Why Is \(\frac{\partial \dot{q}}{ \partial q} = 0\) in Lagrangian Mechanics?

[AJKing]

Question 1

When I take the derivatives of the Lagrangian, specifically of the form:

\frac{\partial L}{ \partial q}

I often find myself saying this:

\frac{\partial \dot{q}}{ \partial q}=0

But why is it true? And is it always true?

 

[AJKing]

[STRIKE]Question 2[/STRIKE] Answer Below

When solving a double pendulum problem I built a Lagrangian of the form:

L(\theta_1, \theta_2, \dot{\theta_1}, \dot{\theta_2})

And found that my Euler Lagrangian equations for each coordinate where coupled to each other, as expected.

But I was a little confused about the direction of motion...

Does

\frac{\partial L}{\partial q_\alpha} = \frac{d}{d t} \frac{\partial L}{\partial \dot{q_\alpha}}

Describe a motion only in the

\hat{q_\alpha}

Direction? Even if the Euler Lagrangian is coupled?

Answer

According to Taylor's text, Classical Mechanics, yes. The \hat{q_\alpha} direction is the only one an Euler Lagrangian equation specifies. But, a true understanding of the motion in that direction comes from:

- \frac{\partial U}{\partial q_\alpha} = \frac{d}{d t} \frac{\partial L}{\partial \dot{q_\alpha}} - \frac{\partial T}{\partial q_\alpha}

Which is used to build:

\vec{F}=-\nabla U

 

[voko]

Question 1

When I take the derivatives of the Lagrangian, specifically of the form:

\frac{\partial L}{ \partial q}

I often find myself saying this:

\frac{\partial \dot{q}}{ \partial q}=0

But why is it true? And is it always true?

Many a student has been confused by this. The short answer is that the Euler-Lagrange equation is derived from a variational principle using the technique of variations, and $q$ and $\dot q$ are merely labels for the arguments of the Lagrangian.

Consider that you have a function $F(x, y)$. At some $(x_0, y_0)$ its value is $z_0$. Now you want to find out its value close to that point. Using the Taylor expansion, you get $F(x_0 + \delta x, y_0 + \delta y) = z_0 + {\partial F \over \partial x} \delta x + {\partial F \over \partial y} \delta y$. Now if you had $x = f(t)$ and $y = \dot f(t)$, you would still use the expression above, where $x$ and $y$ are assumed independent. This is exactly what happens when you derive the Euler-Lagrange equation.