In the Langevin dynamics: neglecting inertia. A mathematician trying to understand physics terminology

Question

If we write the Langevin equation: for a particle with mass $m$, position $x$ and velocity $v$, with some damping coefficient $\gamma$,

$$ m dV(t)=-\gamma V(t)dt+dW(t) ,~~~~~~~dX(t)=V(t)dt.$$

Then as $m\to 0$, we get a first order equation

$$\gamma dX(t)=dW(t),$$

people call this 'neglecting inertia'. If we use the definition of inertia as ``a property of matter by which it continues in its existing state of rest or uniform motion in a straight line, unless that state is changed by an external force''. Why do people say this ? It seems like when $m\neq 0$ the system does not have inertia because friction stops the uniform motion in a straight line?

Answer 1

A physicist would often write these equations as $$ m\ddot{x}=-\gamma\dot{x}+\xi(t)\Leftrightarrow \begin{cases}\dot{x}=v,\\ m\dot{v}=-\gamma v+\xi(t)\end{cases} $$ Admittedly, it lacks mathematical rigor (e.g., equating the derivative with a non-differentiable noise), but it has the advantage of the physical transparency: it is a restatement of the Newton's second law $m\ddot{x}=F$, with two forces: (i) the drag force $f_{drag}=-\gamma v$, and (ii) the random force $\xi(t)$

Overdamped limit (aka "zero mass/inertia")
The equation for velocity can be formally solved as $$ v(t)=v(t')e^{-\frac{\gamma (t-t')}{m}} + \frac{1}{m}\int_{t'}^td\tau e^{-\frac{\gamma (t-\tau)}{m}}\xi(\tau) $$ If the damping is strong, i.e., if the times scales of the behavior of interest are such that we work in the limit $$ \frac{\gamma (t-t')}{m}\gg 1, $$ and the noise varies at scales longer than the one specified before (an admittedly tricky claim for a delta correlated white noise), then the first term in this solution can be neglected, whereas in the second we can take the noise out of the integral, reducing the order of the system of equations to the first-order.

A more hand-waving way of doing this is taking limitr $m\longrightarrow 0$, so that the velocity equation is $$0=-\gamma v(t)+\xi(t)\Rightarrow \gamma v(t)=\gamma\dot{x}=\xi(t).$$ Since mass is the measure of inertia, one speaks of it as neglecting inertia. I suggest reading Reaction-rate theory: fifty years after Kramers for more detailed discussion and references.

Answer 2

Approximations based on "neglecting inertia" occur many times in Physics, and they are always a source of problems. The reason for the name is simple: the mass times acceleration term in the equations of motion is responsible for the validity of the inertia principle, i.e. the fact that in absence of any force (internal, external, random, or with any further specification) a body moves at constant velocity (including the special case of zero-velocity).

Without the random force term (i.e. in the case of ordinary differential equations, ODE), the approximation may look a little puzzling from the mathematical side since it corresponds to neglecting the highest order term in an ODE. The consequence is that apparently there is a sudden change of the related Cauchy problem (initial velocity and position if $m\neq 0$, vs. only position if $m=0$). The presence of the random force does not change in a substantial way the problem.

Actually, in the ODE case, the limit $m \rightarrow 0$ is a singular limit and has to be dealt with some care.

The basic strategy to overcome the difficulties is to understand that for small values of the mass, the motion is the superposition of a "slow" component, due to the "non-inertial terms" plus a fast varying fluctuation around the slow component, due to the mass term. It is the same approximation implied by the adiabatic evolution in the presence of fast and slow degrees of freedom. In the plasma physics context, it is at the base of the "center of guide approximation", or the "force-free" approximation.

In the case of Langevin's equation, things are a little more complicated, due to the random force, but the main ideas of the ODE case may be used.

Answer 3

I suspect the reason for this terminology is that you could think of the stochastic term as a kind of force, such as applied by random particles in a fluid that we often ignore when simplifying the problem. When you send $m$ to zero, the equation you get states that the change in position is entirely dependent on this random force and the particle isn't really continuing in a straight line (i.e. having inertia).

Answer 4

If we use the definition of inertia as "a property of matter by which it continues in its existing state of rest or uniform motion in a straight line, unless that state is changed by an external force".
[...]
It seems like when $m \neq 0$ the system does not have inertia because friction stops the uniform motion in a straight line?

The way you phrased your question makes me think you have a misunderstanding regarding the concept of inertia.

Inertia, when it comes to massive (non-zero mass) objects, is an always-present property that gives rise to the tendency of matter to retain its current state of motion¹, absent a net force. But that definition perhaps misses to communicate a more important fact: inertia is the tendency of an object to resist changes to its current state of motion¹ (i.e. to resist acceleration).

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¹ "State of motion" just means instantaneous speed and direction.

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But, note that mass is a measure of inertia (a way to quantitatively express it), and that therefore Newton's second law, $m\ddot{x}=F$ (or $\ddot{x}=F/m$), has to do with inertia as well. Also note that the idea of inertia being a "tendency to resist acceleration" breaks down at $m = 0$ (the model has no notion of / does not apply to / does not encompass truly massless particles).

So, it doesn't matter if the object in question is being accelerated (speeding up, slowing down, or on a curved path), it still has inertia (for example, if you try to apply some acceleration in order to deflect it from whatever trajectory it's currently on, the force you'll need to supply will depend on the object's mass/inertia). It's just that such an object is not moving inertially - it's not in uniform motion (rest being a special case of it).

Regarding the context of your question, perhaps "neglecting inertia(l effects)" now makes more sense—it just means treating the inertial term (acceleration term) as negligible.