View Full Version : Can energy conservation be derived from Newton's motion laws only?
Poakfield@msn.com
Dec14-04, 05:35 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>Hi! Could someone please tell me if the law of conservation of energy\ncan be derived only from Newton\'s motion laws? Thanks.\n\nPeter\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>Hi! Could someone please tell me if the law of conservation of energy
can be derived only from Newton's motion laws? Thanks.
Peter
Igor Khavkine
Dec15-04, 12:35 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, 14 Dec 2004 11:35:20 +0000, Poakfield wrote:\n\n> Hi! Could someone please tell me if the law of conservation of energy can\n> be derived only from Newton\'s motion laws? Thanks.\n\nTogether with the assumption of conservative forces only (F = -grad V).\nHere is how:\n\nm x\'\' = -grad V\nm x\' x\'\' = - x\' grad V\nd/dt (m x\'^2)/2 = -d/dt V\nd/dt (m x\'^2/2 + V) = 0\nm v^2/2 + V = E,\n\nwhere v = x\' and E is some constant.\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 Tue, 14 Dec 2004 11:35:20 +0000, Poakfield wrote:
> Hi! Could someone please tell me if the law of conservation of energy can
> be derived only from Newton's motion laws? Thanks.
Together with the assumption of conservative forces only (F = -grad V).
Here is how:
m x'' = -grad Vm x' x'' = - x' grad Vd/dt (m x'^2)/2 = -d/dt Vd/dt (m x'^2/2 + V) =m v^2/2 + V = E,
where v = x' and E is some constant.
Igor
Strong_Field
Dec20-04, 10:06 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>"Igor Khavkine" <k_igor_k@lycos.com> wrote in message\nnews:pan.2004.12.14.15.56.52.1093@lycos.c om...\n> On Tue, 14 Dec 2004 11:35:20 +0000, Poakfield wrote:\n>\n> > Hi! Could someone please tell me if the law of conservation of energy\ncan\n> > be derived only from Newton\'s motion laws? Thanks.\n>\n> Together with the assumption of conservative forces only (F = -grad V).\n> Here is how:\n>\n> m x\'\' = -grad V\n> m x\' x\'\' = - x\' grad V\n> d/dt (m x\'^2)/2 = -d/dt V\n> d/dt (m x\'^2/2 + V) = 0\n> m v^2/2 + V = E,\n>\n> where v = x\' and E is some constant.\n>\n\nI don\'t I understand this derivation.\n\n\n\nAs we read in elementary calculus the instantaneous change represented by\nthe derivative notation can always be represented as finite average, or\ndistance/time. How would you write this derivation without calculus\nnotation?\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.2004.12.14.15.56.52.1093@lycos.com...
> On Tue, 14 Dec 2004 11:35:20 +0000, Poakfield wrote:
>
> > Hi! Could someone please tell me if the law of conservation of energy
can
> > be derived only from Newton's motion laws? Thanks.
>
> Together with the assumption of conservative forces only (F = -grad V).
> Here is how:
>
> m x'' = -grad V
> m x' x'' = - x' grad V
> d/dt (m x'^2)/2 = -d/dt V
> d/dt (m x'^2/2 + V) =
> m v^2/2 + V = E,
>
> where v = x' and E is some constant.
>
I don't I understand this derivation.
As we read in elementary calculus the instantaneous change represented by
the derivative notation can always be represented as finite average, or
distance/time. How would you write this derivation without calculus
notation?
Igor Khavkine
Dec22-04, 05:59 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>Strong_Field wrote:\n> "Igor Khavkine" <k_igor_k@lycos.com> wrote in message\n> news:pan.2004.12.14.15.56.52.1093@lycos.com...\n> > On Tue, 14 Dec 2004 11:35:20 +0000, Poakfield wrote:\n> >\n> > > Hi! Could someone please tell me if the law of conservation of energy\n> can\n> > > be derived only from Newton\'s motion laws? Thanks.\n> >\n> > Together with the assumption of conservative forces only (F = -grad V).\n> > Here is how:\n> >\n> > m x\'\' = -grad V Newton\'s 2nd law with\n> > conservative force.\n> > m x\' x\'\' = - x\' grad V Multiplication by x\' on both sides.\n> > d/dt (m x\'^2)/2 = -d/dt V Invert product rule (f^2)\' = 2 f\' f\n> > and chain rule g(f)\' = f\' g\'(f).\n> > d/dt (m x\'^2/2 + V) = 0 Add dV/dt to both sides.\n> > m v^2/2 + V = E, Indefinite integral of both sides\n> > with respect to dt (don\'t forget\n> > the constant!).\n> >\n> > where v = x\' and E is some constant.\n>\n> I don\'t I understand this derivation.\n\nI\'ve added annotations to every step. Anyone with the elementary\nknowledge of differential and integral calculus should be able to\nfollow. If it is still not clear, please ask a more specific question.\n\n> As we read in elementary calculus the instantaneous change represented by\n> the derivative notation can always be represented as finite average, or\n> distance/time. How would you write this derivation without calculus\n> notation?\n\nI am not sure what you mean. If I take your claim at face value, it is\nsimply wrong. A derivative is defined as a limiting ratio:\n\nf\'(t) = lim_{dt -> 0} [f(t+dt)-f(t)]/dt.\n\nThe derivative f\'(t) cannot be expressed as the above ratio for a\nfinite value dt. It can be *approximated* with the above ratio for\nsmall enough dt, but this is completely unnecessary here and would only\ncomplicate things.\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>Strong_Field wrote:
> "Igor Khavkine" <k_{igor_k}@lycos.com> wrote in message
> news:pan.2004.12.14.15.56.52.1093@lycos.com...
> > On Tue, 14 Dec 2004 11:35:20 +0000, Poakfield wrote:
> >
> > > Hi! Could someone please tell me if the law of conservation of energy
> can
> > > be derived only from Newton's motion laws? Thanks.
> >
> > Together with the assumption of conservative forces only (F = -grad V).
> > Here is how:
> >
> > m x'' = -grad V Newton's 2nd law with
> > conservative force.
> > m x' x'' = - x' grad V Multiplication by x' on both sides.
> > d/dt (m x'^2)/2 = -d/dt V Invert product rule (f^2)' = 2 f' f
> > and chain rule g(f)' = f' g'(f).
> > d/dt (m x'^2/2 + V) = Add dV/dt to both sides.
> > m v^2/2 + V = E, Indefinite integral of both sides
> > with respect to dt (don't forget
> > the constant!).
> >
> > where v = x' and E is some constant.
>
> I don't I understand this derivation.
I've added annotations to every step. Anyone with the elementary
knowledge of differential and integral calculus should be able to
follow. If it is still not clear, please ask a more specific question.
> As we read in elementary calculus the instantaneous change represented by
> the derivative notation can always be represented as finite average, or
> distance/time. How would you write this derivation without calculus
> notation?
I am not sure what you mean. If I take your claim at face value, it is
simply wrong. A derivative is defined as a limiting ratio:
f'(t) = lim_{dt -> 0} [f(t+dt)-f(t)]/dt.
The derivative f'(t) cannot be expressed as the above ratio for a
finite value dt. It can be *approximated* with the above ratio for
small enough dt, but this is completely unnecessary here and would only
complicate things.
Igor
Strong_Field
Dec26-04, 02:46 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>"Igor Khavkine" <igor.kh@gmail.com> wrote in message\nnews:1103584469.379730.180660@f14g2000cwb .googlegroups.com...\n> Strong_Field wrote:\n> > "Igor Khavkine" <k_igor_k@lycos.com> wrote in message\n> > news:pan.2004.12.14.15.56.52.1093@lycos.com...\n> > > On Tue, 14 Dec 2004 11:35:20 +0000, Poakfield wrote:\n> > >\n> > > > Hi! Could someone please tell me if the law of conservation of energy\n> > can\n> > > > be derived only from Newton\'s motion laws? Thanks.\n> > >\n> > > Together with the assumption of conservative forces only (F = -grad V).\n> > > Here is how:\n> > >\n> > > m x\'\' = -grad V Newton\'s 2nd law with\n> > > conservative force.\n> > > m x\' x\'\' = - x\' grad V Multiplication by x\' on both sides.\n> > > d/dt (m x\'^2)/2 = -d/dt V Invert product rule (f^2)\' = 2 f\'= f\n> > > and chain rule g(f)\' = f\' g\'(f).\n> > > d/dt (m x\'^2/2 + V) = 0 Add dV/dt to both sides.\n> > > m v^2/2 + V = E, Indefinite integral of both sides\n> > > with respect to dt (don\'t forget\n> > > the constant!).\n> > >\n> > > where v = x\' and E is some constant.\n\nI am trying to understand if this is only a mathematical derivation or\nif= it has physics content. For instance, I would call, writing (A.B =\ndot pro= duct) and then deriving (A*B cos x) from it a mathematical\nderivation. This is only a manipulation of symbols in an identity.\n\nIf there is an identity, we can always break it apart and label its\nparts with various names. Then, we can equate the labels we assigned to\nthe par= ts of the identity and recover the original identity. This is\nthe mathematic= al derivation. As far as the physics is concerned there\nis only one identity. The names mathematicians give to its parts have no\nphysical meaning. (Thi= s is my opinion, it may not be true.)\n\nThen when I see in a derivation, 1) multiplying both sides with the same\nterm; 2) applying on both sides standard calculus transformations, like\nt= he chain rule, I always suspect that there is a mathematical\nderivation goin= g on and the fundamental physics may be hidden under\nmathematics. Again, th= is may not be true. Because I don=92t understand\nthe notation used in this c= ase.\n\nSo let me ask more specific questions.\n\nFor instance, your\n\nF = -grad V\n\nF = m x=92=92\n\nappears to me that the original identity here is mx=92=92 = -grad V. Th=\ne original identity is divided in two parts and each side is labeled F.\nIn this case F is a placeholder because it must be eliminated to recover\nthe identity. And putting the identity back together seems to me only a\nmathematical derivation. I am more interested in physics than in\nmathematical manipulations and I would prefer to start from the\ninvariant= s. In this case, the mx=92=92 = -grad V appears to be the\nmore fundamental expression.\n\nThese are elementary notations, but I believe they are not elementary\nconcepts. To simplify the equations above let=92s set m = 1. Then your\nequation mx=92=92 = -grad V becomes x=92=92 = -grad V.\n\nx=92=92 is d(dx/dt)/dt. Or the second derivative of distance with\nrespect= to time.\n\nWe can write the first derivative in delta notation as the increment in\ndistance x divided by the increment in time t: delta x / delta t. But\nthe distance moved in the first increment of time equals the\nacceleration. Therefore, we can write\n\ndelta x/delta t = [-grad V].\n\ngrad operator is the generalization of the derivative operator into\nthree dimensions. There is no new concept involved. Using the above\nreasoning w= e can simplify grad V into a delta notation and write it as\n[increment in p= / increment in q].\n\nCan you clarify what p and q is? Is it possible to express grad V as the\nratio of increments? What are those increments? Since we are working in\nsimple orthogonal coordinates, this should be possible to do.\n\nIn other words your mx=92=92 = -grad V, appears to say, (in the first u=\nnit of time) the distance moved divided by the unit of time equals the\nnegative = of an increment p divided by an increment q.\n\nI hope all this makes sense. Again I am trying to understand if there is\nonly a mathematical transformation here or if there is a derivation\ninvolving physics. Physical notions can be written by various notations.\n= And it can be useful to look at the problem with different notations.\nLike in the other thread there is a discussion of how physics would look\nif it is expressed in projective geometry.\n\nWell, after writing the above, the other reply by =93Dan Platt=94\nappeare= d in my newsserver. Combined with what he wrote now it appears\nto me that F = g= rad V is nothing more than a definition of a function\nof distance. And this is = the definition of potential. Then already\nmx=92=92 = -grad V is =93change i= n motion = equals change in\nheight,=94 which is the statement of the conservation of energy.\n\nSo it may be true that there is only a mathematical derivation here.\nComments appreciated. 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>"Igor Khavkine" <igor.kh@gmail.com> wrote in message
news:1103584469.379730.180660@f14g2000cwb.googlegr oups.com...
