Jaycall/KillingVectorApproach

Characteristic Vector Approach to Integrals of Motion Problem

Motivated by our recent work in relativistic fluid dynamics, one insightful approach to the integrals of motion problem involves rewriting the equations of motion in terms of an arbitrary vector field, which we're currently calling a characteristic vector. The benefit here is that we can produce a mathematical condition that tells us what combintations of variables should be grouped together in order to produce conserved quantities. The details follow.

Newtonian Equations of Motion

Using index summation notation (wherein repeated indices are summed over), it is possible to write the Newtonian equations of motion for a test particle in a fixed, static potential in an arbitrary orthogonal coordinate system. The equation describing motion in the $i$-dimension is:

$\frac{d}{dt}\left(m{h_i}^2{\dot{\lambda }}_i\right)=m{h_k}^2{\mathrm{\Gamma }}_{ij}^k{\dot{\lambda }}_j{\dot{\lambda }}_k-m{\partial }_i\mathrm{\Phi }$

where $m$ is the mass of the test particle, the $\lambda$s are the chosen coordinates, and the $\dot{\lambda }$s are the coordinate velocities (or the total time-derivatives of the coordinates), the $h$s are the metric scale factors associated with each of the coordinates, $\mathrm{\Phi }$ is the gravitational potential, and the $\mathrm{\Gamma }$s are the Christoffel symbols.

Metric Components, Orthogonality Condition, and Scale Factors

For any coordinate system it is possible to express the components of the metric in terms of the partial derivatives defining the transformation from any other coordinate system. Written in terms of the transformation from cylindrical coordinates, each metric component is

$g_{ij}=\left({\partial }_{i}R\right)\left({\partial }_{j}R\right)+\left({\partial }_{i}z\right)\left({\partial }_{j}z\right)+R^2\left({\partial }_i\varphi \right)\left({\partial }_j\varphi \right).$

The orthogonality condition is that the metric be diagonal. That is, each off-diagonal component of the metric must equal zero.

$\left({\partial }_{i}R\right)\left({\partial }_{j}R\right)+\left({\partial }_{i}z\right)\left({\partial }_{j}z\right)+R^2\left({\partial }_i\varphi \right)\left({\partial }_j\varphi \right)=0_{ij},i\ne j.$

If the coordinates should be orthogonal, then scale factors can also be defined in terms of the diagonal components of the metric such that $g_{ii}={h_i}^2$. This results in the following definition of the scale factors if the orthogonality condition is met.

$h_i={[{\left({\partial }_{i}R\right)}^2+{\left({\partial }_{i}z\right)}^2+R^2{\left({\partial }_i\varphi \right)}^2]}^{1/2}.$

Christoffel Symbols

For an orthogonal coordinate system, the Christoffel symbols can be written (depending on which, if any, indices are repeated) as

Christoffel Symbols (for orthogonal coordinate systems)
${\mathrm{\Gamma }}_{ij}^k=0$	for	$i\ne j\ne k$
${\mathrm{\Gamma }}_{ii}^k=-\frac{h_i}{h_k}\frac{{\partial }_k{h}_i}{h_k}$	for	$i\ne k$
${\mathrm{\Gamma }}_{ij}^i={\mathrm{\Gamma }}_{ji}^i=\frac{{\partial }_j{h}_i}{h_i}$	otherwise (Mistake in original expression corrected 06/10/2010, as discussed.)
${\mathrm{\Gamma }}_{ii}^i=\frac{{\partial }_i{h}_i}{h_i}$	otherwise

Identification of Conserved Quantity

If, through some miracle, the right-hand side (RHS) of the equation of motion should equal zero, then the combination of variables inside the parentheses must by definition be a conserved quantity. But good luck guessing a coordinate system in which this miracle will occur!

