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Is the numerical value of action functional of any importance?

  1. Jun 23, 2008 #1
    While working out a problem I got a result which gave rise to this doubt regarding value of action functional. Suppose I start from an action, obtain the equation of motion and when I try to check if that solution gives a finite value of action, I get, surprisingly, vanishing value. The actual problem I was doing is a bit complicated to describe here, so take a simpler prob. Suppose for a particle,
    [tex]
    S=\int dt \frac{1}{2}m\dot{x}^2
    [/tex]
    One possible solution is, x(t)=constant. This is a trivial solution, but in the problem I am doing this kind trivial solution is enough. Now for this trivial solution, the value of action functional is zero. It looked strange to me. But then I thought may be the actual value of the action is not that important, what is important is its variation. Is that point of view valid? I would like to know your opinion. Of course the action does not show other undesirable features in my case. Do you think in such cases anything else should be carefully investigated? Thanks.
     
  2. jcsd
  3. Jun 23, 2008 #2
    The functional should be defined as

    [tex]S=\int_{t_1}^{t_2} dt \frac{1}{2}m\dot{x}^2[/tex]

    and computed between two points [tex]x_1=x(t_1)[/tex] and [tex]x_2=x(t_2)[/tex] keeping these points fixed. Then, you take all the possible paths between these two points and you uncover that the only one being an extremum of this functional is that solving Newton equation of motion for a free particle. Indeed one will have

    [tex]S=\frac{1}{2}m\frac{(x_2-x_1)^2}{(t_2-t_1)}[/tex]

    but you are taking [tex]x_2=x_1[/tex] and for this particular solution one has indeed S=0.

    The value of the action with respect to its extremum value is not relevant and can also be zero. This also happens to functions in a more mundane world.

    Jon
     
  4. Jun 23, 2008 #3
    Thanks for the reply. I get your point. Actually I din not put the limits of t because in this case even before I worry about the limits, the integrand becomes zero, since [tex]\dot{x}=0[/tex] in this special case. Since you are saying that this (ie, S=0) also happens in more mundane world, can you please cite other such examples briefly? It would be interesting to know.
     
  5. Jun 23, 2008 #4
    The meaning of what I am saying is the following using a more mundane argument. Consider a parabolic function

    [tex]y=x^2+kx[/tex]

    This function has a minimum at [tex]x_m=-\frac{k}{2}[/tex] with the value of the minimum [tex]y_m=-\frac{k^2}{4}[/tex] and this is quite general. But if you insist to demand the minimum in zero you are just selecting the particular case k=0. This is what happens for your case.

    Jon
     
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