View Full Version : Lie derivative
Albert Ng
Jun10-04, 06:47 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>\nCan anyone explain the meaning of Lie derivative in a simple way ?\nThank you\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>Can anyone explain the meaning of Lie derivative in a simple way ?
Thank you
Igor Khavkine
Jun11-04, 06:29 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"Albert Ng" <asgc71@dsl.pipex.com> wrote in message news:<40c7ba93\\$0\\$280\\$cc9e4d1f@news.dial.pipe x.com>...\n> Can anyone explain the meaning of Lie derivative in a simple way ?\n\nI am not sure what your level of mathematical sophistication is, so\nI\'ll try to describe it as lay terms as I can.\n\nSuppose you have a region of space M (the technical term would be\na manifold). At every point x of this region, you attach an object A(x)\nof a particular type (the types that people are usually consider\nare scalars, vectors, tensors, etc.). Now suppose that you have\ntwo exact copies of this region (call them M1 and M2) with the\nobjects defined on the regions copied as well (call them A1(x) and A2(x)).\n\nSince M1 and M2 are exact copies of each other they can be identified\npointwise by associating a point of M1 to the corresponding point of M2.\nBut M1 and M2 can be identified in more than one way. This identification\nis called a diffeomorphism, and the canonical association that we get\nfrom the fact that M1 and M2 are exact copies is called the identity\ndiffeomorphism.\n\nOnce again, since we have two exact copies of the same thing, the objects\nA1(x) and A2(y) defined on top of M1 and M2 (when x and y are associated\nby the identity diffeomorphism) are exactly the same. That is not true,\nhowever, if the identification between M1 and M2 is not trivial, the\nwe\'ll have A2(y) = A1(x) + dA(x).\n\nThe simplest example is the real line identified with a copy of itself\nthat was moved to the right by amount a (x -> x + a). So, if there was a curve\ndefined on the real line with a peak at 0, the peak will have moved\nto a.\n\nNow suppose that we have a family of these associations that we label\nas f_t, where t is a continuous parameter. We require that when t=0,\nf_0 is just the identity diffeomorphism. We also require that f_t changes\nsmoothly with t. That is if y(t) = f_t(x) is the point of M2 associated\nto the point x of M1, y(t) is a smooth curve in M2 as a function of t.\nThrough each point y of M2 should pass exactly one such curve. And to\neach point of every such curve in M2 we can associate a vector tangent\nto it. This is how we can go from a family of diffeomorphisms f_t to\na vector field.\n\nNow we are getting to the part about Lie derivatives. Reversing the\nconstruction from the above paragraph, it should be obvious that given\na vector field, we can construct a family of diffeomorphisms with the\nsame properties that were required for f_t. When t=0, we have\nA2(y_0) = A1(x), where y_0 = f_0(x). If t is non zero we have\nA2(y_0) = A1(x) + dA(x,t), where dA(x,t) -> 0 as t -> 0. So by smoothness\nof f_t, for t very small we can approximate dA(x,t) = t * A\'(x).\n\nThe Lie derivative of the object A1(x) along the vector field we were\ngiven is precisely the the object A\'(x).\n\nI have glossed over some subtleties, but I hope this spirit of the\ndefinition of the Lie derivative is clear.\n\nHope this helps.\n\nIgor\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>"Albert Ng" <asgc71@dsl.pipex.com> wrote in message news:<40c7ba93$0$280$cc9e4d1f@news.dial.pipex.com>...
> Can anyone explain the meaning of Lie derivative in a simple way ?
I am not sure what your level of mathematical sophistication is, so
I'll try to describe it as lay terms as I can.
Suppose you have a region of space M (the technical term would be
a manifold). At every point x of this region, you attach an object A(x)
of a particular type (the types that people are usually consider
are scalars, vectors, tensors, etc.). Now suppose that you have
two exact copies of this region (call them M1 and M2) with the
objects defined on the regions copied as well (call them A1(x) and A2(x)).
Since M1 and M2 are exact copies of each other they can be identified
pointwise by associating a point of M1 to the corresponding point of M2.
But M1 and M2 can be identified in more than one way. This identification
is called a diffeomorphism, and the canonical association that we get
from the fact that M1 and M2 are exact copies is called the identity
diffeomorphism.
Once again, since we have two exact copies of the same thing, the objects
A1(x) and A2(y) defined on top of M1 and M2 (when x and y are associated
by the identity diffeomorphism) are exactly the same. That is not true,
however, if the identification between M1 and M2 is not trivial, the
we'll have A2(y) = A1(x) + dA(x).
The simplest example is the real line identified with a copy of itself
that was moved to the right by amount a (x -> x + a). So, if there was a curve
defined on the real line with a peak at 0, the peak will have moved
to a.
Now suppose that we have a family of these associations that we label
as f_t, where t is a continuous parameter. We require that when t=0,f_0 is just the identity diffeomorphism. We also require that f_t changes
smoothly with t. That is if y(t) = f_t(x) is the point of M2 associated
to the point x of M1, y(t) is a smooth curve in M2 as a function of t.
Through each point y of M2 should pass exactly one such curve. And to
each point of every such curve in M2 we can associate a vector tangent
to it. This is how we can go from a family of diffeomorphisms f_t to
a vector field.
Now we are getting to the part about Lie derivatives. Reversing the
construction from the above paragraph, it should be obvious that given
a vector field, we can construct a family of diffeomorphisms with the
same properties that were required for f_t. When t=0, we have
A2(y_0) = A1(x), where y_0 = f_0(x). If t is non zero we have
A2(y_0) = A1(x) + dA(x,t), where dA(x,t) -> as t -> . So by smoothness
of f_t, for t very small we can approximate dA(x,t) = t * A'(x).
