Maupertuis's principle applied to gravity field

[zhouhao]

Homework Statement

I learned how to deriving Maupertuis's principle from Hamilton's principle under conservative energy condition and feel this interesting.
But when trying to derive a particle's behavior in gravitivity field,stuck...
The direction of grativity is along $x$ axis,the particle is placed at $(0,0,0)$ as well as initial velocity equal to zero.At time $t$,particle's position is $(x(t),y(t),z(t))$.
Because Maupertuis's principle derived from Hamilton's principle here,it is reasonable to fix $t=0$ and $t={\Delta}t$ just like we did with Hamilton's principle.

Homework Equations

$\delta{x,y,z}(0)=\delta{x,y,z}(\Delta{t})=0$
$\delta{\int_0^{\Delta{t}}}p_x\dot{x}+p_y\dot{y}+p_z\dot{z}dt=m\delta{\int_0^{\Delta{t}}}{\dot{x}}^2+{\dot{y}}^2+{\dot{z}}^2dt=0 \ \ \ (1)$
Conservative energy condition:$mgx=\frac{1}{2}m({\dot{x}}^2+{\dot{y}}^2+{\dot{z}}^2)\ \ \ \ (2)$

The Attempt at a Solution

Substituite equation $(2)$ into $(1)$,we got $\delta\int_0^{\Delta{t}}2mgxdt=\int_0^{\Delta{t}}2mg\delta{x}dt=0$ this means $mg$ should equal to zero,something wrong...

 

[TSny]

In Maupertuis' principle, the time t corresponding to the final endpoint is not fixed. Time is not the integration variable in Maupertuis' principle.
https://en.wikipedia.org/wiki/Maupertuis'_principle#Comparison_with_Hamilton.27s_principle

 

[zhouhao]

In Maupertuis' principle, the time t corresponding to the final endpoint is not fixed. Time is not the integration variable in Maupertuis' principle.
https://en.wikipedia.org/wiki/Maupertuis'_principle#Comparison_with_Hamilton.27s_principle

I think in this way:
$W=-L+\sum\limits_ip_i{\dot{q}}_i$.Because for actual path, we have $W=constant$, apply variation method to Lagrange function as well as constrain variation path with energy conservation condition:
$\delta\int_{t_1}^{t_2}Ldt=\delta\int_{t_1}^{t_2}L+Wdt=\delta\int_{t_1}^{t_2}\sum\limits_ip_i{\dot{q}}_idt=0$
so,under energy conservation condition:
$\delta\int_{t_1}^{t_2}\sum\limits_ip_i{\dot{q}}_idt=0$ for actual path.(I apply this to grativity field above,although this is not rigorous Maupertuis' principle)

What's more,in De brogile's article:On the Theory of Quanta,which confusing me as well:
Maupertuis' principle was deduced in this way:$\delta\int_{t_1}^{t_2}\sum\limits_ip_i{\dot{q}}_idt=\delta\int_{A}^{B}\sum\limits_ip_id{q_i}=0$ with a Footnote added to German translation:To make this proof rigorous,it is necessary,as it well known,to also vary $t_1$,$t_2$;but because of the time independence of the result,our argument is not fase.

 

[TSny]

I think in this way:
$W=-L+\sum\limits_ip_i{\dot{q}}_i$.Because for actual path, we have $W=constant$, apply variation method to Lagrange function as well as constrain variation path with energy conservation condition:
$\delta\int_{t_1}^{t_2}Ldt=\delta\int_{t_1}^{t_2}L+Wdt=\delta\int_{t_1}^{t_2}\sum\limits_ip_i{\dot{q}}_idt=0$

This is done too quickly for me. On the far left you have $\delta\int_{t_1}^{t_2}Ldt$ which is an integral between two fixed times and the variation $\delta$ is for paths that generally have different energy $W$. Then you equate this with $\delta\int_{t_1}^{t_2}L+Wdt$ in which you assume $W$ is constant. But when you do the variation at constant energy, the time $t_2$ will vary as you vary the path. So, I don't follow this line of argument.

so,under energy conservation condition:
$\delta\int_{t_1}^{t_2}\sum\limits_ip_i{\dot{q}}_idt=0$ for actual path.

For Mauperuis' principle the variation $\delta$ is done at constant $W$. But this is impossible to do if you keep $t_1$ and $t_2$ fixed during the variation. For a derivation of Maupertuis' principle from Hamilton's principle, see the Landau and Lifshitz Mechanics text, sections 43 and 44.
https://archive.org/stream/Mechanics3eLandauLifshitz/Mechanics 3e-Landau,Lifshitz#page/n163/mode/2up

What's more,in De brogile's article:On the Theory of Quanta,which confusing me as well:
Maupertuis' principle was deduced in this way:$\delta\int_{t_1}^{t_2}\sum\limits_ip_i{\dot{q}}_idt=\delta\int_{A}^{B}\sum\limits_ip_id{q_i}=0$ with a Footnote added to German translation:To make this proof rigorous,it is necessary,as it well known,to also vary $t_1$,$t_2$;but because of the time independence of the result,our argument is not fase.

For me, this is too sketchy.

 

[zhouhao]

This is done too quickly for me. On the far left you have $\delta\int_{t_1}^{t_2}Ldt$ which is an integral between two fixed times and the variation $\delta$ is for paths that generally have different energy $W$. Then you equate this with $\delta\int_{t_1}^{t_2}L+Wdt$ in which you assume $W$ is constant. But when you do the variation at constant energy, the time $t_2$ will vary as you vary the path. So, I don't follow this line of argument.
For Mauperuis' principle the variation $\delta$ is done at constant $W$. But this is impossible to do if you keep $t_1$ and $t_2$ fixed during the variation. For a derivation of Maupertuis' principle from Hamilton's principle, see the Landau and Lifshitz Mechanics text, sections 43 and 44.
https://archive.org/stream/Mechanics3eLandauLifshitz/Mechanics 3e-Landau,Lifshitz#page/n163/mode/2up

For me, this is too sketchy.

