Calculating the moment of inertia [Archive]

jason cooper

Apr15-04, 11:21 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Classical mechanics question... I want to calculate the moment\nof inertia tensor of a molecule. Since (LaTeX notation here):\n\nI_{\\alpha\\beta} = \\frac{\\partial^{2}T}{\\partial\\dot{q}^{\\alpha} \n\\partial\\dot{q}^{\\beta}}\n\nwhere q^{\\alpha} is a rotational coordinate and the dot\nrepresents the time derivative, and since:\n\n2T = \\dot{y}^{i}\\dot{y}^{j}\\delta_{ij}\n\nwhere y^{i} is a 3N-dimensional (N being the number of atoms)\nvector in mass-weighted Cartesians, I\'ve had excellent success\nusing the definition:\n\nI_{\\alpha\\beta} = H_{\\alpha}^{i}H_{\\beta}^{j}\\delta_{ij}\n\nwhere the tensor H is defined by:\n\nH_{\\alpha}^{i} = dy^{i}/dq^{\\alpha}\n\n(notice no more dots) ie, the columns of H give the displacements\nin 3N mass-wieghted Cartesians (MWC) which correspond to each\nrotational displacement dq^{\\alpha}. I won\'t bore you with what\nI do from there, but the point is that this turns out to be a\nvery convenient way for me to express \'I\'.\n\nNow, I can justify why the above might be true, but what I would\n*really* like to know is that I\'m not the first person to do\nthings this way -- and given the long history of classical\nmechanics, I find it exceedingly hard to believe that I am.\n\nHas anybody here seen this sort of notation used before? Can you\nsupply a literature reference or some places where I can look to\nfind this sort of notation mentioned?\n\nAlternatively, can you poke a hole in the above?\n\nThanks!\n\n-----------------------------------------------------------------\n. . . Except when they don\'t,\nBecause sometimes they won\'t. - Dr. Seuss\n-----------------------------------------------------------------\nJason Cooper jcooper@acs.ucalgary.ca\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Classical mechanics question... I want to calculate the moment
of inertia tensor of a molecule. Since (LaTeX notation here):

I_{\alpha\beta} = \frac{\partial^{2}T}{\partial\dot{q}^{\alpha}[/itex]
[itex]\partial\dot{q}^{\beta}}

where q^{\alpha} is a rotational coordinate and the dot
represents the time derivative, and since:

2T = \dot{y}^{i}\dot{y}^{j}\delta_{ij}

where y^{i} is a 3N-dimensional (N being the number of atoms)
vector in mass-weighted Cartesians, I've had excellent success
using the definition:

I_{\alpha\beta} = H_{\alpha}^{i}H_{\beta}^{j}\delta_{ij}

where the tensor H is defined by:

H_{\alpha}^{i} = dy^{i}/dq^{\alpha}

(notice no more dots) ie, the columns of H give the displacements
in 3N mass-wieghted Cartesians (MWC) which correspond to each
rotational displacement dq^{\alpha}. I won't bore you with what
I do from there, but the point is that this turns out to be a
very convenient way for me to express 'I'.

Now, I can justify why the above might be true, but what I would
*really* like to know is that I'm not the first person to do
things this way -- and given the long history of classical
mechanics, I find it exceedingly hard to believe that I am.

Has anybody here seen this sort of notation used before? Can you
supply a literature reference or some places where I can look to
find this sort of notation mentioned?

Alternatively, can you poke a hole in the above?

Thanks!

-----------------------------------------------------------------
. . . Except when they don't,
Because sometimes they won't. - Dr. Seuss
-----------------------------------------------------------------
Jason Cooper jcooper@acs.ucalgary.ca

