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On the principle of least action in classical mechanics 
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#1
Oct711, 10:21 PM

P: 36

The principle of least action applicable in an uniform field can be obtained as follows:
Particle A [tex]\vec{a}_A = \vec{a}_A[/tex] [tex]\int \vec{a}_A \cdot d\vec{r}_A = \int \vec{a}_A \cdot d\vec{r}_A[/tex] [tex]\int \vec{a}_A \cdot d\vec{r}_A = \Delta \; {\textstyle \frac{1}{2}}\vec{v}_A^2[/tex] [tex]\int \vec{a}_A \cdot d\vec{r}_A = \Delta \; \; \vec{a}_A \cdot \vec{r}_A[/tex] [tex]\Delta \; {\textstyle \frac{1}{2}}\vec{v}_A^2 = \Delta \; \; \vec{a}_A \cdot \vec{r}_A[/tex] [tex]\Delta \; {\textstyle \frac{1}{2}}\vec{v}_A^2  \Delta \; \; \vec{a}_A \cdot \vec{r}_A = 0[/tex] [tex] m_A \left( \Delta \; {\textstyle \frac{1}{2}}\vec{v}_A^2  \Delta \; \; \vec{a}_A \cdot \vec{r}_A \right) = 0[/tex] Alex 


#2
Oct711, 10:36 PM

Sci Advisor
P: 2,809

Isn't that simply the workenergy theorem? I don't see an Action in there...where's the principle of least action in that derivation?
Also...isn't that just a tautology... You used the workenergy theorem (getting the 1/2v^2 term), to derive the workenergy theorem. ... 


#3
Oct911, 02:24 PM

P: 36

Now, following from the last equation:
[tex] m_A \left( \Delta \; {\textstyle \frac{1}{2}}\vec{v}_A^2  \Delta \; \; \vec{a}_A \cdot \vec{r}_A \right) = 0[/tex][itex]\rightarrow[/itex] [tex]\Delta \; T_A  \Delta \; V_A = 0[/tex] [tex]\left( T_{Af}  T_{Ai} \right)  \left( V_{Af}  V_{Ai} \right) = 0[/tex] [tex]\left( T_{Af}  V_{Af} \right)  \left( T_{Ai}  V_{Ai} \right) = 0[/tex] [tex]\left(\; L_{Af} \;\right)  \left(\; L_{Ai} \;\right) = 0[/tex] [tex]\Delta \; L_A = 0[/tex] [tex]\int L_A \; \; dt= 0[/tex] Is the principle of least action a tautology? Alex 


#4
Oct1511, 11:50 AM

P: 36

On the principle of least action in classical mechanics
 aat 



#5
Nov611, 06:28 PM

P: 36

Reformulation:
In classical mechanics, if we consider a force field (uniform or nonuniform) in which the acceleration [itex]\vec{a}_A[/itex] of a particle A is constant, then [tex]\vec{a}_A  \, \vec{a}_A = 0[/tex] [tex]\left( \vec{a}_A  \, \vec{a}_A \right) \cdot \delta \vec{r}_A = 0[/tex] [tex]\int_{t_{1}}^{t_{2}} \left( \vec{a}_A  \, \vec{a}_A \right) \cdot \delta \vec{r}_A \; \, dt = 0[/tex] [tex]\delta \int_{t_{1}}^{t_{2}} \left( {\textstyle \frac{1}{2}} \; \vec{v}_A^{\;2} \, + \, \vec{a}_A^{\vphantom{^{\;2}}} \cdot \vec{r}_A^{\vphantom{^{\;2}}} \right) \, dt = 0[/tex] [tex]m_A^{\vphantom{^{\;2}}} \; \delta \int_{t_{1}}^{t_{2}} \left( {\textstyle \frac{1}{2}} \; \vec{v}_A^{\;2} \, + \, \vec{a}_A^{\vphantom{^{\;2}}} \cdot \vec{r}_A^{\vphantom{^{\;2}}} \right) \, dt = 0[/tex] [tex]\delta \int_{t_{1}}^{t_{2}} \left( T_A  \, V_A \right) \, dt = 0[/tex] [tex]\delta \int_{t_{1}}^{t_{2}} L_A \; \, dt = 0[/tex] where: [tex]T_A = {\textstyle \frac{1}{2}} \; m_A^{\vphantom{^{\;2}}}\vec{v}_A^{\;2}[/tex][tex]V_A =  \; m_A^{\vphantom{^{\;2}}} \; \vec{a}_A^{\vphantom{^{\;2}}} \cdot \vec{r}_A^{\vphantom{^{\;2}}}[/tex] If [itex]\vec{a}_A[/itex] is not constant but [itex]\vec{a}_A[/itex] is function of [itex]\vec{r}_A[/itex] then the same result is obtained, even if Newton's second law were not valid.[itex]{\vphantom{aat}}[/itex] 


