# Why is the Lagrangian quadratic in $\dot{q}$? [duplicate]

My teacher said we only consider Lagrangians which are quadratic in $\dot{q}$, and we don't take other Lagrangians. I couldn't understand why. Can anyone please explain this?

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## marked as duplicate by Qmechanic♦Mar 30 '13 at 22:23

Possible duplicate: physics.stackexchange.com/q/23098/2451 –  Qmechanic Feb 28 '13 at 19:07
It's for the same reason the kinetic energy is quadratic in the velocities. –  alexarvanitakis Feb 28 '13 at 19:14
–  dmckee Feb 28 '13 at 21:37

For simplicity, is the short answer. The Lagrangian does not need to be quadratic for physics. My first instinct was some kind of drag force, but general cases were worked out quite some time ago (1955):

http://prola.aps.org/abstract/PR/v99/i2/p587_1

Essentially, since in many physics problems we do not consider non-constant accelerations or time-dependent potentials (nothing like $U(x,v,t)$), $\dot{q}^2$ is associated to the kinetic energy and only the kinetic energy.

EDIT: Concrete example. $L=\frac{1}{2}m\dot{q}^2-\lambda \dot{q}^3$, potential here is $U=\lambda \dot{q}^3$. Euler-Lagrange gives us

$m\ddot{q}-3\lambda (2\dot{q}\ddot{q})=0$

So that potential gives equations of motion with constant velocity. Not exactly what we expected. So, things get strange when we want to associate Lagrangians to 'usual' phyisics problems unless we just say "quadratic in $\dot{q}$!"

EDIT: Fun Generalization! (Inspired by elfmotat's answer) Take generic Lagrangian $L=\sum_n a_n\dot{q}^n+f(q)$ (Putting all velocity in the first term, generic function of position in the second). Then

$\frac{\partial L}{\partial \dot {q}}=\sum_n na_n \dot{q}^{n-1},\qquad\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right)=\sum_n n(n-1)a_n\dot{q}^{n-2}\ddot{q}$

So equations of motion are

$\sum_n n(n-1)a_n\dot{q}^{n-2}\ddot{q}=\frac{\partial f(q)}{\partial q}$

So, only when $n=2$ is this "nice". Then $f(q)$ is something like the potential and this last line is Newton's 2nd Law.

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If $F=\lambda$, then $V=-\lambda q$. Your equation of motion is wrong - it does not contain $\lambda$. –  Vladimir Kalitvianski Feb 28 '13 at 19:34
I did not say $F=\lambda$. I am starting with a potential which is velocity-dependent (a natural place to assign the linear $\dot{q}$ term, right?), and constructing the Lagrangian. Then I'm pointing out that we do not get the equation of motion we expect to get. I'm trying to illustrate the breakdown in our "usual assumptions" about Lagrangians when they are linear in velocities. –  levitopher Feb 28 '13 at 20:46
Linear in $\dot{q}$ term is a full derivative. Also it can be "absorbed" into redefinition of $\dot{q}$ by a shift transformation. That's why $\lambda$ never appears in the equation of motion. –  Vladimir Kalitvianski Mar 1 '13 at 8:43

A simpler answer is that the term in the Euler-Lagrange equations involving $\dot{q}$ is:

$$\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}$$

So $L$ needs to be quadratic in $\dot{q}$ or else the time derivative will be proportional to something other than $\ddot{q}$.

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