Calvin Ritchie

Apr19-04, 01:39 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Jason Cooper wrote, in part,\n\n>Classical mechanics question... I want to calculate the moment\n>of inertia tensor of a molecule. Since (LaTeX notation here):\n> I_{\\alpha\\beta} = \\frac{\\partial^{2}T}{\\partial\\dot{q}^{\\alpha} \n> \\partial\\dot{q}^{\\beta}}\n>where q^{\\alpha} is a rotational coordinate and the dot\n>represents the time derivative, and since:\n> 2T = \\dot{y}^{i}\\dot{y}^{j}\\delta_{ij}\n>where y^{i} is a 3N-dimensional (N being the number of atoms)\n>vector in mass-weighted Cartesians, I\'ve had excellent success\n>using the definition:\n> I_{\\alpha\\beta} = H_{\\alpha}^{i}H_{\\beta}^{j}\\delta_{ij}\n>where the tensor H is defined by:\n> H_{\\alpha}^{i} = dy^{i}/dq^{\\alpha}\n>(notice no more dots) ie, the columns of H give the displacements\n>in 3N mass-wieghted Cartesians (MWC) which correspond to each\n>rotational displacement dq^{\\alpha}. I won\'t bore you with what\n>I do from there, but the point is that this turns out to be a\n>very convenient way for me to express \'I\'.\n>Now, I can justify why the above might be true, but what I would\n>*really* like to know is that I\'m not the first person to do\n>things this way -- and given the long history of classical\n>mechanics, I find it exceedingly hard to believe that I am.\n>Has anybody here seen this sort of notation used before? Can you\n>supply a literature reference or some places where I can look to\n>find this sort of notation mentioned?\n\nI assume that you\'re considering a "rigid" molecule. If so, your\nnotation confuses me a bit, but I have considerable reservations about the\ncorrectness of your final equation, writing the inertial tensor in terms of\nthe H matrix. It looks to me that your off-diagonal I_(ab) have the wrong\nsign (but otherwise ok if I understand your H as, essentially, a rotation\nmatrix) but that your diagonal I_(aa) is incorrect.\nLet\'s start by writing Cartesian center of mass coordinates, j=1,2,3,\nfor the each of your N atoms as r^(nj), with n=1,N. Now, let\'s go to your\n"mass weighted" coordinates, y^(nj)=Sqrt(m_n)r^(nj). We may write the moment\nof inertia tensor, essentially by definition, about these coordinates as:\n\nI^(jk)=-Sum_over_n{y^(nj)y^(nk)}=I^(kj) for j N.E. k,\nand\nI^(jj)=Sum_over_ k N.E. j{Sum over n[y^(nk)y^(nk)]}.\n\nNote that the diagonal element, I^(jj) is just the sum of moments of each\nparticle about the axis (i.e. coordinate) j ..... the mass of the particle\ntimes its perpendicular distance from the j-axis. The j coordinate of each\nparticle MUST be missing.\nThe expression for I is sometimes written as the sum of two terms,\na diagonal one with elements of\nSums_over_ALL_j and n{y^(nj)y(nj)} times the 3x3 identity,\nminus a symmetric matrix, D,\nD^(j,k)=Sum_over_n{y^(nj)y^(nk)}. (for all j and k, including j=k).\nIt appears to me that your expression is just the D part.\nThe I matrix (tensor, if you wish) is symmetric and real, and may be\ndiagonalized by a real unitary (i.e. rotational) matrix. That\'s just a\nrotation of the original Cartesian coordinate axes to new orthogonal ones.\nEach particle\'s coordinates are then just written w.r.t. the new coordinate\naxes, and the off-diagonal elements of I vanish.\nIt looks to me like you\'re writing some sort of rotation of coordinates\nas your H matrix, but the notation confuses me.\n\nI have a "document", written with MS Word and MathType for my own\nedification and quick reference, titled "General Non-relativistic System of\nMany Classical Interacting Point Particles", which goes through the whole\nformalism, including Lagrangians and Hamiltonians. The whole thing is about\n17 pages. If you\'d like to see it and can read Word docs, I\'d be happy to\nsend you a copy by e-mail.\n\nThe rotations of a rigid body are covered in every decent text on Classical\nMechanics. My favorites are:\n\n(1) Fairly elementary; Corben & Stehle, "Classical Mechanics", 2nd Ed.,\nDover Press, (repub.) 1994; see Chapter 9, and particularly pgs. 146-148.\n\n(2) THE classic (IMO, should be on every Physicist\'s and Physical Chemist\'s\nbookshelf) V. I. Arnold, Mathematical Methods of Classical Mechanics, 2nd\nEd., Springer, 1978, 1989; see Chapter 6, and particularly pgs. 133-147.\n\nIf you want to begin to understand gauge theories, btw, rotating coordinates\nis a good place to start.\n\nDon Ritchie\nSubmitted Thursday evening, 15 April 2004\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Jason Cooper wrote, in part,

