A step in the derivation of the Euler-Lagrange equations using Hamilton's Principle

Question

I am going through the derivation of the Euler-Lagrange equations from Hamilton's principle following Landau and Lifshitz Volume 1. We start by writing the variation in the action as,

$$\delta S = \delta\int_{t_1}^{t_2}L(q,\dot{q},t)dt\ $$ as $q \to q+\delta q.$

In the next step, the variation in $L$ is written as $$\delta L= \frac{\partial{L}}{\partial{q}}\delta q+\frac{\partial{L}}{\partial{\dot{q}}}\delta \dot{q}.$$ Shouldn't there be a term $\frac{\partial{L}}{\partial{t}}\delta t$ as well since the Lagrangian $L$ is a function of $(q,\dot{q},t)$?

Answer 1

1. No, in the principle of stationary action/Hamilton's principle we are only varying the generalized coordinates $q$; not the time-parameter $t$, i.e. $\delta t=0$. Similarly, we are not varying the endpoints $t_1$ and $t_2$.

2. However be aware that there do exist other variational principles, such as e.g. Maupertuis' principle, where $\delta t\neq 0$.

Answer 2

In this answer I approach the question along a different angle than the previous answers.

First we take a look at the standard Euler-Lagrange equation:

$$ \frac{\partial L}{\partial q} - \ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}} \right) = 0 \tag{1} $$

Compare that to the newtonian equation for harmonic oscillation:

$$ - \frac{k}{m} q = \frac{d^2q}{dt^2} \tag{2} $$

(k is the spring constant, according to Hooke's law.)

(2) specifies differentiation with respect to the time coordinate, (1) specifies differentiation with respect to the position coordinate

As we know: the reason that (2) specifies differentiation with respect to time: we want to obtain an expression that gives the position as a function of time.

As we know: in order for the solution to give position as a function of time the differential equation must be in terms of derivative with respect to the time.

In the case of the Euler-Lagrange equation the direction of the differentiation is perpendicular to the direction of the time coordinate; the Euler-Lagrange equation specifies taking (partial) derivative with respect to position coordinate.

For simplification I narrow down to the Lagrangian of classical mechanics: $L=(E_k-E_p)$. Since the potential energy is a function of $q$ only, and the kinetic energy is a function of $\dot{q}$ only: specifying partial differentiation is superfluous, and the expression simplifies to:

$$ \frac{d(-E_p)}{dq} - \ \frac{d}{dt} \left( \frac{dE_k}{d\dot{q}} \right) = 0 \tag{3} $$

It looks as if the operation performed on the kinetic energy is different from the operation performed on the potential energy, but that is not actually the case.

To show that we take the following expression:

$$ \frac{d(-E_p)}{dq} - \frac{d(E_k)}{dq} = 0 \tag{4} $$

For both terms in (4): differentiation with respect to the position coordinate $q$is specified.

Evaluating the derivative of the kinetic energy with respect to the position coordinate:

$$ \frac{d(\tfrac{1}{2}mv^2)}{dq} = \tfrac{1}{2}m\left( 2v\frac{dv}{dq} \right) = m\frac{dq}{dt}\frac{dv}{dq} = m\frac{dv}{dt} = ma \tag{5} $$

Taking the derivative of kinetic energy with respect to position coordinate $q$ results in $ma$, the product of mass and acceleration.

The operation specified by the Euler-Lagrange equation:

$$ \frac{d}{dt} \left( \frac{d(\tfrac{1}{2}m\dot{q}^2)}{d\dot{q}} \right) = \frac{d}{dt} (m\dot{q}) = ma \tag{6} $$

(5) and (6) evaluate to the same quantity: $ma$. That demonstrates that (3) and (4) are indeed the same equation.

So:
We see that the Euler-Lagrange equation is an operator that performs a conversion.

It converts an expression in terms of potential energy and kinetic energy to an expression in terms of force and acceleration.

Stated differently:
The Euler-Lagrange equation performs conversion of an expression in terms of energy to an expresssion in terms of newtonian mechanics.

Of course, that raises the question: what if the energy is expressed in terms of generalized coordinates?

Example:
Let's say that we are modeling the oscillation of the balance wheel of a watch, using polar coordinates, and rotational kinetic energy, and potential energy of a spiral shaped spring.

Then the Euler-Lagrange equation does the following conversions:
-potential energy to torque
-rotational kinetic energy to angular acceleration.

The equation that relates torque, moment of inertia, and angular acceleration:

$$ \tau = I \frac{d^2 \phi}{dt^2} \tag{7} $$

$\tau$ => torque
I => $mr^2$; moment of inertia
$\phi$ => angle

Some people may argue that since (7) is not in cartesian coordinates it isn't newtonian mechanics.

I prefer the point of view that the choice of coordinate system is comparitively unimportant. I do regard (7) as newtonian mechanics since it has that characteristic pattern: There is a force (torque), there is opposition to change (moment of inertia) and there is the second time derivative of the position coordinate.

By extension:
If you have a potential energy that is stated in some form of generalized coordinate, and you take the derivative with respect to position coordinate, then the result is the corresponding generalized force

With all of the above established:

To make Hamilton's stationary action yield results what is needed is a conversion tool. A tool for conversion of expressing the physics taking place in terms of energy to expressing the physics taking place in terms of force and acceleration.

For classical mechanics the Euler-Lagrange equation fits the bill. The variation that leads to the Euler-Lagrange equation is restricted to variation of the coordinate(s) that is(are) perpendicular to the time coordinate.

Final remark:
Many introductions to Hamilton's stationary action use the image that is available in the wikipedia article about stationary action. However, that image is highly misleading. Several of the trajectories in that image loop back on themselves. That depiction violates the restrictions of the variation. The variation is exclusively variation of the position coordinate, so that the Euler-Lagrange equation performs differentiation with respect to the position coordinate.

Answer 3

In deriving EL equations from the least action principle, here, we are varying different paths the system takes, i.e., the functional dependence of $q$ and $\dot{q}$ on $t$. So, the variation of Lagrangian is with respect to the different paths.

On a side note: Also check First variation of a functional.

Answer 4

What we're really doing is calculating $\delta S$. S is a functional of q. So we are really varying the q. Since $\dot{q}$ also depends on q, it varies too. Think of $\delta L$ as the first order expansion when varying q, not as some kind of total derivative.