The Lie derivative of the object A1(x) along the vector field we were
given is precisely the the object A'(x).
I have glossed over some subtleties, but I hope this spirit of the
definition of the Lie derivative is clear.
Hope this helps.
Igor
<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>Igor Khavkine wrote:\n\n>\n> "Albert Ng" <asgc71@dsl.pipex.com> wrote in message\n> news:<40c7ba93\\$0\\$280\\$cc9e4d1f@news.dial.pipe x.com>...\n>> Can anyone explain the meaning of Lie derivative in a simple way ?\n>\n> I am not sure what your level of mathematical sophistication is, so\n> I\'ll try to describe it as lay terms as I can.\n\nI think this is the physical intuition one could have for the Lie\nderivative:\n\nThe Lie derivative of V with respect to a vector field X, measures how much\nthe object V (scalar, vector, ...) changes along the flow generated by the\nvector field X.\n\nExample: Take for the vector field X the velocity field of, say, a river.\nFor V we take the temperature scalar field. Now take a light particle and\nlet if flow (litterally) with the current of the river. L_x(V) then\nmeasures the temperature change.\n\nbest,\nJeroen\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>Igor Khavkine wrote:
>
> "Albert Ng" <asgc71@dsl.pipex.com> wrote in message
> news:<40c7ba93$0$280$cc9e4d1f@news.dial.pipex.com>...
>> Can anyone explain the meaning of Lie derivative in a simple way ?
>
> I am not sure what your level of mathematical sophistication is, so
> I'll try to describe it as lay terms as I can.
I think this is the physical intuition one could have for the Lie
derivative:
The Lie derivative of V with respect to a vector field X, measures how much
the object V (scalar, vector, ...) changes along the flow generated by the
vector field X.
Example: Take for the vector field X the velocity field of, say, a river.
For V we take the temperature scalar field. Now take a light particle and
let if flow (litterally) with the current of the river. L_x(V) then
measures the temperature change.
best,
Jeroen
Alfred Einstead
Jun11-04, 12:36 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>\n"Albert Ng" <asgc71@dsl.pipex.com> wrote:\n> Can anyone explain the meaning of Lie derivative in a simple way ?\n\nNot really. But there\'s a way to characterize it:\n\n(1) It\'s Leibnitz with respect to products and arguments:\nSo, for instance\nL_v(wf) = L_v(w) f + w L_v(f),\nwhere v, w are vector fields, f a function\nL_v(W[w]) = L_v(W)[w] + W[L_v(w)]\nwhere W is a 1-form, w a vector field\n\n(2) It gives you the directional derivative when applied to a scalar\nfield:\nL_v(f) = vf.\n\nThus, L_v(w) is the vector field which when applied to f gives you:\nL_v(w) f = L_v(wf) - w L_v(f)\n= v(wf) - w (vf)\n= [v,w] f\nor\nL_v(w) = [v,w].\n\nAnd L_v(W), for a 1-form W, is a 1-form which when contracted with\na vector field w gives you:\nL_v(W)[w] = L_v(W[w]) - W[L_v(w)]\n= v(W[w]) - W[[v,w]].\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>"Albert Ng" <asgc71@dsl.pipex.com> wrote:
> Can anyone explain the meaning of Lie derivative in a simple way ?
Not really. But there's a way to characterize it:
(1) It's Leibnitz with respect to products and arguments:
So, for instance
L_v(wf) = L_v(w) f + w L_v(f),
where v, w are vector fields, f a function
L_v(W[w]) = L_v(W)[w] + W[L_v(w)]
where W is a 1-form, w a vector field
(2) It gives you the directional derivative when applied to a scalar
field:
L_v(f) = vf.
Thus, L_v(w) is the vector field which when applied to f gives you:
L_v(w) f = L_v(wf) - w L_v(f)= v(wf) - w (vf)
= [v,w] f
or
L_v(w) = [v,w].
And L_v(W), for a 1-form W, is a 1-form which when contracted with
a vector field w gives you:
L_v(W)[w] = L_v(W[w]) - W[L_v(w)]= v(W[w]) - W[[v,w]].
Arkadiusz Jadczyk
Jun12-04, 07:21 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>On 10 Jun 2004 07:47:32 -0400, "Albert Ng" <asgc71@dsl.pipex.com> wrote:\n\n>Can anyone explain the meaning of Lie derivative in a simple way ?\n>Thank you\n\nA vector field X defines a local group of diffeomorphisms exp(Xt)\n\ntake point p, let p(t)=exp(Xt)(p) be a moving point on the trajectory\nthrough p.\n\nTake field of geometric objects A defined along the trajectory.\n\nTake A(pt)) and pull it back to p using exp(-Xt). Denote the result\nA_(t).\n\nThen L_X(A)(p)=lim (A_t)-A(p))/t as t ->0\n\nark\n--\n\nArkadiusz Jadczyk\nhttp://www.cassiopaea.org/quantum_future/homepage.htm\n\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>On 10 Jun 2004 07:47:32 -0400, "Albert Ng" <asgc71@dsl.pipex.com> wrote:
>Can anyone explain the meaning of Lie derivative in a simple way ?
>Thank you
A vector field X defines a local group of diffeomorphisms \exp(Xt)
take point p, let p(t)=\exp(Xt)(p) be a moving point on the trajectory
through p.
Take field of geometric objects A defined along the trajectory.
Take A(pt)) and pull it back to p using \exp(-Xt). Denote the result
A_(t).
