Does $\frac{\partial^2 L}{\partial q\partial \dot q}=0$?

[Leo Liu]

Homework Statement

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Relevant Equations

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This is a conceptual question.
Suppose the lagrangian of a system in generalized coordinates is $$L(q,\dot q,t)$$
If the following expression arises in calculations$$\frac{\partial^2 L}{\partial q\partial \dot q}\,,$$
does it equal to 0?
My reasoning is that since $\dot q$ and $q$ are functions of time, the expression above equals 0. Am I right?
Thanks :)

 

[anuttarasammyak]

Say Force
$$\frac{\partial L}{\partial q}=-k \cdot{\dot{q}}$$
the quantity you say is -k, not zero. It is a system with friction.

 

[Leo Liu]

Say Force
$$\frac{\partial L}{\partial q}=-k \cdot{\dot{q}}$$
the quantity you say is -k, not zero. It is a system with friction.

But $$\frac{\partial L}{\partial q}=-k \cdot{\dot{q}}\neq \frac{\partial}{\partial q}\left(\frac{\partial L}{\partial \dot q}\right)$$
Sorry I should have indicated that it was a second derivative wrt a different var.

 

[anuttarasammyak]

Ya, I was wrong. Friction comes from another way.

 

[ergospherical]

Suppose the lagrangian of a system in generalized coordinates is $L(q,\dot q,t)$. If the following expression arises in calculations $\dfrac{\partial^2 L}{\partial q\partial \dot q}\,,$ does it equal to 0? My reasoning is that since $\dot q$ and $q$ are functions of time, the expression above equals 0. Am I right? Thanks :)

In this context $q$ and $\dot{q}$ are not functions of time, rather they are simply the independent variables which make up $q$-$\dot{q}$ space. (They are only functions of time when evaluated on a solution to the equations of motion $\boldsymbol{\gamma}(t) = (q(t), \dot{q}(t))$).

e.g. take the silly example $L = q^2 \dot{q}$, then $\dfrac{\partial^2 L}{\partial q\partial \dot q} = 2q$.

 

[anuttarasammyak]

e.g. take the silly example , then .

we can add the term $2aq\dot{q}=\frac{d}{dt}(aq^2)$ which does not change equation of motion, to Lagrangean. This term results
$$\frac{\partial^2 L}{\partial q \partial \dot{q}}=2a$$

 

[ergospherical]

what is your point? yes an equivalent lagrangian is $\tilde{L} = q^2 \dot{q} + 2a q \dot{q}$ and $\dfrac{\partial^2 \tilde{L}}{\partial q \partial \dot{q}} = 2q + 2a$...

 

[anuttarasammyak]

I meant adding that term to any normal Lagrangean, not to your example.

Anyway
$$q^2 \dot{q}+2aq\dot{q}=\frac{d}{dt}(\frac{q^3}{3}+aq^2)$$
This Lagrangean tells nothing to us.

 

[ergospherical]

I don't know what you are saying because it wasn't supposed to represent any physical system, it was just an arbitrary example to demonstrate the partial derivative

 

[anuttarasammyak]

I say adding any term of
$$\frac{d}{dt}f(q,t)$$
to Lagrangean does not harm it.

 

[ergospherical]

yes, but what has this got to do with the op?

 

[anuttarasammyak]

I show a counter example to his

Homework Statement:: .
Relevant Equations:: .

does it equal to 0?

for physical case.

e.g. a free particle
$$L=\frac{m}{2}\dot{q}^2$$ and,
$$L=\frac{m}{2}\dot{q}^2+2aq\dot{q}$$ or
$$L=\frac{m}{2}\dot{q}^2+q^2\dot{q}+2aq\dot{q}$$
are equivalent in getting a solution. The latter Lanrangeans show a counter example to OP question.