Calculus of Variations problem [Archive]

Blake Winter

Oct8-04, 06:20 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>\nI\'m trying to work out three questions about the theory of calculus of\nvariations:\nFirst, is it possible to have a Lagrangian which is a function of a,b,\nand c, along with the parameter t, and their first t derivatives, such\nthat the resulting differential equations can be written in the form\nA(a,b), B(a,b), C(b,c)\nWhere A, B, and C are differential equations of those functions a, b,\nc?\n\nSecond, what about if the Lagrangian were to depend on the second t\nderivatives as well?\n\nThird, (and this I believe the answer to is no, but I can\'t find a\nproof of it or generate one myself), is it possible to have a\nLagrangian of a,b, and c where all three functions are varied but c\ndoes not influence the evolution of a or b, although it is influenced\nby one or both of them? This seems strictly impossible to me - it\nwouldn\'t make any sense if it were true - but I can\'t prove that it\nhas to be that way.\n\nMaybe I\'m just forgetting my basic calculus of variations...\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'm trying to work out three questions about the theory of calculus of
variations:
First, is it possible to have a Lagrangian which is a function of a,b,
and c, along with the parameter t, and their first t derivatives, such
that the resulting differential equations can be written in the form
A(a,b), B(a,b), C(b,c)
Where A, B, and C are differential equations of those functions a, b,
c?

Second, what about if the Lagrangian were to depend on the second t
derivatives as well?

Third, (and this I believe the answer to is no, but I can't find a
proof of it or generate one myself), is it possible to have a
Lagrangian of a,b, and c where all three functions are varied but c
does not influence the evolution of a or b, although it is influenced
by one or both of them? This seems strictly impossible to me - it
wouldn't make any sense if it were true - but I can't prove that it
has to be that way.

Maybe I'm just forgetting my basic calculus of variations...

Mark Palenik

Oct11-04, 03:52 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\n"Blake Winter" <blake.winter@houghton.edu> wrote in message\nnews:87423d2a.0410071613.6b09c1ca@posting .google.com...\n>\n> I\'m trying to work out three questions about the theory of calculus of\n> variations:\n> First, is it possible to have a Lagrangian which is a function of a,b,\n> and c, along with the parameter t, and their first t derivatives, such\n> that the resulting differential equations can be written in the form\n> A(a,b), B(a,b), C(b,c)\n> Where A, B, and C are differential equations of those functions a, b,\n> c?\n\nI don\'t think so. For a function of several variables, you would just use:\n\ndf/da - d/dt(df/da\') = 0\ndf/db - d/dt(df/db\') = 0\ndf/dc - d/dt(df/dc\') = 0\n\nif df/dc or df/dc\' leave you with a function that includes b, it seems to me\nthat df/db or df/db\' should leave you with a function that includes c.\n\nI don\'t know why you would want A(a,b), B(a,b), C(b,c), though. Ideally you\nwould want A(a, a\', a\'\'), B(b, b\', b\'\'), C(c, c\', c\'\'). It may or may not\nturn out to be the way you\'ve specified, depending on the Lagrangian.\n\n>\n> Second, what about if the Lagrangian were to depend on the second t\n> derivatives as well?\n\nI don\'t think you could use the df/dx - d/dt(df/dx\'), which is Euler\'s\nformula, but I guess you could do it. I\'m not sure what the physical\nimplications would be, though.\n\n>\n> Third, (and this I believe the answer to is no, but I can\'t find a\n> proof of it or generate one myself), is it possible to have a\n> Lagrangian of a,b, and c where all three functions are varied but c\n> does not influence the evolution of a or b, although it is influenced\n> by one or both of them? This seems strictly impossible to me - it\n> wouldn\'t make any sense if it were true - but I can\'t prove that it\n> has to be that way.\n\nI don\'t think this is strictly a calculus of variations question. What, in\nessence, you\'re asking is if you can have a function f(a,b,c(a,b,x),t),\nwhere a and b are not functions of c(a,b,x). As long as you allow c to be\nvaried by an additional parameter, x, the value of c will not necessarily be\ndependant on one particular value of a or b.\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>"Blake Winter" <blake.winter@houghton.edu> wrote in message
news:87423d2a.0410071613.6b09c1ca@posting.google.c om...
>
> I'm trying to work out three questions about the theory of calculus of
> variations:
> First, is it possible to have a Lagrangian which is a function of a,b,
> and c, along with the parameter t, and their first t derivatives, such
> that the resulting differential equations can be written in the form
> A(a,b), B(a,b), C(b,c)
> Where A, B, and C are differential equations of those functions a, b,
> c?