Then L_X(A)(p)=lim (A_t)-A(p))/t as t ->0
ark
--
Arkadiusz Jadczyk
http://www.cassiopaea.org/quantum_future/homepage.htm
--
Stephen Parrott
Jun12-04, 07:21 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>I doubt that the concept of Lie derivative can be properly understood\nwithout previous understanding of the concept of "flow of a vector\nfield", so the latter is what you should study first.\nDetails can be found in most graduate level differential equations\nor differential geometry texts; I can\'t recommend a specific\none offhand.\n\nIn case you already know what is the flow of a vector field,\nthe concept of Lie derivative (of a tensor field, such as a form field\nor vector field) in the direction of a given vector field (on a\nmanifold) is relatively straightforward, though the details can be\nconfusing. In short, the flow of the vector field\nfurnishes a way to "transport"\na tensor based at any given point of the manifold to another point\nrelated to it by the flow. This gives a sensible way to compare\ntensors based at different points related by the flow.\nThe existence of a way to compare tensors based at different points\n(related by the flow) furnishes a natural way to "differentiate"\ntensor fields relative to the flow. This derivative of a tensor field\nis the Lie derivative.\n\nThe best I can do at a simple explanation is to imagine\na time-independent flow of water in a stream. At any point in the\nstream, the velocity of the water at that point is a vector; the\nassignment of that velocity to that point is a vector field\nwhich I\'ll call w below.\n\nThe *flow* f of that vector field w is a mapping\nt,p --> f(t,p)\nwhich assigns to each time t and each point p in the stream, the\nposition at time t of the water particle which was at point p\nat time 0. It is a basic result in (graduate level) differential\nequations that with every vector field on a manifold is associated a\nflow (defined for small t) related as to that vector field\nin the same way that the flow of water is related to its velocity field.\n\nA "vector" at p may be visualized as an infinitesimal arrow\nwith tail at p and head at a point p + dp.\nIf at time 0 we have such a vector, then at time t, the water particles\noriginally (i.e., at time 0) at p and at p+dp will have moved to\nnew positions f(t,p) and f(t, p+dp). The "vector" with tail f(t,p)\nand head f(t, p+dp) is (by definition) the original vector at p\ntransported to f(t,p) by the flow. To make strict mathematical sense\nof this, one needs a more precise definition of "vector"\nthan "arrow from p to p+dp", but the arrow picture gives the intuitive\ncontent.\n\nThe Lie derivative L_w (v) of a given vector field v\nin the direction of the water\'s vector field w is defined as\nthe vector field whose value at a point p is\nobtained by diffentiating the mapping\n\nt --> w at f(t,p) backward transported to p .\n\nHere "backward transported" means the inverse of the above\ntransportation operation; this may also be viewed as forward\ntransporation relative to the original water flow with all velocities\nreversed. Note that for all t,\n"w at f(t,p) backward transported to p" is a vector at p,\nso that it makes good sense to differentiate the above mapping with\nrespect to t (because the mapping\'s range is a fixed vector space,\nthe space of all vectors with tail at p).\n\nIn a similar way, one transports and Lie differentiates\ngeneral tensor fields. Again, the details can be found in\nalmost any graduate level differential geometry text.\n\nIf I had realized how hard it would be to translate the\nmathematics into ASCII,\nI probably wouldn\'t have started.\nSo, the literal answer to your question is probably that\nthe concept of Lie derivative is inherently complicated and somewhat\nsubtle, and hence difficult or impossible to explain simply without\ndrawing pictures.\n\nYour best bets are either to spend a few days thinking hard\nabout the mathematical definitions as presented in graduate texts,\nor to get someone who understands it to spend a couple of hours\nwith you in front of a blackboard.\nThe definition given from scratch (i.e., including the definition of\nflow of a vector field) would require several class periods, maybe\neven several weeks if one also had to properly define the notion\nof "vector" without drawing pictures of little arrows.\n\n\n\nAlbert Ng wrote:\n> Can anyone explain the meaning of Lie derivative in a simple way ?\n> Thank you\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>I doubt that the concept of Lie derivative can be properly understood
without previous understanding of the concept of "flow of a vector
field", so the latter is what you should study first.
Details can be found in most graduate level differential equations
or differential geometry texts; I can't recommend a specific
one offhand.
In case you already know what is the flow of a vector field,
the concept of Lie derivative (of a tensor field, such as a form field
or vector field) in the direction of a given vector field (on a
manifold) is relatively straightforward, though the details can be
confusing. In short, the flow of the vector field
furnishes a way to "transport"
a tensor based at any given point of the manifold to another point
related to it by the flow. This gives a sensible way to compare
tensors based at different points related by the flow.
The existence of a way to compare tensors based at different points
(related by the flow) furnishes a natural way to "differentiate"
tensor fields relative to the flow. This derivative of a tensor field
is the Lie derivative.
The best I can do at a simple explanation is to imagine
a time-independent flow of water in a stream. At any point in the
stream, the velocity of the water at that point is a vector; the
assignment of that velocity to that point is a vector field
which I'll call w below.
The *flow* f of that vector field w is a mapping
t,p --> f(t,p)
which assigns to each time t and each point p in the stream, the
position at time t of the water particle which was at point p
at time . It is a basic result in (graduate level) differential
equations that with every vector field on a manifold is associated a
flow (defined for small t) related as to that vector field
in the same way that the flow of water is related to its velocity field.
A "vector" at p may be visualized as an infinitesimal arrow
with tail at p and head at a point p + dp.
If at time we have such a vector, then at time t, the water particles
originally (i.e., at time 0) at p and at p+dp will have moved to
new positions f(t,p) and f(t, p+dp). The "vector" with tail f(t,p)
and head f(t, p+dp) is (by definition) the original vector at p
transported to f(t,p) by the flow. To make strict mathematical sense
of this, one needs a more precise definition of "vector"
than "arrow from p to p+dp", but the arrow picture gives the intuitive
content.