> Strong_Field wrote:
> > "Igor Khavkine" <k_{igor_k}@lycos.com> wrote in message
> > news:pan.2004.12.14.15.56.52.1093@lycos.com...
> > > On Tue, 14 Dec 2004 11:35:20 +0000, Poakfield wrote:
> > >
> > > > Hi! Could someone please tell me if the law of conservation of energy
> > can
> > > > be derived only from Newton's motion laws? Thanks.
> > >
> > > Together with the assumption of conservative forces only (F = -grad V).
> > > Here is how:
> > >
> > > m x'' = -grad V Newton's 2nd law with
> > > conservative force.
> > > m x' x'' = - x' grad V Multiplication by x' on both sides.
> > > d/dt (m x'^2)/2 = -d/dt V Invert product rule (f^2)' = 2 f'= f
> > > and chain rule g(f)' = f' g'(f).
> > > d/dt (m x'^2/2 + V) = Add dV/dt to both sides.
> > > m v^2/2 + V = E, Indefinite integral of both sides
> > > with respect to dt (don't forget
> > > the constant!).
> > >
> > > where v = x' and E is some constant.
I am trying to understand if this is only a mathematical derivation or
if= it has physics content. For instance, I would call, writing (A.B =
dot pro= duct) and then deriving (A*B cos x) from it a mathematical
derivation. This is only a manipulation of symbols in an identity.
If there is an identity, we can always break it apart and label its
parts with various names. Then, we can equate the labels we assigned to
the par= ts of the identity and recover the original identity. This is
the mathematic= al derivation. As far as the physics is concerned there
is only one identity. The names mathematicians give to its parts have no
physical meaning. (Thi= s is my opinion, it may not be true.)
Then when I see in a derivation, 1) multiplying both sides with the same
term; 2) applying on both sides standard calculus transformations, like
t= he chain rule, I always suspect that there is a mathematical
derivation goin= g on and the fundamental physics may be hidden under
mathematics. Again, th= is may not be true. Because I don=92t understand
the notation used in this c= ase.
So let me ask more specific questions.
For instance, your
F = -grad VF = m x=92=92
appears to me that the original identity here is mx=92=92 = -grad V. Th=
e original identity is divided in two parts and each side is labeled F.
In this case F is a placeholder because it must be eliminated to recover
the identity. And putting the identity back together seems to me only a
mathematical derivation. I am more interested in physics than in
mathematical manipulations and I would prefer to start from the
invariant= s. In this case, the mx=92=92 = -grad V appears to be the
more fundamental expression.
These are elementary notations, but I believe they are not elementary
concepts. To simplify the equations above let=92s set m = 1. Then your
equation mx=92=92 = -grad V becomes x=92=92 = -grad V.x=92=92 is d(dx/dt)/dt. Or the second derivative of distance with
respect= to time.
We can write the first derivative in \delta notation as the increment in
distance x divided by the increment in time t: \delta x / \delta t. But
the distance moved in the first increment of time equals the
acceleration. Therefore, we can write
\delta x/\delta t = [-grad V].
grad operator is the generalization of the derivative operator into
three dimensions. There is no new concept involved. Using the above
reasoning w= e can simplify grad V into a \delta notation and write it as
[increment in p= / increment in q].
Can you clarify what p and q is? Is it possible to express grad V as the
ratio of increments? What are those increments? Since we are working in
simple orthogonal coordinates, this should be possible to do.
In other words your mx=92=92 = -grad V, appears to say, (in the first u=
nit of time) the distance moved divided by the unit of time equals the
negative = of an increment p divided by an increment q.
I hope all this makes sense. Again I am trying to understand if there is
only a mathematical transformation here or if there is a derivation
involving physics. Physical notions can be written by various notations.
= And it can be useful to look at the problem with different notations.
Like in the other thread there is a discussion of how physics would look
if it is expressed in projective geometry.
Well, after writing the above, the other reply by =93Dan Platt=94
appeare= d in my newsserver. Combined with what he wrote now it appears
to me that F = g= rad V is nothing more than a definition of a function
of distance. And this is = the definition of potential. Then already
mx=92=92 = -grad V is =93change i= n motion = equals change in
height,=94 which is the statement of the conservation of energy.
So it may be true that there is only a mathematical derivation here.
Comments appreciated. Thanks.
Igor Khavkine
Dec28-04, 01:51 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>Strong_Field wrote:\n\n[...]\n\n> I am trying to understand if this is only a mathematical derivation\nor\n> if= it has physics content. For instance, I would call, writing (A.B\n=\n> dot pro= duct) and then deriving (A*B cos x) from it a mathematical\n> derivation. This is only a manipulation of symbols in an identity.\n\nAll derivations are a matter of logic. A derivation can no more be\n"mathematical" than "yellow" or "sweet". It is important not to confuse\nnotation with meaning. The whole derivation can be written out in plain\nEnglish without notation from calculus. However, every derivation\nstarts with some given assumptions. Discussion of meaning is usually\nreserved for these assumptions.\n\nThe physics content is contained in the assumptions. They can be\nsummarized as (1) Newton\'s second law and (2) only conservative forces.\nThe mathematica definition of a conservative force has already been\nstated. Are you looking for a justification of this definition?\n\n> Then when I see in a derivation, 1) multiplying both sides with the\nsame\n> term; 2) applying on both sides standard calculus transformations,\nlike\n> t= he chain rule, I always suspect that there is a mathematical\n> derivation goin= g on and the fundamental physics may be hidden under\n> mathematics. Again, th= is may not be true. Because I don=92t\nunderstand\n> the notation used in this c= ase.\n\nYou still haven\'t told us what your level of mathematical training is,\nor whether you are even familiar with basic calculus. I can try to\nexplain in more detail, but I don\'t know what level of detail to aim\nfor.\n\n> So let me ask more specific questions.\n>\n> For instance, your\n>\n> F = -grad V\n>\n> F = m x=92=92\n>\n> appears to me that the original identity here is mx=92=92 = -grad V.\nTh=\n> e original identity is divided in two parts and each side is labeled\nF.\n> In this case F is a placeholder because it must be eliminated to\nrecover\n> the identity. And putting the identity back together seems to me only\na\n> mathematical derivation. I am more interested in physics than in\n> mathematical manipulations and I would prefer to start from the\n> invariant= s. In this case, the mx=92=92 = -grad V appears to be the\n> more fundamental expression.\n>\n> These are elementary notations, but I believe they are not elementary\n> concepts. To simplify the equations above let=92s set m = 1. Then\nyour\n> equation mx=92=92 = -grad V becomes x=92=92 = -grad V.\n>\n> x=92=92 is d(dx/dt)/dt. Or the second derivative of distance with\n> respect= to time.\n>\n> We can write the first derivative in delta notation as the increment\nin\n> distance x divided by the increment in time t: delta x / delta t.\n\nJust keep in mind that this is only an approximation. The exact result\nis recovered only when delta t goes to zero.\n\n> But\n> the distance moved in the first increment of time equals the\n> acceleration. Therefore, we can write\n>\n> delta x/delta t = [-grad V].\n\nThis is completely incorrect. What you should have written is\n\ndelta (delta x/delta t) / delta t = -grad V.\n\n> grad operator is the generalization of the derivative operator into\n> three dimensions. There is no new concept involved. Using the above\n> reasoning w= e can simplify grad V into a delta notation and write it\nas\n> [increment in p= / increment in q].\n>\n> Can you clarify what p and q is? Is it possible to express grad V as\nthe\n> ratio of increments? What are those increments? Since we are working\nin\n> simple orthogonal coordinates, this should be possible to do.\n\ngrad V is a vector attached to every point in space. This vector points\nin the direction of greatest increase of V and its magnitude gives the\nrate of increase. Its x component is given by delta V/delta x as delta\nx -> 0, where delta V is the change in V in the x direction, similarly\nfor its other components.\n\n> In other words your mx=92=92 = -grad V, appears to say, (in the first\nu=\n> nit of time) the distance moved divided by the unit of time equals\nthe\n> negative = of an increment p divided by an increment q.\n>\n> I hope all this makes sense. Again I am trying to understand if there\nis\n> only a mathematical transformation here or if there is a derivation\n> involving physics. Physical notions can be written by various\nnotations.\n> = And it can be useful to look at the problem with different\nnotations.\n> Like in the other thread there is a discussion of how physics would\nlook\n> if it is expressed in projective geometry.\n>\n> Well, after writing the above, the other reply by =93Dan Platt=94\n> appeare= d in my newsserver. Combined with what he wrote now it\nappears\n> to me that F = g= rad V is nothing more than a definition of a\nfunction\n> of distance. And this is = the definition of potential. Then already\n> mx=92=92 = -grad V is =93change i= n motion = equals change in\n> height,=94 which is the statement of the conservation of energy.\n\nV and grad V are functions of position, not distance. I don\'t\nunderstand what you mean by "change in height". Height of what? Perhaps\nwhat you are trying to get at is that the definition of a conservative\nforce is taylored for the proof conservation of energy. This is true,\nas the name "conservative" implies. Further justification can only go\nto experimental observation, and we do observe that energy is conserved\nin simple mechanical systems.\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>Strong_Field wrote:
[...]
> I am trying to understand if this is only a mathematical derivation
or
> if= it has physics content. For instance, I would call, writing (A.B
=
> dot pro= duct) and then deriving (A*B cos x) from it a mathematical
> derivation. This is only a manipulation of symbols in an identity.
All derivations are a matter of logic. A derivation can no more be
"mathematical" than "yellow" or "sweet". It is important not to confuse
notation with meaning. The whole derivation can be written out in plain
English without notation from calculus. However, every derivation
starts with some given assumptions. Discussion of meaning is usually
reserved for these assumptions.
The physics content is contained in the assumptions. They can be
summarized as (1) Newton's second law and (2) only conservative forces.
The mathematica definition of a conservative force has already been
stated. Are you looking for a justification of this definition?
> Then when I see in a derivation, 1) multiplying both sides with the
same
> term; 2) applying on both sides standard calculus transformations,
like
> t= he chain rule, I always suspect that there is a mathematical
> derivation goin= g on and the fundamental physics may be hidden under
> mathematics. Again, th= is may not be true. Because I don=92t
understand
> the notation used in this c= ase.
You still haven't told us what your level of mathematical training is,
or whether you are even familiar with basic calculus. I can try to
explain in more detail, but I don't know what level of detail to aim
for.
> So let me ask more specific questions.
