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On the principle of least action in classical mechanics

  1. Oct 7, 2011 #1
    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. jcsd
  3. Oct 7, 2011 #2

    Matterwave

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    Isn't that simply the work-energy 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 work-energy theorem (getting the 1/2v^2 term), to derive the work-energy theorem.

    ...
     
  4. Oct 9, 2011 #3
    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
     
    Last edited: Oct 9, 2011
  5. Oct 15, 2011 #4
    - aat -
     
  6. Nov 6, 2011 #5
    Reformulation:

    In classical mechanics, if we consider a force field (uniform or non-uniform) 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]
     
  7. Nov 7, 2011 #6
    The last line of your first proof seems to imply that the area of a rectangle is always zero.
     
  8. Nov 25, 2011 #7

    [itex]\delta[/itex] is a variation.
     
  9. Nov 25, 2011 #8

    Ken G

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    No. The reason we know this is that it can be used to predict motions under conservative fields, but we only know this by testing those predictions, not by writing them down. So this means we have to find where, in your argument, you inserted something that had dynamical content-- something that was true in our world but was not a general truism. That appears when you substitute the integral of the acceleration times dr by a potential change. You might claim that this is just your definition of a potential change, but that would not make sense in general-- the potential change must be a function only of change in position for your argument to hold, so acceleration must be a function of position that can be recast that way (the acceleration field must be curl free and a function only of location). That's why it is not a tautology. But what it does show is that any situation in which acceleration obeys that rule will yield a principle of least action. Which is the cause of which is not answered by classical physics.
    That is Newton's second law, the insertion of the concept of "force" is unnecessary in that law, but what is essential is having some means of saying what the acceleration field is, i.e., you still have to specify V for the principle of least action to mean anything. You are merely claiming that it would be true for some undefined V, but what gives least action its teeth is that we can know V independently of the motion. That's more or less the whole point of Newton's laws, at least when applied to conservative forces as it is here.
     
    Last edited: Nov 25, 2011
  10. Nov 27, 2011 #9
    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.
     
    Last edited: Nov 27, 2011
  11. Nov 27, 2011 #10

    Ken G

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    Yes, if the acceleration is constant and known, then you don't need the principle of least action. So what? That's not what the principle is for-- the principle is for finding the acceleration when you don't know the acceleration but you do know the potential energy function (it is especially useful if you also have complicated constraints on the motion, like strings and pulleys and whatnot). That is the whole point of the principle of least action, it is pretty useless in any other situation. It is not tautological, however, because it uses potential energy functions that have to be determined separately from the principle of least action. All you are saying is that the principle of least action can be recast in the statement that the dot product of acceleration times displacement equals the potential energy function (per gram), and this potential energy function can then be included explicitly in "the action", the minimization of which can generate the equations of motion. That is already well known.
     
    Last edited: Nov 27, 2011
  12. Nov 27, 2011 #11
    These equations are tautological statements even if you don't know the value of the acceleration.
     
  13. Nov 28, 2011 #12

    Ken G

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    That is quite demonstrably false. If you don't know the acceleration, and you also don't know the potential energy function, you can't do squat. If you don't know the acceleration, and you do know the potential energy function, then you can find the acceleration-- if you also know Newton's laws (either in their standard form, or in the form of a principle of least action, or any other mathematical equivalent like the statement you wrote). If you can't know A unless you already know B, then B is clearly not tautological. The error in your conclusion is that you think the acceleration dotted into the displacement goes into the action. It does not. What goes into the action is the potential energy function, which (per gram) equals the acceleration dotted into the displacement, but you infer the latter from the former, not the other way around. So the real physical content of Newton's laws (for conservative forces) is the statement that the potential energy function (per gram) gives the acceleration dotted into the displacement. This can be derived from the principle of least action, or taken as known (in which case it can derive the principle of least action-- it is not unusual that the order of a derivation can be reversed if a different assumption is made about what is known).

    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 Euler-Lagrange equations makes it quite useful for dealing with complicated constraints on the motion.
     
    Last edited: Nov 28, 2011
  14. Nov 28, 2011 #13

    """"if the acceleration is constant"""" then


    In addition, sorry Ken G, but your comments are very boring.
     
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