The Lie derivative L_w (v) of a given vector field v
in the direction of the water's vector field w is defined as
the vector field whose value at a point p is
obtained by diffentiating the mapping
t --> w at f(t,p) backward transported to p .
Here "backward transported" means the inverse of the above
transportation operation; this may also be viewed as forward
transporation relative to the original water flow with all velocities
reversed. Note that for all t,
"w at f(t,p) backward transported to p" is a vector at p,
so that it makes good sense to differentiate the above mapping with
respect to t (because the mapping's range is a fixed vector space,
the space of all vectors with tail at p).
In a similar way, one transports and Lie differentiates
general tensor fields. Again, the details can be found in
almost any graduate level differential geometry text.
If I had realized how hard it would be to translate the
mathematics into ASCII,
I probably wouldn't have started.
So, the literal answer to your question is probably that
the concept of Lie derivative is inherently complicated and somewhat
subtle, and hence difficult or impossible to explain simply without
drawing pictures.
Your best bets are either to spend a few days thinking hard
about the mathematical definitions as presented in graduate texts,
or to get someone who understands it to spend a couple of hours
with you in front of a blackboard.
The definition given from scratch (i.e., including the definition of
flow of a vector field) would require several class periods, maybe
even several weeks if one also had to properly define the notion
of "vector" without drawing pictures of little arrows.
Albert Ng wrote:
> Can anyone explain the meaning of Lie derivative in a simple way ?
> Thank you
Stephen Parrott
Jun14-04, 03:08 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>In my original reply, there was a bad typo in the definition\nof the Lie derivative L_w(v) of a vector field v in the direction\nof a vector field w. The definition should have read as follows:\n\n> The Lie derivative L_w (v) of a given vector field v\n> in the direction of the water\'s vector field w is defined as\n> the vector field whose value at a point p is\n> obtained by diffentiating the mapping\n>\n> t --> v at f(t,p) backward transported to p .\n>\n> Here "backward transported" means the inverse of the above\n> transportation operation; this may also be viewed as forward\n> transporation relative to the original water flow with all velocities\n> reversed. Note that for all t,\n> "v at f(t,p) backward transported to p" is a vector at p,\n> so that it makes good sense to differentiate the above mapping with\n> respect to t (because the mapping\'s range is a fixed vector space,\n> the space of all vectors with tail at p).\n\n(In the original, v was mistakenly replaced by w in the displayed\nequation and the sentence 5 lines below it.)\n\nStephen Parrott wrote:\n> I doubt that the concept of Lie derivative can be properly understood\n> without previous understanding of the concept of "flow of a vector\n> field", so the latter is what you should study first.\n> Details can be found in most graduate level differential equations\n> or differential geometry texts; I can\'t recommend a specific\n> one offhand.\n>\n> In case you already know what is the flow of a vector field,\n> the concept of Lie derivative (of a tensor field, such as a form field\n> or vector field) in the direction of a given vector field (on a\n> manifold) is relatively straightforward, though the details can be\n> confusing. In short, the flow of the vector field\n> furnishes a way to "transport"\n> a tensor based at any given point of the manifold to another point\n> related to it by the flow. This gives a sensible way to compare\n> tensors based at different points related by the flow.\n> The existence of a way to compare tensors based at different points\n> (related by the flow) furnishes a natural way to "differentiate"\n> tensor fields relative to the flow. This derivative of a tensor field\n> is the Lie derivative.\n>\n> The best I can do at a simple explanation is to imagine\n> a time-independent flow of water in a stream. At any point in the\n> stream, the velocity of the water at that point is a vector; the\n> assignment of that velocity to that point is a vector field\n> which I\'ll call w below.\n>\n> The *flow* f of that vector field w is a mapping\n> t,p --> f(t,p)\n> which assigns to each time t and each point p in the stream, the\n> position at time t of the water particle which was at point p\n> at time 0. It is a basic result in (graduate level) differential\n> equations that with every vector field on a manifold is associated a\n> flow (defined for small t) related as to that vector field\n> in the same way that the flow of water is related to its velocity field.\n>\n> A "vector" at p may be visualized as an infinitesimal arrow\n> with tail at p and head at a point p + dp.\n> If at time 0 we have such a vector, then at time t, the water particles\n> originally (i.e., at time 0) at p and at p+dp will have moved to\n> new positions f(t,p) and f(t, p+dp). The "vector" with tail f(t,p)\n> and head f(t, p+dp) is (by definition) the original vector at p\n> transported to f(t,p) by the flow. To make strict mathematical sense\n> of this, one needs a more precise definition of "vector"\n> than "arrow from p to p+dp", but the arrow picture gives the intuitive\n> content.\n>\n> The Lie derivative L_w (v) of a given vector field v\n> in the direction of the water\'s vector field w is defined as\n> the vector field whose value at a point p is\n> obtained by diffentiating the mapping\n>\n> t --> w at f(t,p) backward transported to p .\n>\n> Here "backward transported" means the inverse of the above\n> transportation operation; this may also be viewed as forward\n> transporation relative to the original water flow with all velocities\n> reversed. Note that for all t,\n> "w at f(t,p) backward transported to p" is a vector at p,\n> so that it makes good sense to differentiate the above mapping with\n> respect to t (because the mapping\'s range is a fixed vector space,\n> the space of all vectors with tail at p).\n>\n> In a similar way, one transports and Lie differentiates\n> general tensor fields. Again, the details can be found in\n> almost any graduate level differential geometry text.\n>\n> If I had realized how hard it would be to translate the\n> mathematics into ASCII,\n> I probably wouldn\'t have started.\n> So, the literal answer to your question is probably that\n> the concept of Lie derivative is inherently complicated and somewhat\n> subtle, and hence difficult or impossible to explain simply without\n> drawing pictures.\n>\n> Your best bets are either to spend a few days thinking hard\n> about the mathematical definitions as presented in graduate texts,\n> or to get someone who understands it to spend a couple of hours\n> with you in front of a blackboard.\n> The definition given from scratch (i.e., including the definition of\n> flow of a vector field) would require several class periods, maybe\n> even several weeks if one also had to properly define the notion\n> of "vector" without drawing pictures of little arrows.\n>\n>\n>\n> Albert Ng wrote:\n>\n>>Can anyone explain the meaning of Lie derivative in a simple way ?\n>>Thank you\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>In my original reply, there was a bad typo in the definition
of the Lie derivative L_w(v) of a vector field v in the direction
of a vector field w. The definition should have read as follows:
> The Lie derivative L_w (v) of a given vector field v
> in the direction of the water's vector field w is defined as
> the vector field whose value at a point p is
> obtained by diffentiating the mapping
>
> t --> v at f(t,p) backward transported to p .
