# Question on Reynolds number using Newton's definition of inertial force and how to obtain the common expression for Reynolds number?

The concept of inertial force is traced back to Newton’s Philosophiae naturalis principia mathematica, The Principia, Newton. At the beginning of his book, Newton presents eight definitions forming the basis for his three Laws. The two most important definitions for us are definition no. 2, or the definition of quantity of motion, and no. 3, or the definition of inertial force or the force that measures how difficult it is to change the quantity of motion of a body. Together with Newton’s second law, we can state that the change in quantity of motion (or time rate of momentum change) is identical to the inertial force that is equal to the resultant of all forces acting upon the body: $$\tag{1} \text{Inertial Force}=d\vec p/dt=f_1+f_2+\cdots+f_n$$

The Navier-Stokes equation is a restatement of Newton's second law on a per unit volume basis: $$\tag{2} \rho \vec a = \vec f_{\text{press}}+\vec f_{\text{visc}}+\vec f_{\text{grav}}$$

$$\rho \left( \frac{\partial \vec{v}}{\partial t} + \vec{v} \cdot \nabla \vec{v} \right) = -\nabla p + \mu \nabla^2 \vec{v} + \rho \vec{g}$$

It is said that the Reynolds number, $$Re$$, is the ratio of the inertial force ($$\equiv$$ resultant force) to the viscous force on the fluid body:$$\tag{3} Re=\frac{\text{inertial force}}{\text{viscous force}}$$

Based on (3) would it be correct to interpret the Reynolds number as the following two equalities, Eqns(4) and (5)?: $$\tag{4} Re=\frac{d\vec p/dt}{\vec f_{\text{visc}}}=\frac{\left(\vec f_{\text{press}}+\vec f_{\text{visc}}+\vec f_{\text{grav}}\right)}{\vec f_{\text{visc}}}$$

$$\tag{5} Re=\frac{\rho \left( \frac{\partial \vec{v}}{\partial t} + \vec{v} \cdot \nabla \vec{v} \right)}{\mu \nabla^2 \vec{v}}$$

Or is the "inertial force" in the Reynolds number thought of in a different way? For clarity, let me ask this: Consider the case of steady fully developed flow through a pipe of constant cross-sectional area. The local acceleration is zero $$(\partial \vec v/\partial t=0)$$ because the flow is steady, and the convective acceleration is zero $$(\vec v \cdot \nabla \vec v=u \partial u/\partial x \hat i=0)$$ because the flow is fully developed and is not converging/diverging. Since $$|\vec v|>0$$, and $$Re=\rho |\vec v| L/\mu$$, do you say the inertial force is some value other than zero?

Lastly, I usually see the Reynolds number expressed as: $$\tag{6} Re=\frac{\rho |\vec v| L}{\mu}$$

How does one go from (5) to (6)?

One way I know to get from Eqn(5) to (6) is as follows, but requires some simplifying assumptions and therefore does not answer my actual question. Consider steady flow $$(\partial \vec v/\partial t=0)$$ in the vicinity of a sphere. For fluid motion parallel to the x-coordinate direction, the inertial force per unit volume at a given position in the flow is $$\rho u \partial u/\partial x$$. Thus we have for Eqn(5):

$$\tag{7} Re=\frac{\rho u \frac{\partial u}{\partial x}}{\mu \frac{\partial^2 u}{\partial y^2}}$$

Let us assume that the local velocity $$u$$, the velocity gradient $$\partial u/\partial x$$, and the second derivative of velocity $$\partial^2 u/\partial y^2$$ vary in proportion to the magnitudes of characteristic quantities of the flow. These include, in addition to the density $$\rho$$ and the viscosity $$\mu$$, a characteristic velocity $$U$$, and a characteristic length, the sphere radius $$R$$. Formally applying dimensional analysis to the quantities $$u\partial u/\partial x$$ and $$\partial^2 u/\partial y^2$$, one obtains the result that $$u\partial u/\partial x$$ is proportional to $$U^2/R$$ and $$\partial^2 u/\partial y^2$$ is proportional to $$U/R^2$$. Thus, comparing with (7), $$\tag{8} Re=\frac{\rho \frac{U^2}{R}}{\mu \frac{U}{R^2}}=\frac{\rho U R}{\mu}$$

• You might find this handy physics.stackexchange.com/questions/138598/… – joseph h Sep 15 '20 at 2:17
• Does this answer your question? How Reynolds number was derived? – BioPhysicist Sep 15 '20 at 3:08
• @BioPhysicist I don’t see anything in that post that addresses my two questions (in bold). – Armadillo Sep 15 '20 at 3:32
• @Drjh I don’t see anything in that post that addresses my two questions (in bold). – Armadillo Sep 15 '20 at 3:33

The characterization of the Reynolds number as a ratio of "inertial effects" to "viscous effects" is only approximate, and is based on the nondimensionalization of the Navier-Stokes equations. If you construct nondimensional variables $$x^* = x/L$$, $$\vec{u}^* = \vec{u}^*/U$$, $$t^* = t/(L/U)$$, and $$p^* = p/(U\mu/L)$$, the Navier-Stokes equation (with no gravity) transforms into:

$$Re\left(\frac{\partial \vec{v^*}}{\partial t} + \vec{v^*} \cdot \nabla \vec{v^*} \right) = -\nabla p^* + \mu \nabla^2 \vec{v^*}$$

with the traditional definition of the Reynolds number, $$Re = \frac{\rho U L}{\mu}$$.

If (and this is a big if) the nondimensional variables were scaled in such a way that they are always approximately $$1$$, then the relevance of each side of the equation (and by extension the inertial/viscous effects) is modulated by the magnitude of the scalar $$Re$$.

The challenge is that, because of the nature of even simple solutions to Navier-Stokes, it is effectively impossible to construct such a scaling for any nontrivial flow. As a result, the Reynolds number often fails to consistently predict the ratio of inertial to viscous effects in flows; whereas the object you constructed in (5), which I call the Reynolds tensor, does. (It's a very unwieldy object and of not much practical use, but it does what it says on the tin.)

The Reynolds number does modulate transitions to turbulence, but only as a result of its presence as the single nondimensional parameter that has any relevance in most flows, which from dimensional analysis means it must be an indicator of turbulence transitions. You can read more about this in Chapters 8-10 of my book: https://wp.me/P7zu73-Of

• I enjoyed reading your book. As you say, Re is an approximation to the ratio of convective to viscous effects, which is described by your Reynolds tensor (which is my Eqn(5) with the local acceleration term excluded. Side question, why do you exclude the local acceleration from your Reynolds tensor?). I am in agreement with your conclusion: Re has no physical meaning, only a mathematical one. – Armadillo Sep 16 '20 at 17:08
• Good question, and I’m glad you enjoyed the book! I defined the Reynolds tensor without the time rate-of-change term for two reasons; 1) because I used it as a metric of convective to viscous effects, not as a metric of inertial to viscous effects, and 2) I never really discussed flows that changed in time that weren’t turbulent, so I never saw the needed to define it in such a way. That being said, your definition is absolutely the right one if you want to calculate the ratio of inertial to viscous effects in general flows! – aghostinthefigures Sep 16 '20 at 17:33