- #1
michael879
- 698
- 7
Let me begin by saying I know I'm doing something wrong here, but I'm having trouble seeing what it is. This is a reformulation of https://www.physicsforums.com/threads/plugging-eom-into-lagrangian.905099/, where I've reduced the issue to a much simpler problem. Moderators, feel free to close my original post
Start with a 1-dimension particle at x(t) undergoing some force F(t), with the action,
[itex]S[x] \equiv \int{\left(\dfrac{1}{2}m\dot{x}^2 + Fx\right)dt}[/itex].
Under arbitrary variations [itex]x(t)\rightarrow x(t)+\delta{x}(t)[/itex], the variation in the action is given by,
[itex]\delta{S} = \int{\left(-m\ddot{x}+F\right)\delta{x}dt}[/itex] + 'surface' terms.
The surface terms can be eliminated by choosing [itex]\delta{x}[/itex] to vanish on the boundary, and by requiring [itex]\delta{S}=0[/itex] we get the Euler-Lagrange equation [itex]F = m\ddot{x}[/itex].
Now, if in addition to our initial assumption about the action we add in [itex]F = m\ddot{x}[/itex] as a postulate, we arrive at a paradox. The action can be reduced to,
[itex]S'[x] = \int{\left(\dfrac{1}{2}m\dot{x}^2 + m\ddot{x}x\right)dt} = \int{\left(-\dfrac{1}{2}m\dot{x}^2 + m\dfrac{d}{dt}\left(\dot{x}x\right)\right)dt}[/itex].
The right-hand term can be dropped and we're left with just,
[itex]S'[x] = \int{\left(-\dfrac{1}{2}m\dot{x}^2 \right)dt}[/itex].
This results in the Euler-Lagrange equation [itex]\ddot{x}=0[/itex], meaning that [itex]F=0[/itex]. Clearly plugging the EOM back into the Lagrangian is a tricky operation, but if done carefully it should still produce valid results, which it doesn't seem to be doing here
*note* Interestingly, repeating the same procedure for a spring-type force with action,
[itex]S[x] \equiv \int \left(\dfrac{1}{2}m\dot{x}^2 - \dfrac{1}{2} kx^2\right)dt[/itex],
results in [itex]S'[x] = 0[/itex]. This is exactly what I would expect. If there is only a single EOM, and you make it one of your postulates, the resulting action is trivial. Arriving at a conflicting action just has me confused
Start with a 1-dimension particle at x(t) undergoing some force F(t), with the action,
[itex]S[x] \equiv \int{\left(\dfrac{1}{2}m\dot{x}^2 + Fx\right)dt}[/itex].
Under arbitrary variations [itex]x(t)\rightarrow x(t)+\delta{x}(t)[/itex], the variation in the action is given by,
[itex]\delta{S} = \int{\left(-m\ddot{x}+F\right)\delta{x}dt}[/itex] + 'surface' terms.
The surface terms can be eliminated by choosing [itex]\delta{x}[/itex] to vanish on the boundary, and by requiring [itex]\delta{S}=0[/itex] we get the Euler-Lagrange equation [itex]F = m\ddot{x}[/itex].
Now, if in addition to our initial assumption about the action we add in [itex]F = m\ddot{x}[/itex] as a postulate, we arrive at a paradox. The action can be reduced to,
[itex]S'[x] = \int{\left(\dfrac{1}{2}m\dot{x}^2 + m\ddot{x}x\right)dt} = \int{\left(-\dfrac{1}{2}m\dot{x}^2 + m\dfrac{d}{dt}\left(\dot{x}x\right)\right)dt}[/itex].
The right-hand term can be dropped and we're left with just,
[itex]S'[x] = \int{\left(-\dfrac{1}{2}m\dot{x}^2 \right)dt}[/itex].
This results in the Euler-Lagrange equation [itex]\ddot{x}=0[/itex], meaning that [itex]F=0[/itex]. Clearly plugging the EOM back into the Lagrangian is a tricky operation, but if done carefully it should still produce valid results, which it doesn't seem to be doing here
*note* Interestingly, repeating the same procedure for a spring-type force with action,
[itex]S[x] \equiv \int \left(\dfrac{1}{2}m\dot{x}^2 - \dfrac{1}{2} kx^2\right)dt[/itex],
results in [itex]S'[x] = 0[/itex]. This is exactly what I would expect. If there is only a single EOM, and you make it one of your postulates, the resulting action is trivial. Arriving at a conflicting action just has me confused