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

  • Thread starter Thread starter Leo Liu
  • Start date Start date
AI Thread Summary
The discussion revolves around whether the expression \(\frac{\partial^2 L}{\partial q \partial \dot{q}} = 0\) for a Lagrangian \(L(q, \dot{q}, t)\). It is clarified that \(q\) and \(\dot{q}\) are independent variables in the context of Lagrangian mechanics, not functions of time, which impacts the evaluation of the partial derivatives. A counterexample is provided where a Lagrangian of the form \(L = q^2 \dot{q}\) results in \(\frac{\partial^2 L}{\partial q \partial \dot{q}} = 2q\), demonstrating that the expression does not equal zero. Additionally, the discussion highlights that adding total time derivatives to the Lagrangian does not affect the equations of motion, further complicating the original question. Thus, the assertion that the expression equals zero is incorrect in general cases.
Leo Liu
Messages
353
Reaction score
156
Homework Statement
.
Relevant Equations
.
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 :)
 
Last edited:
Physics news on Phys.org
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.
 
  • Skeptical
Likes Leo Liu
anuttarasammyak said:
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.
 
  • Like
Likes anuttarasammyak
Ya, I was wrong. Friction comes from another way.
 
  • Informative
Likes Leo Liu
Leo Liu said:
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##.
 
  • Informative
Likes Leo Liu
ergospherical said:
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
 
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##...
 
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.
 
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
 
  • #10
I say adding any term of
\frac{d}{dt}f(q,t)
to Lagrangean does not harm it.
 
  • Like
Likes Leo Liu
  • #11
yes, but what has this got to do with the op?
 
  • Skeptical
Likes Leo Liu
  • #12
I show a counter example to his
Leo Liu said:
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.
 
Last edited:
  • Like
Likes Leo Liu
Back
Top