I don't think so. For a function of several variables, you would just use:

df/da - d/dt(df/da') =df/db - d/dt(df/db') =df/dc - d/dt(df/dc') =

if df/dc or df/dc' leave you with a function that includes b, it seems to me
that df/db or df/db' should leave you with a function that includes c.

I don't know why you would want A(a,b), B(a,b), C(b,c), though. Ideally you
would want A(a, a', a''), B(b, b', b''), C(c, c', c''). It may or may not
turn out to be the way you've specified, depending on the Lagrangian.

>
> Second, what about if the Lagrangian were to depend on the second t
> derivatives as well?

I don't think you could use the df/dx - d/dt(df/dx'), which is Euler's
formula, but I guess you could do it. I'm not sure what the physical
implications would be, though.

>
> Third, (and this I believe the answer to is no, but I can't find a
> proof of it or generate one myself), is it possible to have a
> Lagrangian of a,b, and c where all three functions are varied but c
> does not influence the evolution of a or b, although it is influenced
> by one or both of them? This seems strictly impossible to me - it
> wouldn't make any sense if it were true - but I can't prove that it
> has to be that way.

I don't think this is strictly a calculus of variations question. What, in
essence, you're asking is if you can have a function f(a,b,c(a,b,x),t),
where a and b are not functions of c(a,b,x). As long as you allow c to be
varied by an additional parameter, x, the value of c will not necessarily be
dependant on one particular value of a or b.

Igor Khavkine

Oct11-04, 03:52 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>\nOn Fri, 08 Oct 2004 11:20:40 +0000, Blake Winter wrote:\n\n>\n> I\'m trying to work out three questions about the theory of calculus of\n> variations:\n> First, is it possible to have a Lagrangian which is a function of a,b, and\n> c, along with the parameter t, and their first t derivatives, such that\n> the resulting differential equations can be written in the form A(a,b),\n> B(a,b), C(b,c)\n> Where A, B, and C are differential equations of those functions a, b, c?\n\nThree coupled harmonic oscillators (let a_t, ... denote time derivatives):\n\nL = (a_t)^2 + (b_t)^2 + (c_t)^2 - a^2 - b^2 - c^2 + ab + bc + ca.\n\n> Second, what about if the Lagrangian were to depend on the second t\n> derivatives as well?\n\nThere is no problem with including higher derivatives of the dynamical\nvariables in the Lagrangian. The resulting equations will be higher than\nsecond order though. For example, you could easily add terms of the form\n(a_tt)^2 to the Lagrangian above.\n\n> Third, (and this I believe the answer to is no, but I can\'t find a proof\n> of it or generate one myself), is it possible to have a Lagrangian of a,b,\n> and c where all three functions are varied but c does not influence the\n> evolution of a or b, although it is influenced by one or both of them?\n> This seems strictly impossible to me - it wouldn\'t make any sense if it\n> were true - but I can\'t prove that it has to be that way.\n\nI don\'t believe so. This would violate Newton\'s third law, which is built\ninto the Lagrangian formalism. Thinking strictly mathematically, if the\ndifferential equation for c has terms depending on a and b, then the\nLagrangian must have had terms depending on a and d, and b and c,\nrespectively. But these terms imply that the differential equations for a\nand b also have terms depending on c.\n\nHope this helps.\n\nIgor\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 Fri, 08 Oct 2004 11:20:40 +0000, Blake Winter wrote:

>
> I'm trying to work out three questions about the theory of calculus of
> variations:
> First, is it possible to have a Lagrangian which is a function of a,b, and
> c, along with the parameter t, and their first t derivatives, such that
> the resulting differential equations can be written in the form A(a,b),
> B(a,b), C(b,c)
> Where A, B, and C are differential equations of those functions a, b, c?

