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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.

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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