When does exponentially damping term in Lagrangian make physical sense

Question

I have a question on adding dissipation into the Lagrangian. On this page: https://profoundphysics.com/friction-in-lagrangian-mechanics/ the approach with adding an exponential factor to the Lagrangian is descibed. Starting with a lagrangian $L(\dot{q},q) = T(\dot{q}) - V(q)$, one can add an exponential damping factor: $L(\dot{q},q,t) = e^{ct}\left( T(\dot{q}) - V(q) \right)$. However, they claim that this only makes physical sense if one has "linear velocity dependent friction". I'm not sure of why this is the case though. From the Legendre transform $H(q,p) + L(\dot{q},q) = p \cdot \dot{q}$, I can take the gradient w.r.t. $p$, to get that $\dot{q} = \nabla_p H(q,p)$, with $H(q,p) = V(q) + T^*(p)$. (Here $ T^{\ast}(p) $ is the convex conjugate of $T$). In the case when $T(\dot{q}) = \frac{\lVert \dot{q} \rVert_2^2}{2}$, I think $ T^{\ast}(p) = \frac{\lVert p \rVert_2^2}{2}$, and in this case $\dot{q} = p$, and the friction term is linear and velocity dependent. For other kinetic energy functions $T$ this might not hold though. Does the approach with the exponential damping term only make sense "physically" when the kinetic energy is on this form (quadratic in $\dot{q}$)?

Answer 1

Starting with a lagrangian $L(\dot{q},q) = T(\dot{q}) - V(q)$, one can add an exponential damping factor: $L(\dot{q},q,t) = e^{ct}\left( T(\dot{q}) - V(q) \right)$. However, they claim that this only makes physical sense if one has "linear velocity dependent friction".

I'm not sure of why this is the case though.

This could be the case, but only when equations of motion bear out such an interpretation.

In general, the equations of motion generated by your Lagrangian are: $$ c\frac{\partial T}{\partial \dot q}+ \frac{d}{dt}\frac{\partial T}{\partial \dot q}=-\frac{\partial V}{\partial q}\;.\tag{A} $$

Now consider a simple example case where the kinetic energy is a homogeneous function of degree 2 in the velocity: $$ T = \frac{1}{2}m\dot q^2 \tag{B} $$

Inserting Eq. (B) into Eq. (A) gives: $$ cm\dot q + m\ddot q = -\frac{\partial V}{\partial q}\;. $$

The term $$ -\frac{\partial V}{\partial q}\equiv F_{cons} $$ represents the conservative forces, by definition.

So, we have: $$ ma = F_{cons} - cm\dot q\;, $$ where the last term on the right is a linear function of the velocity or, if you prefer, a "linear velocity dependent friction."

For other kinetic energy functions T this might not hold though.

In general, for an arbitrary function $T(\dot q)$, you are correct.

However, in classical non-relativistic mechanics, the function $T$ is not completely arbitrary, but is defined in terms of cartesian coordinates as a homogeneous function of degree 2 in the cartesian velocity components: $$ T = \sum_{i=1}^N \frac{1}{2}m_i (\dot x_i^2+\dot y_i^2 + \dot z_i^2)\;, $$ where the sum is over the $N$ particles in the system. Therefore, we can say that, in cartesian coordinates, generally: $$ m \ddot x_i = -\frac{\partial V}{\partial x_i} - cm_i \dot x_i\;, $$ etc. for $y_i$ and $z_i$. So, that we can write: $$ m \vec a_i = -\vec \nabla_i V - cm_i\vec v_i\;, $$ for all of the $N$ particles indexed by $i$.

Does the approach with the exponential damping term only make sense "physically" when the kinetic energy is on this form (quadratic in $\dot{q}$)?

For any given physical system, you should just work out the equations of motion. For example, this might be the wrong interpretation when there is a velocity dependent potential. But, it seems to be a pretty good interpretation otherwise.