>
> For instance, your
>
> F = -grad V
>
> F = m x=92=92
>
> appears to me that the original identity here is mx=92=92 = -grad V.
Th=
> e original identity is divided in two parts and each side is labeled
F.
> In this case F is a placeholder because it must be eliminated to
recover
> the identity. And putting the identity back together seems to me only
a
> mathematical derivation. I am more interested in physics than in
> mathematical manipulations and I would prefer to start from the
> invariant= s. In this case, the mx=92=92 = -grad V appears to be the
> more fundamental expression.
>
> These are elementary notations, but I believe they are not elementary
> concepts. To simplify the equations above let=92s set m = 1. Then
your
> equation mx=92=92 = -grad V becomes x=92=92 = -grad V.
>
> x=92=92 is d(dx/dt)/dt. Or the second derivative of distance with
> respect= to time.
>
> We can write the first derivative in \delta notation as the increment
in
> distance x divided by the increment in time t: \delta x / \delta t.
Just keep in mind that this is only an approximation. The exact result
is recovered only when \delta t goes to zero.
> But
> the distance moved in the first increment of time equals the
> acceleration. Therefore, we can write
>
> \delta x/\delta t = [-grad V].
This is completely incorrect. What you should have written is
\delta (\delta x/\delta t) / \delta t = -grad V.
> grad operator is the generalization of the derivative operator into
> three dimensions. There is no new concept involved. Using the above
> reasoning w= e can simplify grad V into a \delta notation and write it
as
> [increment in p= / increment in q].
>
> Can you clarify what p and q is? Is it possible to express grad V as
the
> ratio of increments? What are those increments? Since we are working
in
> simple orthogonal coordinates, this should be possible to do.
grad V is a vector attached to every point in space. This vector points
in the direction of greatest increase of V and its magnitude gives the
rate of increase. Its x component is given by \delta V/\delta x as \deltax -> 0, where \delta V is the change in V in the x direction, similarly
for its other components.
> In other words your mx=92=92 = -grad V, appears to say, (in the first
u=
> nit of time) the distance moved divided by the unit of time equals
the
> negative = of an increment p divided by an increment q.
>
> I hope all this makes sense. Again I am trying to understand if there
is
> only a mathematical transformation here or if there is a derivation
> involving physics. Physical notions can be written by various
notations.
> = And it can be useful to look at the problem with different
notations.
> Like in the other thread there is a discussion of how physics would
look
> if it is expressed in projective geometry.
>
> Well, after writing the above, the other reply by =93Dan Platt=94
> appeare= d in my newsserver. Combined with what he wrote now it
appears
> to me that F = g= rad V is nothing more than a definition of a
function
> of distance. And this is = the definition of potential. Then already
> mx=92=92 = -grad V is =93change i= n motion = equals change in
> height,=94 which is the statement of the conservation of energy.
V and grad V are functions of position, not distance. I don't
understand what you mean by "change in height". Height of what? Perhaps
what you are trying to get at is that the definition of a conservative
force is taylored for the proof conservation of energy. This is true,
as the name "conservative" implies. Further justification can only go
to experimental observation, and we do observe that energy is conserved
in simple mechanical systems.
Hope this helps.
Igor
whopkins@csd.uwm.edu
Dec30-04, 11:46 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>Poakfield@msn.com wrote:\n> Hi! Could someone please tell me if the law of conservation of energy\n> can be derived only from Newton\'s motion laws? Thanks.\n\nThey can if you make some additional assumptions.\n\nSuppose the system can be described by a set of N coordinates\n(q^1,q^2,...q^N), with a force law given by a mass matrix\nm = (m_{ab}: a,b=1,...,N), and force vector F = (F_a: a=1,...,N)\nwith the equations of motion (in matrix form):\ndq/dt = v; m dv/dt = F = F(q,v).\nWe\'ll even allow for the N^2 mass coefficients to be q and v\ndependent.\n\nIf the following conditions occur:\nHelmholz\'s Conditions:\nm is non-singular (i.e., no 0 eigenvalues).\nm = m^T; dm_{ab}/dv^c = dm_{ac}/dv^b\nD m = -1/2 (dF/dv + (dF/dv)^T)\nD (dF/dv - (dF/dv)^T) = 2 (dF/dq - (dF/dq)^T)\nwhere\nD = d/dt + sum v^c d/dq^c,\nwhere ()^T denotes matrix transpose, then the system has a\nHamiltonian. If the m\'s and F\'s have no explicit\ntime-dependence then the Hamiltonian can be made time-independent\nand becomes a constant of motion. This, then, can be taken\nas the conservation of energy.\n\nSo, that gives you the criteria for what kind of force (and\nmass) laws enable Hamiltonians and energy conservation in\nterms of the Hamiltonian.\n\nIf the force is velocity-independent, then the last equation\nbecomes dF/dq = (dF/dq)^T, which is the condition that\ndetermines the derivability of F from a potential: F = -dU/dq.\nSo, forces derivable from potentials are covered as a special\ncase.\n\nMore generally, the first 2 equations for the mass are\nexactly what determine the dervability of the mass matrix\nfrom a vector p, with m = dp/dv; and the vector p, in turn,\nfrom a function T with p = dT/dv. These would define,\nrespectively, the momentum and kinetic energy. Substituting\nthis into the first equation involving m, then you get\nD (dp_a/dv^b) = -1/2 (dF_a/dv^b + dF_b/dv^a).\nManipulating the left-hand side, you get\n(d/dt + sum v^c d/dq^c) (dp_a/dv^b)\n= d/dv^b (d/dt + sum v^c d/dq^c)p_a - dp_a/dq^b\n= d/dv^b (Dp_a) - d/dv^a (dT/dq^b).\nSince dp_a/dv^b = dp_b/dv^a, this can be written symmetrically\nin terms of the a and b indices as\n1/2 d/dv^b (Dp_a - dT/dq^a) + 1/2 d/dv^a (Dp_b - dT/dq^b).\nSo, collecting terms, you have:\ndQ_a/dv^b + dQ_b/dv^a = 0\nwith\nQ_a = Dp_a - dT/dq^a + F_a.\n\nThis implies that Q is linear in v, with\nQ_a = sum H_{ab}(q) v^b + E_a(q)\nand that the coefficients H are anti-symmetric\nH_{ab} = -H_{ba}.\nThus, the force assumes the general form:\nF_a = dT/dq^a - Dp_a + E_a + sum H_{ab} v^b.\n\nSubstituting into the last equation for F:\nD (dF/dv - (dF/dv)^T) = 2 (dF/dq - (dF/dq)^T)\nyou will find that E and H are derivable from potentials\nA, phi such that\nH_{ab} = dA_b/dq^a - dA_a/dq^b\nE_a = -dA^a/dt - d(phi)/dq^a.\n\nOut of these, the force law\nF_a = sum (m_{ab} dv^b/dt)\ntransforms into Euler equations for a Lagrangian expressible\nin terms of T, A and phi. You can work out the details for\nyourself. The p\'s will be the conjugate momenta, and the\nHamiltonian\'s existence will be assured by the non-singularity\nof the mass matrix m. All these quantities can be made\nexplicitly time-independent if the mass matrix m and force\nvector F were.\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>Poakfield@msn.com wrote:
> Hi! Could someone please tell me if the law of conservation of energy
> can be derived only from Newton's motion laws? Thanks.
They can if you make some additional assumptions.
Suppose the system can be described by a set of N coordinates
(q^1,q^2,...q^N), with a force law given by a mass matrix
m = (m_{ab}: a,b=1,...,N), and force vector F = (F_a: a=1,...,N)
with the equations of motion (in matrix form):
dq/dt = v; m dv/dt = F = F(q,v).
We'll even allow for the N^2 mass coefficients to be q and v
dependent.
If the following conditions occur:
Helmholz's Conditions:
m is non-singular (i.e., no eigenvalues).
m = m^T; dm_{ab}/dv^c = dm_{ac}/dv^bD m = -1/2 (dF/dv + (dF/dv)^T)D (dF/dv - (dF/dv)^T) = 2 (dF/dq - (dF/dq)^T)
where
D = d/dt + sum v^c d/dq^c,
where ()^T denotes matrix transpose, then the system has a
Hamiltonian. If the m's and F's have no explicit
time-dependence then the Hamiltonian can be made time-independent
and becomes a constant of motion. This, then, can be taken
as the conservation of energy.
So, that gives you the criteria for what kind of force (and
mass) laws enable Hamiltonians and energy conservation in
terms of the Hamiltonian.
If the force is velocity-independent, then the last equation
becomes dF/dq = (dF/dq)^T, which is the condition that
determines the derivability of F from a potential: F = -dU/dq.
So, forces derivable from potentials are covered as a special
case.
More generally, the first 2 equations for the mass are
exactly what determine the dervability of the mass matrix
from a vector p, with m = dp/dv; and the vector p, in turn,
from a function T with p = dT/dv. These would define,
respectively, the momentum and kinetic energy. Substituting
this into the first equation involving m, then you get
D (dp_a/dv^b) = -1/2 (dF_a/dv^b + dF_b/dv^a).
Manipulating the left-hand side, you get
(d/dt + sum v^c d/dq^c) (dp_a/dv^b)= d/dv^b (d/dt + sum v^c d/dq^c)p_a - dp_a/dq^b= d/dv^b (Dp_a) - d/dv^a (dT/dq^b).
Since dp_a/dv^b = dp_b/dv^a, this can be written symmetrically
in terms of the a and b indices as
1/2 d/dv^b (Dp_a - dT/dq^a) + 1/2 d/dv^a (Dp_b - dT/dq^b).
So, collecting terms, you have:
dQ_a/dv^b + dQ_b/dv^a =
with
Q_a = Dp_a - dT/dq^a + F_a.
This implies that Q is linear in v, with
Q_a = sum H_{ab}(q) v^b + E_a(q)
and that the coefficients H are anti-symmetric
H_{ab} = -H_{ba}.
Thus, the force assumes the general form:
F_a = dT/dq^a - Dp_a + E_a + sum H_{ab} v^b.
Substituting into the last equation for F:
D (dF/dv - (dF/dv)^T) = 2 (dF/dq - (dF/dq)^T)
you will find that E and H are derivable from potentials
A, \phi such that
H_{ab} = dA_b/dq^a - dA_a/dq^bE_a = -dA^a/dt - d(\phi)/dq^a.
Out of these, the force law
F_a = sum (m_{ab} dv^b/dt)
transforms into Euler equations for a Lagrangian expressible
in terms of T, A and \phi. You can work out the details for
yourself. The p's will be the conjugate momenta, and the
Hamiltonian's existence will be assured by the non-singularity
of the mass matrix m. All these quantities can be made
explicitly time-independent if the mass matrix m and force
vector F were.
