| Topic: |
Science > Physics |
| User: |
"proton" |
| Date: |
28 Mar 2005 06:50:14 AM |
| Object: |
Question about Least action principle |
I have seen a derivation of Newton's Laws from Lagrange's equation and
the least action principle (Landau and Lifshitz, most interesting, by
the way).
In this derivation it states that the mass of a body must be positive
because otherwise the integral of the action:
S = intg (m v^2 / 2) dt would not have a minimum, which is a
prerequisite for the principle of least action.
However, this is for a free particle, with no acting field.
If there is a field, then the integral becomes:
S = intg (m v^2 / 2 - U (r, t)), where U is the potential energy of the
field.
Since U can be positive or negative, does this mean that in this case
there might not be a minimum? Does it mean that in some cases the mass
might be negative?
Thanks for any help.
.
|
|
| User: "richard miller" |
|
| Title: Re: Question about Least action principle |
31 Mar 2005 01:40:53 PM |
|
|
"proton" <leosarasua@gmail.com> wrote in message
news:1112014214.117008.321040@f14g2000cwb.googlegroups.com...
I have seen a derivation of Newton's Laws from Lagrange's equation and
the least action principle (Landau and Lifshitz, most interesting, by
the way).
In this derivation it states that the mass of a body must be positive
because otherwise the integral of the action:
S = intg (m v^2 / 2) dt would not have a minimum, which is a
prerequisite for the principle of least action.
However, this is for a free particle, with no acting field.
If there is a field, then the integral becomes:
S = intg (m v^2 / 2 - U (r, t)), where U is the potential energy of the
field.
Since U can be positive or negative, does this mean that in this case
there might not be a minimum? Does it mean that in some cases the mass
might be negative?
Thanks for any help.
It is interesting in that the 'first variation of the functional L', where L
is the Lagrangian, which gives Lagrange's eqations, gives a stationary, not
specifically a max or min. You need to go to higher order terms todetermine
max/min.
I haven't seen the Landau and Lifshitz derivation of Newton's Laws but, as I
posted on sci.physics.research a couple of days ago, Newton's laws are
really a special case of the Lagragian formulation with the Lagrangian
suitably chosen (L=T-V). There are many Lagrangians. Furthermore, I would
prefer on sci.math to stick to the pure math element which does not state
that the first variation is necessarily a minimum, merely a stationary
point.
Richard Miller
.
|
|
|
| User: "richard miller" |
|
| Title: Re: Question about Least action principle |
31 Mar 2005 03:05:30 PM |
|
|
"richard miller" <richard@microscitech.freeserve.co.uk> wrote in message
news:d2hjme$j4d$1@newsg3.svr.pol.co.uk...
"proton" <leosarasua@gmail.com> wrote in message
news:1112014214.117008.321040@f14g2000cwb.googlegroups.com...
I have seen a derivation of Newton's Laws from Lagrange's equation and
the least action principle (Landau and Lifshitz, most interesting, by
the way).
snip for brevity
Thanks for any help.
then snip again for more brevity
suitably chosen (L=T-V). There are many Lagrangians. Furthermore, I would
prefer on sci.math to stick to the pure math element which does not
state...
make that sci.physics too!
Richard Miller
.
|
|
|
|
|
| User: "" |
|
| Title: Re: Question about Least action principle |
28 Mar 2005 01:56:34 PM |
|
|
In article <1112014214.117008.321040@f14g2000cwb.googlegroups.com>, "proton" <leosarasua@gmail.com> writes:
I have seen a derivation of Newton's Laws from Lagrange's equation and
the least action principle (Landau and Lifshitz, most interesting, by
the way).
In this derivation it states that the mass of a body must be positive
because otherwise the integral of the action:
S = intg (m v^2 / 2) dt would not have a minimum, which is a
prerequisite for the principle of least action.
However, this is for a free particle, with no acting field.
If there is a field, then the integral becomes:
S = intg (m v^2 / 2 - U (r, t)), where U is the potential energy of the
field.
Since U can be positive or negative, does this mean that in this case
there might not be a minimum?
Since you've both positive (from the mv^2/2 part) and negative
contributions, in general there'll be a minimum.
Does it mean that in some cases the mass
might be negative?
Theidea is that the mass is an intrinsic property of the body, not a
function of external fields. Thus, whatever it is in the absence of
any field, that's what it remains.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
.
|
|
|
|
|
Related Articles |
|
|