>
> Here "backward transported" means the inverse of the above
> transportation operation; this may also be viewed as forward
> transporation relative to the original water flow with all velocities
> reversed. Note that for all t,
> "v at f(t,p) backward transported to p" is a vector at p,
> so that it makes good sense to differentiate the above mapping with
> respect to t (because the mapping's range is a fixed vector space,
> the space of all vectors with tail at p).
(In the original, v was mistakenly replaced by w in the displayed
equation and the sentence 5 lines below it.)
Stephen Parrott wrote:
> I doubt that the concept of Lie derivative can be properly understood
> without previous understanding of the concept of "flow of a vector
> field", so the latter is what you should study first.
> Details can be found in most graduate level differential equations
> or differential geometry texts; I can't recommend a specific
> one offhand.
>
> In case you already know what is the flow of a vector field,
> the concept of Lie derivative (of a tensor field, such as a form field
> or vector field) in the direction of a given vector field (on a
> manifold) is relatively straightforward, though the details can be
> confusing. In short, the flow of the vector field
> furnishes a way to "transport"
> a tensor based at any given point of the manifold to another point
> related to it by the flow. This gives a sensible way to compare
> tensors based at different points related by the flow.
> The existence of a way to compare tensors based at different points
> (related by the flow) furnishes a natural way to "differentiate"
> tensor fields relative to the flow. This derivative of a tensor field
> is the Lie derivative.
>
> The best I can do at a simple explanation is to imagine
> a time-independent flow of water in a stream. At any point in the
> stream, the velocity of the water at that point is a vector; the
> assignment of that velocity to that point is a vector field
> which I'll call w below.
>
> The *flow* f of that vector field w is a mapping
> t,p --> f(t,p)
> which assigns to each time t and each point p in the stream, the
> position at time t of the water particle which was at point p
> at time . It is a basic result in (graduate level) differential
> equations that with every vector field on a manifold is associated a
> flow (defined for small t) related as to that vector field
> in the same way that the flow of water is related to its velocity field.
>
> A "vector" at p may be visualized as an infinitesimal arrow
> with tail at p and head at a point p + dp.
> If at time we have such a vector, then at time t, the water particles
> originally (i.e., at time 0) at p and at p+dp will have moved to
> new positions f(t,p) and f(t, p+dp). The "vector" with tail f(t,p)
> and head f(t, p+dp) is (by definition) the original vector at p
> transported to f(t,p) by the flow. To make strict mathematical sense
> of this, one needs a more precise definition of "vector"
> than "arrow from p to p+dp", but the arrow picture gives the intuitive
> content.
>
> The Lie derivative L_w (v) of a given vector field v
> in the direction of the water's vector field w is defined as
> the vector field whose value at a point p is
> obtained by diffentiating the mapping
>
> t --> w at f(t,p) backward transported to p .
>
> Here "backward transported" means the inverse of the above
> transportation operation; this may also be viewed as forward
> transporation relative to the original water flow with all velocities
> reversed. Note that for all t,
> "w at f(t,p) backward transported to p" is a vector at p,
> so that it makes good sense to differentiate the above mapping with
> respect to t (because the mapping's range is a fixed vector space,
> the space of all vectors with tail at p).
>
> In a similar way, one transports and Lie differentiates
> general tensor fields. Again, the details can be found in
> almost any graduate level differential geometry text.
>
> If I had realized how hard it would be to translate the
> mathematics into ASCII,
> I probably wouldn't have started.
> So, the literal answer to your question is probably that
> the concept of Lie derivative is inherently complicated and somewhat
> subtle, and hence difficult or impossible to explain simply without
> drawing pictures.
>
> Your best bets are either to spend a few days thinking hard
> about the mathematical definitions as presented in graduate texts,
> or to get someone who understands it to spend a couple of hours
> with you in front of a blackboard.
> The definition given from scratch (i.e., including the definition of
> flow of a vector field) would require several class periods, maybe
> even several weeks if one also had to properly define the notion
> of "vector" without drawing pictures of little arrows.
>
>
>
> Albert Ng wrote:
>
>>Can anyone explain the meaning of Lie derivative in a simple way ?
