I Derivation of Hamilton's Principle: Questions

[PhysicsLama]

TL;DR Summary

I want to know how the term $m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0$ can be derived, because it is a key part in understanding how hamiltons principle can be derived.

I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term
$m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0$ (1) one may derivate $\delta \int (T - V) dt = 0$ (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the action-integral can be derived.
If needed I can add the steps from how to get from (1) to (2).

The paper from where I took (1) is the following: https://www.researchgate.net/public...ple_for_the_Derivation_of_Equations_of_Motion

Thanks for your help

 

[wrobel]

because it is a key part in understanding how hamiltons principle can be derived.

No it is just a part of some little-known text

 

[PhysicsLama]

Do you know where that equation came from?

 

[JimWhoKnew]

Welcome to PF.

I confess I didn't have the patience to read the linked paper. Yet, here is my take. Tell me whether it answers your question.

We don't have$$m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0$$as you wrote in post #1. But we note that the LHS is a total derivative, so in the absence of $~\mathbf{F}_i~$ we get from equations (1)-(4) in the referenced paper:$$\delta \int_{t_1}^{t_2}L~dt=m\int_{t_1}^{t_2}\frac{d}{dt} (\frac{dr}{dt}\delta r)~dt = m\left[\frac{dr}{dt} \delta r\right]_{t_1}^{t_2} \quad.$$If we impose$$\delta r(t_1)=0=\delta r(t_2)$$(which we do!), the RHS is zero. This is just what we need in order to proceed.

edit:
The vanishing of $~\delta r~$ at the endpoints is certainly a key ingredient in Hamilton's principle.

 

[PhysicsLama]

Welcome to PF.

I confess I didn't have the patience to read the linked paper. Yet, here is my take. Tell me whether it answers your question.

We don't have$$m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0$$as you wrote in post #1. But we note that the LHS is a total derivative, so in the absence of $~\mathbf{F}_i~$ we get from equations (1)-(4) in the referenced paper:$$\delta \int_{t_1}^{t_2}L~dt=m\int_{t_1}^{t_2}\frac{d}{dt} (\frac{dr}{dt}\delta r)~dt = m\left[\frac{dr}{dt} \delta r\right]_{t_1}^{t_2} \quad.$$If we impose$$\delta r(t_1)=0=\delta r(t_2)$$(which we do!), the RHS is zero. This is just what we need in order to proceed.

edit:
The vanishing of $~\delta r~$ at the endpoints is certainly a key ingredient in Hamilton's principle.

Thanks for your answer.

Now the thing I do not understand is how to even derive the LHS from equation (4) (in the paper). Even if we look at your explanation we still take the Hamilton Principle

$\delta \int L \; dt = 0$

for granted and just put in equation (4). But to derive the Hamilton Principle with the logic of this paper we need the term on the LHS of (4). My Problem is that I don't understand where the LHS of equation (4) in the paper is taken from, without one knowing the form of Hamiltons Principle (like you did), which we can't use because we are deriving it from (4). For me it's like trying to proof a theorem by using the theorem.

Thanks for your help already in advance :)

 

[wrobel]

Spoiler

I confess I didn't have the patience to read the linked paper.

So am I. Actually I looked through it and found it turbid. Heaps of excellent classical texts are devoted to the topic.

 

[PhysicsLama]

Okay thanks for the time. Do you have any names for those classical texts?

 

[wrobel]

Paul Appell: Traité de mécanique rationnelle vol 1,2
Greenwood D.T. Classical dynamics
Jorge V. José, Eugene J. Saletan Classical Dynamics A Contemporary Approach
Pars, L. A A treatise on the analytical dynamics

 

[TSny]

Now the thing I do not understand is how to even derive the LHS from equation (4) (in the paper). Even if we look at your explanation we still take the Hamilton Principle

$\delta \int L \; dt = 0$

for granted and just put in equation (4).

At which point of the paper do you believe the author takes $\delta \int L \;dt = 0$ for granted?

The paper starts with d'Alembert's principle in the form of equation (1) of the paper. Equation (4) is used to make a substitution in (1) to derive the "extended Hamilton's principle", equation (8): $\int_{t_1}^{t_2} (\delta L + \delta W) \;dt = 0$.
I don't see any circular reasoning by the author in getting to (8).

