- #1
TheDragon
- 10
- 0
In Newton's problem,and other central force problems in Classical Mechanics, you can get with decreasing the center of mass movement to the lagrangian:
L=1/2m(r' ^2+r^2 \varphi'^2)-V(r)
because \varphi is cyclic, you can write:
\frac{d}{dt}(mr^2 \varphi')=0
or, defining the angular momentum:
mr^2 \varphi'=l
In order for you to understand my question we will go through two ways:
Way no.1) Now if you go ahead and use euler-lagrange equation to r you will get:
\frac{d}{dt}(mr')=mr\varphi'^2-\frac{∂V(r)}{∂r}
and exchanging for l you can get:
\frac{d}{dt}(mr')=\frac{l^2}{mr^3}-\frac{∂V(r)}{∂r}
Way no.2) But if you first excange phi and then use euler-lagrange to r you get:
L=1/2m(r' ^2+\frac{l^2}{m^2r^2})-V(r)=1/2mr' ^2+\frac{l^2}{2mr^2}-V(r)
and then euler lagrange for r:
\frac{d}{dt}(mr')=\frac{∂}{∂r}(\frac{l^2}{2mr^2})-\frac{∂V(r)}{∂r}=-\frac{l^2}{mr^3}-\frac{∂V(r)}{∂r}
The plus in way no.1 became a minus in way no.2. How can you explain it?
If you can, don't show me a solution using energy because I already know that one. I want to use only lagrangian formalism.
L=1/2m(r' ^2+r^2 \varphi'^2)-V(r)
because \varphi is cyclic, you can write:
\frac{d}{dt}(mr^2 \varphi')=0
or, defining the angular momentum:
mr^2 \varphi'=l
In order for you to understand my question we will go through two ways:
Way no.1) Now if you go ahead and use euler-lagrange equation to r you will get:
\frac{d}{dt}(mr')=mr\varphi'^2-\frac{∂V(r)}{∂r}
and exchanging for l you can get:
\frac{d}{dt}(mr')=\frac{l^2}{mr^3}-\frac{∂V(r)}{∂r}
Way no.2) But if you first excange phi and then use euler-lagrange to r you get:
L=1/2m(r' ^2+\frac{l^2}{m^2r^2})-V(r)=1/2mr' ^2+\frac{l^2}{2mr^2}-V(r)
and then euler lagrange for r:
\frac{d}{dt}(mr')=\frac{∂}{∂r}(\frac{l^2}{2mr^2})-\frac{∂V(r)}{∂r}=-\frac{l^2}{mr^3}-\frac{∂V(r)}{∂r}
The plus in way no.1 became a minus in way no.2. How can you explain it?
If you can, don't show me a solution using energy because I already know that one. I want to use only lagrangian formalism.