- #1
gulsen
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I've recently started Feynman & Gibbs. I was sure exercises will be fun, but i can't enjoy myself when i fail solving the first one! Exercise 1-1 says: show that free particle action is
[tex]\frac{m}{2} \frac{x_b^2 - x_a^2}{t_b-t_a}[/tex]
I tried finding anti-derivative of [itex]\dot x^2[/itex], ended up with [itex]x\dot x - \int x d(\dot x)[/itex] via integration by parts. Couldn't do much about the integral. Of course, I know the solution, and can show it using Euler-Lagrange equations, [itex]x=vt[/itex] where [itex]v[/itex] is a constant (taking [itex]x_0=0[/itex]). I can "solve" the question by substituting [itex]x[/itex] in the action integral
[tex]S = \int_{t_a}^{t_b} \frac{m}{2}v^2 dt[/tex]
but i suppose this counts as cheating --kinda solving backwards.
Any ideas on how this kind of stuff can be solved?
[tex]\frac{m}{2} \frac{x_b^2 - x_a^2}{t_b-t_a}[/tex]
I tried finding anti-derivative of [itex]\dot x^2[/itex], ended up with [itex]x\dot x - \int x d(\dot x)[/itex] via integration by parts. Couldn't do much about the integral. Of course, I know the solution, and can show it using Euler-Lagrange equations, [itex]x=vt[/itex] where [itex]v[/itex] is a constant (taking [itex]x_0=0[/itex]). I can "solve" the question by substituting [itex]x[/itex] in the action integral
[tex]S = \int_{t_a}^{t_b} \frac{m}{2}v^2 dt[/tex]
but i suppose this counts as cheating --kinda solving backwards.
Any ideas on how this kind of stuff can be solved?