Study the action in a one-dimensional movement (Hamilton Principle)

AI Thread Summary
The discussion revolves around proving that the real trajectory of a particle, derived from the Lagrangian, minimizes the action according to the Hamilton Principle. The trajectory is expressed as x(t) = (1/2)gt^2 + ut, and alternative trajectories are considered through the function α(t). The user initially struggles with an integral that does not yield zero, which is necessary for the proof. Feedback from other users highlights a sign error in the Lagrangian and suggests using integration by parts to resolve the issue. After correcting the error, the user successfully demonstrates that the action is minimized, confirming the validity of their approach.
alex23
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Homework Statement
Study the action (in the sense of Hamilton) for the one-dimensional movement of a particle of mass 1 in a field of constant forces and prove that it reaches a minimum for the real trajectory of the particle.
Relevant Equations
##L=T-V##
##T=\frac{1}{2}m\dot x^2##
##V=-mgx##
Action: ##S=\int{L(t,x,\dot x)dt}##
Using the Lagrangian of the system I reached that ##x(t)=\frac{1}{2} gt^2+ut ## is the real trajectory of the particle.
After that, I consider different trajectories: ##x(\alpha,t) = x(t) + \alpha(t)## with ##\alpha(t)## being an arbitrary function of t expect for the conditions ##\alpha(0)=\alpha(T)=0##, where ##T## is the final time of the movement. This way ##x(\alpha,t)## shows a family of trajectories that start and end in the same points as the real trajectory ##x(t)##.
Now, the problem asks me to study the action and prove that the real trajectory makes the action an extremal (in this case a minimum). For that, I calculated the variance of ##S: \delta S= S(\alpha) - S(0)##.
To do that I calculated the Lagrangian of both ##x(\alpha,t)## and ##x(0,t)## and introduced them both to calculate ##S(\alpha)## and ##S(0)## and introduce them in the formula above. Doing that I reached this:
$$ \displaystyle\int_{0}^{T}{[\frac{1}{2}(gt+u+\dot \alpha(t))^2 -g(\frac{1}{2}gt^2+ut+\alpha(t))]dt} - \displaystyle\int_{0}^{T}{[\frac{1}{2}(g^2t^2+u^2+2ugt)-(\frac{1}{2}g^2t^2+ugt)]dt} $$
Where ##\dot\alpha(t)## is used to denote the total time derivate of ##\alpha(t)##, so ##\dot\alpha(t)=\frac{d\alpha(t)}{dt}##.
After expanding the integral and combining them I reached to this:
$$ \displaystyle\int_{0}^{T}{(\frac{1}{2}\dot\alpha^2(t)+gt\dot\alpha(t)+u\dot\alpha(t)-g\dot\alpha(t))dt} $$
This leaves me with 3 integrals, the first:
##\frac{1}{2}\displaystyle\int_{0}^{T}{\dot\alpha^2(t)dt}##, which is always positive, which would make it so it had a minimum in ##\alpha=0## and hence proving that the real trajectory makes the action S an extremal.
The second integral:
##u\displaystyle\int_{0}^{T}{\dot\alpha(t)dt}=u[\alpha(T)-\alpha(0)]=0##, which would make it so it didn't contribute.
And the third integral and the reason for my problems:
##g\displaystyle\int_{0}^{T}{(t\dot\alpha(t)-\alpha(t))dt}##, which I can't seem to make 0 since I think it would not be 0 in the general case and I can't see where I was mistaken before to make this integral appear.
I know for a fact that the objectives is to reach that ##\delta S=S(\alpha)-S(0)=\frac{1}{2}\displaystyle\int_{0}^{T}{\dot\alpha^2(t)}##. And making the argument I made above would be sufficient to solve the problem.
I don't think the mistakes is after expanding and simplifying the integrals, so the error should come from before that, but I can't seem to locate where I was mistaken. Any help pointing where the mistake is would be greatly appreciated.
Thanks in advance!

Edit: I edited the post to properly format everything in LaTeX since I didn't know it was supported by the forum, hopefully it makes it easier to read! And sorry for the inconvenience.
 
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I think you just have a careless sign error (easy to do). There are two negative signs that contribute to the ##mgx## term in the Lagrangian. One negative sign is the one in ##L = T-V## and the other is in ##V = -mgx##.

Otherwise, your work looks good to me.

You might think about doing integration by parts for ##\int t \dot \alpha dt##

Thanks for using LaTex!
 
TSny said:
I think you just have a careless sign error (easy to do). There are two negative signs that contribute to the ##mgx## term in the Lagrangian. One negative sign is the one in ##L = T-V## and the other is in ##V = -mgx##.

Otherwise, your work looks good to me.

You might think about doing integration by parts for ##\int t \dot \alpha dt##

Thanks for using LaTex!
You are absolutely right, I had a sign error in the Lagrangian. After fixing that and doing the integration by parts as you suggested I managed to get ##\delta S=\frac{1}{2}\displaystyle\int_{0}^{T}{\dot\alpha^2(t)dt}## as I as supposed to.
Thanks!
 
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