# Maupertuis's principle applied to gravity field

• zhouhao
In summary, the conversation discusses the derivation of Maupertuis's principle from Hamilton's principle under the conservative energy condition. The person mentions their interest in this topic but is stuck when trying to apply it to a particle's behavior in a gravitational field. They mention that the time t corresponding to the final endpoint is not fixed in Maupertuis's principle. They also bring up a footnote in de Broglie's article which they find confusing. The expert summarizer notes that the conversation is too sketchy and mentions the equations and explanations provided by the person. They also mention that the energy conservation condition may affect the calculation of the integral for Maupertuis's principle.
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 said:
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 neccessary,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.

zhouhao said:
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 neccessary,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.

TSny said:
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?

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zhouhao said:
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.

Last edited:
zhouhao

## 1. What is Maupertuis's principle applied to gravity field?

Maupertuis's principle, also known as the principle of least action, states that the path taken by a system between two points in time is the one that minimizes the action, or the integral of the Lagrangian, of the system. When applied to the gravity field, it means that the path taken by an object between two points is the one that minimizes the gravitational potential energy.

## 2. How does Maupertuis's principle relate to Newton's laws of motion?

Maupertuis's principle is essentially a reformulation of Newton's laws of motion. It provides an alternative way of describing the motion of a system by minimizing the action instead of solving differential equations. However, the principle is equivalent to Newton's laws and can be derived from them.

## 3. Can Maupertuis's principle be applied to other forces besides gravity?

Yes, Maupertuis's principle can be applied to any conservative force, meaning a force that conserves energy. This includes forces such as electric and magnetic forces, as well as forces that arise from potential energy functions, such as the spring force.

## 4. How does Maupertuis's principle apply to celestial bodies?

Maupertuis's principle can be used to determine the orbits of celestial bodies, such as planets and satellites, in the presence of a gravitational field. By minimizing the action, we can determine the most likely path of a celestial body, which is often an elliptical orbit around a larger body.

## 5. What are the practical applications of Maupertuis's principle applied to gravity field?

Maupertuis's principle has practical applications in many fields, including astrophysics, aerospace engineering, and celestial mechanics. It is used to calculate the trajectories of spacecraft and satellites, as well as to predict the motion of celestial bodies in the solar system. It also has applications in the study of black holes and other objects with strong gravitational fields.

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