You are right.I must misunderstand something.
Lagrangian function:$L=\frac{1}{2}m{\dot{q}_x}^2+\frac{1}{2}m{\dot{q}_y}^2+\frac{1}{2}m{\dot{q}_z}^2+mgq_x\ \ \ \ \ (1)$

Energy conservation condition:$mgq_x=\frac{1}{2}m{\dot{q}_x}^2+\frac{1}{2}m{\dot{q}_y}^2+\frac{1}{2}m{\dot{q}_z}^2\ \ \ (2.1)$ and $mg\delta{q_x}=m\dot{q}_x\delta{\dot{q}_x}+m\dot{q}_y\delta{\dot{q}_y}+m\dot{q}_z\delta{\dot{q}_z} \ \ \ \ (2.2)$

In my imagination,$\frac{\partial{L}}{\partial{q}_i}-\frac{d}{dt}\frac{\partial{L}}{\partial{\dot{q}_i}}=0$ make $\delta\int_{t_1}^{t_2}Ldt=0$,This could be right for equation $(1)$ and equavalent to $\ddot{q}_x=g,\ddot{q}_y=\ddot{q}_z=0$

Equation $(2.2)$ make $\delta\int_{t_1}^{t_2}Wdt=0$
So $\delta\int_{t_1}^{t_2}(L+W)dt=0$,then we have $\delta\int_{t_1}^{t_2}\sum\limits_ip_i\dot{q}_idt=0$

But equation $(2.1)$ would make $L=2mgq_x \ \ \ (3)$,
$\frac{\partial{L}}{\partial{q}_i}-\frac{d}{dt}\frac{\partial{L}}{\partial{\dot{q}_i}}=0$ is not right argument for equation (3) now.
It seems that energy conservation condition make $\delta\int_{t_1}^{t_2}Ldt=\int_{t_1}^{t_2}2mg\delta{q}_xdt\ne0$.

If varying $t_2$,the $\delta\int_{t_1}^{t_2}Ldt$ would equal zero under energy conservation condition?

 

[TSny]

You are right.I must misunderstand something.
Lagrangian function:$L=\frac{1}{2}m{\dot{q}_x}^2+\frac{1}{2}m{\dot{q}_y}^2+\frac{1}{2}m{\dot{q}_z}^2+mgq_x\ \ \ \ \ (1)$

Energy conservation condition:$mgq_x=\frac{1}{2}m{\dot{q}_x}^2+\frac{1}{2}m{\dot{q}_y}^2+\frac{1}{2}m{\dot{q}_z}^2\ \ \ (2.1)$ and $mg\delta{q_x}=m\dot{q}_x\delta{\dot{q}_x}+m\dot{q}_y\delta{\dot{q}_y}+m\dot{q}_z\delta{\dot{q}_z} \ \ \ \ (2.2)$

In my imagination,$\frac{\partial{L}}{\partial{q}_i}-\frac{d}{dt}\frac{\partial{L}}{\partial{\dot{q}_i}}=0$ make $\delta\int_{t_1}^{t_2}Ldt=0$

Yes, this is correct if the $\delta$ symbol means to vary the paths such that the initial and final times ($t_1$ and $t_2$) are kept fixed for all paths. For this type of path variation, the energy $W$ generally will not be conserved (except for the true path).

Equation $(2.2)$ make $\delta\int_{t_1}^{t_2}Wdt=0$

Here, you are assuming the energy $W$ remains constant for the varied paths. So, here you are using a different type of path variation than before. So, it might be good to use a different symbol $\Delta$ for this type of path variation. For this type of variation, you must allow the endpoint times $t_1$ and/or $t_2$ to vary as you vary the path.

[EDIT: So, $\Delta\int_{t_1}^{t_2}Wdt \neq 0$ if you are varying paths keeping $W$ constant. Rather, you can see that you would get $\Delta\int_{t_1}^{t_2}Wdt = W \Delta t_2 - W \Delta t_1$, where $\Delta t_2$ is the variation in the upper time limit when varying the path. Similarly for $\Delta t_1$.]

So $\delta\int_{t_1}^{t_2}(L+W)dt=0$

Here it looks like you are not considering that the two types of variation are different. [EDIT: From what I wrote above, this equation does not follow.]

But equation $(2.1)$ would make $L=2mgq_x \ \ \ (3)$,
$\frac{\partial{L}}{\partial{q}_i}-\frac{d}{dt}\frac{\partial{L}}{\partial{\dot{q}_i}}=0$ is not right argument for equation (3) now.

Lagrange's equations $\frac{\partial{L}}{\partial{q}_i}-\frac{d}{dt}\frac{\partial{L}}{\partial{\dot{q}_i}}=0$ assume that you are expressing $L$ as a function of both the $q_i$ and the $\dot{q}_i$ according to $L = T - V$.

It seems that energy conservation condition make $\delta\int_{t_1}^{t_2}Ldt=\int_{t_1}^{t_2}2mg\delta{q}_xdt\ne0$

The variation of the paths on the left side of the equation are the $\delta$ type of variation where energy is not conserved for the varied paths. But the integral involving $2mg$ assumes energy is conserved in the variation.

A careful discussion of going from Hamilton's principle to Maupertuis' principle can be found in Goldstein's Mechanics, 3rd edition, starting on page 356. The text uses the $\delta$ versus $\Delta$ notation to distinguish the types of variation of path.