Introduce Characteristic Vector

The trick is to introduce an arbitrary vector field, which will be used to build a location-dependent weighted linear combination of the equations of motion. Since the equations of motion associated with each of the coordinates collectively form a single vector equation, we will just form an inner product of our characteristic vector with each side of the vector equation of motion. The result is

CV.01

$C_i\frac{d}{dt}\left(m{h_i}^2{\dot{\lambda }}_i\right)=m{h_k}^2{\mathrm{\Gamma }}_{ij}^k{\dot{\lambda }}_j{\dot{\lambda }}_k{C}_i-m{C}_i{\partial }_i\mathrm{\Phi }$

Next, we bring $C_i$ inside the total time-derivative on the left-hand side (LHS). This produces an additional term, which we promptly move over to the RHS and include as part of what is commonly referred to as the source (since it's the source of any change in the quantity in parentheses).

$\frac{d}{dt}\left(m{h_i}^2{\dot{\lambda }}_i{C}_i\right)=m{h_i}^2{\dot{\lambda }}_i{\dot{C}}_i+m{h_k}^2{\mathrm{\Gamma }}_{ij}^k{\dot{\lambda }}_j{\dot{\lambda }}_k{C}_i-m{C}_i{\partial }_i\mathrm{\Phi }$

By doing this, we have formed a new conservative quantity; that is, a new quantity in the parentheses which will be conserved if and only if the source is zero. The utility is in the fact that we should be able to force the source associated with this new conservative quantity to go to zero by choosing the right characteristic vector. Once we find the right characteristic vector, we'll be able to use it directly to build the conserved quantity $m{h_i}^2{\dot{\lambda }}_i{C}_i$.

Strategic Choice of Characteristic Vector

So we're looking for a vector $\overrightarrow{C}$ such that

${h_i}^2{\dot{\lambda }}_i{\dot{C}}_i+{h_k}^2{\mathrm{\Gamma }}_{ij}^k{\dot{\lambda }}_j{\dot{\lambda }}_k{C}_i-C_i{\partial }_i\mathrm{\Phi }=0.$

The good news is that this is a single scalar equation constraining the three independent components of $\overrightarrow{C}$, so several families of solutions should exist. Consequently, we can afford to be choosey!

In the problem we're exploring using T3 coordinates, we already know a conserved quantity associated with the azimuthal coordinate — angular momentum. We're looking for an additional (independent) conserved quantity associated with the two radial coordinates. So it makes sense to look for one of the solutions that has $C_3=0$. Furthermore, we'd like to avoid dealing with the unknown potential function (which varies only with ${\lambda }_1$) as much as possible, so as long as we're being choosey, let's look for a solution that has $C_1=0$. We now have three conditions on three components. The condition constraining $C_2$ is

${h_2}^2{\dot{\lambda }}_2{\dot{C}}_2+{h_k}^2{\mathrm{\Gamma }}_{2j}^k{\dot{\lambda }}_j{\dot{\lambda }}_k{C}_2=0.$

This can be rewritten more simply as

${\dot{C}}_2=-\left(\frac{{h_k}^2}{{h_2}^2}{\mathrm{\Gamma }}_{2j}^k\frac{{\dot{\lambda }}_j{\dot{\lambda }}_k}{{\dot{\lambda }}_2}\right)C_2.$

Perhaps the best approach to solving this condition is separation of variables.

	$\int \frac{d{C}_2}{C_2}$	$=$	$-\int \left(\frac{{h_k}^2}{{h_2}^2}{\mathrm{\Gamma }}_{2j}^k\frac{{\dot{\lambda }}_j{\dot{\lambda }}_k}{{\dot{\lambda }}_2}\right)dt$
$\Rightarrow$	$\ln C_2$	$=$	$-\int \left(\frac{{h_k}^2}{{h_2}^2}{\mathrm{\Gamma }}_{2j}^k\frac{{\dot{\lambda }}_j{\dot{\lambda }}_k}{{\dot{\lambda }}_2}\right)dt$
$\Rightarrow$	$C_2$	$=$	$\exp \{-\int \left(\frac{{h_k}^2}{{h_2}^2}{\mathrm{\Gamma }}_{2j}^k\frac{{\dot{\lambda }}_j{\dot{\lambda }}_k}{{\dot{\lambda }}_2}\right)dt\}.$

The final necessary step will be to write the quantity in parentheses as the exact derivative of some other quantity, call it $\mathrm{\Xi }$. The long-sought conserved quantity will then be $m{h_2}^2{\dot{\lambda }}_2\exp (-\mathrm{\Xi })$ , where

$\mathrm{\Xi }\equiv \int \left(\frac{{h_k}^2}{{h_2}^2}{\mathrm{\Gamma }}_{2j}^k\frac{{\dot{\lambda }}_j{\dot{\lambda }}_k}{{\dot{\lambda }}_2}\right)dt.$

Question from Joel

Is the following logic correct?