>Classical mechanics question... I want to calculate the moment
>of inertia tensor of a molecule. Since (LaTeX notation here):
> I_{\alpha\beta} = \frac{\partial^{2}T}{\partial\dot{q}^{\alpha}
> \partial\dot{q}^{\beta}}
>where q^{\alpha} is a rotational coordinate and the dot
>represents the time derivative, and since:
> 2T = \dot{y}^{i}\dot{y}^{j}\delta_{ij}
>where y^{i} is a 3N-dimensional (N being the number of atoms)
>vector in mass-weighted Cartesians, I've had excellent success
>using the definition:
> I_{\alpha\beta} = H_{\alpha}^{i}H_{\beta}^{j}\delta_{ij}
>where the tensor H is defined by:
> H_{\alpha}^{i} = dy^{i}/dq^{\alpha}
>(notice no more dots) ie, the columns of H give the displacements
>in 3N mass-wieghted Cartesians (MWC) which correspond to each
>rotational displacement dq^{\alpha}. I won't bore you with what
>I do from there, but the point is that this turns out to be a
>very convenient way for me to express 'I'.
>Now, I can justify why the above might be true, but what I would
>*really* like to know is that I'm not the first person to do
>things this way -- and given the long history of classical
>mechanics, I find it exceedingly hard to believe that I am.
>Has anybody here seen this sort of notation used before? Can you
>supply a literature reference or some places where I can look to
>find this sort of notation mentioned?

I assume that you're considering a "rigid" molecule. If so, your
notation confuses me a bit, but I have considerable reservations about the
correctness of your final equation, writing the inertial tensor in terms of
the H matrix. It looks to me that your off-diagonal I_(ab) have the wrong
sign (but otherwise ok if I understand your H as, essentially, a rotation
matrix) but that your diagonal I_(aa) is incorrect.
Let's start by writing Cartesian center of mass coordinates, j=1,2,3,
for the each of your N atoms as r^(nj), with n=1,N. Now, let's go to your
"mass weighted" coordinates, y^(nj)=\Sqrt(m_n)r^(nj). We may write the moment
of inertia tensor, essentially by definition, about these coordinates as:

I^(jk)=-Sum_over_n{y^(nj)y^(nk)}=I^(kj)[/itex] for j N.E. k,
and
I^(jj)=Sum_over_ k N.E. j{Sum over [itex]n[y^(nk)y^(nk)]}.

Note that the diagonal element, I^(jj) is just the sum of moments of each
particle about the axis (i.e. coordinate) j ..... the mass of the particle
times its perpendicular distance from the j-axis. The j coordinate of each
particle MUST be missing.
The expression for I is sometimes written as the sum of two terms,
a diagonal one with elements of
Sums_over_ALL_j and n{y^(nj)y(nj)} times the 3x3 identity,
minus a symmetric matrix, D,
D^(j,k)=Sum_over_n{y^(nj)y^(nk)}. (for all j and k, including j=k).
It appears to me that your expression is just the D part.
The I matrix (tensor, if you wish) is symmetric and real, and may be
diagonalized by a real unitary (i.e. rotational) matrix. That's just a
rotation of the original Cartesian coordinate axes to new orthogonal ones.
Each particle's coordinates are then just written w.r.t. the new coordinate
axes, and the off-diagonal elements of I vanish.
It looks to me like you're writing some sort of rotation of coordinates
as your H matrix, but the notation confuses me.

I have a "document", written with MS Word and MathType for my own
edification and quick reference, titled "General Non-relativistic System of
Many Classical Interacting Point Particles", which goes through the whole
formalism, including Lagrangians and Hamiltonians. The whole thing is about
17 pages. If you'd like to see it and can read Word docs, I'd be happy to
send you a copy by e-mail.

The rotations of a rigid body are covered in every decent text on Classical
Mechanics. My favorites are:

(1) Fairly elementary; Corben & Stehle, "Classical Mechanics", 2nd Ed.,
Dover Press, (repub.) 1994; see Chapter 9, and particularly pgs. 146-148.

(2) THE classic (IMO, should be on every Physicist's and Physical Chemist's
bookshelf) V. I. Arnold, Mathematical Methods of Classical Mechanics, 2nd
Ed., Springer, 1978, 1989; see Chapter 6, and particularly pgs. 133-147.

