- #1
arroy_0205
- 129
- 0
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.
[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.