, but anyone who wants to check (I'm not perfect either) would have to go through the math, computing the Christoffel symbols and writing down the geodesic equation from them (the Christoffel symbols.)

The meta-point I want to make is that anyone without tensor analysis will already be thoroughly lost at this point - they won't be able to find the geodesics, so they won't be able to calculate anything, and I don't think they will even be able to tell in a qualitiative manner how the geodesics will act (whether they will diverge or converge) without some fairly sophisticaed math.

I was hoping there might be some easier way to determine the characteristics of the geodesic without the math, but I'm not aware of it, and unless someone else can come up with something, I'll have to assume that it doesn't exist.

If one does have tensor analysis, it's much better to skip on to the Geodesic deviation equation

http://math.ucr.edu/home/baez/gr/geodesic.deviation.html
http://scholar.uwinnipeg.ca/courses/38/4500.6-001/Cosmology/GeodesicDeviation.htm [Broken]

than to go through all the pain of finding the geodesics IMO. This will lead one more or less directly to the internal forces.

What disturbs me a little bit is that some posters to this thread don't seem to grasp the seemingly obvious points that both Hurkyl and I are making about the fact that a string exerts forces. Thus, they wouldn't even understand why we care wheter or not the geodesics converge or diverge. I think perhaps a Newtonian analogy might help. More in another post.

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pervect
Staff Emeritus
Why threads exert forces - a Newtonian problem

I'm going to pose the following purely Newtonian problem in the hope that the solution to it will clarify some of the aspects of the relativistic problem. Of course, I can't actually make Zanket, for example, solve it, I can just offer it as a (hopefully simple!) challenge.

Let's start by imagining a single point-mass particle of mass m orbiting a massive planet with a much larger mass M, so that M >> m. To make life really easy, we will put m into a circular orbit around M or radius R (M is assumed to be stationary since M >> m), and we will assume that M is a sphere of uniform density, so that it in essence acts like a point particle. Then the force between the orbiting particle and the central mass will be

F = g M m / R^2

and we can find the orbital period T by saying that F = m*r*(2*pi/T)^2

Now, let's imagine that we have two point mass particles, m1 and m2, each of mass m, connected by a string, orbiting the same planet.

The string is a very short string, of length l (l << R), and the center of mass of both particles remains at a distance R away from the planet. Furthermore, the string always points directly towards the planet.

The problem is simply stated:

1) draw a free-body diagram of the forces on both point particles
2) compute the tension in the string.
3) Does m1 follow a geodesic (the path a body would follow with no applied forces) Why or why not?
4) Does m2 follow a geodesic?

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pervect
Staff Emeritus
OK, I think I've figured out the problem (or one of the problems, possibly there are more) with the FRW metric where a(t) = H t is. It's non-causal, because it's got closed timelike curves. "Not globally hyperbolic" to use some of the buzzwords from the textbooks.

Note again that a(t) = H t has not been shown to be the solution of any possible matter distribution. It's definitely not the solution of any matter distribution that satisfies the weak energy condition, because otherwise it would not have CTC's.

Here's a plot of a sample solutions to the geodesic, with C=2. As proper time advances, the olution starts out near x = +3 on the graph (numerically x goes to infinity at t=0), progresses up until coordinate time reaches a maximum of t=2, then coordinate time starts running backwards as proper time advances, until the solution reaches the singularity again at t=0 and x=-3 (on the graph) or - infinity in the complete solution.

We haven't "closed the loop yet", but clearly by symmetry, closing timelike geodesics (which start out at -x instead of +x) exist.

The equations plotted are

x := k + (1/(2*H))*(ln(H*C+lambda)-ln(H*C-lambda));
t := sqrt(H^2*C^2-lambda^2);

lambda is an affine parameter which is some scaled version of proper time, and starts out at lambda=-HC and progresses to lambda=+HC to generate the plot.

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Hurkyl
Staff Emeritus
Gold Member
OK, I think I've figured out the problem (or one of the problems, possibly there are more) with the FRW metric where a(t) = constant is. It's non-causal, because it's got closed timelike curves.
That can't be right. a(t) = constant is simply Minowski space -- the only geodesics are straight lines. (Do null paths count as geodesics?)

I guess you meant to talk about the a(t) = Ht case again, though. You should check the domains on which your solutions are valid -- there might be a singularity through which you can't naively continue the geodesics.