Three coupled harmonic oscillators (let a_t, ... denote time derivatives):

L = (a_t)^2 + (b_t)^2 + (c_t)^2 - a^2 - b^2 - c^2 + ab + bc + ca.

> Second, what about if the Lagrangian were to depend on the second t
> derivatives as well?

There is no problem with including higher derivatives of the dynamical
variables in the Lagrangian. The resulting equations will be higher than
second order though. For example, you could easily add terms of the form
(a_{tt})^2 to the Lagrangian above.

> Third, (and this I believe the answer to is no, but I can't find a proof
> of it or generate one myself), is it possible to have a Lagrangian of a,b,
> and c where all three functions are varied but c does not influence the
> evolution of a or b, although it is influenced by one or both of them?
> This seems strictly impossible to me - it wouldn't make any sense if it
> were true - but I can't prove that it has to be that way.

I don't believe so. This would violate Newton's third law, which is built
into the Lagrangian formalism. Thinking strictly mathematically, if the
differential equation for c has terms depending on a and b, then the
Lagrangian must have had terms depending on a and d, and b and c,
respectively. But these terms imply that the differential equations for a
and b also have terms depending on c.

Hope this helps.

Igor

Blake Winter

Oct11-04, 01:06 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\nIgor Khavkine <k_igor_k@lycos.com> wrote\n> I don\'t believe so. This would violate Newton\'s third law, which is built\n> into the Lagrangian formalism. Thinking strictly mathematically, if the\n> differential equation for c has terms depending on a and b, then the\n> Lagrangian must have had terms depending on a and d, and b and c,\n> respectively. But these terms imply that the differential equations for a\n> and b also have terms depending on c.\n>\n> Hope this helps.\n>\n> Igor\n\nThat does help. The reasoning with Newton\'s law was what I had\noriginally considered in deciding the answer is negative.\n\nBut I think that the question is related to my first question: If the\nsolution to an extremization problem can be written as three\ndifferential equations A(a,b), B(a,b), C(b,c) then A and B will fully\ndetermine variables a and b, and so while variable b influences\nvariable c there will not be a corresponding effect on b. Hence I\nsuspect that even if the Lagrangian can be seperated out, the\ndifferential equations which one will get as the solution from\nEuler-Lagrange will not be able to be so seperated.\nLooking at the Euler-Lagrange equations, I think this can be seen by\nthe fact that if dL/dc or dL/dc_t involves b or b_t, then dL/db or\ndL/db_t must involve c, so the differential equation from b will\ninvolve c terms as well. Since the higher derivative analogues of the\nEuler-Lagrange equation work the same way, I suspect it is true for\nany number of derivatives placed in the Lagrangian.\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 <k_{igor_k}@lycos.com> wrote
> I don't believe so. This would violate Newton's third law, which is built
> into the Lagrangian formalism. Thinking strictly mathematically, if the
> differential equation for c has terms depending on a and b, then the
> Lagrangian must have had terms depending on a and d, and b and c,
> respectively. But these terms imply that the differential equations for a
> and b also have terms depending on c.
>
> Hope this helps.
>
> Igor

That does help. The reasoning with Newton's law was what I had
originally considered in deciding the answer is negative.

But I think that the question is related to my first question: If the
solution to an extremization problem can be written as three
differential equations A(a,b), B(a,b), C(b,c) then A and B will fully
determine variables a and b, and so while variable b influences
variable c there will not be a corresponding effect on b. Hence I
suspect that even if the Lagrangian can be seperated out, the
differential equations which one will get as the solution from
Euler-Lagrange will not be able to be so seperated.
Looking at the Euler-Lagrange equations, I think this can be seen by
the fact that if dL/dc or dL/dc_t involves b or b_t, then dL/db or
dL/db_t must involve c, so the differential equation from b will
involve c terms as well. Since the higher derivative analogues of the
Euler-Lagrange equation work the same way, I suspect it is true for
any number of derivatives placed in the Lagrangian.

