# $\gamma$ in Newton's Second Law of Motion in Differential Form

I am teaching myself Differential Equations from a website. In the website I am up to Direction Fields and an example of a differential equation is Newton's Second Law of Motion. It is written on the website like this:

$$m\frac{dv}{dt}=mg-\gamma v$$

I know that $m$ is the mass, $g$ is the gravitational acceleration, $v$ is the velocity, and $t$ is the time, but what does $\gamma$ stand for?

• Looks like it has something to do with viscosity. – Ruslan Oct 3 '13 at 9:52
• Comment to the question (v1): A force of the type $\vec{F}=-\gamma \vec{v}$ is known as kinetic friction, damping, or Stokes' drag, depending on the context. – Qmechanic Oct 3 '13 at 10:18

The $\gamma$ here is some coefficient/constant (which has the units $\mathrm{kg\,s^{-1}}$). This formula is saying that the resultant force $F = \mathrm{d}v/\mathrm{d}t$ is equal to the force on the mass from gravity minus some force $\gamma v$ which is dependent on velocity.

To me, you are right, this does not make too much sense from a Newtonian standpoint (IMO) as this is representing a drag linearly dependent on velocity which is never the case in my experience. It would make more sense if the resistive force was written as $\lambda v^{2}$ (and the units of $\lambda$ amended accordingly for some arbitrary example case), this could then represent an air/fluid resistance where $\lambda$ is some type of 'drag coefficient'.

I hope this helps.

• In Stokes' law, e.g., one has drag force proportional to first power of velocity, not second one. – Ruslan Oct 3 '13 at 9:57
• This is a Newtonian equation taken at face value. A classic introduction to aerodynamic drag for basic Newtonian physics is to use $\lambda v^{2}$ and there is nothing wrong with doing so. This clearly is not an advanced equation and all I am doing it answering the question, not second guessing what the author of the equation was thinking at the time of writing. – MoonKnight Oct 3 '13 at 10:15
• The down vote is non-sense. Tell me the reason. This site makes a mockery of StackExchange. – MoonKnight Oct 9 '13 at 13:27

The general form of Newton's Law for constant mass is

$$m \frac{dv}{dt} = F$$

so in your case, $F = mg - \gamma v$ is the provided force law. In your case your force happens to depend on the velocity; the greater the velocity, the more negative the force, so it is a kind of friction or drag. $\gamma$ is just the proportionality constant between the friction force $F_{fric} = -\gamma v$ and the velocity, just like $k$ is the proportionality constant between the position and the force $F_{spring} = -kx$ for a spring.