#6
Nov711, 07:39 AM

P: 389

The last line of your first proof seems to imply that the area of a rectangle is always zero.



#7
Nov2511, 08:50 AM

P: 36

[itex]\delta[/itex] is a variation. 


#8
Nov2511, 09:43 PM

PF Gold
P: 3,086




#9
Nov2711, 01:26 PM

P: 36

key p:
I don't need of the definition of potential energy. If the acceleration [itex]\vec{a}_A[/itex] and the mass [itex]m_A[/itex] of a particle A are constant, then [tex]\vec{a}_A  \, \vec{a}_A = 0[/tex] [tex]\left( \vec{a}_A  \, \vec{a}_A \right) \cdot \delta \vec{r}_A = 0[/tex] [tex]\int_{t_{1}}^{t_{2}} \left( \vec{a}_A  \, \vec{a}_A \right) \cdot \delta \vec{r}_A \; \, dt = 0[/tex] [tex]\delta \int_{t_{1}}^{t_{2}} \left( {\textstyle \frac{1}{2}} \; \vec{v}_A^{\;2} \, + \, \vec{a}_A^{\vphantom{^{\;2}}} \cdot \vec{r}_A^{\vphantom{^{\;2}}} \right) \, dt = 0[/tex] [tex]m_A^{\vphantom{^{\;2}}} \; \delta \int_{t_{1}}^{t_{2}} \left( {\textstyle \frac{1}{2}} \; \vec{v}_A^{\;2} \, + \, \vec{a}_A^{\vphantom{^{\;2}}} \cdot \vec{r}_A^{\vphantom{^{\;2}}} \right) \, dt = 0[/tex] [tex] \; \delta \int_{t_{1}}^{t_{2}} \left( {\textstyle \frac{1}{2}} \; m_A^{\vphantom{^{\;2}}} \; \vec{v}_A^{\;2} \, + \, m_A^{\vphantom{^{\;2}}} \; \vec{a}_A^{\vphantom{^{\;2}}} \cdot \vec{r}_A^{\vphantom{^{\;2}}} \right) \, dt = 0[/tex] Therefore, if the acceleration [itex]\vec{a}_A[/itex] and the mass [itex]m_A[/itex] of a particle A are constant, then the above equations are tautological statements. 


#10
Nov2711, 03:13 PM

PF Gold
P: 3,086




#11
Nov2711, 03:23 PM

P: 36




#12
Nov2811, 04:06 AM

PF Gold
P: 3,086

What makes the principle of least action an incredibly powerful principle of physics, and therefore not a tautology, is just one thing: prior knowledge of the potential energy function. Once you have that, then the principle of least action is just one of many equivalent ways to say one thing: how to use that prior knowledge. The reason it gets used in some problems, rather than Newton's laws or any of the other equivalent mathematical statements, is that it is in the form of an extremum principle, which via the EulerLagrange equations makes it quite useful for dealing with complicated constraints on the motion. 


#13
Nov2811, 08:07 AM

P: 36

""""if the acceleration is constant"""" then In addition, sorry Ken G, but your comments are very boring. 


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