Strong_Field
Jan4-05, 02:06 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>"Igor Khavkine" <igor.kh@gmail.com> wrote in message\nnews:1104169210.571937.84550@f14g2000cwb. googlegroups.com...\n> Strong_Field wrote:\n>\n> It is important not to confuse\n> notation with meaning. The whole derivation can be written out in plain\n> English without notation from calculus. However, every derivation\n> starts with some given assumptions. Discussion of meaning is usually\n> reserved for these assumptions.\n>\n“Newton’s laws” and “Energy conservation” appears to be two different names\nfor the same thing. In any derivation, there is the first statement X, and\nthere is the last stament Y. If your derivation starts from X and ends with\nthe expression Y, then the expression Y must be hidden in the expression X.\nIf Y is not hidden in X and you could find Y starting from X you would be\nviolating conservation laws. You would be creating something from nothing.\nEquations themselves obey the rules of thermodynamics. An equation is\nnothing more than a definiton of conservation rules. Otherwise you cannot\nexpress the laws of thermodynamics with physics equations. So, in my\nopinion, what physicists call a “derivation” is nothing more than an\nalgebraic simplification process. To claim otherwise would be against the\nlaws of physics. Corrections, comments welcome.\n\n\n> The physics content is contained in the assumptions. They can be\n> summarized as (1) Newton\'s second law and (2) only conservative forces.\n> The mathematica definition of a conservative force has already been\n> stated. Are you looking for a justification of this definition?\n\nTo a disinterested observer who does not know the conventions and traditions\nof physics but is familiar with mathematical notation, what you call “Newton\n’s second law” is simply a definition. “Newton’s second law” makes the two\nsymbols “F” and “a” synonyms. There is no reason to write the superluous\nsymbol “m” which will cancel at the next step. (Or if you wish we can set it\nto unity.)\n\n\n\nF == a is in itself nothing more than saying that in your equations you may\nwrite “F” or if you think it looks better you may write “a” instead. There\nis no physics content in F == a.\n\n\n\nIt is an old physics tradition to call “acceleration” “force” and I have\nnothing to argue with that. Physicists also like to call old definitions\ncoming down from the 18th century “laws of nature,” and this is\nunderstandable too. Like in any other professional business, physics\nprofessionals must be able to communicate with each other and they must have\na standard language that all agree upon. In this language words such as\n“observer” or “law” do not have their colloquial meanings; they have\ntechnical meanings. For a non physicist it is important to see through this\nhigh level technical language to understand the fundamental physics content\nthat physicists already know.\n\n\n\nIt is always possible to put the prose words into precise mathematical\nstatements. So when a physicist says “Newton’s laws” he has precise\nmathematical definition in mind. It does not matter if it is called a\n“law, ”a definition,” whatever. As you say, physicists never confuse\nnotation with meaning. That’s why a non-physicist must always be clear about\nthe meanings.\n>\n> > We can write the first derivative in delta notation as the increment in\n> > distance x divided by the increment in time t: delta x / delta t.\n>\n> Just keep in mind that this is only an approximation. The exact result\n> is recovered only when delta t goes to zero.\n\nWhat do you mean by exact? To me the distance the moon moves in delta t =\n1/1000 seconds is an exact observation. Delta t is infinitely distant from\n0, but it is exact. What I mean is that no new insight is gained by\nconsidering the case t-->0. In practice you choose the smallest unit you can\nmeasure.\n>\n> > But\n> > the distance moved in the first increment of time equals the\n> > acceleration. Therefore, we can write\n> >\n> > delta x/delta t = [-grad V].\n>\n> This is completely incorrect. What you should have written is\n>\n> delta (delta x/delta t) / delta t = -grad V.\n>\nWhat I wrote is elementary physics. I am saying that velocity in the first\nunit of time equals acceleration. You are saying this is completely wrong.\n\n> grad V is a vector attached to every point in space...\n\nFirst, thanks for this explanation. Instead of considering the most general\ncase, let’s look at the specific case of orbits. I believe grad V could be\napplied to orbital situation. There is no doubt that the method of studying\nmotion with the gradient vector point of view may be useful in some\nsituations. But in this case, it appears to me, the complications that ensue\nby introducing this notation is unnecessarily high for physics returns we\nget. Take the case of circular orbits. You divide the circle into points.\nThese points are spaced with equal intervals. You attach an arrow to each\npoint. Each arrow is perpendicular to the radius of the orbit. Each arrow\nhas the same length.... All this cluttering of the figure is done in order\nto say that the orbit has uniform motion. What is gained here by talking\nabout grad V? It seems superfluous to me.\n\n> Perhaps\n> what you are trying to get at is that the definition of a conservative\n> force is taylored for the proof conservation of energy.\n\nYes. Something like that. What does it mean when you assume “only\nconservative forces?” As I understand it, by tradition physicists prefer to\ncall “acceleration” “force.” Substituting “acceleration” in “only\nconservative force” I get:\n\n\n\n“Only conservative [acceleration].”\n\n\n\n“Only Conservative accelerations” may be physicists’ way of saying “let’s\nconsider motions where net acceleration is zero.” The trivial case of which\nis the uniform motion. From a naïve point of view, it appears that your\nderivation amounts to writing “acceleration” in two different notations,\nthen setting, acceleration to zero, and calling this situation “gradient\nforce.” I am not sure this is correct though.\n\n\n> This is true,\n> as the name "conservative" implies. Further justification can only go\n> to experimental observation, and we do observe that energy is conserved\n> in simple mechanical systems.\n\nThis seems not to be relevant to the derivation question. The question is\nnot about the observation of energy conservation but about the derivation of\nit from the Newtonian laws of motion.\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" <igor.kh@gmail.com> wrote in message
news:1104169210.571937.84550@f14g2000cwb.googlegro ups.com...
> Strong_Field wrote:
>
> It is important not to confuse
> notation with meaning. The whole derivation can be written out in plain
> English without notation from calculus. However, every derivation
> starts with some given assumptions. Discussion of meaning is usually
> reserved for these assumptions.
>
“Newton’s laws” and “Energy conservation” appears to be two different names
for the same thing. In any derivation, there is the first statement X, and
there is the last stament Y. If your derivation starts from X and ends with
the expression Y, then the expression Y must be hidden in the expression X.
If Y is not hidden in X and you could find Y starting from X you would be
violating conservation laws. You would be creating something from nothing.
Equations themselves obey the rules of thermodynamics. An equation is
nothing more than a definiton of conservation rules. Otherwise you cannot
express the laws of thermodynamics with physics equations. So, in my
opinion, what physicists call a “derivation” is nothing more than an
algebraic simplification process. To claim otherwise would be against the
laws of physics. Corrections, comments welcome.
> The physics content is contained in the assumptions. They can be
> summarized as (1) Newton's second law and (2) only conservative forces.
> The mathematica definition of a conservative force has already been
> stated. Are you looking for a justification of this definition?
To a disinterested observer who does not know the conventions and traditions
of physics but is familiar with mathematical notation, what you call “Newton
’s second law” is simply a definition. “Newton’s second law” makes the two
symbols “F” and “a” synonyms. There is no reason to write the superluous
symbol “m” which will cancel at the next step. (Or if you wish we can set it
to unity.)
F == a is in itself nothing more than saying that in your equations you may
write “F” or if you think it looks better you may write “a” instead. There
is no physics content in F == a.
It is an old physics tradition to call “acceleration” “force” and I have
nothing to argue with that. Physicists also like to call old definitions
coming down from the 18th century “laws of nature,” and this is
understandable too. Like in any other professional business, physics
professionals must be able to communicate with each other and they must have
a standard language that all agree upon. In this language words such as
“observer” or “law” do not have their colloquial meanings; they have
technical meanings. For a non physicist it is important to see through this
high level technical language to understand the fundamental physics content
that physicists already know.
It is always possible to put the prose words into precise mathematical
statements. So when a physicist says “Newton’s laws” he has precise
mathematical definition in mind. It does not matter if it is called a
“law, ”a definition,” whatever. As you say, physicists never confuse
notation with meaning. That’s why a non-physicist must always be clear about
the meanings.
>
> > We can write the first derivative in \delta notation as the increment in
> > distance x divided by the increment in time t: \delta x / \delta t.
>
> Just keep in mind that this is only an approximation. The exact result
> is recovered only when \delta t goes to zero.
What do you mean by exact? To me the distance the moon moves in \delta t =1/1000 seconds is an exact observation. \Delta t is infinitely distant from
0, but it is exact. What I mean is that no new insight is gained by
considering the case t-->0. In practice you choose the smallest unit you can
measure.
>
> > But
> > the distance moved in the first increment of time equals the
> > acceleration. Therefore, we can write
> >
> > \delta x/\delta t = [-grad V].
>
> This is completely incorrect. What you should have written is
>
> \delta (\delta x/\delta t) / \delta t = -grad V.
>
What I wrote is elementary physics. I am saying that velocity in the first
unit of time equals acceleration. You are saying this is completely wrong.
> grad V is a vector attached to every point in space...
First, thanks for this explanation. Instead of considering the most general
case, let’s look at the specific case of orbits. I believe grad V could be
applied to orbital situation. There is no doubt that the method of studying
motion with the gradient vector point of view may be useful in some
situations. But in this case, it appears to me, the complications that ensue
by introducing this notation is unnecessarily high for physics returns we
get. Take the case of circular orbits. You divide the circle into points.
These points are spaced with equal intervals. You attach an arrow to each
point. Each arrow is perpendicular to the radius of the orbit. Each arrow
has the same length.... All this cluttering of the figure is done in order
to say that the orbit has uniform motion. What is gained here by talking
about grad V? It seems superfluous to me.
> Perhaps
> what you are trying to get at is that the definition of a conservative
> force is taylored for the proof conservation of energy.
Yes. Something like that. What does it mean when you assume “only
conservative forces?” As I understand it, by tradition physicists prefer to
call “acceleration” “force.” Substituting “acceleration” in “only
conservative force” I get:
“Only conservative [acceleration].”
“Only Conservative accelerations” may be physicists’ way of saying “let’s
consider motions where net acceleration is zero.” The trivial case of which
is the uniform motion. From a naïve point of view, it appears that your
derivation amounts to writing “acceleration” in two different notations,
then setting, acceleration to zero, and calling this situation “gradient
force.” I am not sure this is correct though.
> This is true,
> as the name "conservative" implies. Further justification can only go
> to experimental observation, and we do observe that energy is conserved
> in simple mechanical systems.
This seems not to be relevant to the derivation question. The question is
not about the observation of energy conservation but about the derivation of
it from the Newtonian laws of motion.
