- #1
JohanL
- 158
- 0
If you have an action integral
[tex]
\int_{A}^{B} \sqrt{F\mathbf{(r)}} dr
[/tex]
and
F=a-bz^2 , b>0, a-bd^2>0
the minimum of the action integral is equivalent to
[tex]
\frac{d}{dt}\frac{dG}{\dot{z}}-\frac{dG}{z}=0
[/tex]
where
[tex]
G=\sqrt{F}
[/tex]
or am i doing this in a completeley wrong way?
[tex]
\int_{A}^{B} \sqrt{F\mathbf{(r)}} dr
[/tex]
and
F=a-bz^2 , b>0, a-bd^2>0
the minimum of the action integral is equivalent to
[tex]
\frac{d}{dt}\frac{dG}{\dot{z}}-\frac{dG}{z}=0
[/tex]
where
[tex]
G=\sqrt{F}
[/tex]
or am i doing this in a completeley wrong way?