Suppose we examine the term in which $j=2$ and $k=1$. The integral becomes,

$\mathrm{\Xi }{|}_{j=2,k=1}=\int \left(\frac{{h_1}^2}{{h_2}^2}{\mathrm{\Gamma }}_{22}^1\frac{{\dot{\lambda }}_2{\dot{\lambda }}_1}{{\dot{\lambda }}_2}\right)dt=\int \left(\frac{{h_1}^2}{{h_2}^2}{\mathrm{\Gamma }}_{22}^1\right)d{\lambda }_1$

Given that the scale factors and the Christoffel symbols are expressible entirely in terms of ${\lambda }_1$ and ${\lambda }_2$, one could imagine a situation — for example in the quadratic case of $q^2=2$ — in which this integral could be completed analytically. (I presume that wherever ${\lambda }_2$ appears inside this integral, it can be treated as a constant because the two coordinates are independent of one another.)

Response from Jay

Yes, I believe you've worked this term out correctly, but as you can see in what follows, it cancels out with the other cross term. Writing out each of the terms in $\mathrm{\Xi }$ gives,

$\frac{d\mathrm{\Xi }}{dt}=\frac{{h_1}^2}{{h_2}^2}{\mathrm{\Gamma }}_{21}^1\frac{{\dot{\lambda }}_1{\dot{\lambda }}_1}{{\dot{\lambda }}_2}+\frac{{h_1}^2}{{h_2}^2}{\mathrm{\Gamma }}_{22}^1\frac{{\dot{\lambda }}_2{\dot{\lambda }}_1}{{\dot{\lambda }}_2}+\frac{{h_1}^2}{{h_2}^2}{\mathrm{\Gamma }}_{23}^1\frac{{\dot{\lambda }}_3{\dot{\lambda }}_1}{{\dot{\lambda }}_2}+\frac{{h_2}^2}{{h_2}^2}{\mathrm{\Gamma }}_{21}^2\frac{{\dot{\lambda }}_1{\dot{\lambda }}_2}{{\dot{\lambda }}_2}+\frac{{h_2}^2}{{h_2}^2}{\mathrm{\Gamma }}_{22}^2\frac{{\dot{\lambda }}_2{\dot{\lambda }}_2}{{\dot{\lambda }}_2}+\frac{{h_2}^2}{{h_2}^2}{\mathrm{\Gamma }}_{23}^2\frac{{\dot{\lambda }}_3{\dot{\lambda }}_2}{{\dot{\lambda }}_2}+\frac{{h_3}^2}{{h_2}^2}{\mathrm{\Gamma }}_{21}^3\frac{{\dot{\lambda }}_1{\dot{\lambda }}_3}{{\dot{\lambda }}_2}+\frac{{h_3}^2}{{h_2}^2}{\mathrm{\Gamma }}_{22}^3\frac{{\dot{\lambda }}_2{\dot{\lambda }}_3}{{\dot{\lambda }}_2}+\frac{{h_3}^2}{{h_2}^2}{\mathrm{\Gamma }}_{23}^3\frac{{\dot{\lambda }}_3{\dot{\lambda }}_3}{{\dot{\lambda }}_2}.$