 

[JimWhoKnew]

Now the thing I do not understand is how to even derive the LHS from equation (4) (in the paper). Even if we look at your explanation we still take the Hamilton Principle

$\delta \int L \; dt = 0$

for granted and just put in equation (4). But to derive the Hamilton Principle with the logic of this paper we need the term on the LHS of (4). My Problem is that I don't understand where the LHS of equation (4) in the paper is taken from, without one knowing the form of Hamiltons Principle (like you did), which we can't use because we are deriving it from (4). For me it's like trying to proof a theorem by using the theorem.

In addition to @TSny's explanation in #9: the usual practice is to assume $\delta ~\int L~dt = 0~$ , and derive the equations of motion from it. In the paper, they try to motivate this practice (I think), so they go in the opposite direction (as it developed historically): they start from d'Alembert's principle (equation (1) ) and work their way to show that it yields $\delta ~\int L~dt = 0~$ . Note that equation (1) is satisfied for all $~\delta \mathbf{r}_i~$ only if all the terms in brackets are zero, and this happens only when Newton's second law is satisfied. So we may say that the starting point is Newton's law.
Equation (4) is nothing but a straightforward mathematical manipulation. You don't need to "derive" the LHS, you only have to show that it equals the RHS, which they do. Now it can be used, along with the endpoints condition, to get $$\delta \int_{t_1}^{t_2}L~dt=m\int_{t_1}^{t_2}\frac{d}{dt} (\frac{dr}{dt}\delta r)~dt = m\left[\frac{dr}{dt} \delta r\right]_{t_1}^{t_2}=0 $$as intended.

Clearer?

There are plenty of good books on this subject.
Feynman's lecture is accessible for free.
Goldstein's "Classical Mechanics" has been a canonical text for decades.

Edit: rephrased a little.

 

[wrobel]

Formula (4) from the article merely does not make sense. Indeed,
in the text above $\delta\boldsymbol {r}_i$ stands for virtual displacements. By definition a virtual displacement is a vector from some vector space and we know how this vector space is defined, that is ok.
In the notation $\delta\boldsymbol {r}_i$ the letter $\delta$ is not an operation it is a part of the symbol. For example we can use $\boldsymbol \xi_i$ to denote virtual displacements.
In formula (4) $\boldsymbol r_i(t)$ is a real motion of the system.
The term
$$\delta\frac{d\boldsymbol r_i}{dt}$$ is a nonsense.
Formulas like that (which the author uses implicitly):
$$\delta\Big(\frac{d\boldsymbol r_i}{dt}\cdot \frac{d\boldsymbol r_i}{dt}\Big)=2\frac{d\boldsymbol r_i}{dt}\cdot \delta\frac{d\boldsymbol r_i}{dt}$$
are the nonsense as well.

 

[PhysicsLama]

Okay thanks guys I think I have undertstood better now. But do you have any Papers or Books where I specificely could read something directly about how to derive Hamiltons Principle, so how to derive that $\delta \int L \; dt$ (A), not how to get from Hamiltons Principle to the equations of motion but how to basically get to the Hamilton Principle. In all the books and papers you suggested like the Feynman lectures or Greenwoods classical dynamics for instance the Hamilton Principle (A) was just given to be true and never derived somehow mathematically. The paper I linked was the only one I found online specific explaining how to derive (A) but it seems to be a not very good written one and even an at least partly wrong one.

 

[wrobel]

how to basically get to the Hamilton Principle

to get to the Hamilton Principle from what?

The standard scheme is as follows.

(the Lagrange-D'Alambert principle+certain assumptions) $\Longleftrightarrow$ (the Lagrange equations) $\Longleftrightarrow$ (the Hamilton principle)

 

[JimWhoKnew]

Formula (4) from the article merely does not make sense. Indeed,
in the text above $\delta\boldsymbol {r}_i$ stands for virtual displacements. By definition a virtual displacement is a vector from some vector space and we know how this vector space is defined, that is ok.
In the notation $\delta\boldsymbol {r}_i$ the letter $\delta$ is not an operation it is a part of the symbol. For example we can use $\boldsymbol \xi_i$ to denote virtual displacements.
In formula (4) $\boldsymbol r_i(t)$ is a real motion of the system.
The term
$$\delta\frac{d\boldsymbol r_i}{dt}$$ is a nonsense.
Formulas like that (which the author uses implicitly):
$$\delta\Big(\frac{d\boldsymbol r_i}{dt}\cdot \frac{d\boldsymbol r_i}{dt}\Big)=2\frac{d\boldsymbol r_i}{dt}\cdot \delta\frac{d\boldsymbol r_i}{dt}$$
are the nonsense as well.