>>Thank you
Thomas Larsson
Jun14-04, 03:09 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>"Albert Ng" <asgc71@dsl.pipex.com> wrote in message news:<40c7ba93\\$0\\$280\\$cc9e4d1f@news.dial.pipe x.com>...\n> Can anyone explain the meaning of Lie derivative in a simple way ?\n> Thank you\n\n\nA nice but unconventional definition of the Lie derivative is\nas a representation of the diffeomorphism group. Let\n\nX = X^i(x) d_i\n\nbe a vector field (= infinitesimal diffeomorphism) in N\ndimensions, where x = (x^i) are the coordinates and d_i = d/dx^i\nthe corresponding partial derivatives. The structure of the\ndiffeomorphism algebra is encoded in the bracket\n\n[X, Y] = XY - YX = X^i d_i Y^j d_j - Y^j d_j X^i d_i.\n\nWe may now define the Lie derivative as an operator L_X that\nsatisfies\n\n[L_X, L_Y] = L_[X,Y].\n\nIf T^i_j are some matrices satisfying gl(N), then\n\nL_X = X^i d_i + d_j X^i T^j_i\n\nsatisfies the diffeomorphism algebra. By substituting a matrix\nrepresentation of gl(N) for T^i_j, we get the action of the Lie\nderivative on the corresponding tensor density. In this way we\nrecover the Lie derivative acting on functions, vectors, forms,\netc.\n\nOne nice thing about this algebraic viewpoint is that it allows\nfor many generalizations. Assume e.g. that we are interested in\nflows on a N-dimensional symplectic manifold. Then X and Y are\nHamiltonian vector fields (infinitesimal canonical\ntransformations), so L_X should yield a representation of this\nalgebra instead. There is a similar explicit formula for L_X in\nthis case, with T^i_j replaced by matrices satisfying sp(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>"Albert Ng" <asgc71@dsl.pipex.com> wrote in message news:<40c7ba93$0$280$cc9e4d1f@news.dial.pipex.com>...
> Can anyone explain the meaning of Lie derivative in a simple way ?
> Thank you
A nice but unconventional definition of the Lie derivative is
as a representation of the diffeomorphism group. Let
X = X^i(x) d_i
be a vector field (= infinitesimal diffeomorphism) in N
dimensions, where x = (x^i) are the coordinates and d_i = d/dx^i
the corresponding partial derivatives. The structure of the
diffeomorphism algebra is encoded in the bracket
[X, Y] = XY - YX = X^i d_i Y^j d_j - Y^j d_j X^i d_i.
We may now define the Lie derivative as an operator L_X that
satisfies
[L_X, L_Y] = L_[X,Y][/itex].
If T^{i_j} are some matrices satisfying gl(N), then
[itex]L_X = X^i d_i + d_j X^i T^{j_i}
satisfies the diffeomorphism algebra. By substituting a matrix
representation of gl(N) for T^{i_j}, we get the action of the Lie
derivative on the corresponding tensor density. In this way we
recover the Lie derivative acting on functions, vectors, forms,
etc.
One nice thing about this algebraic viewpoint is that it allows
for many generalizations. Assume e.g. that we are interested in
flows on a N-dimensional symplectic manifold. Then X and Y are
Hamiltonian vector fields (infinitesimal canonical
transformations), so L_X should yield a representation of this
algebra instead. There is a similar explicit formula for L_X in
this case, with T^{i_j} replaced by matrices satisfying sp(N).
Arkadiusz Jadczyk
Jun16-04, 03:55 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>On Sat, 12 Jun 2004 12:21:34 +0000 (UTC), Stephen Parrott\n<spardr@netscape.net> wrote:\n\n>Details can be found in most graduate level differential equations\n>or differential geometry texts; I can\'t recommend a specific\n>one offhand.\n\nBishop and Crittenden, "Geometry of Manifolds", Academic Press 1964,\npages 14-17 has flows of vector fields and Lie derivative explained,\neven with illustrations.\n\nark\n--\n\nArkadiusz Jadczyk\nhttp://www.cassiopaea.org/quantum_future/homepage.htm\n\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>On Sat, 12 Jun 2004 12:21:34 +0000 (UTC), Stephen Parrott
<spardr@netscape.net> wrote:
>Details can be found in most graduate level differential equations
>or differential geometry texts; I can't recommend a specific
>one offhand.
Bishop and Crittenden, "Geometry of Manifolds", Academic Press 1964,
pages 14-17 has flows of vector fields and Lie derivative explained,
even with illustrations.
ark
--
Arkadiusz Jadczyk
http://www.cassiopaea.org/quantum_future/homepage.htm
--
Alfred Einstead
Jun16-04, 04:00 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>Arkadiusz Jadczyk <arkREMOVETHIS@ANDTHIScassiopaea.org> wrote:\n> let p(t)=exp(Xt)(p) be a moving point on the trajectory\n> through p.\n> Take A(pt)) and pull it back to p using exp(-Xt). Denote the result\n> A_(t).\n> Then L_X(A)(p)=lim (A_t)-A(p))/t as t ->0\n\nYou could just say it directly:\nL_X(A)(p) = d/dt ((exp(Xt)_{*} A exp(-Xt) p)\nat t = 0, removing parentheses for brevity.\n\nIn comparison, the commutator for a Lie group G is defined by:\n[X, A] = d/dt (exp(Xt) A exp(-Xt)).\n[with obvious notation to denote the left and right actions of\nG on TG].\n\nBut all that\'s applicable only to vector fields. The point\nof the construction is to effect the property laid out in\na previous article, that\nL_X(Af) = L_X(A) f + A (L_X f)\nor\nX(Af) = L_X(A) f + A (Xf)\n(from which it follows that L_X(A) = [X,A]).\n\nIn order to realize the characterization provided in the article\nbefore for 1-forms W:\nL_X(W[A]) = L_X(W)[A] + W[L_X(A)]\nor\nX(W[A]) = L_X(W)[A] + W[ [X,A] ],\nyou need a different analytic explication. So, the analytic\nexplications can\'t really be considered as fundamental\ncharacterizations of what a Lie derivative is, but only as a\nconstruction that realises the more fundamental (algebraic)\ncharacterization, such as that provided in the previous\narticle.\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>Arkadiusz Jadczyk <arkREMOVETHIS@ANDTHIScassiopaea.org> wrote:
> let p(t)=\exp(Xt)(p) be a moving point on the trajectory
> through p.