Plugging in values for the Christoffel symbols leads to the expression

$\frac{d\mathrm{\Xi }}{dt}=\frac{{h_1}^2}{{h_2}^2}\frac{{\partial }_2{h}_1}{h_1}\frac{{\dot{\lambda }}_1{\dot{\lambda }}_1}{{\dot{\lambda }}_2}-\frac{{h_1}^2}{{h_2}^2}\frac{h_2}{h_1}\frac{{\partial }_1{h}_2}{h_1}\frac{{\dot{\lambda }}_2{\dot{\lambda }}_1}{{\dot{\lambda }}_2}+\frac{{h_2}^2}{{h_2}^2}\frac{{\partial }_1{h}_2}{h_2}\frac{{\dot{\lambda }}_1{\dot{\lambda }}_2}{{\dot{\lambda }}_2}+\frac{{h_2}^2}{{h_2}^2}\frac{{\partial }_2{h}_2}{h_2}\frac{{\dot{\lambda }}_2{\dot{\lambda }}_2}{{\dot{\lambda }}_2}+\frac{{h_3}^2}{{h_2}^2}\frac{{\partial }_2{h}_3}{h_3}\frac{{\dot{\lambda }}_3{\dot{\lambda }}_3}{{\dot{\lambda }}_2}.$

And limiting our interest to motion within the meridional plane (setting ${\dot{\lambda }}_3=0$) and simplifying

CV.02

$\frac{d\mathrm{\Xi }}{dt}\equiv -\frac{d\ln C_2}{dt}$	$=$	$\frac{h_1}{h_2}\frac{{\partial }_2{h}_1}{h_2}\frac{{{\dot{\lambda }}_1}^2}{{\dot{\lambda }}_2}-\cancel{\frac{{\partial }_1{h}_2}{h_2}{\dot{\lambda }}_1}+\cancel{\frac{{\partial }_1{h}_2}{h_2}{\dot{\lambda }}_1}+\frac{{\partial }_2{h}_2}{h_2}{\dot{\lambda }}_2$
	$=$	$\left[\right(\frac{h_1{\dot{\lambda }}_1}{h_2{\dot{\lambda }}_2}{)}^2\frac{\partial \ln h_1}{\partial \ln {\lambda }_2}+\frac{\partial \ln h_2}{\partial \ln {\lambda }_2}]\frac{d\ln {\lambda }_2}{dt}$
NOTE: Sign error fixed on 15 July 2010. Specifically, $d\mathrm{\Xi }/dt\equiv d\ln C_2/dt$ changed to $d\mathrm{\Xi }/dt\equiv -d\ln C_2/dt$.

Application to T3 Coordinates

On a separate page, we have applied this "characteristic vector" approach specifically to T3 coordinates.

Related Talk Session

Joel's First Talk Message to Jay

Jay: In the email message that you sent to me today, you indicated that you had added the 3rd component to the dx/dt equation. I don't see that modification in the current T3Coordinates page. --Tohline 15:05, 29 May 2010 (MDT)

Joel: It's at the bottom of the section entitled "Time-Derivative of Position and Velocity Vectors". Under the history tab, can you see the edits that I made? It was the very first edit and was fairly minor. I only added a couple terms. --Jaycall 20:17, 29 May 2010 (MDT)

The mediaWiki "talk" page recommends that when you reply to a query, you should not start a new subsection but, rather, just indent (using one or more colons) immediately following the query. Also, it recommends that you "sign" each "talk" message. You do this by typing 2 dashes followed by 4 tildes! I'm going to edit your "reply" (and this additional one from me) to put it in this recommended format. But you have to "sign" your own remark. --Tohline 16:23, 29 May 2010 (MDT)

By the way, I can see the edits that you made to add the 3rd component to the dx/dt equation. I did not spot the changes earlier, but via the history "diff" function, I'm able to see the edits clearly. --Tohline 16:41, 29 May 2010 (MDT)

Killing Vector Approach

Thanks for writing out all the terms in the expression for $d\mathrm{\Xi }/dt$ in your "Killing Vector Approach" discussion. On the one hand, I'm happy that the term I mentioned cancels with another one because that means we have fewer terms to integrate. On the other hand, it was one term that I thought we might actually be able to manipulate analytically. Of the two terms that remain, one is relatively simple — the one that is proportional to ${\dot{\lambda }}_2$ — but the first term is a bear! That will take some thinking! --Tohline 10:50, 30 May 2010 (MDT)