Also, you mentioned that the closing timelike geodesic should exist "by symmetry" -- isn't it exactly the same geodesic, but traced in the opposite direction?

Anyways, I rereviewed my solution:
x = K + (log |&lambda;|) / (2H)
t = &radic;(2 H C &lambda;)

And once again, they seem to work as solutions to the differential equations:
x'' + (2/t) t' x' = 0
t'' + H2 t (x')2 = 0

Are you sure my work is wrong?

pervect
Staff Emeritus
Hurkyl said:
That can't be right. a(t) = constant is simply Minowski space -- the only geodesics are straight lines. (Do null paths count as geodesics?)
A silly typo - I meant a(t)=Ht

I guess you meant to talk about the a(t) = Ht case again, though. You should check the domains on which your solutions are valid -- there might be a singularity through which you can't naively continue the geodesics.
I don't see any singularities but the one at t=0, but I'm open for alternate explanations for the behavior of the solution. By varying lambda from -HC to +HC, the solution starts and terminates at t=0, the only singularity I'm aware of.

Also, you mentioned that the closing timelike geodesic should exist "by symmetry" -- isn't it exactly the same geodesic, but traced in the opposite direction?
Yep, that's the simplest way of closing it.

Anyways, I rereviewed my solution:
x = K + (log |λ|) / (2H)
t = √(2 H C λ)

And once again, they seem to work as solutions to the differential equations:
x'' + (2/t) t' x' = 0
t'' + H2 t (x')2 = 0
You're right, I owe you an apology.

BIG APOLOGY TO HURKYL, HIS SOLUTION WAS CORRECT!

My solution seems to work, too, though. This suggests that the solutions must be equivalent. My solution does not look like yours for the case plotted (C=2), but it does look like yours when C approaches inifinity. (This is my constant C, different from your constant C).

Thus suggests that your solution is a special case of mine. Let's see if we can make them equivalent via a linear transform (the only sort allowed for affine transformation).

If we take

lambda' = a+lambda/b

and substitute

t = sqrt(H^2C^2 -(a+lambda'/b)^2)

and let a = -HC we get

t = sqrt( 2HC/b lambda' - lambda'^2/b^2)

So if we let C = k b, and take the limit as b->infinity, we retrive your solution

t = sqrt(2 H K lambda' - lambda'^2 / b^2), but as b-> infinity this reduces to

t = sqrt(2 H k lamba')

The solution for x is also the same, except for an infinte offset, which is annoying but expected since your solution for x asymptotically approaches infinity for infinite time, while mine asymptotically approaches zero.

BTW, if

t'' + k/t^3 = 0

t = sqrt(p lambda^2 + q lambda + r)

should be a solution when 4k + 4pr - q^2 = 0

Hurkyl
Staff Emeritus
Gold Member
Ah, I found the problem -- your geodesics are space-like, not time-like.

a(t)2 x'2 > t'2 for all &lambda; in the interval (-HC, HC)

By the way: how did you arise at that form for the solution to the differential equation?

pervect
Staff Emeritus
Hurkyl said:
Ah, I found the problem -- your geodesics are space-like, not time-like.

a(t)2 x'2 > t'2 for all λ in the interval (-HC, HC)
<Sigh> Back to the drawing board. I see now that your geodesics were all null geodesics.

I think (one of) the correct solutions for timelike geodesics should be

t = sqrt(lambda^2 - H^2 C^2), but I'll have to work on it more to be sure.

By the way: how did you arise at that form for the solution to the differential equation?
I used symbolic algebra package (maple) to solve the differential equation - it couldn't produce a usable solution for the original problem, but it could solve it once the problem was simplified.

Basically you input the following two lines into the program

deq5 := diff(t(lambda),lambda,lambda)+H^2*C^2/t(lambda)^3;
dsolve(deq5);

and out pops

$$t \left( \lambda \right) ={\frac {\sqrt {-{\it \_C1}\, \left( {H}^{2}{ C}^{2}-{\lambda}^{2}{{\it \_C1}}^{2}-2\,\lambda\,{{\it \_C1}}^{2}{\it \_C2}-{{\it \_C2}}^{2}{{\it \_C1}}^{2} \right) }}{{\it \_C1}}}$$

and

$$t \left( \lambda \right) =-{\frac {\sqrt {-{\it \_C1}\, \left( {H}^{2} {C}^{2}-{\lambda}^{2}{{\it \_C1}}^{2}-2\,\lambda\,{{\it \_C1}}^{2}{ \it \_C2}-{{\it \_C2}}^{2}{{\it \_C1}}^{2} \right) }}{{\it \_C1}}}$$

(It will also output the latex code to display this for you, if you ask it nicely enough).