How do you derive Euler-Lagrange equations from Hamilton's principle without using$$\delta\frac{d\mathbf{r}_i}{dt}$$and without replacing it by$$\frac{d}{dt}\delta\mathbf{r}_i$$?

 

[wrobel]

How do you derive Euler-Lagrange equations from Hamilton's principle without using

$$L=L(t,x,\dot x);\quad\frac{d}{ds}\Big|_{s=0}\int_{t_1}^{t_2}L\big(t,x(t)+s\xi(t),\dot x(t)+s\dot \xi(t)\big)dt$$
$$=\int_{t_1}^{t_2}(L_x\xi+L_{\dot x}\dot\xi) dt=
L_{\dot x}\xi\Big|_{t_1}^{t_2}+\int_{t_1}^{t_2}\Big(L_x-\frac{d L_{\dot x}}{dt}\Big)\xi dt$$

 

[JimWhoKnew]

$$L=L(t,x,\dot x);\quad\frac{d}{ds}\Big|_{s=0}\int_{t_1}^{t_2}L\big(\tau,x(\tau)+s\xi(\tau),\dot x(\tau)+s\dot \xi(\tau)\big)d\tau$$
$$=\int_{t_1}^{t_2}(L_x\xi+L_{\dot x}\dot\xi) d\tau=
L_{\dot x}\xi\Big|_{t_1}^{t_2}+\int_{t_1}^{t_2}\Big(L_x-\frac{d L_{\dot x}}{dt}\Big)\xi d\tau$$

Take a look at Greenwood's equations (4-51), (4-52), (4-53). Still nonsense?

Substituting $~\delta \mathbf{r}=s\cdot\mathbf{\xi}(t)~$ is just doing the same thing with a different notation (Greenwood's (4-46) ).

 

[wrobel]

Take a look at Greenwood's equations (4-51), (4-52), (4-53). Still nonsense?

Sure. Because a definition of $\delta\boldsymbol {\dot r}_i$ is missing.

 

[PhysicsLama]

to get to the Hamilton Principle from what?

The standard scheme is as follows.

(the Lagrange-D'Alambert principle+certain assumptions) $\Longleftrightarrow$ (the Lagrange equations) $\Longleftrightarrow$ (the Hamilton principle)

Ah okay, so basically Lagranges Equations originally come from d'alemberts Principle and not from the Hamilton Principle, the Hamilton Principle is only a consequence of Lagranges Interpretation of Mechanics? Have I understood that right, therefore the Hamilton Principle is always just given without any explenation of its mathematical derivation.

 

[JimWhoKnew]

Ah okay, so basically Lagranges Equations originally come from d'alemberts Principle and not from the Hamilton Principle, the Hamilton Principle is only a consequence of Lagranges Interpretation of Mechanics? Have I understood that right, therefore the Hamilton Principle is always just given without any explenation of its mathematical derivation.

I refreshed my memory using Goldstein's book. I'll consider only one generalized coordinate $q$ compatible with the constraints. The generalization to any number of GC is straightforward. Start from Newton's second law in Cartesian coordinates$$\mathbf{F}_i-\dot{\mathbf{p}}_i=0\quad.$$Noting that no virtual work is done by the constraint forces when the virtual displacement is $\delta q$ , one gets d'Alembert principle$$\delta K:=\left[\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right)-\frac{\partial L}{\partial q}\right]~\delta q=0\quad.$$Now you can see that mathematically$$\delta\int_{t_1}^{t_2}L(q,\dot{q},t)~dt= \int_{t_1}^{t_2}\delta K~dt +\left[\frac{\partial L}{\partial \dot{q}} \delta q\right]_{t_1}^{t_2}$$where the equal time distance $~\delta q(t)~$ to the virtually varied path is the same on both sides of the equation. So to guarantee ("derive") $~\delta\int L~dt=0~$ , you must impose the endpoints condition. Once you've done that, Hamilton's principle is equivalent to Newton's law or d'Alembert principle, and can replace them as the starting point. It contains all the physical content they do (which is basically what @wrobel wrote in #13).

But Hamilton's principle can also be used to derive the generalization of Euler-Lagrange equations for non-mechanical systems (such as fields, for example). Newton's law can't go that far.

Remember that Physics is an empirical science. We can't derive everything. Newton has postulated his second law because experiments/observations suggested it.

 

[PhysicsLama]

Thanks for your explanation I have finally understood it now :)