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This is in reference to the question: Deriving the Lagrangian for a free particle

My question is specifically in regards to QMechanic's answer to this question, and I have quoted the relevant part of this answer below:

In physics, it is often implicitly assumed that the Lagrangian L=L($\overrightarrow{q}$ ,$\overrightarrow{v}$ ,t) depends smoothly on the (generalized) positions $q_i$, velocities $v_i$, and time t, i.e. that the Lagrangian L is a differentiable function}

Why should we assume the Lagrangian be a differentiable function of positions and velocities?

For example, every potential of the form $1/r^n$ is non-differentiable at the origin, for $n>0$ and we do encounter such potentials all the time (for $n=1$, we recover the usual central force problem).

Can we not have any situations where the Lagrangian is not a differentiable function of the velocities?

This is relevant because later in the answer, they say:

This is differentiable wrt. the speed v=$|\overrightarrow{v}|$, but it is not differentiable wrt. the velocity $\overrightarrow{v}$ at $\overrightarrow{v}=\overrightarrow{0}$ if $\alpha \neq 0$. Therefore the second branch (6) is discarded.

That is, they are using the non-differentiability of a candidate Lagrangian with respect to velocities to rule it out.

TVina
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2 Answers2

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Why should we assume the Lagrangian be a differentiable function of positions and velocities?

Because it makes things easier and is often true.

Can we not have any situations where the Lagrangian is not a differentiable function of the velocities?

Double negative notwithstanding, I guess you are asking for examples where the Lagrangian is not a differentiable function of velocity.

One example is the Lagrangian of a relativistic point particle: $$ L = -mc^2\sqrt{1 - \frac{v^2}{c^2}} $$

Clearly, the derivative with respect to $v^2$ has some issues at $v^2=c^2$.

Update: OP is asking more question in the comments...

Thanks! Is there any reason why a Lagrangian of the form L=av+b (where v is the magnitude of velocity) would not be an acceptable Lagrangian

Acceptable to whom? You can certainly consider it...

It doesn't really generate an interesting Hamiltonian: $$ H = -b\;. $$

And, as explained in the linked-answer, it generates silly equations of motion.

Maybe it would be helpful to explain that, in classical mechanics, the kinetic energy (in its usual form as $\frac{1}{2}\sum_i m_i v_i^2$) is more primal than the Lagrangian. The total kinetic energy (in its usual form) results as the quantity that is changed by the total work. If you can write this work as resulting from conservative forces then the Lagrangian appears and becomes useful.

hft
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The point is, the function you mentioned is non differentiable at a single point which we can disregard by just restricting our domain of interest to be the complement of that point.

The lagrangian in classical mechanics is $K-U$, the kinetic energy is quadratic in velocities and hence is a polynomial (and hence trivially differentiable); the potential may depend on the velocity but that's rare and even in the physical cases where it does (electromagnetism) it's linear so no problem there as well. Also, it may be that you choose a coordinate system which has a singularity (e.g. polar coordinates), but that is just an artifact of the coordinate system. Meanwhile regarding the potential energy, ask yourself: What kind of potential energies have you encountered that model real life interactions? the gravitational and coulomb are both differentiable apart from one point (and so is the generalized coulomb). Perhaps it's useful to point out that when we say differentiable we mean differentiable almost everywhere (in the intuitive not the measure-theoretic sense); because a few points of discontinuity or differentiability isn't a huge deal. To convince yourself of the need of "differentiability almost everywhere" you may be interested to look at a continuous nowhere differentiable function; do you honestly think that a force of nature would behave this way, for example? Not to mention that if the potential is discontinuous at a point this would imply that its gradient (force) is infinite at that point; as I said, that's okay if we have a point or two where this happens (like gravitational and coulomb), but becomes highly problematic and unrealistic otherwise; we simply don't observe entire portions of space where the force is infinite.

Leonid
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