Igor Khavkine
Jan6-05, 04:10 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>Strong_Field wrote:\n\n> "Newton\'s laws" and "Energy conservation" appears to be two different names\n> for the same thing. In any derivation, there is the first statement X, and\n> there is the last stament Y. If your derivation starts from X and ends with\n> the expression Y, then the expression Y must be hidden in the expression X.\n> If Y is not hidden in X and you could find Y starting from X you would be\n> violating conservation laws. You would be creating something from nothing.\n> Equations themselves obey the rules of thermodynamics. An equation is\n> nothing more than a definiton of conservation rules. Otherwise you cannot\n> express the laws of thermodynamics with physics equations. So, in my\n> opinion, what physicists call a "derivation" is nothing more than an\n> algebraic simplification process. To claim otherwise would be against the\n> laws of physics. Corrections, comments welcome.\n\nThe above paragraph makes very little sense. From what I understand, you\nare trying to apply something very vague to something very precise. It\nis unlikely that you can apply thermodynamics to mathematical logic\nunless propositional statements suddenly turn into an ideal gas.\n\nAnd, as I\'ve already mentioned, one does not discriminate against\nderivations or proofs, whether one is a physicist, mathematician,\nchemist, psychologist, bus driver, etc. One only discriminates against\nassumptions.\n\n> To a disinterested observer who does not know the conventions and\ntraditions\n> of physics but is familiar with mathematical notation, what you call\n"Newton\n> \'s second law" is simply a definition. "Newton\'s second\nlaw" makes the two\n> symbols "F" and "a" synonyms. There is no reason to write the\nsuperluous\n> symbol "m" which will cancel at the next step. (Or if you wish we\ncan set it\n> to unity.)\n>\n> F =3D=3D a is in itself nothing more than saying that in your equations\nyou may\n> write "F" or if you think it looks better you may write "a"\ninstead. There\n> is no physics content in F =3D=3D a.\n\nThis is not true. Both F and a have independent meaning and can be\nmeasured\nindependently. The latter with a ruler and a stop watch and the former\nwith a standard, like a spring of a certain stiffness. Equating them\ngives\na non-trivial statement about how interactions between objects affect\ntheir\nmotion. Also, the mass does not cancel from the equation unless we have\nonly gravitational forces. But since it\'s a dimensionful quantity, you\ncould\nset its numerical value to be 1 in the appropriate units.\n\n> > > We can write the first derivative in delta notation as the\nincrement in\n> > > distance x divided by the increment in time t: delta x / delta t.\n> >\n> > Just keep in mind that this is only an approximation. The exact\nresult\n> > is recovered only when delta t goes to zero.\n>\n> What do you mean by exact? To me the distance the moon moves in delta\nt =3D\n> 1/1000 seconds is an exact observation. Delta t is infinitely distant\nfrom\n> 0, but it is exact. What I mean is that no new insight is gained by\n> considering the case t-->0. In practice you choose the smallest unit\nyou can\n> measure.\n\nYou must keep in mind that Newtonian mechanics is a simplified model of\nthe real world. In this model time, space, and other quantities are\ncontinuous and even smooth. That is why one can speak of derivatives\nand\nhow they can be evaluated exactly. There is no garantee that numbers\nfrom\na theoretical calculation will predict exactly numbers obtained from\nexperiment, only that they\'ll be close.\n\n> > > But\n> > > the distance moved in the first increment of time equals the\n> > > acceleration. Therefore, we can write\n> > >\n> > > delta x/delta t =3D [-grad V].\n> >\n> > This is completely incorrect. What you should have written is\n> >\n> > delta (delta x/delta t) / delta t =3D -grad V.\n> >\n> What I wrote is elementary physics. I am saying that velocity in the\nfirst\n> unit of time equals acceleration. You are saying this is completely\nwrong.\n\nIf you believe this to be elementary physics, you clearly do not\nunderstand\nit. In words, your statement is "velocity is equal to minus the\ngradient\nof the potential," keeping in mind that your "=3D" sign actually means\n"approximately equal when delta t is small". The correct statement\nis "acceleration is equal to minus the gradient of the potential". If\nyou\nreally want to stick to the case where a particle starts accelerating\nfrom rest, then you could write something like\n\nv(at time delta t) =3D (-grad V)*(delta t),\n\nwhich reads "if a particle starts from rest (velocity=3D0 at time 0),\nafter\na time delta t its velocity is (-grad V)*(delta t)". The devil is in\nthe details, as you can see.\n\n>\n> > grad V is a vector attached to every point in space...\n>\n> First, thanks for this explanation. Instead of considering the most\ngeneral\n> case, let\'s look at the specific case of orbits. I believe grad V\ncould be\n> applied to orbital situation. There is no doubt that the method of\nstudying\n> motion with the gradient vector point of view may be useful in some\n> situations. But in this case, it appears to me, the complications\nthat ensue\n> by introducing this notation is unnecessarily high for physics\nreturns we\n> get. Take the case of circular orbits. You divide the circle into\npoints.\n> These points are spaced with equal intervals. You attach an arrow to\neach\n> point. Each arrow is perpendicular to the radius of the orbit. Each\narrow\n> has the same length.... All this cluttering of the figure is done in\norder\n> to say that the orbit has uniform motion. What is gained here by\ntalking\n> about grad V? It seems superfluous to me.\n\nOnce again, I am lost in what you are trying to say. Force is a vector,\nit is applied to a particle when it\'s at a given position, say x. This\nvector at position x could for example come from the gradient of a\nscalar potential V, namely grad V. I fail to see what orbits,\nespecially\ncircular ones have to do with this. Yes, a circular orbit is a\ntrajectory\nof a particle in a certain potential, the Newtonian gravitational\npotential.\nNo, your example does not say anything about superfluousness of grad V,\nwhatever you mean by that.\n\n> > Perhaps\n> > what you are trying to get at is that the definition of a\nconservative\n> > force is taylored for the proof conservation of energy.\n>\n> Yes. Something like that. What does it mean when you assume "only\n> conservative forces?" As I understand it, by tradition physicists\nprefer to\n> call "acceleration" "force." Substituting "acceleration"\nin "only\n> conservative force" I get:\n\nNo.\n\n> "Only conservative [acceleration]."\n\n???\n\n> "Only Conservative accelerations" may be physicists\' way of\nsaying "let\'s\n> consider motions where net acceleration is zero." The trivial case\nof which\n> is the uniform motion. From a na=EFve point of view, it appears that\nyour\n> derivation amounts to writing "acceleration" in two different\nnotations,\n> then setting, acceleration to zero, and calling this situation\n"gradient\n> force." I am not sure this is correct though.\n\nThis is correct in no shape or form.\n\n> > This is true,\n> > as the name "conservative" implies. Further justification can only\ngo\n> > to experimental observation, and we do observe that energy is\nconserved\n> > in simple mechanical systems.\n>\n> This seems not to be relevant to the derivation question. The\nquestion is\n> not about the observation of energy conservation but about the\nderivation of\n> it from the Newtonian laws of motion.\n\nThis question has been answered. Newton\'s laws + assumption of\nconservative\nforces give conservation of enerty. Not it is not a vaccuous statement.\nBoth, Newton\'s laws and assumption of conservative force are justified\nby observing experiments; nature is the highest authority.\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>Strong_Field wrote:
> "Newton's laws" and "Energy conservation" appears to be two different names
> for the same thing. In any derivation, there is the first statement X, and
> there is the last stament Y. If your derivation starts from X and ends with
> the expression Y, then the expression Y must be hidden in the expression X.
> If Y is not hidden in X and you could find Y starting from X you would be
> violating conservation laws. You would be creating something from nothing.
> Equations themselves obey the rules of thermodynamics. An equation is
> nothing more than a definiton of conservation rules. Otherwise you cannot
> express the laws of thermodynamics with physics equations. So, in my
> opinion, what physicists call a "derivation" is nothing more than an
> algebraic simplification process. To claim otherwise would be against the
> laws of physics. Corrections, comments welcome.
The above paragraph makes very little sense. From what I understand, you
are trying to apply something very vague to something very precise. It
is unlikely that you can apply thermodynamics to mathematical logic
unless propositional statements suddenly turn into an ideal gas.
And, as I've already mentioned, one does not discriminate against
derivations or proofs, whether one is a physicist, mathematician,
chemist, psychologist, bus driver, etc. One only discriminates against
assumptions.
> To a disinterested observer who does not know the conventions and
traditions
> of physics but is familiar with mathematical notation, what you call
"Newton
> 's second law" is simply a definition. "Newton's second
law" makes the two
> symbols "F" and "a" synonyms. There is no reason to write the
superluous
> symbol "m" which will cancel at the next step. (Or if you wish we
can set it
> to unity.)
>
> F =3D=3D a is in itself nothing more than saying that in your equations
you may
> write "F" or if you think it looks better you may write "a"
instead. There
> is no physics content in F =3D=3D a.
This is not true. Both F and a have independent meaning and can be
measured
independently. The latter with a ruler and a stop watch and the former
with a standard, like a spring of a certain stiffness. Equating them
gives
a non-trivial statement about how interactions between objects affect
their
motion. Also, the mass does not cancel from the equation unless we have
only gravitational forces. But since it's a dimensionful quantity, you
could
set its numerical value to be 1 in the appropriate units.
> > > We can write the first derivative in \delta notation as the
increment in
> > > distance x divided by the increment in time t: \delta x / \delta t.
> >
> > Just keep in mind that this is only an approximation. The exact
result
> > is recovered only when \delta t goes to zero.
>
> What do you mean by exact? To me the distance the moon moves in \deltat =3D
> 1/1000 seconds is an exact observation. \Delta t is infinitely distant
from
> 0, but it is exact. What I mean is that no new insight is gained by
> considering the case t-->0. In practice you choose the smallest unit
you can
> measure.
You must keep in mind that Newtonian mechanics is a simplified model of
the real world. In this model time, space, and other quantities are
continuous and even smooth. That is why one can speak of derivatives
and
how they can be evaluated exactly. There is no garantee that numbers
from
a theoretical calculation will predict exactly numbers obtained from
experiment, only that they'll be close.
> > > But
> > > the distance moved in the first increment of time equals the
> > > acceleration. Therefore, we can write
> > >
> > > \delta x/\delta t =3D [-grad V].
> >
> > This is completely incorrect. What you should have written is
> >
> > \delta (\delta x/\delta t) / \delta t =3D -grad V.
> >
> What I wrote is elementary physics. I am saying that velocity in the
first
> unit of time equals acceleration. You are saying this is completely
wrong.
If you believe this to be elementary physics, you clearly do not
understand
it. In words, your statement is "velocity is equal to minus the
gradient
of the potential," keeping in mind that your "=3D" sign actually means
"approximately equal when \delta t is small". The correct statement
is "acceleration is equal to minus the gradient of the potential". If
you
really want to stick to the case where a particle starts accelerating
from rest, then you could write something like
v(at time \delta t) =3D (-grad V)*(\delta t),
which reads "if a particle starts from rest (velocity=3D0 at time 0),
after
a time \delta t its velocity is (-grad V)*(\delta t)". The devil is in
the details, as you can see.
>
> > grad V is a vector attached to every point in space...
>
> First, thanks for this explanation. Instead of considering the most
general
> case, let's look at the specific case of orbits. I believe grad V
could be
> applied to orbital situation. There is no doubt that the method of
studying
> motion with the gradient vector point of view may be useful in some
> situations. But in this case, it appears to me, the complications
that ensue
> by introducing this notation is unnecessarily high for physics
returns we
> get. Take the case of circular orbits. You divide the circle into
points.
> These points are spaced with equal intervals. You attach an arrow to
each
> point. Each arrow is perpendicular to the radius of the orbit. Each
arrow
> has the same length.... All this cluttering of the figure is done in
order
> to say that the orbit has uniform motion. What is gained here by
talking
> about grad V? It seems superfluous to me.
Once again, I am lost in what you are trying to say. Force is a vector,
it is applied to a particle when it's at a given position, say x. This
vector at position x could for example come from the gradient of a
scalar potential V, namely grad V. I fail to see what orbits,
especially
circular ones have to do with this. Yes, a circular orbit is a
trajectory
of a particle in a certain potential, the Newtonian gravitational
potential.
No, your example does not say anything about superfluousness of grad V,
whatever you mean by that.
> > Perhaps
> > what you are trying to get at is that the definition of a
conservative
> > force is taylored for the proof conservation of energy.
>
> Yes. Something like that. What does it mean when you assume "only
> conservative forces?" As I understand it, by tradition physicists
prefer to
> call "acceleration" "force." Substituting "acceleration"
in "only
> conservative force" I get:
No.