The above expression for t from the computer was sort of unwieldly, so I simplified it - a little too much. Actually, I think throwing out C2 didn't matter, but I chose the wrong sign for C1.

My version of maple is quite outdated - I got it secondhand so I could run GRTensor II. Still I've found it wildly useful for almost all sorts of mathematical problems.

Hurkyl said:
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.
Sorry for the delay in replying (holiday intrudes). It has to make sense to me; how it is typically done is a lower priority to me (it’s a chicken & egg thing—I adopt what is typically done when it makes sense).

I thought a lot about your posts. I think I see what you’re getting at. Thanks for your input so far.

As a layman, my terminology is often lacking. When you say there’s a nonzero net force on a particle, to me that means that the particle is noninertially accelerated (it feels a push). I think what you mean (and your terminology could be fine for all I know) about the short thread is that, unlike the infinitely long thread, there’s no balanced force on the thread to keep its particles from falling toward each other to their closest state, precluding breakage.

Let 5 individually floating particles be in a line, equally spaced, and let gravity coalesce them. It seems that in your viewpoint the distance between the middle particle and its adjacent particles will shrink slower than will the distance between the outer pairs. That makes sense to me now; there is a higher gravitational attraction toward the middle for the outer particles than for the inner particles, and there is no net gravity for the middle particle (i.e. the forces on the middle particle are balanced).

If I have that right, then I think I understand intuitively why a very long thread can become taut and break due to cosmic expansion as described by the paradigm: it eventually breaks when the binding forces between particles are on average balanced to a degree that there is not enough time for the thread to maintain its length. The binding forces work to coalesce the particles of the thread but the net attraction of particles to their neighbors tends to zero toward the middle of the thread, so the rate of coalescing tends to zero toward the middle. If that rate is near zero all along a billion light years of thread, say, then even a seemingly small expansion rate along that length of thread could break it. The particles of a finitely long thread are not stationary in the comoving space—except maybe the middle particle—but their peculiar velocity may be small compared to the rate of cosmic expansion, in which case the thread becomes tauter and tauter until it breaks.

Please tell me if I’ve got this right or am at least on the right track.

Hurkyl
Staff Emeritus
Gold Member
As a layman, my terminology is often lacking. When you say there’s a nonzero net force on a particle, to me that means that the particle is noninertially accelerated (it feels a push)
Yes.

If I have that right, then I think I understand intuitively why a very long thread can become taut and break due to cosmic expansion as described by the paradigm: it eventually breaks when the binding forces between particles are on average balanced to a degree that there is not enough time for the thread to maintain its length. The binding forces work to coalesce the particles of the thread but the net attraction of particles to their neighbors tends to zero toward the middle of the thread, so the rate of coalescing tends to zero toward the middle. If that rate is near zero all along a billion light years of thread, say, then even a seemingly small expansion rate along that length of thread could break it. The particles of a finitely long thread are not stationary in the comoving space—except maybe the middle particle—but their peculiar velocity may be small compared to the rate of cosmic expansion, in which case the thread becomes tauter and tauter until it breaks.
This sounds right, or at least close to right.

I understand that a floating particle can be gravitationally accelerated with no force felt by it, but not how it can feel a push. Maybe the other binding forces can exert a push?

In any case, given enough mass to collapse, how would the universe collapse if, hypothetically, it were perfectly symmetrical and hence all forces were balanced everywhere? In that case, wouldn't every particle be comoving with the expanding space forever (i.e. no peculiar velocity), regardless of the density of the universe? Does collapse depend upon asymmetry? I thought it did not depend on that.

pervect
Staff Emeritus
OK, it looks like the most convenient form for the geodesics are

x := k+1/2/H*ln(1+H*C/lambda)
t := sqrt(lambda^2+H*C*lambda)

which are reasonably well behaved. k and C are arbitrary constants, H is Hubble's constant for the uniform expansion.