Igor Khavkine

Oct12-04, 10:50 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\nOn Mon, 11 Oct 2004 18:06:31 +0000, Blake Winter wrote:\n\n>\n>\n> Igor Khavkine <k_igor_k@lycos.com> wrote\n>> I don\'t believe so. This would violate Newton\'s third law, which is\n>> built into the Lagrangian formalism. Thinking strictly mathematically,\n>> if the differential equation for c has terms depending on a and b, then\n>> the Lagrangian must have had terms depending on a and d, and b and c,\n>> respectively. But these terms imply that the differential equations for\n>> a and b also have terms depending on c.\n>>\n>> Hope this helps.\n>>\n>> Igor\n>\n> That does help. The reasoning with Newton\'s law was what I had originally\n> considered in deciding the answer is negative.\n>\n> But I think that the question is related to my first question: If the\n> solution to an extremization problem can be written as three differential\n> equations A(a,b), B(a,b), C(b,c) then A and B will fully determine\n> variables a and b, and so while variable b influences variable c there\n> will not be a corresponding effect on b. Hence I suspect that even if the\n> Lagrangian can be seperated out, the differential equations which one will\n> get as the solution from Euler-Lagrange will not be able to be so\n> seperated. Looking at the Euler-Lagrange equations, I think this can be\n> seen by the fact that if dL/dc or dL/dc_t involves b or b_t, then dL/db or\n> dL/db_t must involve c, so the differential equation from b will involve c\n> terms as well. Since the higher derivative analogues of the\n> Euler-Lagrange equation work the same way, I suspect it is true for any\n> number of derivatives placed in the Lagrangian.\n\nYou are right. When I saw the first part of your question, I originally\nsupposed that you meant something like a cyclic set of equations A(a,b),\nB(b,c), C(c,a). But I see now that it is not what you meant, and the\nexample I gave is irrelevant. Each equation obtained from the\nEuler-Lagrange procedure is labeled by the variable that was varied in\norder to obtain it. So I\'ll suppose that A was obtained from the variation\nof a, etc. So if we have A(a,b) and B(a,b), then it should be clear by now\nthat since neither A nor B depend on c, then C must also be independent of\na and b. This conclusion is independent of the number of derivatives that\nare included in your Lagrangian.\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>On Mon, 11 Oct 2004 18:06:31 +0000, Blake Winter wrote:

>
>
> Igor Khavkine <k_{igor_k}@lycos.com> wrote
>> I don't believe so. This would violate Newton's third law, which is
>> built into the Lagrangian formalism. Thinking strictly mathematically,
>> if the differential equation for c has terms depending on a and b, then
>> the Lagrangian must have had terms depending on a and d, and b and c,
>> respectively. But these terms imply that the differential equations for
>> a and b also have terms depending on c.
>>
>> Hope this helps.
>>
>> Igor
>
> That does help. The reasoning with Newton's law was what I had originally
> considered in deciding the answer is negative.
>
> But I think that the question is related to my first question: If the
> solution to an extremization problem can be written as three differential
> equations A(a,b), B(a,b), C(b,c) then A and B will fully determine
> variables a and b, and so while variable b influences variable c there
> will not be a corresponding effect on b. Hence I suspect that even if the
> Lagrangian can be seperated out, the differential equations which one will
> get as the solution from Euler-Lagrange will not be able to be so
> seperated. Looking at the Euler-Lagrange equations, I think this can be
> seen by the fact that if dL/dc or dL/dc_t involves b or b_t, then dL/db or
> dL/db_t must involve c, so the differential equation from b will involve c
> terms as well. Since the higher derivative analogues of the
> Euler-Lagrange equation work the same way, I suspect it is true for any
> number of derivatives placed in the Lagrangian.

You are right. When I saw the first part of your question, I originally
supposed that you meant something like a cyclic set of equations A(a,b),
B(b,c), C(c,a). But I see now that it is not what you meant, and the
example I gave is irrelevant. Each equation obtained from the
Euler-Lagrange procedure is labeled by the variable that was varied in
order to obtain it. So I'll suppose that A was obtained from the variation
of a, etc. So if we have A(a,b) and B(a,b), then it should be clear by now
that since neither A nor B depend on c, then C must also be independent of
a and b. This conclusion is independent of the number of derivatives that
are included in your Lagrangian.