> "Only conservative [acceleration]."
???
> "Only Conservative accelerations" may be physicists' way of
saying "let's
> consider motions where net acceleration is zero." The trivial case
of which
> is the uniform motion. From a na=EFve point of view, it appears that
your
> derivation amounts to writing "acceleration" in two different
notations,
> then setting, acceleration to zero, and calling this situation
"gradient
> force." I am not sure this is correct though.
This is correct in no shape or form.
> > This is true,
> > as the name "conservative" implies. Further justification can only
go
> > to experimental observation, and we do observe that energy is
conserved
> > in simple mechanical systems.
>
> This seems not to be relevant to the derivation question. The
question is
> not about the observation of energy conservation but about the
derivation of
> it from the Newtonian laws of motion.
This question has been answered. Newton's laws + assumption of
conservative
forces give conservation of enerty. Not it is not a vaccuous statement.
Both, Newton's laws and assumption of conservative force are justified
by observing experiments; nature is the highest authority.
Igor
Maarten van Reeuwijk
Jan6-05, 04:12 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>> =93Newton=92s laws=94 and =93Energy conservation=94 appearsacceleratesi=\nfferent\n> names for the same thing.\n\nAs explained in another post: Newton\'s laws and energy conservation are o=\nnly\nidentical when the force-field is conservative and in absence of other\nenergy-conversion processes, such as electro-magnetism and heat. There is\n*no* general law of conservation of *mechanical* energy. If internal ener=\ngy\nis negligible the energy equation can be=20\n\n> There is no physics content in F == a.\n\na is not a coefficient, but represents a second derivative of position! I=\nf F\n== m d2 x / dt2 does not contain physics to you I suggest reading up =\non\nyour basic calculus.\n\n> What I wrote is elementary physics. I am saying that velocity in the fi=\nrst\n> unit of time equals acceleration. You are saying this is completely wro=\nng.\n\nIt is wrong. Let\'s see how it works for a car at rest that is subjected t=\no a\nlinear force in time F = A t.=20\n\nIntegrating Newton\'s second law once yields the velocity of the car:\n\ndv / dt = F/m <=>\nv = int F/m = A t^2 / 2m + B and B = 0 as the initial velocity is 0\n\nIt is easy to see that your argument is flawed, as your estimate of veloc=\nity\nwould be v = F/m = A t / m, which would be only valid if the force wa=\ns\nconstant in time (and therefore in space, when conservative fields are\nconcerned).=20\n\n> figure is done in order to say that the orbit has uniform motion. What =\nis\n> gained here by talking about grad V? It seems superfluous to me.\n\nAt first it can seem overly complicated but you learn to love it as all\nproblems (with conservative forces) can be treated similarly and there is=\na\nwhole bunch of theorems that are valid for those fields (circle integral\nzero etc). I would really recommend reading up on them. The Feynman\nLectures could be a very good start, very intuitive and not too\nmathematical.\n\n> =93Only Conservative accelerations=94 may be physicists=92 way of sayin=\ng =93let=92s\n> consider motions where net acceleration is zero.=94\n\nForget F==a, because F= m d2 x / dt2 ! There are conservative *forc=\nes*, not\nconservative accelerations.\n\nHTH,\nMaarten\n\n--=20\n==========================\n================= =========\n=================\nMaarten van Reeuwijk Thermal and Fluids Sciences\nPhd student dept. of Multiscale Physics\nwww.ws.tn.tudelft.nl Delft University of Technology\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>> =93Newton=92s laws=94 and =93Energy conservation=94 appearsacceleratesi=
fferent
> names for the same thing.
As explained in another post: Newton's laws and energy conservation are o=
nly
identical when the force-field is conservative and in absence of other
energy-conversion processes, such as electro-magnetism and heat. There is
*no* general law of conservation of *mechanical* energy. If internal ener=
gy
is negligible the energy equation can be=20
> There is no physics content in F == a.
a is not a coefficient, but represents a second derivative of position! I=
f F
== m d2 x / dt2 does not contain physics to you I suggest reading up =
on
your basic calculus.
> What I wrote is elementary physics. I am saying that velocity in the fi=
rst
> unit of time equals acceleration. You are saying this is completely wro=
ng.
It is wrong. Let's see how it works for a car at rest that is subjected t=
o a
linear force in time F = A t.=20
Integrating Newton's second law once yields the velocity of the car:
dv / dt = F/m <=>v = \int F/m = A t^2 / 2m + B and B = as the initial velocity is
It is easy to see that your argument is flawed, as your estimate of veloc=
ity
would be v = F/m = A t / m, which would be only valid if the force wa=
s
constant in time (and therefore in space, when conservative fields are
concerned).=20
> figure is done in order to say that the orbit has uniform motion. What =
is
> gained here by talking about grad V? It seems superfluous to me.
At first it can seem overly complicated but you learn to love it as all
problems (with conservative forces) can be treated similarly and there is=
a
whole bunch of theorems that are valid for those fields (circle integral
zero etc). I would really recommend reading up on them. The Feynman
Lectures could be a very good start, very intuitive and not too
mathematical.
> =93Only Conservative accelerations=94 may be physicists=92 way of sayin=
g =93let=92s
> consider motions where net acceleration is zero.=94
Forget F==a, because F= m d2 x / dt2 ! There are conservative *forc=es*, not
conservative accelerations.
HTH,
Maarten
--=20
==========================
==========================
=================
Maarten van Reeuwijk Thermal and Fluids Sciences
Phd student dept. of Multiscale Physics
www.ws.tn.tudelft.nl Delft University of Technology
Maarten van Reeuwijk
Jan6-05, 04:12 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>Just had a look at the Feynman Lectures and it seems to answer your\nquestions. Interesting chapters of Part I are\n\n4 - Conservation of Energy\n\n7 - The theory of gravitation\n\n9 - Newton\'s law of dynamics, and in 9.7 Planetary motions are dealt with,\nin exactly the way you\'re trying (discretized).\n\n14 - Work and potential energy, including the definition of conservative\nforces.\n\nHTH, Maarten\n--\n================================================ ===================\nMaarten van Reeuwijk Thermal and Fluids Sciences\nPhd student dept. of Multiscale Physics\nwww.ws.tn.tudelft.nl Delft University of Technology\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>Just had a look at the Feynman Lectures and it seems to answer your
questions. Interesting chapters of Part I are
4 - Conservation of Energy
7 - The theory of gravitation
9 - Newton's law of dynamics, and in 9.7 Planetary motions are dealt with,
in exactly the way you're trying (discretized).
14 - Work and potential energy, including the definition of conservative
forces.
HTH, Maarten
--
================================================== =================
Maarten van Reeuwijk Thermal and Fluids Sciences
Phd student dept. of Multiscale Physics
www.ws.tn.tudelft.nl Delft University of Technology
Strong_Field
Jan11-05, 03:48 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>"Igor Khavkine" <igor.kh@gmail.com> wrote in message\nnews:1104911484.419196.39400@c13g2000cwb. googlegroups.com...\n\n> ... One only discriminates against assumptions.\n\nOk. You are right. I went back to your derivation and I looked at it again.\nLet me ask about the assumptions you make. The "Newton\'s laws" assumption\nis the left hand side of your first line:\n\nmx\'\' = -grad V\n\nThe "conservative forces" assumption is the right hand side of the equation.\nFrom the writings of the other posters I got the impression that in physics\nthere is not a definite law of conservation for mechanical systems. Then,\nwhen you say "conservative forces" do you mean only gravity? Are there other\nconservative forces in physics other than gravity? So, let me understand if\nthe assumption of conservative forces reduces the problem only to orbital\nmotion. In other words, is this derivation limited with orbital motion?\n\nAssuming that this is the case, I would say that the function V is 1/r. Can\nV be any scalar function or are we working strictly with V = 1/r?\n\nIf we are working with V = 1/r or V = 1/x, then, when I write the grad\nexplicitly as dV/dx, dV/dy, dV/dz, I believe that dV/dy = 0 and dV/dz = 0.\nAgain limiting the derivation to orbital case only.\n\nIf this is true so far, then, I may write, mx\'\' = -dV/dx. So please let me\nknow if I am on the right track here. If so I\'ll look at the rest of the\nderivation.\n\n> This is not true. Both F and a have independent meaning and can be\n> measured independently. The latter with a ruler and a stop watch and\n> the former with a standard, like a spring of a certain stiffness.\n> Equating them gives a non-trivial statement about how interactions\n> between objects affect their motion.\n\nI am looking at the expression F = ma as a literal observer. It may be\npossible to measure F and a separately and then equate them. But this is not\nwhat F = ma says if we look at it as is. Because I believe that there are\nnot two measurable quantities in this equation. Assume that I am totally ign\norant of the conventions of physics. This would make me a good disinterested\nobserver. You gave me this equation and told me to measure F with it. Fine.\nI take a spring, hang a weight m on it and measure the distance it travels\nin unit time. I say I measured force. What did I measure in reality? I\nmeasured acceleration and I called it force. If we restrict our discussion\nto this equation only, I see only one measurable quantity.\n\nAs you say at the end of your post "nature is the highest authority." I\nagree. As I understand it we communicate with nature through measurement. If\nwe accept that nature is the highest authority, then I would not impose on\nnature quantities which I am unable to measure, such as that force in the\nabove definition. Again, this may be my ignorance about how to measure\nforce. But otherwise, in terms of measurable quantities I only see distance\nand time in F = ma.\n\n\n> You must keep in mind that Newtonian mechanics is a simplified model of\n> the real world.\n\nAs far as I know this is true for any theory. I don\'t think there are any\nexperiments in physics which are stated as absolute measurements without\nmeasurement errors.\n\n> If you believe this to be elementary physics, you clearly do not\n> understand it. In words, your statement is "velocity is equal to\nminus the gradient of the potential....\n\nYou left out my "in the first increment of time." My statement is "In the\nfirst increment of time delta x = (-grad V) (delta t). This is equivalent to\nyour statement. But "acceleration is equal to minus the gradient of the\npotential" is fine with me.\n\n> This question has been answered. Newton\'s laws + assumption of\n> conservative forces give conservation of enerty. Not it is not\n> a vaccuous statement. Both, Newton\'s laws and assumption of\n> conservative force are justified by observing experiments;\n> nature is the highest authority.\n\nAgreed. But if I understand you correctly you are saying that because both\nNewton\'s laws and assumption of conservative forces are justified by\nexperiments they can be derived from each other. I don\'t agree with this\nreasoning.\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" <igor.kh@gmail.com> wrote in message
news:1104911484.419196.39400@c13g2000cwb.googlegro ups.com...
> ... One only discriminates against assumptions.
Ok. You are right. I went back to your derivation and I looked at it again.
Let me ask about the assumptions you make. The "Newton's laws" assumption
is the left hand side of your first line:
mx'' = -grad V
The "conservative forces" assumption is the right hand side of the equation.
From the writings of the other posters I got the impression that in physics
there is not a definite law of conservation for mechanical systems. Then,
when you say "conservative forces" do you mean only gravity? Are there other
conservative forces in physics other than gravity? So, let me understand if
the assumption of conservative forces reduces the problem only to orbital
motion. In other words, is this derivation limited with orbital motion?
Assuming that this is the case, I would say that the function V is 1/r. Can
V be any scalar function or are we working strictly with V = 1/r?