If you want to begin to understand gauge theories, btw, rotating coordinates
is a good place to start.

Don Ritchie
Submitted Thursday evening, 15 April 2004

Calvin Ritchie

Apr21-04, 10:53 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nJason:\nI awoke during the night last night and realized that I had completely\nmissed your point when I replied to your query about moments of inertia. As\na result, my reply was overly pedantic and condescending. I apologize.\nYou wrote:\n>........ I\'ve had excellent success\n>using the definition:\n> I_{\\alpha\\beta} = H_{\\alpha}^{i}H_{\\beta}^{j}\\delta_{ij}\n>where the tensor H is defined by:\n> H_{\\alpha}^{i} = dy^{i}/dq^{\\alpha}\n>(notice no more dots) ie, the columns of H give the displacements\n>in 3N mass-wieghted Cartesians (MWC) which correspond to each\n>rotational displacement dq^{\\alpha}. .......\n\nYour q^{\\alpha} are (now) obviously the angles of rotation about the\nCartesian axis alpha. That\'s the point that I missed.\nAlthough I\'ve never seen precisely that expression for the moment of\ninertia tensor, it is certainly implied throughout any development.\nNevertheless, it\'s a clever and economical expression. It\'s easiest to see\nin the vector cross product expressions, for example pg. 136 ff of the\nArnold reference that I gave in my last message.\n\nDon Ritchie\nsubmitted morning Friday 16 April\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Jason:
I awoke during the night last night and realized that I had completely
missed your point when I replied to your query about moments of inertia. As
a result, my reply was overly pedantic and condescending. I apologize.
You wrote:
>........ I've had excellent success
>using the definition:
> I_{\alpha\beta} = H_{\alpha}^{i}H_{\beta}^{j}\delta_{ij}
>where the tensor H is defined by:
> H_{\alpha}^{i} = dy^{i}/dq^{\alpha}
>(notice no more dots) ie, the columns of H give the displacements
>in 3N mass-wieghted Cartesians (MWC) which correspond to each
>rotational displacement dq^{\alpha}. .......

Your q^{\alpha} are (now) obviously the angles of rotation about the
Cartesian axis \alpha. That's the point that I missed.
Although I've never seen precisely that expression for the moment of
inertia tensor, it is certainly implied throughout any development.
Nevertheless, it's a clever and economical expression. It's easiest to see
in the vector cross product expressions, for example pg. 136 ff of the
Arnold reference that I gave in my last message.

Don Ritchie
submitted morning Friday 16 April

jason cooper

Apr22-04, 04:17 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Calvin Ritchie (DonRitchie870@csWebmail.com) wrote:\n\n: Your q^{\\alpha} are (now) obviously the angles of rotation about the\n: Cartesian axis alpha. That\'s the point that I missed.\n: Although I\'ve never seen precisely that expression for the moment of\n: inertia tensor, it is certainly implied throughout any development.\n: Nevertheless, it\'s a clever and economical expression. It\'s easiest to see\n: in the vector cross product expressions, for example pg. 136 ff of the\n: Arnold reference that I gave in my last message.\n\nThanks for the tip. Actually, your last message spurred me to\ncheck the formula I gave directly against the cross products.\nThe reference should be helpful.\n\n-----------------------------------------------------------------\n. . . Except when they don\'t,\nBecause sometimes they won\'t. - Dr. Seuss\n-----------------------------------------------------------------\nJason Cooper jcooper@acs.ucalgary.ca\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Calvin Ritchie (DonRitchie870@csWebmail.com) wrote:

: Your q^{\alpha} are (now) obviously the angles of rotation about the
: Cartesian axis \alpha. That's the point that I missed.
: Although I've never seen precisely that expression for the moment of
: inertia tensor, it is certainly implied throughout any development.
: Nevertheless, it's a clever and economical expression. It's easiest to see
: in the vector cross product expressions, for example pg. 136 ff of the
: Arnold reference that I gave in my last message.

Thanks for the tip. Actually, your last message spurred me to
check the formula I gave directly against the cross products.
The reference should be helpful.

-----------------------------------------------------------------
. . . Except when they don't,
Because sometimes they won't. - Dr. Seuss
-----------------------------------------------------------------
Jason Cooper jcooper@acs.ucalgary.ca