Igor

Igor Khavkine

Oct13-04, 02:13 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 Tue, 12 Oct 2004 15:49:58 +0000, Mark Palenik wrote:\n\n> "Igor Khavkine" <k_igor_k@lycos.com> wrote in message\n> news:pan.2004.10.08.15.24.17.940897@lycos.com...\n >>\n>> On Fri, 08 Oct 2004 11:20:40 +0000, Blake Winter wrote:\n\n>> > I\'m trying to work out three questions about the theory of calculus of\n>> > variations:\n>> > First, is it possible to have a Lagrangian which is a function of a,b,\n> and\n>> > c, along with the parameter t, and their first t derivatives, such\n>> > that the resulting differential equations can be written in the form\n>> > A(a,b), B(a,b), C(b,c)\n>> > Where A, B, and C are differential equations of those functions a, b,\n>> > c?\n>>\n>> Three coupled harmonic oscillators (let a_t, ... denote time\n>> derivatives):\n>>\n>> L = (a_t)^2 + (b_t)^2 + (c_t)^2 - a^2 - b^2 - c^2 + ab + bc + ca.\n>>\n> Hehe - silly me, I forgot about linear combinations.\n>\n> I was ready to reply with a question when I realized that\n>\n> -2a + b + c - 2a_tt = 0\n> -2b + a + c - 2b_tt = 0\n> -2c + b + a - 2c_tt = 0\n>\n> could make\n>\n> -2a + b + c - 2a_tt = 0\n> -2b + a + c - 2b_tt = 0\n> -3c + 3b - 2c_tt + 2b_tt = 0\n>\n> but why is this form desirable?\n\nThere is nothing special about this form. I misread the original question,\nso I came up with an example Lagrangian that produces equations of the\nform A(a,b), B(b,c), C(c,a) (note the cyclicity). But the actual answer\nto the original question is negative. I already addressed this in another\npost in this thread.\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>On Tue, 12 Oct 2004 15:49:58 +0000, Mark Palenik wrote:

> "Igor Khavkine" <k_{igor_k}@lycos.com> wrote in message
> news:pan.2004.10.08.15.24.17.940897@lycos.com...
>>
>> On Fri, 08 Oct 2004 11:20:40 +0000, Blake Winter wrote:

>> > I'm trying to work out three questions about the theory of calculus of
>> > variations:
>> > First, is it possible to have a Lagrangian which is a function of a,b,
> and
>> > c, along with the parameter t, and their first t derivatives, such
>> > that the resulting differential equations can be written in the form
>> > A(a,b), B(a,b), C(b,c)
>> > Where A, B, and C are differential equations of those functions a, b,
>> > c?
>>
>> Three coupled harmonic oscillators (let a_t, ... denote time
>> derivatives):
>>
>> L = (a_t)^2 + (b_t)^2 + (c_t)^2 - a^2 - b^2 - c^2 + ab + bc + ca.
>>
> Hehe - silly me, I forgot about linear combinations.
>
> I was ready to reply with a question when I realized that
>
> -2a + b + c - 2a_tt =
> -2b + a + c - 2b_tt =
> -2c + b + a - 2c_tt =
>
> could make
>
> -2a + b + c - 2a_tt =
> -2b + a + c - 2b_tt =
> -3c + 3b - 2c_tt + 2b_tt =
>
> but why is this form desirable?

There is nothing special about this form. I misread the original question,
so I came up with an example Lagrangian that produces equations of the
form A(a,b), B(b,c), C(c,a) (note the cyclicity). But the actual answer
to the original question is negative. I already addressed this in another
post in this thread.