If we are working with V = 1/r or V = 1/x, then, when I write the grad
explicitly as dV/dx, dV/dy, dV/dz, I believe that dV/dy = and dV/dz = .
Again limiting the derivation to orbital case only.
If this is true so far, then, I may write, mx'' = -dV/dx. So please let me
know if I am on the right track here. If so I'll look at the rest of the
derivation.
> This is not true. Both F and a have independent meaning and can be
> measured independently. The latter with a ruler and a stop watch and
> the former with a standard, like a spring of a certain stiffness.
> Equating them gives a non-trivial statement about how interactions
> between objects affect their motion.
I am looking at the expression F = ma as a literal observer. It may be
possible to measure F and a separately and then equate them. But this is not
what F = ma says if we look at it as is. Because I believe that there are
not two measurable quantities in this equation. Assume that I am totally ign
orant of the conventions of physics. This would make me a good disinterested
observer. You gave me this equation and told me to measure F with it. Fine.
I take a spring, hang a weight m on it and measure the distance it travels
in unit time. I say I measured force. What did I measure in reality? I
measured acceleration and I called it force. If we restrict our discussion
to this equation only, I see only one measurable quantity.
As you say at the end of your post "nature is the highest authority." I
agree. As I understand it we communicate with nature through measurement. If
we accept that nature is the highest authority, then I would not impose on
nature quantities which I am unable to measure, such as that force in the
above definition. Again, this may be my ignorance about how to measure
force. But otherwise, in terms of measurable quantities I only see distance
and time in F = ma.
> You must keep in mind that Newtonian mechanics is a simplified model of
> the real world.
As far as I know this is true for any theory. I don't think there are any
experiments in physics which are stated as absolute measurements without
measurement errors.
> If you believe this to be elementary physics, you clearly do not
> understand it. In words, your statement is "velocity is equal to
minus the gradient of the potential....
You left out my "in the first increment of time." My statement is "In the
first increment of time \delta x = (-grad V) (\delta t). This is equivalent to
your statement. But "acceleration is equal to minus the gradient of the
potential" is fine with me.
> This question has been answered. Newton's laws + assumption of
> conservative forces give conservation of enerty. Not it is not
> a vaccuous statement. Both, Newton's laws and assumption of
> conservative force are justified by observing experiments;
> nature is the highest authority.
Agreed. But if I understand you correctly you are saying that because both
Newton's laws and assumption of conservative forces are justified by
experiments they can be derived from each other. I don't agree with this
reasoning.
Strong_Field
Jan11-05, 03:48 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>"Maarten van Reeuwijk" <maarten@remove_this_ws.tn.tudelft.nl> wrote in\nmessage news:crhq3n\\$buo\\$1@news.tudelft.nl...\n\n> As explained in another post: Newton\'s laws and energy conservation are only\n> identical when the force-field is conservative and in absence of other\n> energy-conversion processes, such as electro-magnetism and heat. There is\n> *no* general law of conservation of *mechanical* energy. ...\n\nI think I am trying to understand the situation. As I wrote on another post,\nnow it appears that "force field is conservative" reduces the derivation to\nan orbital problem. Because you say that there is no general law of\nconservation of mechanical energy.\n>\n>\n> ... If F == m d2 x / dt2 does not contain physics to you I suggest\n> reading up on your basic calculus.\n\nWhy calculus and not physics :)\n\n> It is easy to see that your argument is flawed, as your estimate of velocity\n> would be v = F/m = A t / m, which would be only valid if the force was\n> constant in time (and therefore in space, when conservative fields are\n> concerned).\n\nSo, constant force assumption is the assumption of conservative field. Is\nthere any force which creates constant acceleration other than gravity?\n\nIt seems to me that conservative field = constant force = constant\nacceleration = gravity = gravitational field. They all mean the same thing.\n\nIn other words if we limit our derivation to application of Newton\'s laws to\norbital problems we can say that Newton\'s laws are equivalent to energy\nconservation. Is this a true statement?\n>\n> Forget F==a, because F= m d2 x / dt2 ! There are conservative *forces*, not\n> conservative accelerations.\n>\nI think that in physics "constant" and "conservation" is used more or less\nas synonyms. "Constant acceleration" means that acceleration is conserved.\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>"Maarten van Reeuwijk" <maarten@remove_this_ws.tn.tudelft.nl> wrote in
message news:crhq3n$buo$1@news.tudelft.nl...
> As explained in another post: Newton's laws and energy conservation are only
> identical when the force-field is conservative and in absence of other
> energy-conversion processes, such as electro-magnetism and heat. There is
> *no* general law of conservation of *mechanical* energy. ...
I think I am trying to understand the situation. As I wrote on another post,
now it appears that "force field is conservative" reduces the derivation to
an orbital problem. Because you say that there is no general law of
conservation of mechanical energy.
>
>
> ... If F == m d2 x / dt2 does not contain physics to you I suggest
> reading up on your basic calculus.
Why calculus and not physics :)
> It is easy to see that your argument is flawed, as your estimate of velocity
> would be v = F/m = A t / m, which would be only valid if the force was
> constant in time (and therefore in space, when conservative fields are
> concerned).
So, constant force assumption is the assumption of conservative field. Is
there any force which creates constant acceleration other than gravity?
It seems to me that conservative field = constant force = constant
acceleration = gravity = gravitational field. They all mean the same thing.
In other words if we limit our derivation to application of Newton's laws to
orbital problems we can say that Newton's laws are equivalent to energy
conservation. Is this a true statement?
>
> Forget F==a, because F= m d2 x / dt2 ! There are conservative *forces*, not
> conservative accelerations.
>
I think that in physics "constant" and "conservation" is used more or less
as synonyms. "Constant acceleration" means that acceleration is conserved.
Igor Khavkine
Jan13-05, 01:29 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>\nOn Tue, 11 Jan 2005 21:48:27 +0000, Strong_Field wrote:\n\n> "Igor Khavkine" <igor.kh@gmail.com> wrote in message\n> news:1104911484.419196.39400@c13g2000cwb.googlegro ups.com...\n>\n>> ... One only discriminates against assumptions.\n>\n> Ok. You are right. I went back to your derivation and I looked at it\n> again. Let me ask about the assumptions you make. The "Newton\'s laws"\n> assumption is the left hand side of your first line:\n\nNo, the assumption that Newton\'s Laws hold is that the equation below\nholds (i.e. the left hand side is equal to the right hand side).\n\n> mx\'\' = -grad V\n>\n> The "conservative forces" assumption is the right hand side of the\n> equation.\n\nNo, the assumption of conservative forces fixes the particular form of the\nright hand side of the above equation. In general, the RHS could be any\nvector valued function of x, while in the case of conservative forces it\nmust be the gradient of some scalar potential.\n\n> From the writings of the other posters I got the impression that\n> in physics there is not a definite law of conservation for mechanical\n> systems. Then, when you say "conservative forces" do you mean only\n> gravity? Are there other conservative forces in physics other than\n> gravity? So, let me understand if the assumption of conservative forces\n> reduces the problem only to orbital motion. In other words, is this\n> derivation limited with orbital motion?\n\nOrbital motion has nothing to do with this. Newtonian gravity provides a\nconservative force, the potential is basically 1/r for each point mass.\nThe electrostatic force is conservative, it is derived from a similar\npotential. The elastic force exerted by a stretched string is conservative,\nthe potential for a spring is basically x^2 where x is the displacement\nfrom equilibrium. Orbital motion is a particular class of solutions of the\ntwo body problem in Newtonian gravity, but there are other classes of\nsolutions and there are other systems to consider examples of which I gave\nabove.\n\n> Assuming that this is the case, I would say that the function V is 1/r.\n> Can V be any scalar function or are we working strictly with V = 1/r?\n>\n> If we are working with V = 1/r or V = 1/x, then, when I write the grad\n> explicitly as dV/dx, dV/dy, dV/dz, I believe that dV/dy = 0 and dV/dz = 0.\n> Again limiting the derivation to orbital case only.\n>\n> If this is true so far, then, I may write, mx\'\' = -dV/dx. So please let me\n> know if I am on the right track here. If so I\'ll look at the rest of the\n> derivation.\n\nCareful! 1/r = 1/sqrt(x^2+y^2+z^2) in three dimensions is not the same as\n1/x.\n\n>> This is not true. Both F and a have independent meaning and can be\n>> measured independently. The latter with a ruler and a stop watch and the\n>> former with a standard, like a spring of a certain stiffness. Equating\n>> them gives a non-trivial statement about how interactions between\n>> objects affect their motion.\n>\n> I am looking at the expression F = ma as a literal observer. It may be\n> possible to measure F and a separately and then equate them. But this is\n> not what F = ma says if we look at it as is.\n\nI do not see what you are trying to achieve by trying to be a "literal\nobserver". The way I see it, you are either trying to understand what is\nmeant when physicists write F = ma, or you are trying to find your own\ninterpretation of that equation. In the latter case you\'re on your own. In\nthe former case, it matters not what kind of observer you are. You should\nbe making an effort to understand what is meant by written formulas (which\nare admittedly ambiguous without explanation) instead of trying to fit\ntheir meaning into your own presuppositions.\n\n> Because I believe that\n> there are not two measurable quantities in this equation. Assume that I\n> am totally ign orant of the conventions of physics. This would make me a\n> good disinterested observer. You gave me this equation and told me to\n> measure F with it. Fine. I take a spring, hang a weight m on it and\n> measure the distance it travels in unit time. I say I measured force.\n> What did I measure in reality? I measured acceleration and I called it\n> force. If we restrict our discussion to this equation only, I see only\n> one measurable quantity.\n\nAnd you are wrong. First of all, if you indeed measure "the distance it\ntravels in unit time" you measure the velocity of the spring. What you\nneed to measure is the displacement of the spring from equilibrium when it\ncomes to rest (it\'s not moving, velocity is zero).\n\n>> If you believe this to be elementary physics, you clearly do not\n>> understand it. In words, your statement is "velocity is equal to\n> minus the gradient of the potential....\n>\n> You left out my "in the first increment of time." My statement is "In\n> the first increment of time delta x = (-grad V) (delta t). This is\n> equivalent to your statement. But "acceleration is equal to minus the\n> gradient of the potential" is fine with me.\n\nAs I\'ve mentioned your statement is at best approximately true, and only\nwhen the particle in question starts from rest (zero velocity) if at time\nt=0 it is already moving then equation is completely wrong. The two\nstatements from the previous paragraph, yours and mine, are\ndefinitely not equivalent. Also, you are confusing predictions from theory\nwith (in principle infinite precision, derivatives exist, etc.) with\nmeasurements (error bars, estimation of derivatives by finite differences,\netc.). The two can only be compared with approximate equal signs, not\nequal signs.\n\nI\'ve noticed from this discussion that you seem to lack a basic\nunderstanding of elementary physics. Unfortunately, the task of correcting\nthis situation and your misconceptions is a bit too ambitious for me to\nundertake. Instead I recommend consulting a good book on elementary\nphysics. For example, take a look at the first volume of Feynman\'s\nlectures on physics, especially chapters 4, 8, and 9. I believe it will\nanswer some of your questions in detail.\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, 11 Jan 2005 21:48:27 +0000, Strong_Field wrote:
> "Igor Khavkine" <igor.kh@gmail.com> wrote in message
> news:1104911484.419196.39400@c13g2000cwb.googlegro ups.com...