I know it seems like the wrong direction, but unless we can express this summary integral entirely in terms of ${\lambda }_1$, ${\lambda }_2$, ${\dot{\lambda }}_1$ and ${\dot{\lambda }}_2$, I think that in order to make progress we're going to have to write things out in terms of $\varpi$, $z$, $\dot{\varpi }$ and $\dot{z}$. Of course, that will make the expression much messier, but I think it will be very difficult to recognize this quantity as an exact derivative unless everything's in terms of the same coordinates. --Jaycall 12:55, 30 May 2010 (MDT)

I agree. It is for similar reasons that I am planning on focusing on the case where $q^2=2$. At least in this special case we can invert the coordinate expressions to give $\varpi$ and $z$ in terms of the $\lambda$s. Incidentally, I have added a link to your "Killing Vector" page as well as a link to this associated "Talk" page on the page where I itemize my various "ramblings". --Tohline 16:16, 30 May 2010 (MDT)

Good point. That would be a good reason to focus on one specific case. I haven't looked closely at the inverted coordinate transformation for the $q^2=2$ case for several days. Since the relation is quadratic, is it obvious which root should be used?

The proper physical root was obvious to me when I performed the coordinate inversion in the case of T1 Coordinates, so I presume it will be obvious for the T3 Coordinate system. But the inversion needs to be redone for the case of T3 Coordinates; do you want to take care of this and type up the result? In the (quadratic) case of $q^2=2$, it may be as simple as replacing ${\chi }_2$ with $1/{\lambda }_2$. --Tohline 11:21, 31 May 2010 (MDT)

Sure. --Jaycall 13:34, 31 May 2010 (MDT)

Mistake? Please check the next to last row in the tabulated expressions for Christoffel symbols; shouldn't the $i$ and $j$ indexes be swapped? --Tohline 09:39, 6 June 2010 (MDT)

You are correct. Good eye.--Jaycall 09:42, 9 June 2010 (MDT)

Actually, it wasn't my good eye. I discovered this from my inelegant brute-force derivations. In particular, when I was trying to derive the same form of the equation of motion from two different points of view, I had to plug in some of your Christoffel symbol expressions. At first I could not get my EOM to match yours; it was while trying to understand this mismatch that I realized that your general expression for ${\mathrm{\Gamma }}_{ij}^i$ did not match the specific expression for ${\mathrm{\Gamma }}_{21}^2$ that you typed on the "T3 Coordinates" page. In my effort to get the EOMs to match, I decided that your general expression was the incorrect one. --Tohline 14:58, 9 June 2010 (MDT)

For the record, on 10 June 2010, I modified the table that contains the general Christoffel symbol expressions in order to correct the order of these indexes.--Tohline 18:55, 11 June 2010 (MDT)

Logarithmic Derivatives of T3 Scale Factors

Check out my new subsection (under T3 Coordinates) entitled Logarithmic Derivatives of Scale Factors. First, see if you agree that there is a mistake (typo) in one of your tables of partial derivatives. Specifically, I think that the correct expression is:

$\frac{\partial z}{\partial {\lambda }_2}=-(q^2-1)\frac{{\varpi }^{2}z{\ell }^2}{{\lambda }_2}.$

Second, see if you agree with my derived expressions for $\partial \ln h_i/\partial \ln {\lambda }_j$. --Tohline 22:28, 31 May 2010 (MDT)

You're certainly right about the sign error. I have corrected the expression in the relevant table. --Jaycall 14:28, 1 June 2010 (MDT)

I have not yet been able to confirm your expressions for logarithmic derivatives of the scale factors. I'm not sure what approach you took in deriving them, but since I had already calculated partials of the scale factors (${\partial }_i{h}_j$ and so forth), I derived a little trick to help simplify the work. Do you agree that

$\frac{\partial \ln h_i}{\partial \ln {\lambda }_j}\equiv \frac{{\lambda }_j}{h_i}{\partial }_j{h}_i?$

Yes, this is precisely how I defined the logarithmic derivatives. And I used your tables of partial derivatives (e.g., $\partial \varpi /\partial {\lambda }_j$ and $\partial z/\partial {\lambda }_j$) to obtain ${\partial }_i{h}_j$ and so forth. --Tohline 17:30, 1 June 2010 (MDT)