Igor

Mark Palenik

Oct14-04, 11:07 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\n\n"Igor Khavkine" <k_igor_k@lycos.com> wrote in message\nnews:pan.2004.10.13.15.40.28.134828@lycos .com...\n> On Tue, 12 Oct 2004 15:49:58 +0000, Mark Palenik wrote:\n>\n> > "Igor Khavkine" <k_igor_k@lycos.com> wrote in message\n> > news:pan.2004.10.08.15.24.17.940897@lycos.com...\n > >>\n> >> On Fri, 08 Oct 2004 11:20:40 +0000, Blake Winter wrote:\n>\n> >> > I\'m trying to work out three questions about the theory of calculus\nof\n> >> > variations:\n> >> > First, is it possible to have a Lagrangian which is a function of\na,b,\n> > and\n> >> > c, along with the parameter t, and their first t derivatives, such\n> >> > that the resulting differential equations can be written in the form\n> >> > A(a,b), B(a,b), C(b,c)\n> >> > Where A, B, and C are differential equations of those functions a, b,\n> >> > c?\n> >>\n> >> Three coupled harmonic oscillators (let a_t, ... denote time\n> >> derivatives):\n> >>\n> >> L = (a_t)^2 + (b_t)^2 + (c_t)^2 - a^2 - b^2 - c^2 + ab + bc + ca.\n> >>\n> > Hehe - silly me, I forgot about linear combinations.\n> >\n> > I was ready to reply with a question when I realized that\n> >\n> > -2a + b + c - 2a_tt = 0\n> > -2b + a + c - 2b_tt = 0\n> > -2c + b + a - 2c_tt = 0\n> >\n> > could make\n> >\n> > -2a + b + c - 2a_tt = 0\n> > -2b + a + c - 2b_tt = 0\n> > -3c + 3b - 2c_tt + 2b_tt = 0\n> >\n> > but why is this form desirable?\n>\n> There is nothing special about this form. I misread the original question,\n> so I came up with an example Lagrangian that produces equations of the\n> form A(a,b), B(b,c), C(c,a) (note the cyclicity). But the actual answer\n> to the original question is negative. I already addressed this in another\n> post in this thread.\n>\n> Igor\n\nOk, thanks. I\'ve only been doing this stuff for half a semester (actually\nwe\'re moving on to other stuff now), so I just wanted to be sure there\nwasn\'t something interesting that I hadn\'t learned yet - and it\'s easy to\nget into trouble when you try to explain things you\'ve just learned (maybe I\nshould stop doing that here? :) ).\n\nAs for your other post, sorry, I either didn\'t see it at the time, or I just\nsent this before it appeared on the group.\n\nThis might not go through unless I say something physically relevant, so let\nme just ask\n\nWhen doing a mechanics problem, does anyone have a way of deciding whether\nto use the Hamiltonian or the Lagrangian? They seem about equal in terms of\namount of number crunching involved (although, I suppose the Hamiltonian\nmight require a bit less - at least when H = T+U). Is the Hamiltonian\nsimply more useful because its applications to other areas of physics, or is\nthere ever a reason that it makes mechanics problems easier than minimizing\nthe action? I guess that\'s a bit off topic, but I\'ve been curious.\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>"Igor Khavkine" <k_{igor_k}@lycos.com> wrote in message
news:pan.2004.10.13.15.40.28.134828@lycos.com...
> On Tue, 12 Oct 2004 15:49:58 +0000, Mark Palenik wrote:
>
> > "Igor Khavkine" <k_{igor_k}@lycos.com> wrote in message
> > news:pan.2004.10.08.15.24.17.940897@lycos.com...
> >>
> >> On Fri, 08 Oct 2004 11:20:40 +0000, Blake Winter wrote:
>
> >> > I'm trying to work out three questions about the theory of calculus
of
> >> > variations:
> >> > First, is it possible to have a Lagrangian which is a function of
a,b,
> > and
> >> > c, along with the parameter t, and their first t derivatives, such
> >> > that the resulting differential equations can be written in the form
> >> > A(a,b), B(a,b), C(b,c)
> >> > Where A, B, and C are differential equations of those functions a, b,
> >> > c?
> >>
> >> Three coupled harmonic oscillators (let a_t, ... denote time
> >> derivatives):
> >>
> >> L = (a_t)^2 + (b_t)^2 + (c_t)^2 - a^2 - b^2 - c^2 + ab + bc + ca.
> >>
> > Hehe - silly me, I forgot about linear combinations.
> >
> > I was ready to reply with a question when I realized that
> >
> > -2a + b + c - 2a_tt =
> > -2b + a + c - 2b_tt =
> > -2c + b + a - 2c_tt =
> >
> > could make
> >
> > -2a + b + c - 2a_tt =
> > -2b + a + c - 2b_tt =
> > -3c + 3b - 2c_tt + 2b_tt =
> >
> > but why is this form desirable?
>
> There is nothing special about this form. I misread the original question,
> so I came up with an example Lagrangian that produces equations of the
> form A(a,b), B(b,c), C(c,a) (note the cyclicity). But the actual answer
> to the original question is negative. I already addressed this in another
> post in this thread.
>
> Igor

Ok, thanks. I've only been doing this stuff for half a semester (actually
we're moving on to other stuff now), so I just wanted to be sure there
wasn't something interesting that I hadn't learned yet - and it's easy to
get into trouble when you try to explain things you've just learned (maybe I
should stop doing that here? :) ).

