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

In summary, the conversation discussed the use of the Lagrangian to find the real trajectory of a particle, as well as the consideration of different trajectories starting and ending at the same points. The problem at hand was to study the action and prove that the real trajectory minimizes it. After some calculations, a sign error was discovered in the Lagrangian, which was then corrected. Integration by parts was also suggested to simplify the calculations. Ultimately, the correct result was obtained, showing that the real trajectory indeed minimizes the action.
  • #1
alex23
3
0
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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  • #2
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!
 
  • #3
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!
 

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

1. What is the Hamilton Principle?

The Hamilton Principle, also known as the principle of least action, is a fundamental principle in classical mechanics that states that the path taken by a system between two points in time is the one that minimizes the action integral.

2. How is the Hamilton Principle used in one-dimensional movement?

In one-dimensional movement, the Hamilton Principle is used to determine the equations of motion for a system by minimizing the action integral along the path of motion. This allows for the prediction of the position, velocity, and acceleration of a system at any given time.

3. What is the significance of studying one-dimensional movement using the Hamilton Principle?

Studying one-dimensional movement using the Hamilton Principle allows for a deeper understanding of the underlying principles and laws governing the motion of a system. It also provides a powerful mathematical tool for predicting the behavior of systems in a wide range of scientific fields.

4. What are the limitations of using the Hamilton Principle in one-dimensional movement?

The Hamilton Principle is limited in its applicability to systems that can be described by a single generalized coordinate. It also assumes that the system is conservative, meaning that there are no external forces acting on the system.

5. Can the Hamilton Principle be extended to higher dimensions?

Yes, the Hamilton Principle can be extended to higher dimensions using the Hamiltonian formalism. This involves introducing additional generalized coordinates and momenta to describe the motion of a system in three-dimensional space. However, the principle remains the same - minimizing the action integral along the path of motion.

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