- #1
Alex Cros
- 28
- 1
Hello everyone,
Reading Landau and Lifshitz Course of Theoretical Physics Volume 1: Mechanics (page 3) I got suck in the following step (and I cite in italics):
The change in S when q is replaced by q+δq is
[tex]\int_{t_1}^{t_2} L(q+δq, \dot q +δ\dot q, t)dt - \int_{t_1}^{t_2} L(q, \dot q, t)dt[/tex]
(So far so good)
When this difference is expanded in powers of δq and [itex]δ\dot q[/itex] in the integrand, the leading terms are of the first order.
How do you expand that? Could anyone show me how explicitly if you don't know the explicit form of the Lagrangian?
The necessary condition for S (where S is the action) to have a minimum is that these terms (called the first variation, or simply the variation, of the integral) should be zero. Thus the principle of least action may be written in the form
[tex]δS = δ \int_{t_1}^{t_2} L(q, \dot q, t)dt = 0[/tex]
(Which I'm fine with the above expression)
Or, effecting the variation,
[tex]\int_{t_1}^{t_2} (
\frac{\partial L}{\partial q}δq+
\frac{\partial L}{\partial \dot q}δ\dot q)
dt = 0[/tex]
Now my guess would have included [itex]\frac{\partial L}{\partial t}[/itex] like:
[tex]\int_{t_1}^{t_2} (
\frac{\partial L}{\partial q}δq+
\frac{\partial L}{\partial \dot q}δ\dot q + \frac{\partial L}{\partial t}dt)
dt = 0[/tex]
To perform the total differential of all variables.
Explain me like if I'm five why this guess is wrong.
Thanks so much in advance! And sorry for my lack of elemental knowledge.
-Alex
Reading Landau and Lifshitz Course of Theoretical Physics Volume 1: Mechanics (page 3) I got suck in the following step (and I cite in italics):
The change in S when q is replaced by q+δq is
[tex]\int_{t_1}^{t_2} L(q+δq, \dot q +δ\dot q, t)dt - \int_{t_1}^{t_2} L(q, \dot q, t)dt[/tex]
(So far so good)
When this difference is expanded in powers of δq and [itex]δ\dot q[/itex] in the integrand, the leading terms are of the first order.
How do you expand that? Could anyone show me how explicitly if you don't know the explicit form of the Lagrangian?
The necessary condition for S (where S is the action) to have a minimum is that these terms (called the first variation, or simply the variation, of the integral) should be zero. Thus the principle of least action may be written in the form
[tex]δS = δ \int_{t_1}^{t_2} L(q, \dot q, t)dt = 0[/tex]
(Which I'm fine with the above expression)
Or, effecting the variation,
[tex]\int_{t_1}^{t_2} (
\frac{\partial L}{\partial q}δq+
\frac{\partial L}{\partial \dot q}δ\dot q)
dt = 0[/tex]
Now my guess would have included [itex]\frac{\partial L}{\partial t}[/itex] like:
[tex]\int_{t_1}^{t_2} (
\frac{\partial L}{\partial q}δq+
\frac{\partial L}{\partial \dot q}δ\dot q + \frac{\partial L}{\partial t}dt)
dt = 0[/tex]
To perform the total differential of all variables.
Explain me like if I'm five why this guess is wrong.
Thanks so much in advance! And sorry for my lack of elemental knowledge.
-Alex