As for your other post, sorry, I either didn't see it at the time, or I just
sent this before it appeared on the group.

This might not go through unless I say something physically relevant, so let
me just ask

When doing a mechanics problem, does anyone have a way of deciding whether
to use the Hamiltonian or the Lagrangian? They seem about equal in terms of
amount of number crunching involved (although, I suppose the Hamiltonian
might require a bit less - at least when H = T+U). Is the Hamiltonian
simply more useful because its applications to other areas of physics, or is
there ever a reason that it makes mechanics problems easier than minimizing
the action? I guess that's a bit off topic, but I've been curious.

Frank Hellmann

Oct15-04, 01:55 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>\n\n"Mark Palenik" <markpalenik@wideopenwest.com> wrote in message news:<Rvadnd7In-uRKfDcRVn-uQ@wideopenwest.com>...\n\n> When doing a mechanics problem, does anyone have a way of deciding whether\n> to use the Hamiltonian or the Lagrangian? They seem about equal in terms of\n> amount of number crunching involved (although, I suppose the Hamiltonian\n> might require a bit less - at least when H = T+U). Is the Hamiltonian\n> simply more useful because its applications to other areas of physics, or is\n> there ever a reason that it makes mechanics problems easier than minimizing\n> the action? I guess that\'s a bit off topic, but I\'ve been curious.\n\nIt really depends not only on the problem but also on the question you\nare asking.\nIn the Hamiltonian formalism you can use Hamilton Jacobi Theory, which\nis the most powerfull and generall way to obtain closed sollutions to\na physical system. Then there are some nice geometric phase space ways\nof looking at things in the Hamiltonian formalism that can be used for\nanalysis of the system. KAM is formulated in a Hamiltonian context:\nhttp://mathworld.wolfram.com/Kolmogorov-Arnold-MoserTheorem.html\n\nA lot of the advantages come down to explicitly splitting momenta and\nposition, which allows for more generall coordinate transformations.\n\nBasically the general thrust seems to me to be that Lagrangians are\nusefull to build theories with certain symmetries (and that\'s\nvirtually all of them today), and Hamiltonians are usefull to actually\ntry and work out what those theories predict.\n\nIt\'s trivial to write down a relativistically covariant Lagrangian\nfrom nothing, it\'s quite hard to do so for a Hamiltonian.\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>"Mark Palenik" <markpalenik@wideopenwest.com> wrote in message news:<Rvadnd7In-uRKfDcRVn-uQ@wideopenwest.com>...

> When doing a mechanics problem, does anyone have a way of deciding whether
> to use the Hamiltonian or the Lagrangian? They seem about equal in terms of
> amount of number crunching involved (although, I suppose the Hamiltonian
> might require a bit less - at least when H = T+U). Is the Hamiltonian
> simply more useful because its applications to other areas of physics, or is
> there ever a reason that it makes mechanics problems easier than minimizing
> the action? I guess that's a bit off topic, but I've been curious.

It really depends not only on the problem but also on the question you
are asking.
In the Hamiltonian formalism you can use Hamilton Jacobi Theory, which
is the most powerfull and generall way to obtain closed sollutions to
a physical system. Then there are some nice geometric phase space ways
of looking at things in the Hamiltonian formalism that can be used for
analysis of the system. KAM is formulated in a Hamiltonian context:
http://mathworld.wolfram.com/Kolmogorov-Arnold-MoserTheorem.html

A lot of the advantages come down to explicitly splitting momenta and
position, which allows for more generall coordinate transformations.

Basically the general thrust seems to me to be that Lagrangians are
usefull to build theories with certain symmetries (and that's
virtually all of them today), and Hamiltonians are usefull to actually
try and work out what those theories predict.

It's trivial to write down a relativistically covariant Lagrangian
from nothing, it's quite hard to do so for a Hamiltonian.