> Take A(pt)) and pull it back to p using \exp(-Xt). Denote the result
> A_(t).
> Then L_X(A)(p)=lim (A_t)-A(p))/t as t ->0
You could just say it directly:
L_X(A)(p) = d/dt ((\exp(Xt)_{*} A \exp(-Xt) p)at t = 0, removing parentheses for brevity.
In comparison, the commutator for a Lie group G is defined by:
[X, A] = d/dt (\exp(Xt) A \exp(-Xt)).
[with obvious notation to denote the left and right actions of
G on TG].
But all that's applicable only to vector fields. The point
of the construction is to effect the property laid out in
a previous article, that
L_X(Af) = L_X(A) f + A (L_X f)
or
X(Af) = L_X(A) f + A (Xf)
(from which it follows that L_X(A) = [X,A]).
In order to realize the characterization provided in the article
before for 1-forms W:
L_X(W[A]) = L_X(W)[A] + W[L_X(A)]
or
X(W[A]) = L_X(W)[A] + W[ [X,A] ],
you need a different analytic explication. So, the analytic
explications can't really be considered as fundamental
characterizations of what a Lie derivative is, but only as a
construction that realises the more fundamental (algebraic)
characterization, such as that provided in the previous
article.
<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>k_igor_k@lycos.com (Igor Khavkine) wrote in message news:<f1ac2e6e.0406101436.24873a5c@posting.google. com>...\n> "Albert Ng" <asgc71@dsl.pipex.com> wrote in message news:<40c7ba93\\$0\\$280\\$cc9e4d1f@news.dial.pipe x.com>...\n> > Can anyone explain the meaning of Lie derivative in a simple way ?\n>\n> I am not sure what your level of mathematical sophistication is, so\n> I\'ll try to describe it as lay terms as I can.\n>\n> Suppose you have a region of space M (the technical term would be\n> a manifold). At every point x of this region, you attach an object A(x)\n> of a particular type (the types that people are usually consider\n> are scalars, vectors, tensors, etc.). Now suppose that you have\n> two exact copies of this region (call them M1 and M2) with the\n> objects defined on the regions copied as well (call them A1(x) and A2(x)).\n>\nIn " Geometry of Physics" Frankel descibes a flow with a tangent\nvector to the congruence moved forward via the differential to a new\ntangent vector corrosponding to a increase in a parameter ,say s. This\ndifferential is the Jacobian of the transformation ( in this case a\none parameter transformation). Another vector field is described that\ncrosses the original one with its own congruence and\nparameterization.At the point where the original congruence is given\nby parameter s there is also a "crossing vector " from the second\ncongruence. Do we take the inverse of the original Jacobian to move\nthe "crossing vector" back to the original tangent vector of the first\ncongruence? Or do i have this wrong? After we back transport this\nvector along the original congruence ,what do we compare it to,the\ncrossing vector that was at the origin ( s=0 ) of the original\ncongruence? Thanks\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>k_{igor_k}@lycos.com (Igor Khavkine) wrote in message news:<f1ac2e6e.0406101436.24873a5c@posting.google.com>...
> "Albert Ng" <asgc71@dsl.pipex.com> wrote in message news:<40c7ba93$0$280$cc9e4d1f@news.dial.pipex.com>...
> > Can anyone explain the meaning of Lie derivative in a simple way ?
>
> I am not sure what your level of mathematical sophistication is, so
> I'll try to describe it as lay terms as I can.
>
> Suppose you have a region of space M (the technical term would be
> a manifold). At every point x of this region, you attach an object A(x)
> of a particular type (the types that people are usually consider
> are scalars, vectors, tensors, etc.). Now suppose that you have
> two exact copies of this region (call them M1 and M2) with the
> objects defined on the regions copied as well (call them A1(x) and A2(x)).
>
In " Geometry of Physics" Frankel descibes a flow with a tangent
vector to the congruence moved forward via the differential to a new
tangent vector corrosponding to a increase in a parameter ,say s. This
differential is the Jacobian of the transformation ( in this case a
one parameter transformation). Another vector field is described that
crosses the original one with its own congruence and
parameterization.At the point where the original congruence is given
by parameter s there is also a "crossing vector " from the second
congruence. Do we take the inverse of the original Jacobian to move
the "crossing vector" back to the original tangent vector of the first
congruence? Or do i have this wrong? After we back transport this
vector along the original congruence ,what do we compare it to,the
crossing vector that was at the origin ( s=0 ) of the original
congruence? Thanks
<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>k_igor_k@lycos.com (Igor Khavkine) wrote in message news:<f1ac2e6e.0406101436.24873a5c@posting.google. com>...\n> "Albert Ng" <asgc71@dsl.pipex.com> wrote in message news:<40c7ba93\\$0\\$280\\$cc9e4d1f@news.dial.pipe x.com>...\n> > Can anyone explain the meaning of Lie derivative in a simple way ?\n>\n> I am not sure what your level of mathematical sophistication is, so\n> I\'ll try to describe it as lay terms as I can.\n>\n> Suppose you have a region of space M (the technical term would be\n> a manifold). At every point x of this region, you attach an object A(x)\n> of a particular type (the types that people are usually consider\n> are scalars, vectors, tensors, etc.). Now suppose that you have\n> two exact copies of this region (call them M1 and M2) with the\n> objects defined on the regions copied as well (call them A1(x) and A2(x)).\n>\nIn " Geometry of Physics" Frankel descibes a flow with a tangent\nvector to the congruence moved forward via the differential to a new\ntangent vector corrosponding to a increase in a parameter ,say s. This\ndifferential is the Jacobian of the transformation ( in this case a\none parameter transformation). Another vector field is described that\ncrosses the original one with its own congruence and\nparameterization.At the point where the original congruence is given\nby parameter s there is also a "crossing vector " from the second\ncongruence. Do we take the inverse of the original Jacobian to move\nthe "crossing vector" back to the original tangent vector of the first\ncongruence? Or do i have this wrong? After we back transport this\nvector along the original congruence ,what do we compare it to,the\ncrossing vector that was at the origin ( s=0 ) of the original\ncongruence? Thanks\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>k_{igor_k}@lycos.com (Igor Khavkine) wrote in message news:<f1ac2e6e.0406101436.24873a5c@posting.google.com>...