Getting the relative acceleration of two neighboring geodesics out of these equations is messy, though. I think I'm getting the result I expect from the geodesic deviation equation (acceleration / unit length = 0), but it's hard to have a great deal of confidence at this point.

The procedure I used is to set t=1, which implies

(1) lambda = -1/2*H*C+1/2*sqrt(H^2*C^2+4)

The distance in "physical" coordiantes from x=0 is

H*t*x evaluated at the lambda in (1)

dis:= k*H+1/2*ln(-(H*C+sqrt(H^2*C^2+4))/(H*C-sqrt(H^2*C^2+4)))

which can be series expanded assuming C is a small number as

k*H+1/2*H*C+(-1/48*H^3)*C^3+3/1280*H^5*C^5+Order(C^6)

A "velocity" in physical coordiantes is
d(H t x)/ d lambda evaluated at the lambda in (1)

This is actually velocity scaled by some factor dt/dlambda, but we require velocity to be zero because the endpoints of the string are stationary so the scale factor isn't important.

The series expansion for this is

k*H+1/8*k*H^3*C^2+1/24*H^3*C^3+(-1/128*k*H^5)*C^4+(-1/240*H^5)*C^5+Order(C^6)

From the requirement that the velocity be zero, we see that k is a constant of order C^3, as the lowest order terms in k and C are

k*H + 1/24 H^3 C^3 = 0

Finally we evaluate acceleration, and find

d^2 (H t x) / dlambda^2 =
(-1/4*k*H^3)*C^2+(-1/8*H^3)*C^3+(-1/4*k*H^5-1/2*(-1/2*k*H^4+1/4*H^3)*H+1/8*H^4)*C^4+(-11/192*H^5-1/4*(-1/2*k*H^4+1/4*H^3)*H^2-1/2*(1/4*k*H^5-1/4*H^4)*H)*C^5+Order(C^6)

This will again have to be scaled by some factor

Since acceleration is proportional to C^3 in the lowest order term, while distance is linear in C, the acceleration / unit distance approaches zero as the distance (proportional to C) approaches zero.

Since the acceleration is zero, the scale factor doesn't matter.

pervect
Staff Emeritus
Question for today:

What are the geodesics of a piece of string?

This has meaning only if a string follows a geodesic - but I still think that the tension in a string in the a(t)=HT FRW metric is zero, so the string as a whole should follow a geodesic.

It therfore seems to me that any of the space-like geodesics that I previously found earlier (more or less by accident in trying to find a timelike one) should then represent a piece of string (possibly a moving piece of string).

Corollary: assuming this is correct, (I think all the bugs have been worked out - there certainly were enough of them!)

there is a maximum possible length that a string can be at any given age of the universe, t

The previously derived geodesics are

x := k + (1/(2*H))*(ln(H*C+lambda)-ln(H*C-lambda));
> t := sqrt(H^2*C^2-lambda^2);

this represents a string with a midpoint at x=0, t=CH

The string does not extend past the time coordinate CH because that is the age of the string.

The length of this string is

$$\int (\sqrt{(H t \frac{dx}{d\lambda})^2 - (\frac{dt}{d\lambda})^2}) d \lambda$$

for lambda = -HC to HC

which is a finite number (I get 2 H C - the e quantity under the intergal works out to be equal to 1 * dlambda!)

So inifinte strings don't exist -- (at least not straight infinite strings)

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pervect
Staff Emeritus
I just found the easy way to find geodesics for any a(t).

http://panic.berkeley.edu/~wangfa/solutions/sol8.pdf

Solution:
Orient coordinates so that the particle is moving along the x-axis and restrict
attention to the two relevant dimensions (t; x). The metric is [cf. (18.1)]
ds^2 = -dt^2 + a^2(t)dx^2 (1)

This is unchanged under displacements in x. There is thus a Killing vector
E^a = (0; 1) and a conserved quantity

a^2(t) dx/dtau

Another intergal can be found by the normilization condition

a^2(t) (dx/dtau)^2 - (dt/dtau)^2 = -1
The conserved quantity is indeed conserved in my solution for the specific case a(t) = Ht (required for the solution to be correct). This is a generalziation of Hurkyl's observation that dx/dlambda = C/t^2 for the specific geodesic a(t) = H t.

The normalization condition is also satisfied, though this is more or less a happy accident (the normalization condition insures that lambda is scaled 1 for 1 with proper time tau).

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