Mistake! Jay: I have found a mistake in my original derivation of the logarithmic derivatives of the $h_1$ scale factor. Perhaps now they will match yours. Check it out and let me know! --Tohline 12:41, 4 June 2010 (MDT)

Joel: I am finally able to confirm your derivation of the logarithmic derivatives of the scale factors. I came at them from a different angle with a fresh head, and I am now confident that what you have is correct. Next I will look into the general case. --Jaycall 14:29, 10 July 2010 (MDT)

Two Views of Equation of Motion

Jay: I have created a new wiki page to explicitly analyze how your Characteristic Vector formalism applies to the T3 Coordinate system. On this page is a subsection entitled "Two Views of Equation of Motion" in which I have shown that the $2^{\mathrm{n}\mathrm{d}}$ component of the equation of motion can be written in the following form,

$\frac{d(h_2^2{\dot{\lambda }}_2)}{dt}=\left(h_1\frac{\partial h_1}{\partial {\lambda }_2}\right){\dot{\lambda }}_1^2+\left(h_2\frac{\partial h_2}{\partial {\lambda }_2}\right){\dot{\lambda }}_2^2$ ,

assuming that $\partial \mathrm{\Phi }/\partial {\lambda }_2=0$. More to the point, I have shown that I can derive this form of this component of the equation of motion using either (A) your Christoffel symbol formalism, or (B) the more classical formalism that I have been using, which I obtained from Appendix 1.B of Binney & Tremaine (1987) and Morse & Feshbach (1953).

Now, I shouldn't be shocked that both derivations lead to the same form of the equation — after all, we're doing physics aren't we? But here is what surprised me: The Christoffel symbol formalism produces this form of the equation without ever assuming a particular coordinate system whereas, in order to derive the equation using the more classical formalism, I plugged in some relations between the partial derivatives of the scale factors that I thought were specific to T3 coordinates. This leads me to ask, "How many of the T3 Coordinate relations derived earlier are generic relations that apply to any orthogonal, axisymmetric coordinate system, and how many are unique to the T3 Coordinate System?" I'm particularly interested in knowing how generalizable the relations are that appear inside boxes labeled "T3 Coordinates" under the subsections I have entitled, "Derived Identity for T3 Corodinates" and "Logarithmic Derivatives of Scale Factors". Can you answer this? --Tohline 10:46, 7 June 2010 (MDT)

Joel: Yes, this is the power of using a covariant formulation. The only assumptions we made were that the coordinates were orthogonal and axisymmetric. I believe that the relation under "Derived Identity for T3 Coordinates" would apply to any coordinate system meeting these criteria. I will double check, though. (How did you determine that the position vector could be written $\overrightarrow{x}={\hat{e}}_1(h_1{\lambda }_1)+{\hat{e}}_2(h_2{\lambda }_2)$?)

The logarithmic derivatives of the scale factors cannot be general in their current form due to the appearance of factors of $q$, but I suspect they can be generalized and molded into a form that closely resembles what you have. I will work on this, too. --Jaycall 10:39, 9 June 2010 (MDT)

Jay: You asked how I determined the expression for the position vector. I don't have my detailed notes with me right now, but I can outline the steps. In Morse & Feshbach's classical presentation of orthogonal curvilinear coordinates, they define a set of so-called direction cosines, ${\gamma }_{ij}$. (I don't have the generalized definition with me at this time, but they are expressible in terms of derivatives of the scale factors with respect to the various coordinates; I'm certain direction cosines are definable in terms of Christoffel symbols.) Each of the Cartesian unit vectors ($\hat{ı},\hat{ȷ},\hat{k}$) can be written in terms of products of these direction cosines and the three unit vectors of your chosen curvilinear coordinate system (${\hat{e}}_1,{\hat{e}}_2,{\hat{e}}_3$). According to Morse & Feshbach, for example, $\hat{ı}=({\hat{e}}_1{\gamma }_{11}+{\hat{e}}_2{\gamma }_{21}+{\hat{e}}_3{\gamma }_{31})$. So, all I did to obtain an expression for the position vector in T3 coordinates was define $\overrightarrow{x}=(\hat{ı}x+\hat{ȷ}y+\hat{k}z)$, then replace each Cartesian unit vector by its expression in terms of products of the T3-coordinate unit vectors & the appropriate direct cosine, then gather together and simplify terms. Does this make sense? --Tohline 14:59, 10 June 2010 (MDT)