> "Albert Ng" <asgc71@dsl.pipex.com> wrote in message news:<40c7ba93$0$280$cc9e4d1f@news.dial.pipex.com>...
> > Can anyone explain the meaning of Lie derivative in a simple way ?
>
> I am not sure what your level of mathematical sophistication is, so
> I'll try to describe it as lay terms as I can.
>
> Suppose you have a region of space M (the technical term would be
> a manifold). At every point x of this region, you attach an object A(x)
> of a particular type (the types that people are usually consider
> are scalars, vectors, tensors, etc.). Now suppose that you have
> two exact copies of this region (call them M1 and M2) with the
> objects defined on the regions copied as well (call them A1(x) and A2(x)).
>
In " Geometry of Physics" Frankel descibes a flow with a tangent
vector to the congruence moved forward via the differential to a new
tangent vector corrosponding to a increase in a parameter ,say s. This
differential is the Jacobian of the transformation ( in this case a
one parameter transformation). Another vector field is described that
crosses the original one with its own congruence and
parameterization.At the point where the original congruence is given
by parameter s there is also a "crossing vector " from the second
congruence. Do we take the inverse of the original Jacobian to move
the "crossing vector" back to the original tangent vector of the first
congruence? Or do i have this wrong? After we back transport this
vector along the original congruence ,what do we compare it to,the
crossing vector that was at the origin ( s=0 ) of the original
congruence? Thanks
Michael Murray
Jun30-04, 05:38 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>> >\n> In " Geometry of Physics" Frankel descibes a flow with a tangent\n> vector to the congruence moved forward via the differential to a new\n> tangent vector corrosponding to a increase in a parameter ,say s. This\n> differential is the Jacobian of the transformation ( in this case a\n> one parameter transformation). Another vector field is described that\n> crosses the original one with its own congruence and\n> parameterization.At the point where the original congruence is given\n> by parameter s there is also a "crossing vector " from the second\n> congruence. Do we take the inverse of the original Jacobian to move\n> the "crossing vector" back to the original tangent vector of the first\n> congruence? Or do i have this wrong? After we back transport this\n> vector along the original congruence ,what do we compare it to,the\n> crossing vector that was at the origin ( s=0 ) of the original\n> congruence? Thanks\n\nIf X is a vector field then it gives rise to a one-parameter\nfamily of transformations f : R x M -> M (actually knowing\nyou can define it for all of R rather than some open set\naround 0 requires some other conditions like M compact).\n\nLet f_t : M -> M be the function that sends y to f(t,y).\n\nIf Y is another vector field then you can pick a point x in M and then\nY_x is in the tangent space at x. Now look at Y_{f_t(x)} which\nis in the tangent space at f_t(x) for every t. Move this\nback with the differential to the inverse of the function f_t.\nThen its in Y_x. Call the result something like (f_t^{-1})_*(Y_f_t(x)).\nThen do the usual type of limit for a derivative\n\nlim_{t \\to 0} [ (f_t^{-1})_*(Y_f_t(x)) - Y_x ] / t\n\nthis is [X, Y](x) or L_X(Y)(x) or whatever your notation for\nLie derivative is. [Its quite possible I have got the sign wrong and\nits actually the negative of this. ]\n\n\nMichael\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>> >
> In " Geometry of Physics" Frankel descibes a flow with a tangent
> vector to the congruence moved forward via the differential to a new
> tangent vector corrosponding to a increase in a parameter ,say s. This
> differential is the Jacobian of the transformation ( in this case a
> one parameter transformation). Another vector field is described that
> crosses the original one with its own congruence and
> parameterization.At the point where the original congruence is given
> by parameter s there is also a "crossing vector " from the second
> congruence. Do we take the inverse of the original Jacobian to move
> the "crossing vector" back to the original tangent vector of the first
> congruence? Or do i have this wrong? After we back transport this
> vector along the original congruence ,what do we compare it to,the
> crossing vector that was at the origin ( s=0 ) of the original
> congruence? Thanks
If X is a vector field then it gives rise to a one-parameter
family of transformations f : R x M -> M (actually knowing
you can define it for all of R rather than some open set
around requires some other conditions like M compact).
Let f_t : M -> M be the function that sends y to f(t,y).
If Y is another vector field then you can pick a point x in M and then
Y_x is in the tangent space at x. Now look at Y_{f_t(x)} which
is in the tangent space at f_t(x) for every t. Move this
back with the differential to the inverse of the function f_t.
Then its in Y_x. Call the result something like (f_t^{-1})_*(Y_{f_t}(x)).
Then do the usual type of limit for a derivative
lim_{t \to 0}[/itex] [ [itex](f_t^{-1})_*(Y_{f_t}(x)) - Y_x ] / t
this is [X, Y](x) or L_X(Y)(x) or whatever your notation for
Lie derivative is. [Its quite possible I have got the sign wrong and
its actually the negative of this. ]
Michael
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