>
>> ... One only discriminates against assumptions.
>
> Ok. You are right. I went back to your derivation and I looked at it
> again. Let me ask about the assumptions you make. The "Newton's laws"
> assumption is the left hand side of your first line:
No, the assumption that Newton's Laws hold is that the equation below
holds (i.e. the left hand side is equal to the right hand side).
> mx'' = -grad V
>
> The "conservative forces" assumption is the right hand side of the
> equation.
No, the assumption of conservative forces fixes the particular form of the
right hand side of the above equation. In general, the RHS could be any
vector valued function of x, while in the case of conservative forces it
must be the gradient of some scalar potential.
> From the writings of the other posters I got the impression that
> in physics there is not a definite law of conservation for mechanical
> systems. Then, when you say "conservative forces" do you mean only
> gravity? Are there other conservative forces in physics other than
> gravity? So, let me understand if the assumption of conservative forces
> reduces the problem only to orbital motion. In other words, is this
> derivation limited with orbital motion?
Orbital motion has nothing to do with this. Newtonian gravity provides a
conservative force, the potential is basically 1/r for each point mass.
The electrostatic force is conservative, it is derived from a similar
potential. The elastic force exerted by a stretched string is conservative,
the potential for a spring is basically x^2 where x is the displacement
from equilibrium. Orbital motion is a particular class of solutions of the
two body problem in Newtonian gravity, but there are other classes of
solutions and there are other systems to consider examples of which I gave
above.
> Assuming that this is the case, I would say that the function V is 1/r.
> Can V be any scalar function or are we working strictly with V = 1/r?
>
> If we are working with V = 1/r or V = 1/x, then, when I write the grad
> explicitly as dV/dx, dV/dy, dV/dz, I believe that dV/dy = and dV/dz = .
> Again limiting the derivation to orbital case only.
>
> If this is true so far, then, I may write, mx'' = -dV/dx. So please let me
> know if I am on the right track here. If so I'll look at the rest of the
> derivation.
Careful! 1/r = 1/\sqrt(x^2+y^2+z^2) in three dimensions is not the same as
1/x.
>> This is not true. Both F and a have independent meaning and can be
>> measured independently. The latter with a ruler and a stop watch and the
>> former with a standard, like a spring of a certain stiffness. Equating
>> them gives a non-trivial statement about how interactions between
>> objects affect their motion.
>
> I am looking at the expression F = ma as a literal observer. It may be
> possible to measure F and a separately and then equate them. But this is
> not what F = ma says if we look at it as is.
I do not see what you are trying to achieve by trying to be a "literal
observer". The way I see it, you are either trying to understand what is
meant when physicists write F = ma, or you are trying to find your own
interpretation of that equation. In the latter case you're on your own. In
the former case, it matters not what kind of observer you are. You should
be making an effort to understand what is meant by written formulas (which
are admittedly ambiguous without explanation) instead of trying to fit
their meaning into your own presuppositions.
> Because I believe that
> there are not two measurable quantities in this equation. Assume that I
> am totally ign orant of the conventions of physics. This would make me a
> good disinterested observer. You gave me this equation and told me to
> measure F with it. Fine. I take a spring, hang a weight m on it and
> measure the distance it travels in unit time. I say I measured force.
> What did I measure in reality? I measured acceleration and I called it
> force. If we restrict our discussion to this equation only, I see only
> one measurable quantity.
And you are wrong. First of all, if you indeed measure "the distance it
travels in unit time" you measure the velocity of the spring. What you
need to measure is the displacement of the spring from equilibrium when it
comes to rest (it's not moving, velocity is zero).
>> If you believe this to be elementary physics, you clearly do not
>> understand it. In words, your statement is "velocity is equal to
> minus the gradient of the potential....
>
> You left out my "in the first increment of time." My statement is "In
> the first increment of time \delta x = (-grad V) (\delta t). This is
> equivalent to your statement. But "acceleration is equal to minus the
> gradient of the potential" is fine with me.
As I've mentioned your statement is at best approximately true, and only
when the particle in question starts from rest (zero velocity) if at time
t=0 it is already moving then equation is completely wrong. The two
statements from the previous paragraph, yours and mine, are
definitely not equivalent. Also, you are confusing predictions from theory
with (in principle infinite precision, derivatives exist, etc.) with
measurements (error bars, estimation of derivatives by finite differences,
etc.). The two can only be compared with approximate equal signs, not
equal signs.
I've noticed from this discussion that you seem to lack a basic
understanding of elementary physics. Unfortunately, the task of correcting
this situation and your misconceptions is a bit too ambitious for me to
undertake. Instead I recommend consulting a good book on elementary
physics. For example, take a look at the first volume of Feynman's
lectures on physics, especially chapters 4, 8, and 9. I believe it will
answer some of your questions in detail.
Igor
Maarten van Reeuwijk
Jan14-05, 11:03 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>Strong_Field wrote:\n\n> "Maarten van Reeuwijk" <maarten@remove_this_ws.tn.tudelft.nl> wrote in\n> message news:crhq3n\\$buo\\$1@news.tudelft.nl...\n>\n>> As explained in another post: Newton\'s laws and energy conservation are\n>> only identical when the force-field is conservative and in absence of\n>> other energy-conversion processes, such as electro-magnetism and heat.\n>> There is *no* general law of conservation of *mechanical* energy. ...\n>\n> I think I am trying to understand the situation. As I wrote on another\n> post, now it appears that "force field is conservative" reduces the\n> derivation to an orbital problem. Because you say that there is no general\n> law of conservation of mechanical energy.\n>>\n>>\n>> ... If F == m d2 x / dt2 does not contain physics to you I suggest\n>> reading up on your basic calculus.\n>\n> Why calculus and not physics :)\n\ngood point :), but you know what I mean\n\n>> It is easy to see that your argument is flawed, as your estimate of\n>> velocity would be v = F/m = A t / m, which would be only valid if the\n>> force was constant in time (and therefore in space, when conservative\n>> fields are concerned).\n>\n> So, constant force assumption is the assumption of conservative field.\n\nNo, this was just an example to show you that your reasoning is only valid\nfor constant forces, which normally is not the case. Conservative force\nfields are very special fields where it does not matter how you travel from\nA to B, the energy cost is identical. This is very different from normal\nlife-experience, because normally there is friction involved which makes\nenergy-losses path-dependent. Explanations on conservative force-fields,\nsee for example:\n\nhttp://mathworld.wolfram.com/ConservativeField.html\nhttp://theory.uwinnipeg.ca/physics/work/node8.html\n\n> It seems to me that conservative field = constant force = constant\n> acceleration = gravity = gravitational field. They all mean the same\n> thing.\n\nNope, see above. I would really recommend reading the chapters of the\nFeynman-lectures I recommended in a previous post. Paragraph 9.7 deals with\nplanetary motion, quite similar to what you\'re trying.\n\n> In other words if we limit our derivation to application of Newton\'s laws\n> to orbital problems we can say that Newton\'s laws are equivalent to energy\n> conservation. Is this a true statement?\n\nNo, they are not equivalent, from Newton\'s laws we can derive an evolution\nequation of mechanical energy. Only because heat losses can be neglected,\nis mechanical energy conserved.\n\n> I think that in physics "constant" and "conservation" is used more or less\n> as synonyms. "Constant acceleration" means that acceleration is conserved.\n\nNope, see above\n\nHTH, Maarten\n\n--\n================================================ ===================\nMaarten van Reeuwijk Thermal and Fluids Sciences\nPhd student dept. of Multiscale Physics\nwww.ws.tn.tudelft.nl Delft University of Technology\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>Strong_Field wrote:
> "Maarten van Reeuwijk" <maarten@remove_this_ws.tn.tudelft.nl> wrote in
> message news:crhq3n$buo$1@news.tudelft.nl...
>
>> As explained in another post: Newton's laws and energy conservation are
>> only identical when the force-field is conservative and in absence of
>> other energy-conversion processes, such as electro-magnetism and heat.
>> There is *no* general law of conservation of *mechanical* energy. ...
>
> I think I am trying to understand the situation. As I wrote on another
> post, now it appears that "force field is conservative" reduces the
> derivation to an orbital problem. Because you say that there is no general
> law of conservation of mechanical energy.
>>
>>
>> ... If F == m d2 x / dt2 does not contain physics to you I suggest
>> reading up on your basic calculus.
>
> Why calculus and not physics :)
good point :), but you know what I mean
>> It is easy to see that your argument is flawed, as your estimate of
>> velocity would be v = F/m = A t / m, which would be only valid if the
>> force was constant in time (and therefore in space, when conservative
>> fields are concerned).
>
> So, constant force assumption is the assumption of conservative field.
No, this was just an example to show you that your reasoning is only valid
for constant forces, which normally is not the case. Conservative force
fields are very special fields where it does not matter how you travel from
A to B, the energy cost is identical. This is very different from normal
life-experience, because normally there is friction involved which makes
energy-losses path-dependent. Explanations on conservative force-fields,
see for example:
http://mathworld.wolfram.com/ConservativeField.html
http://theory.uwinnipeg.ca/physics/work/node8.html
> It seems to me that conservative field = constant force = constant
> acceleration = gravity = gravitational field. They all mean the same
> thing.
Nope, see above. I would really recommend reading the chapters of the
Feynman-lectures I recommended in a previous post. Paragraph 9.7 deals with
planetary motion, quite similar to what you're trying.
> In other words if we limit our derivation to application of Newton's laws
> to orbital problems we can say that Newton's laws are equivalent to energy
> conservation. Is this a true statement?
No, they are not equivalent, from Newton's laws we can derive an evolution
equation of mechanical energy. Only because heat losses can be neglected,
is mechanical energy conserved.
> I think that in physics "constant" and "conservation" is used more or less
> as synonyms. "Constant acceleration" means that acceleration is conserved.
Nope, see above
HTH, Maarten
--
================================================== =================
Maarten van Reeuwijk Thermal and Fluids Sciences
Phd student dept. of Multiscale Physics
www.ws.tn.tudelft.nl Delft University of Technology
J. J. Lodder
Jan18-05, 01:21 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<Poakfield@msn.com> wrote:\n\n> Hi! Could someone please tell me if the law of conservation of energy\n> can be derived only from Newton\'s motion laws? Thanks.\n\nNo, for the law of conservation of energy cannot be derived at all.\nThe reason is that we do not know what energy is,\nwithout using the law of conservation of energy.\n\nThe only correct definition of energy I am aware of is:\nEnergy is 1/2 m v^2, and everything else that is missing\nfrom the law of conservation of energy.\n\nHowever, once we have used a few particular cases\nto identify all relevant forms of energy\nthe law must hold (and has been found to hold)\nin all further cases.\n\nBest,\n\nJan\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><Poakfield@msn.com> wrote:
> Hi! Could someone please tell me if the law of conservation of energy
> can be derived only from Newton's motion laws? Thanks.
No, for the law of conservation of energy cannot be derived at all.
The reason is that we do not know what energy is,
without using the law of conservation of energy.
The only correct definition of energy I am aware of is:
Energy is 1/2 m v^2, and everything else that is missing
from the law of conservation of energy.
However, once we have used a few particular cases
to identify all relevant forms of energy
the law must hold (and has been found to hold)
in all further cases.
Best,
Jan
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