Yes, it does. That's very helpful. --Jaycall 09:49, 11 June 2010 (MDT)

Jay: For the record, I have put together a wiki page that discusses the definition and utility of Direction Cosines.--Tohline 18:47, 3 July 2010 (MDT)

Joel, I finally figured out what was going on this afternoon with the strange conserved quantities we were getting, like ${h_2}^4{\dot{{\lambda }_2}}^2$ and, consequently, ${\lambda }_1{\lambda }_2$. There is a mistake in the derivation stemming all the way back from Eq. CV.02 on my page on the characteristic vector approach. My derivation of the brute force condition on $C_2$ was correct, but you can see from above that $\mathrm{\Xi }$ does not equal $\ln C_2$, it equals $-\ln C_2$. Consequently, the conclusion is no longer that $C_2$ must equal ${h_2}^2\dot{{\lambda }_2}$, but rather ${\left({h_2}^2\dot{{\lambda }_2}\right)}^{-1}$. So in fact you did find a conserved quantity--$1$!  : ) I guess it was a tautology after all; your intuition was right. --Jaycall 18:50, 14 July 2010 (MDT)

Jay, very good catch! Looking back through the history of that wiki page, it is clear that I am the one who introduced this sign error. I mistakingly set $\mathrm{\Xi }=\ln C_2$ in equation CV.02 at the same time that I "Added a couple of right-justified equation numbers." I'll march through the relevent wiki pages today and attempt to correct this mistake wherever it appears. --Tohline 08:30, 15 July 2010 (MDT)

Joel, I'm sorry yesterday's excitement didn't pan out. It would have been really neat. Well, at least I learned something. On my drive home yesterday I was thinking about what it meant for there to be a constraint that involved only the coordinates (and not their rates of change). We were uncomfortable with this idea, but now I understand why in general no conserved quantity can take this form. If there were a constraint on only the coordinates, then this would limit a particle to a curve or a surface...a subspace of physical space (as opposed to a subspace of phase space). Yet it would definitely be possible to get a particle to leave the subspace by giving it an initial velocity in any direction normal to the subspace. Any conserved quantity must definitely involve the coordinate velocities (unless you're artificially restricting the motion). --Jaycall 14:15, 15 July 2010 (MDT)

Good observation. --Tohline 16:28, 15 July 2010 (MDT)

Special Quadratic Case

Jay: On a new accompanying wiki page I have begun to investigate how our system behaves in the special case of $q^2=2$. I have added a pointer/link to this new page — and a separate pointer/link to the parallel investigation that you have told me you are conducting — on the page containing a table of contents to my various research ramblings. I have not yet actually gone to the page that details your analysis because, at least for a while, I prefer for our trains of thought to remain independent of one another. However, in an effort to simplify the comparison that we ultimately will need to make between our independent analyses, I strongly recommend that you write your expressions for the scale factors, etc. for the special case of $q^2=2$ in terms of a variable, $\mathrm{\Lambda }$, defined such that,

$\mathrm{\Lambda }\equiv [1+(\frac{2{\lambda }_1}{{\lambda }_2}{)}^2{]}^{1/2}=\cosh (2\mathrm{Z}).$

By the way, if you don't agree that $\sqrt{1+(2{\lambda }_1/{\lambda }_2{)}^2}$ equals $\cosh (2\mathrm{Z})$, please let me know!--Tohline 19:49, 11 June 2010 (MDT)

Joel: I have checked and agree that in the $q^2=2$ case only, the above relation among $\mathrm{\Lambda }$, $Z$ and the T3 coordinates is correct. --Jaycall 14:58, 10 July 2010 (MDT)